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Technical Mathematics


Author: Todd Simeone (Technology Faculty, Heartland Community College)

Volume I

Updated November 18, 2016

2


Table of Contents

Chapter 1 – Prerequisite material ................................................................................................................. 5

1.1 – Arithmetic with whole numbers ...................................................................................................... 5

1.2 – Fractions ........................................................................................................................................... 7

1.3 – Mixed numbers and Decimals .......................................................................................................... 8

1.4 – Order of Operations ....................................................................................................................... 10

1.5 – Solving Linear Equations ................................................................................................................ 12

1.6 – Ratio and Proportion ...................................................................................................................... 13

1.7 – Scientific Notation .......................................................................................................................... 14

Chapter 1 Homework problems.............................................................................................................. 17

Chapter 2 – Measurement Systems ............................................................................................................ 21

2.1 – Standard Systems ........................................................................................................................... 21

2.2 – Areas and Volumes ........................................................................................................................ 25

2.3 – Temperature .................................................................................................................................. 27

2.4 – Converting between systems ......................................................................................................... 29


Chapter 3 – Percentages, precision, error, and tolerance intervals ........................................................... 32

3.1 – Percentages: basic conversions ..................................................................................................... 32

3.2 – Percentages: base, part, and rate .................................................................................................. 34

3.3 – Precision and error ......................................................................................................................... 36

3.4 – Tolerance intervals ......................................................................................................................... 38


Chapter 4 – Number systems ...................................................................................................................... 42

4.1 – Decimal number system ................................................................................................................ 42

4.2 – Electronic machines ....................................................................................................................... 43

4.3 – Base-2 (Binary) ............................................................................................................................... 44

4.4 – Base – 16 (Hexadecimal) ................................................................................................................ 46


Chapter 5 – Formulas and data substitution .............................................................................................. 53

5.1 – Substituting data into formulas ..................................................................................................... 53

3


5.2 – Simple algebraic manipulation of formulas ................................................................................... 55


Chapter 6 – Basic Geometry ....................................................................................................................... 58

6.1 – Angles ............................................................................................................................................. 58

6.2 – Triangles ......................................................................................................................................... 60

The Pythagorean Theorem ................................................................................................................. 61

Area ..................................................................................................................................................... 64

6.3 – Quadrilaterals ................................................................................................................................. 67

6.4 – Circles ............................................................................................................................................. 69

Circle terminology and basic formulas................................................................................................ 69

Arcs and associate angles ................................................................................................................... 70


Chapter 7 – Basic Trigonometry.................................................................................................................. 80

7.1 – The Trigonometric ratios ................................................................................................................ 80

7.2 – Finding trigonometric values on a calculator ................................................................................. 83

Using the calculator to determine trigonometric values of angles. ................................................... 84

Using the calculator to determine angles given their trigonometric values. ..................................... 85

7.3 – Using trigonometry to find unknown components of a triangle ................................................... 87


Chapter 8 – Interpreting Charts and Graphs............................................................................................... 93

8.1 – Charts and Graphs .......................................................................................................................... 93

Bar charts ............................................................................................................................................ 93

Pie charts ............................................................................................................................................. 96

Coordinate system graphs .................................................................................................................. 97

Chapter 8 Homework problems............................................................................................................ 100

Chapter 9 – Basic Statistics ....................................................................................................................... 105

9.1 – Data presentation ........................................................................................................................ 105

9.2 – Measures of central tendency ..................................................................................................... 109

9.3 – Measures of dispersion ................................................................................................................ 112

9.4 – Normally distributed data sets ..................................................................................................... 116

9.5 – Controlled and capable processes ............................................................................................... 119

Chapter 9 Homework problems............................................................................................................ 121

4


Chapter 10 – Basic Logic ........................................................................................................................... 125

10.1 – Logical expressions ..................................................................................................................... 125

The ‘And’ logical operator ................................................................................................................. 126

The ‘Or’ logical operator ................................................................................................................... 127

The ‘Not’ logical operator ................................................................................................................. 127

10.2 – Logic truth tables ....................................................................................................................... 128

10.3 – Schematics ................................................................................................................................. 131

Chapter 10 Homework problems.......................................................................................................... 133

Supplemental Chapter – Systems of Linear Equations ............................................................................. 135

S.1 – Solving systems of 2 linear equations using a graphing tool ....................................................... 135

S.2 – Solving systems of 2 equations using the Addition-Subtraction method .................................... 138

S.3 – Solving systems of 2 equations using the Substitution method .................................................. 142

S.4 – Applications associated with systems of equations ..................................................................... 146

S.5 – Systems of 3 linear equations ...................................................................................................... 148

Chapter S Homework problems ............................................................................................................ 151

Answers to Homework Problems ............................................................................................................. 155

Chapter 1 Answers ................................................................................................................................ 155

Chapter 2 Answers ................................................................................................................................ 158

Chapter 3 Answers ................................................................................................................................ 160

Chapter 4 Answers ................................................................................................................................ 162

Chapter 5 Answers ................................................................................................................................ 163

Chapter 6 Answers ................................................................................................................................ 164

Chapter 7 Answers ................................................................................................................................ 166

Chapter 8 Answers ................................................................................................................................ 167

Chapter 9 Answers ................................................................................................................................ 169

Chapter 10 Answers .............................................................................................................................. 172

Supplemental Chapter Answers............................................................................................................ 174

5


Chapter 1 – Prerequisite material

It is assumed that readers of this text have been exposed to the majority of the Chapter 1 content (if not

all of it). As such much of the content is provided with brevity.

1.1 – Arithmetic with whole numbers

It is assumed that the reader is comfortable with the content of section 1.1 more than any other section

in this chapter. A small sampling of examples is provided with no formal discussion.

Adding whole numbers:

Example 1

17)14()3(

11143

11)14(3

17143

Subtracting whole numbers:

Example 2

11)14()3(

17143

17)14(3

11143

Multiplying whole numbers:

Example 3

42)14()3(

42143

42)14(3

42143

6


Dividing whole numbers:

Example 4

5)3()15(

53)15(

5)3(15

5315

Exponentiation:

Example 5

211 11 11 121

Example 6

32 2 2 2 8

Example 7

68 8 8 8 8 8 8 262,144

7


1.2 – Fractions

Adding and subtracting fractions

When adding fractions, the denominator must be the same. This is referred to as a ‘common

denominator’. As in the first section, only a limited discussion is provided for this review material.

Example 1

6

5

6

1

6

4 Since the two fractions had 6 as the common denominator, the two numerators

are added and placed over the common denominator

A similar process is used to subtract fractions:

Example 2

8

1

8

4

8

5

When adding fractions that do not have the same denominator, a common denominator must be found.

Example 3

3

1

7

4 21 can be used as the common denominator since 7 and 3 are both

divisors of 21

4 3 12

7 3 21

1 7 7

3 7 21

The fractions in the original problem are replaced with equivalent

fractions over that have a common denominator

21

19

21

7

21

12

3

1

7

4 The answer to the original problem is

19

21.

8


A similar process is used when subtracting fractions with different denominators.

Example 4

3

2

6

11

2 2 4

3 2 6

6

7

6

4

6

11

3

2

6

11 The answer to the original problem is

7

6.

1.3 – Mixed numbers and Decimals

Mixed numbers are numbers that contain both an integer and a fraction. They always have an

equivalent fractional representation knows as an improper fraction.

Example 1

Write the improper fraction 23

7 as a mixed number.

Solution: When 23 is divided by 7, the result is 3 with a remainder of two. 3 becomes the integer in the

mixed number, 2 is the numerator of the fraction, and 7 is the denominator as follows:

23 23

7 7

Example 2

Write the mixed number 11

413

as an improper fraction.

Solution: To determine the numerator, the following calculation is used: (4 13) 11 63 . As a result,

63 becomes the numerator and the denominator is 13 as follows:

11 634

13 13

9


In addition to improper fractions and mixed numbers, decimal numbers are another way to represent

these values.

Example 3

Write the fraction 3

8 as a decimal.

Solution: Using a calculator, 3 8 0.375 , and so 3

0.3758 .

Example 4

Write the mixed number 7

516

as a decimal.

Solution: Using a calculator, 5 7 16 5.4375 , and so 7

5 5.437516

.

Example 5

Write the decimal 0.2 as a fraction.

Solution: Since the 2 is in the tenths place, 2 1

0.210 5

Example 6

Write the decimal 3.21 as a mixed number.

Solution: Since the 1 is in the hundredths place, 21

3.21 3100

10


1.4 – Order of Operations

When multiple operations are to be performed, there is an accepted order in which the operations must

take place. Many students remember the acronym PEMDAS to help keep track of this order. PEMDAS

stands for the following:

1. P = Parenthesis

Any operations within parentheses are done first. Parentheses are used to override the normal

order of operations.

2. E = Exponents

3. MD = Multiplication and division

In step 3, all multiplication and division operations are done. If there is more than one, they are

done in the order they appear from left to right.

4. AS = Addition and subtraction

In step 4, all addition and subtraction operations are done. If there are more than one, they are

done in the order they appear from left to right.

Example 1

Calculate: 333

Solution:

333 33 is calculated first since multiplication is done before addition

93 93 is calculated since it is the lone remaining calculation to perform

12 The resulting answer is 12

11


Example 2

Calculate: 210 2 7 42 15 (3 4)

Solution:

210 2 7 42 15 (3 4) )43( is calculated first since it is within parentheses

210 2 7 42 15 7 27 is calculated next since it is an exponent

10 2 49 42 15 7 210 is calculated next since it is the left-most

multiplication

20 49 42 15 7 715 is calculated next since it is multiplication

20 49 42 105 4920 is calculated next since all remaining

operations have the same precedence and it is the

left-most operation remaining

29 42 105 4229 is calculated next since all remaining

operations have the same precedence and it is the

left-most operation remaining

13 105 10513 is the lone remaining calculation to perform

92 The resulting answer is 92

12


1.5 – Solving Linear Equations

Linear equations are equations that have only one variable that has an exponent of 1. For example, the

following is a linear equation:

5 14x

The goal when solving a linear equation is to determine all of the numbers that can be substituted in

place of the variable which result in a true statement. 9 is a solution to the above equation because

when 9 is substituted in place of x, a true statement results:

Example 1a

5 14

9 5 14

14 14

x

Following a similar line of thinking, 10 is NOT a solution because when 10 is substituted in place of x, a

false statement results:

Example 1b

5 14

10 5 14

15 14

x

Additional values can be substituted in for x, but only 9 will result in a true statement (and hence only 9

is a solution to this equation).

As equations become more complex, there are a few rules that can be applied to help determine the

solutions:

1. Any value can be added to or subtracted from both sides of an equation.

2. Both sides of the equation can be multiplied or divided by any non-zero value.

13


Example 2

Find the solution of the equation: 2173 x

Solution:

2173 x

17217173 x 17 is added to both sides of the equation

153 x

3

15

3

3

x Both sides of the equation are divided by 3

5x x has been isolated, and the solution is 5

1.6 – Ratio and Proportion

Ratios are ways to compare one measurement to another and are represented as fractions. For

example, if the wall in a room is 30m, and the opposite wall is 100m, the values can be represented as:

30

100which can be simplified to

3

10.

2 ratios are in proportion if they are equal to each other. For example, 7

11and

21

33are in proportion

because7 21

11 33 .

Sometimes it is necessary to determine a specific value which will put two ratios in proportion. Cross-

multiplication can be used to accomplish this task as follows:

14


Example 1

Find the value for x which solves the following proportion: 5

217 31

x

Solution:

5

217 31

x

31 217 5x Cross-multiplication is performed on the proportion

31 1085x

31

1085

31

31

x Both sides of the equation are divided by 31

35x x has been isolated, and the solution is 35

1.7 – Scientific Notation

Scientific notation provides another way to represent numbers. This notation is often used in a technical

setting to provide a short-hand notation for numbers that might otherwise be very long. Scientific

notation depends on powers of 10 to represent numbers. If you consider the following calculation:

133.6 10

the calculated value is:

36,000,000,000,000 .

Since these are just two different representations of the same value, either can be used. However, since

the first one is shorter and more concise it is often used in technical settings, and is in fact known as the

scientific notation representation. The second representation is in what is known as decimal notation.

It can be seen that the second representation was generated from the first number by simply moving

the decimal point 13 places to the right. Scientific notation representations always work this way. The

only rules to remember when converting from scientific notation are:

1. If the exponent is positive, move the decimal point to the right

2. If the exponent is negative, move the decimal point to the left

15


The following table demonstrates this process:

Scientific notation Rule to be applied Converted to decimal notation 74.7 10 Positive exponent – move decimal to right 47,000,000 101.26 10 Positive exponent – move decimal to right 12,600,000,000 117.3 10 Negative exponent – move decimal to left 0.000000000073

52.514 10 Negative exponent – move decimal to left 0.00002514

To convert from decimal notation to scientific notation, the process is reversed. Furthermore, the

decimal point is always moved so that the remaining number only has one digit to the left of the decimal

point. The rules to convert from decimal notation to scientific notation are as follows:

1. Move the decimal point so the number has exactly one digit to the left of the decimal

2. The exponent is the number of places the decimal point was moved

a. If the decimal was moved to the left, the exponent is positive

b. If the decimal was moved to the right, the exponent is negative

The following table demonstrates this process:

Decimal notation Rule to be applied Converted to scientific

notation 231,000,000 Move decimal to left – positive exponent 82.31 10

4,127,000,000,000,000 Move decimal to left – positive exponent 154.127 10

0.000061 Move decimal to right – negative exponent 56.1 10

0.0003 Move decimal to right – negative exponent 43 10

When performing calculations with numbers in scientific notation, the most straightforward way to

handle it is to perform the calculations on the numbers separately from the calculations on the powers

of 10 as in the following example:

16


Example 1

Calculate the following:

8 5 (3.2 10 ) (2.11 10 )

Solution:

8 5

8 5

13

(3.2 10 ) (2.11 10 )

(3.2 2.11) (10 10 )

6.752 10

Example 2


11 3

4

(6.21 10 ) (4.11 10 )

9.1 10

Solution:

11 3

4

11 3

4

4

(6.21 10 ) (4.11 10 )

9.1 10

6.21 4.11 10 10

9.1 10

2.8 10

As demonstrated by the next example, care must be taken to ensure the answer is in scientific notation.

Example 2


5 3 (9.2 10 ) (8.5 10 )

Solution:

5 3

5 3

8

9

(9.2 10 ) (8.5 10 )

(9.2 8.5) (10 10 )

78.2 10

7.82 10

17


Chapter 1 Homework problems

Section 1.2 Homework

Calculate the following.

1. 3 5

7 7

2. 4 12

11 11

3. 3 13

14 7

4. 41 5

6 6

5. 2 5

3 9

6. 1 4

2 7

7. 9 3

13 11

8. 2 4

3 17

9. 12 1

13 2

10. 33

113

18



Fill in the following chart.

Mixed number Improper Fraction Decimal

47

7

1311

41

24

9

1582

14

–6.62

74.6



1. 7 2 11 8 2

2. (7 2) 11 8 2

3. (7 2) (11 8) 2

4. 2 3 2(8 3 )

5. 2 49 (3 7)2 14 2 4

6.

4

2 140.3 17 0.01

11

19



In each of the equations, solve for the variable.

1. 11 28x

2. 2 7 71x

3. 5 4 3 16x x

4. 3 7 1 2

2 3 6 3x x

5. 0.007 0.314 2.16x


In each of the following, find the value of x which solves the proportion.

1. 22

7 14

x

2. 4

3 5

x

3. 9

27 33

x

4. 23 1035

4833 1611

x

5. 0.023 1.07

4.69 0.943

x

20



Fill in the following chart:

Scientific Notation Decimal Notation 89.21 10

58.7 10 145 10

509,000,000,000

0.00007 0.0000000314

Calculate the following and express the answer in scientific notation.

1. 7 5

(2 10 ) (4 10 )

2. 7 3

(2.01 10 ) (3.72 10 )

3. 17 53

(3 10 ) (3 10 )

4. 9 4

(5.1 10 ) (3.2 10 )

5.

14

5

8.21 10

9.1 10

6. 11 3 5

4 6

(7.21 10 ) (3.94 10 ) (1.7 10 )

(9.1 10 ) (4 10 )

21


Chapter 2 – Measurement Systems

2.1 – Standard Systems

While the U.S. typically uses the Imperial System for measurement, the rest of the world uses the Metric

System. For this reason, the metric system is also known as the international system, which in other

languages is referenced as the System International, or SI system. The Metric System is easier to use

because everything is based on powers of 10. The following table provides the base values in the metric

system:

Prefix Symbol Power of 10 Decimal Equivalent

Tera T 12 1,000,000,000,000

Giga G 9 1,000,000,000

Mega M 6 1,000,000

Kilo k 3 1,000

Hecto h 2 100

Deka da 1 10

Standard Unit

Deci d –1 0.1

Centi c –2 0.01

Milli m –3 0.001

Micro u –6 0.000001

Nano n –9 0.000000001

Pico p –12 0.000000000001

The standard unit is dictated by the type of measurement that is being done according to the following

table:

Measurement Unit Symbol

Length Meter m

Mass Gram g

Volume (liquid) Liter L

Current Ampere (amp) A

Power Watt W

Resistance Ohm Ω

Time Second s

Temperature Celsius C

Using these charts, measurements can be converted from one prefix to another as in the following

examples.

22


Example 1

Convert 10 km to dm.

Solution:

Looking at the charts, it can be determined that km is the abbreviation for kilometers (k = kilo

and m = meters). Similarly, dm is the abbreviation for decimeters

The exponent for km is 3, and the exponent for dm is –1

The exponents are subtracted to find how many places the decimal point needs to be moved:

3 – (-1) = 4

The decimal point is moved 4 places. Since decimeters are smaller than kilometers, it is moved

to the right

10 km = 100,000 dm

Example 2

Convert 0.0032 hm to m.

Solution:

Looking at the charts, it can be determined that hm is the abbreviation for hectometers and m is

the abbreviation for meters (since it has no prefix, it is just the standard unit)

The exponent for hm is 2, and the exponent for m is 0 (the standard unit always has an

exponent of 0)


2 – 0 = 2

The decimal point is moved 2 places. Since meters are smaller than hectometers, it is moved to

the right

0.0032 hm = 0.32 m

23


Example 3

Convert 432,000 cm to dam.

Solution:

Looking at the charts, it can be determined that cm is the abbreviation for centimeters and dam

is the abbreviation for dekameters

The exponent for dam is 1, and the exponent for cm is –2


1 – (-2) = 3

The decimal point is moved 3 places. Since dekameters are larger than centimeters, it is moved

to the left

432,000 cm = 432 dam

One of the powerful aspects of the metric system is that the standard unit does not affect the way the

conversions are done. In the following examples, notice how the standard unit changes, but the process

is the same as for the above examples in meters (and would be the same for watts, liters, seconds, etc.).

Example 4

Convert 0.0013 A to mA.

Solution:

Looking at the charts, it can be determined that A is the abbreviation for Amp and mA is the

abbreviation for milliamps

The exponent for A is 0, and the exponent for mA is –3


0 – (-3) = 3

The decimal point is moved 3 places. Since mA are smaller than A, it is moved to the right

0.0013 A = 1.3 mA

24


Example 5

Convert 1,000,000,000,000 ug to kg.

Solution:

Looking at the charts, it can be determined that ug is the abbreviation for microgram and kg is

the abbreviation for kilogram

The exponent for ug is –6, and the exponent for k is 3


3 – (-6) = 9

The decimal point is moved 9 places. Since kg are larger than ug, it is moved to the left

1,000,000,000,000 ug = 1,000 kg

25


2.2 – Areas and Volumes

Care must be taken when areas or volumes are involved. The following example demonstrates how to

convert square cm (cm2) to square dam (dam2). Again, the standard unit itself (i.e. m2) is inconsequential

to the process.

Example 1

Convert 25,000 cm2 to dam2.

Solution:

Looking at the charts, it can be determined that cm2 is the abbreviation for square centimeters,

and dam2 is the abbreviation for square dekameters

The exponent for cm is –2, and the exponent for dam is 1

The exponents are subtracted and then multiplied by 2 because this is an area measurement to

find how many places the decimal point needs to be moved:

1 – (-2) = 3, and then 3 x 2 = 6

The decimal point is moved 6 places. Since dam2 are larger than cm2, it is moved to the left

25,000 cm2 = 0.025 dam2

26


Similarly, volumes will use a factor of 3. The following demonstrates a volume conversion.

Example 2

Convert 3.8 Mm3 to m3.

Solution:

Looking at the charts, it can be determined that Mm3 is the abbreviation for cubic megameters,

and m3 is the abbreviation for cubic meters

The exponent for Mm is 6, and the exponent for m is 0

The exponents are subtracted and then multiplied by 3 because this is a volume measurement

to find how many places the decimal point needs to be moved:

6 – (0) = 6, and then 6 x 3 = 18

The decimal point is moved 18 places. Since m3 are smaller than Mm3, it is moved to the right

3.8 Mm3 = 3,800,000,000,000,000,000 m3 or 3.8 x 1018 m3

27


2.3 – Temperature

In the metric system, temperatures are handled in a completely different way than other

measurements. The Celsius scale provides the unit of measurement (you may hear about the Kelvin

scale which is also used in the metric system but it is used primarily for more scientific and engineering

applications). In the English system, the Fahrenheit system is used. The following table represents 3

important temperatures on the scales:

Celsius temperature Fahrenheit temperature Significance of measurement

100° 212° Boiling point of water

0° 32° Freezing point of water

–273° –459.4° Absolute zero

The following formulas provide the means to convert between these two temperature systems:

Example 1

Convert 22.3° C to F.

Solution:

Since we are converting to Fahrenheit, we will use the second formula listed above

95

32

9(22.3) 32

5

40.14 32

72.14

F C

F

F

F

So 22.3° C = 72.14° F

)32(95 FC

3259 CF

28


Example 2

Convert –13.67° F to C.

Solution:

Since we are converting to Celsius, we will use the first formula listed above

59

59

59

( 32)

( 13.67 32)

( 45.67)

25.37

C F

C

C

C

So –13.67° F = –25.37° C

29


2.4 – Converting between systems

It is often necessary to convert from one number system to the other. Because the systems are so

different and have so many units to deal with, many times some research must be done to determine

how one unit relates to another.

Example 1

Chicago, IL and New York City, NY are approximately 800 miles apart. How far is this distance in km?

Solution:

A quick internet search reveals that 1 mi = 1.609 km. In order to convert mi to km, this

conversion factor must be used as follows:

800 800(1.609 ) 1287.2km

mi kmmi

Example 2

What is the weight capacity in kg of a 0.5 ton truck?

Solution:

A quick internet search reveals that 1 ton = 907.185 kg. In order to convert tons to kg, this

conversion factor must be used as follows:

0.5 0.5(907.185 ) 453.593kg

tons kgton

30




1. Convert 46,120 cm to:

a. mm

b. m

c. dam

d. km

2. Convert 17 MW to:

a. mW

b. W

c. kW

d. GW

3. Convert 0.0005 L to:

a. dL

b. cL

c. kL

d. daL


1. Convert 1,000,000 mm2 to:

a. um2

b. m2

c. km2

2. Convert 133.21 10 km3 to:

a. cm3

b. m3

c. Tm3

31



1. Fill in the following chart:

Degrees Fahrenheit Degrees Celsius

–48°

111°

3.2°

27°

2. A third temperature scale that is often used in Science and Engineering applications is the Kelvin

scale. The units on the Kelvin scale are the same as those on Celsius, except the scale is shifted

so that zero on the Kelvin scale is the coldest possible temperature (known as absolute zero).

The following equation relates Kelvin and Celsius: 273K C .

Use this formula to verify the values of absolute zero on the Celsius and Fahrenheit scales that

were given earlier in the text.

Section 2.4 Homework (an internet search may be required to obtain the necessary conversion

factors).

1. Convert 7.2 gallons to liters.

2. Convert 10,000 feet to meters.

3. Convert 727 grams to pounds.

4. Convert 10 hectares to acres.

5. Convert 2800 cm3 to in3.

32


Chapter 3 – Percentages, precision, error, and tolerance intervals

3.1 – Percentages: basic conversions

Percentages are another way to numbers. If one half of a quantity is being discussed, that quantity can

be represented as a fraction or a decimal as follows:

10.5

2

Using a percentage, the decimal representation is multiplied by 100. So a third way to represent this

quantity is illustrated as follows:

1

0.5 50%2

In general, conversions are done according to the rules below:

To convert a decimal to a percentage, move the decimal point 2 places to the right

To convert a percentage to a decimal, move the decimal point 2 places to the left

To convert a fraction to a percentage, first convert it to a decimal and then move the decimal

point 2 places to the right

The following table provides examples of these relationships:

Fractional representation Decimal representation Percentage representation

2

5 0.4 40%

3

10 0.3 30%

13

500 0.026 2.6

4

3000 0.0013 0.13%

27

2 13.5 1350%

33


Example 1

Convert 0.31 to a percentage

Solution:

To convert a decimal to a percentage, the decimal point is moved 2 places to the right:

0.31 = 31%

Example 2

Convert 43.21% to a decimal

Solution:

To convert a percentage to a decimal, the decimal point is moved 2 places to the left:

43.21% = 0.4321

Example 3

Convert 4

11 to a percentage

Solution:

To convert a fraction to a percent, the fraction is converted to a decimal and then the decimal is

converted to a percent:

4

0.363611

0.3636 36.36%

So 4

36.36%11

34


3.2 – Percentages: base, part, and rate

There are times when what percentage one number is of another needs to be determined. For example,

700 is 50% of 1400.

In this case, we have the following:

1400 is called the ‘base’ (B in the formulas below)

700 is called the ‘part’ (P in the formulas below)

50% is called the ‘rate’ (R in the formulas below)

Note: the decimal version of R is used in the formulas below

The formula that relates these components together are:

P BR used when the ‘part’ needs to be found

PB

R used when the ‘base’ needs to be found

PR

B used when the ‘rate’ needs to be found

Example 1

What is 38% of 654?

Solution:

Here 38% is the rate (0.38 is used in the formula), and 654 is the base. Since we need to find the

part, the problem is solved as follows:

654 0.38

248.52

P BR

P

P

So 248.52 is 38% of 654

35


Example 2

29 is what percent of 514?

Solution:

Here 29 is the part and 514 is the base. Since we need to find the rate, the problem is solved as

follows:

29

514

0.0564

PR

B

R

R

So 29 is 5.64% of 514

Example 3

63 is 0.03% of what number?

Solution:

Here 63 is the part and 0.03% is the rate. Since we need to find the base, the problem is solved

as follows:

63

0.0003

210,000

PB

R

B

B

So 63 is 0.03% of 210,000

36


3.3 – Precision and error

The precision of a measurement is defined to be the place of the digit that the measurement was

rounded to. The following table represents the precisions of sample measurements:

Measurement Precision Explanation

170 m 10 m Measurement was rounded to

the tens place

420,000 mi 10,000 mi Measurement was rounded to

the ten thousands places

0.023 A 0.001 A Measurement was rounded to

the thousandths place

0.10 ft 0.01 ft Measurement was rounded to

the hundredths place (as indicated by the last zero)

4250 cm 1 cm Measurement was rounded to the ones place (as indicated by

the tagged zero)

The precision of a measurement allows for different analytical values to be calculated related to error.

The following definitions are associated with this concept:

Greatest possible error – one half of the precision

Relative error – greatest possible error divided by the actual measurement

Percent of error – relative error expressed as a percent

Example 1

Find the precision, greatest possible error, relative error, and percent of error in the measurement

7200 m.

Solution:

The precision is 100 m since the measurement was rounded to the hundreds place

The greatest possible error is 50 m since that is one half of the precision

The relative error is 0.00694 since 50

0.006947200

The percent of error is 0.694% since that is the percentage representation of the relative error

37


Example 2

Find the precision, greatest possible error, relative error, and percent of error in the measurement

0.0012 A.

Solution:

The precision is 0.0001 A since the measurement was rounded to the ten-thousandths place

The greatest possible error is 0.00005 A since that is one half of the precision

The relative error is 0.0417 since 0.00005

0.04170.0012

The percent of error is 4.17% since that is the percentage representation of the relative error

38


3.4 – Tolerance intervals

When manufacturing items that are to meet a specific measurement, no process can be perfect due to

what is called common cause variation. As a result, manufacturers must be presented with tolerance

intervals that indicate how much a particular measurement can differ from the specifications. The

following definitions are associated with this concept:

Tolerance – acceptable amount a measurement may vary from the specification

Upper tolerance limit – the largest measurement that is still within tolerance

Lower tolerance limit – the smallest measurement that is still within tolerance

Tolerance interval – twice the tolerance

The following table represents sample tolerance values:

Specification Tolerance Upper limit Lower limit Tolerance Interval

0.001 m 0.0002 m 0.0012 m 0.0008 m 0.0004 m

60 g 0.05 g 60.05 g 59.95 g 0.1 g

110 lb 2 lb 112 lb 108 lb 4 lb

Example 1

If an order is placed for a widget with a length specification of 2.12 mm and a tolerance of 0.0001

mm, find the upper limit, lower limit, and tolerance interval.

Solution:

The upper limit is 2.12 + 0.0001 = 2.1201 mm

The lower limit is 2.12 – 0.0001 = 2.1199 mm

The tolerance interval is 2 x 0.0001 = 0.0002

39


Example 2

If an order is placed for a widget with a weight of 170 kg and a tolerance interval of 15, find the

tolerance, the upper limit, and the lower limit.

Solution:

The tolerance is 15/2 = 7.5 kg

The upper limit is 170 + 7.5 = 177.5 kg

The lower limit is 170 – 7.5 = 162.5 kg

40




Fill in the blank spaces in the following table:

Fractional representation Decimal representation Percentage representation

1

10

13

74

213

84

0.05

0.0023

17.44

71%

17.33%

0.0056%


1. What is 14% of 150?

2. What is 11.35% of 720?

3. What is 0.001% of 4410?

4. 15 is what percent of 22?

41


5. 400 is what percent of 100?

6. 0.62 is what percent of 96?

7. 29 is 32% of what number?

8. 1.35 is 0.37% of what number?

9. 18 is 300% of what number?

10. If Julie received a 7% raise and her new salary is $33,560, what was her original salary?



Measurement Precision Greatest Possible

Error Relative Error Percent of Error

30 V

21.5 m

0.0003 g

105,000,000 mi



Specification Tolerance Upper limit Lower limit Tolerance Interval

17 m 1 m

60.01 mL 0.005 mL

0.5 oz 1.5 oz.

0.0002 W 59.9999

2.5 cm 0.0003 cm

4.5 m 1 m

42


Chapter 4 – Number systems

4.1 – Decimal number system

The typical system that is used in everyday mathematics is the decimal number system, also known as

the base-10 system. The system is structured based on powers of 10 (hence the name of base-10). For

example, the number 50,413 can be written in expanded form as follows:

4 3 2 1 0

50,413 5(10000) 0(1000) 4(100) 1(10) 3(1)

or more specifically:

50,413 5(10 ) 0(10 ) 4(10 ) 1(10 ) 3(10 )

The place of every digit is important, as it indicates the power of ten that number is associated with.

Another aspect of the base-10 number system is that there are 10 digits necessary to represent

numbers in the system. These are the familiar digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Every number in base-10

can be represented by using these 10 digits and the concept regarding their place in the number.

Furthermore, all of the familiar operations (+, –, x, ÷) can be performed on these numbers.

Since electronic systems (such as computers) only store data in terms of ‘on’ and ‘off’, they are unable

to directly store a 10-state system. This concept will be discussed in the following sections.

43


4.2 – Electronic machines

Computers, calculators, and other electronic devices are sometimes known as finite state machines. This

references the fact that these devices have memory which can be in any of a number of different states.

More importantly to this discussion, because these machines are electronic they ultimately can only

store on and off sequences. Because there are only two different options, machines do not have the

ability to store a 10-digit sequence like the numbers in a base-10 system. To simulate this system, we

need to use a binary number system (or base-2) in place of the more familiar base-10 system. However,

different number systems are all structured the same way. The only thing that changes is the base that is

used for each place where the digit resides.

One example of how this manifests itself in technology is how a computer stores a symbol such as a

letter ‘A’. Ultimately, this letter must be stored as a sequence of ons and offs so each letter (and in fact

all letters, digits, symbols, etc.) is given a numeric code. Most personal computers use the Unicode

system to encode these characters. In Unicode, the letter ‘A’ has a code of 65. However, even the

number 65 is one step removed from what the machine can store since 65 is in the decimal (10-state)

base. 65 is converted to binary: 65 = 01000001 in base 2 (groupings on computers are in multiples of 8).

Hence, the number 65 actually looks like this sequence:

off, on, off, off, off, off, off, on

Numbers can be written in expanded form using the same technique as that which was used for base-10

as follows:

7 6 5 4 3 2 1 0

20(2 ) 1(2 ) 0(2 ) 1(2 ) 0(2 ) 1(20100 ) 1(2 ) 0(2 )0001

Note the subscript on the number 201000001 indicates that it is in base-2 as opposed to base-10.

Furthermore, if you add the values on the right side of the equation, you get the original value of 65.

While any positive integer can be used as a base, the bases relevant to a technological discussion are the

following:

Base Name Valid digits

2 Binary 0 and 1

8 Octal 0, 1, 2, 3, 4, 5, 6, and 7

10 Decimal 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

16 Hexadecimal (Hex) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F

Binary numbers are important because they provide a means of simulating the on/off aspect of

electronic devices. Hexadecimal numbers are used in different technological applications such as

networking and design (among others). Since octal numbers are no longer as prevalent as they used to

be, they are not covered in this text.

44


4.3 – Base-2 (Binary)

As mentioned above, the binary number system has only 2 valid digits: 0 and 1. Furthermore, the place

of each digit in a number indicates the associated power of 2.

Any number in any non-decimal number base can be converted to base 10 by writing the number in

expanded form and then calculating the products and sums on the right-hand side.

Example 1

Convert 101102 to decimal

Solution:

101102 is a binary number as indicated by the subscript. Writing this number in expanded form

yields the following:

4 3 2 1 0

2

4 3 2 1 0

10110 1(2 ) 0(2 ) 1(2 ) 1(2 ) 0(2 )

1(2 ) 0(2 ) 1(2 ) 1(2 ) 0(2 ) 22

So the binary number 101102 is equivalent to the decimal number 22

101102 = 22

Example 2

Convert 11100102 to decimal

Solution:

6 5 4 1

2

6 5 4 1

1110010 1(2 ) 1(2 ) 1(2 ) 1(2 ) (Note the zeroes were ignored)

1(2 ) 1(2 ) 1(2 ) 1(2 ) 114

So 11100102 = 114

45


When converting numbers from base-10 to base-2, a different approach is used as follows:

Example 3

Convert 152 to binary

Solution:

2152

2 76 0 76 represents the quotient and 0 represents the remainder when 153 is divided by 2

2 38 0

219 0

2 9 1

2 4 1

2 2 0

21 0

2 0 1 The process stops when a 0 is obtained in the bracket

Reading the remainders from the bottom to the top gives the result.

152 = 100110002

Example 4

Convert 37 to binary

Solution:

2 37

2 18 1

2 9 0

2 4 1

2 2 0

2 1 0

2 0 1

37 = 1001012

46


4.4 – Base – 16 (Hexadecimal)

Because the hexadecimal number system (often abbreviated to ‘hex’) is base-16, it needs 16 valid digits.

This is problematic because there are only 10 standard digits (0 through 9). Letters are commonly used

for the remaining 6 digits according to the following chart:

Digit in Hex Base-10 value

0 – 9 Standard meaning

A 10

B 11

C 12

D 13

E 14

F 15

As in the other bases, the place of each digit is important. In hex, the placement indicates the associated

power of 16. Converting base-16 numbers to decimal can be done by putting the number in expanded

form and then calculating the products and sums on the right-hand side.

Example 1

Convert 3CD216 to decimal

Solution:

3CD216 is a hexadecimal number as indicated by the subscript. Writing this number in expanded

form yields the following:

3 2 1 0

163 2 3(16 ) 12(16 ) 13(16 ) 2(16 ) 15,570CD

So 3CD216 = 15,570

47


Example 2

Convert A141E216 to decimal

Solution:

A141E216 is a hexadecimal number as indicated by the subscript. Writing this number in expanded

form yields the following:

5 4 3 2 1 0

16141 2 10(16 ) 1(16 ) 4(16 ) 1(16 ) 14(16 ) 2(16 ) 10,568,162A E

So A141E216 = 10,568,162

When converting numbers from base-10 to base-16, the same division/remainder approach is used as

that which was used for binary:

Example 3

Convert 5038 to hexadecimal

Solution:

16 5038

16 314 14 (which is E in base-16)

1619 10 (which is A in base-16)

161 3

16 0 1

5038 = 13AE16

48


Example 4

Convert 13020 to hexadecimal

Solution:

1613020

16 813 12 (which is C in base-16)

16 50 13 (which is D in base-16)

16 3 2

16 0 3

13020 = 32DC16

Base-16 numbers can be converted to base-2 numbers using the following process:

1. Convert the base-16 number to base-10

2. Convert the base-10 number to base-2

However, base-16 and base-2 numbers have a unique relationship whereby every 4 digits in a base-2

number equates to 1 digit in a base-16 number. This can be useful to provide a shortcut to converting

numbers between these two bases.

Example 5

Convert 1011000101102 to hexadecimal.

Solution:

The digits in 101100010110 is broken up into groupings of 4:

1011 0001 0110

The groupings are now associated with their base-16 number using expanded form:

10112 = 1(23) + 1(22) + 1(20) = 13 = D16

00012 = 1(20) = 116

01102 = 1(22) + 1(21) = 616

Placing the values in their respective locations gives 1011000101102 = D1616

49


Example 6

Convert 11000110101012 to hexadecimal.

Solution:

The digits in 101100010110 is broken up into groupings of 4 and replaced with their hex equivalents:

1 1000 1101 0101

1 8 D 5

So 11000110101012 = 18D516

The reverse process is used to convert base-16 to base-2.

Example 7

Convert A316 to binary.

Solution:

The digits in A3 are associated with their corresponding groupings of 4:

A 3

1010 0011

So A316 = 101000112

Example 8

Convert 1CC16 to binary.

Solution:

The digits in 1CC are associated with their corresponding groupings of 4:

1 C C

0001 1100 1100

So 1CC16 = 1110011002

50


Number Bases Chart

The integers 0 – 32 in Base 10, 2, 8, and 16

Decimal (Base 10) Binary (Base 2) Octal (Base 8) Hexadecimal (Base 16)

0 0 0 0

1 1 1 1

2 10 2 2

3 11 3 3

4 100 4 4

5 101 5 5

6 110 6 6

7 111 7 7

8 1000 10 8

9 1001 11 9

10 1010 12 A

11 1011 13 B

12 1100 14 C

13 1101 15 D

14 1110 16 E

15 1111 17 F

16 10000 20 10

17 10001 21 11

18 10010 22 12

19 10011 23 13

20 10100 24 14

21 10101 25 15

22 10110 26 16

23 10111 27 17

24 11000 30 18

25 11001 31 19

26 11010 32 1A

27 11011 33 1B

28 11100 34 1C

29 11101 35 1D

30 11110 36 1E

31 11111 37 1F

32 100000 40 20

51




1. Write 6192 in expanded form (the absence of a subscript indicates this is a base 10 number).

2. Write 1012 in expanded form



5. Write AE316 in expanded form

6. Write 13C2D016 in expanded form


1. Convert 1012 to Base 10




5. Convert 547 to binary

6. Convert 1024 to binary


1. Convert AE316 to Base 10

2. Convert 13C2D016 to Base 10


4. Convert 501,322 to Base 16

52



6. Convert 110011010010112 to Base 16

7. Convert A316 to Base 2

8. Convert 19CD16 to Base 2

53


Chapter 5 – Formulas and data substitution

5.1 – Substituting data into formulas

Many formulas are encountered in technical settings to show relationships among the values in a given

situation. A common simple formula in electronics is Ohm’s law which states:

E IR

where E is the voltage (measured in volts), I is the current (measured in amps) and R is the resistance

(measured in ohms). Hence, the voltage in a circuit is equal to the current multiplied by the resistance.

Example 1a

Find the voltage in a circuit if the current is 0.0009A and the resistance is 10,000Ω.

Solution:

0.0009 10,000

9

E IR

E A

E V

So the voltage in this circuit is 9V.

Example 1b

It is not uncommon for current and resistance values to be given in uA (microamps) and kΩ

(kiloohms or k-ohms). As a result, it would be common to see the problem in Example 1a phrased

this way:

Find the voltage in a circuit if the current is 900uA and the resistance is 10kΩ.

Solution:

0.0009 10,000 (since 900uA=0.0009A and 10k 10,000 )

9

E IR

E A

E V

Using different (but equivalent) units does not change the situation, and hence the voltage is still 9V.

54


There are many formulas which will be encountered in technical settings. A small subset is provided

here.

Formula Description f pa force = (pressure)(area)

P IE power = (current)(voltage) 2P I R power = (square of the current)(resistance)

QI

t

Electric chargecurrent=

time

20.7854E b SN Engine displacement = 0.7854(square of the bore)(Stroke)(number of cylinders)

932

5F C

Fahrenheit temp = 9/5(Celsius temp) + 32

Substituting data into formulas always uses essentially the same process. The values are substituted in

and the calculations are performed to determine the unknown quantity.

Example 2

Find the engine displacement given that the bore is 3.63 in, the stroke is 3.65 in, and the engine has

8 cylinders.

Solution:

2

3

0.7854(3.63 )(3.65)(8)

302

E

E in

Example 3

Water boils at 100°C. What is the equivalent Fahrenheit temperature?

Solution:

932

5

9(100) 32

5

180 32

212

F C

F

F

F

So water boils at 212°F.

55


5.2 – Simple algebraic manipulation of formulas

Problems encountered in technology are often not always in terms of the isolated variable. In one of the

situations above, the formula was written in terms of V and the situation required that the value of V be

found. However, it is quite possible that V would be a known quantity and one of the other two values

would need to be found. In this case, minor algebraic manipulation would have to be performed to

isolate the required variable.

Example 1

A light bulb in a circuit has a voltage of 12V and current of 24mA. What is its resistance?

Solution:

Both sides of the equation are divided by I

The equation is flipped to place R on the left

12 The kno

0.024

E IR

ER

I

ER

I

VR

A

wn values are substitued in for E and I

500 The value of R is calculated (and left in since k is typically

used for values over 100

R

0)

So the resistance is 500Ω.

Example 2

What is the wattage of the bulb in example 1?

Solution:

(12 )(0.024 ) The known values are substitued in for I and E

0.288 The value for P is calculated and converted to mW

288

P IE

P V A

P W

P mW

So the power is 288mW.

56


Example 3 As of the end of 2013, the highest recorded temperature in Normal, IL was 107.6°F. What is this temperature in Celsius? Solution:

932

5

932 C 32 is subtracted from both sides of the equation

5

5 5( 32) C both sides of the equation are multiplied by

9 9

5( 32)

9

F C

F

F

C F

the equation is flipped to place C on the left

5(107.6 32) the known value is substituted for F

9

42 the value of C is calculated

C

C

So as of the end of 2013, the highest recorded temperature in Normal, IL was 42°C.

57




1. The pressure between a boy and the ground he stands on is 1.875 N/cm2. If the boy’s feet cover

an area of 240 cm2, what is his weight (in N)?

2. If the current in a circuit is 4A, and the voltage is 12V, what is the power?

3. If the current in a circuit is 2000mA, and the voltage is 9V, what is the power?

4. If the current in a circuit is 3A, and the resistance is 2 Ω, what is the power?

5. Find the engine displacement of a twin cylinder engine in cm3 if the bore is 5.4 cm and the

stroke is 4.0 cm.


1. If the Power in a circuit is 50W, and the voltage is 12V, what is the current?

2. If the Power in a circuit is 50W, and the current is 9A, what is the voltage?

3. If the Power in a circuit is 50W, and the current is 9A, what is the resistance?

4. If the Power in a circuit is 50W, and the resistance is 5 Ω, what is the current?

5. In the late 1960s, Toyota produced a small 1988 cm3 6-cylinder engine. If the bore of the engine

was 75 mm, what was the stroke?

58


Chapter 6 – Basic Geometry

6.1 – Angles

Most shapes are in some way related to line segments and angles. There are many different

relationships inherent in geometry, so it is important to first understand what angles are. An angle is

formed when two line segments meet in a point. This point is known as the vertex of the angle. In the

following image, O is the angle vertex:

Angles can be measured using a few different approaches. One measurement systems uses degrees. Any

angle that is formed by two perpendicular segments as in the image below is known as a right angle and

has a measurement of 90°.

The question can reasonably be asked “What is a degree?” If a circle is ‘cut’ into 360 equal pie-shaped

wedges, each of these wedges is exactly 1°, regardless of the size of the circle:

Angles can now be classified as follows:

An ‘acute’ angle measures less than 90°

A ‘right’ angle measures exactly 90°

An obtuse angle measures more than 90°

Two angles are complements if the sum of their measures is 90°

Two angles are supplements if the sum of their measures is 180°

59


There are many angular relationships that exist when angles are formed by intersecting lines as follows.

In the following image, angles A and B are called vertical angles and will always have the same

measurement. Angles C and D are also vertical angles.

A somewhat more complicated example occurs when two parallel lines are intersected by a third line

(called a transversal). The image and table below show the various relationships that result in this

scenario (the table does not show all possible relationships).

Angle pair Relationship Reason

A and D Equal Vertical angles

D and E Equal Alternate-interior

A and C Supplements Form a straight line

C and E Supplements E = A and A and C are supplements

E and G Supplements Form a straight line

A and E Equal A = D = E

A and G Supplements A = E and E and G are supplements

60


6.2 – Triangles

Polygons are closed figures that are made up of line segments. They are categorized by the number of

sides. Triangles are the 3-sided polygons where the sum of the interior angles is always 180°. There are

two different ways to categorize triangles as follows:

Categorization 1 – based on the interior angles

A triangle is ‘acute’ if all three angles measure less than 90°

A triangle is ‘right’ if one of its angles measures 90° (the other two angles will always be less

than 90° and will be complements of each other). The longest side of a right triangle is called the

hypotenuse, and the two shorter sides are called the legs.

A triangle is obtuse if one of the angles measures more than 90° (the other two angles will

always be less than 90°)

Categorization 2 – based on the lengths of the sides

A triangle is ‘scalene’ if all three sides have different lengths

A triangle is ‘isosceles’ if two of the sides have the same length

A triangle is ‘equilateral’ if all three sides have the same length

Example 1

What is the measure of angle A in the image below?

Solution

Since A is a right triangle (indicated by the small square in the lower-left corner), the other two

angles are complements of each other. The measure of the angle A can be found by subtracting the

other angle from 90°:

A = 90° - 37° = 53°.

61


Example 2

An isosceles triangle has a perimeter of 22m and the shorter side is 4m. What is the length of the

two equal sides?

Solution

The perimeter of a figure is the distance around the figure. In the case of a triangle, it is the sum of

the 3 sides. Using ‘L’ as a variable for the length of the equal sides gives the following:

4 + L + L = 22

4 + 2L = 22

2L = 18

L = 9

So the length of the two equal sides is 9m.

The Pythagorean Theorem

When the lengths of two sides of a right triangle are known, the third side can always be calculated

using the Pythagorean Theorem. This theorem is demonstrated as follows:

In the image below, 2 2 2c a b

62


Example 3

Find the length of the unknown side in the following figure:

Solution

This is a right triangle so the Pythagorean Theorem is used. Since x represents the hypotenuse, the

solution is calculated as follows:

2 2 2

2 2

27 16

27 16

729 256

985

31.4

x

x

x

x

x

So the length of x is 31.4 cm.

Care needs to be taken because the variable c here is always indicating the hypotenuse. The following

problem indicates how to solve for one of the legs.

63


Example 4

Find the length of the unknown side in the following figure:

Solution

This is a right triangle so the Pythagorean Theorem is used. Since x represents one of the legs, the

solution is calculated as follows:

2 2 2

2 2 2

2 2 2

2 2

100 37

100 37

100 37

100 37

10000 1369

8631

92.9

x

x

x

x

x

x

x

So the length of x is 92.9 mm.

64


Area

The formula for the area of a triangle is given by 12

A bh where b is the base and h is the height. The

following example applies this formula.

Example 5

Find the area of the triangle.

Solution

The height of this triangle is 19 in and the base is 42 in.

12

12

212

2

(42 )(19 )

(798 )

399

A bh

A in in

A in

A in

So the area of the triangle is 399 in2.

65


In a right triangle, the two legs represent the base and the height.

Example 6


Solution

The height of this triangle is 14.3 m and the base is 7.1 m.

12

12

212

2

(14.3 )(7.1 )

(101.53 )

50.765

A bh

A m m

A m

A m

So the area of the triangle is 50.765 m2.

66


Example 7


Solution

Since only the length of one of the legs is known, the other must be found using the Pythagorean

Theorem.

2 2 2

2 2 2

2 2 2

2 2

27.5 20.3

27.5 20.3

27.5 20.3

27.5 20.3

756.25 412.09

344.16

18.55 cm

x

x

x

x

x

x

x

Now that the 2 legs are known to be 20.3 cm and 18.55 cm, the area can be found.

12

12

212

2

(20.3 )(18.55 )

(376.565 )

188.3

A bh

A cm cm

A cm

A cm

So the area of the triangle is 188.3cm2.

67


6.3 – Quadrilaterals

Quadrilaterals are the four-sided polygons. Two of the common types found in a technical setting are

parallelograms and trapezoids.

Parallelograms have two pairs of parallel sides. Examples are rectangle and squares, although

not all parallelograms have 4 right angles.

Trapezoids have one pair of parallel sides.

Name Area formula Example

General parallelogram A bh

Rectangle A lw

Trapezoid 12

( )A a b h

Example 1

Find the area of the parallelogram.

Solution

The base of the parallelogram is 9.5 in and the height is 7 in.

2

(7in)(9.5in)

66.5

A bh

A

A in

So the area of the parallelogram is 66.5 in2.

68


Example 2

Find the area of the trapezoid.

Solution

The height of the trapezoid is 5.1 ft and the two bases are 13.2 and 16.8 ft.

12

12

12

2

( )

(13.2 16.8)(5.1)

(30)(5.1)

76.5

A a b h

A

A

A ft

So the area of the trapezoid is 76.5 ft2.

69


6.4 – Circles

Circle terminology and basic formulas

Important terminology associated with circles is discussed in the following table. The most important of

these are the radius and diameter.

Line / Line segment Description

Chord Line segment that begins on one point of the circle and ends on the other

Diameter Chord that goes through the center of the circle

Radius Line segment starting at the center of the circle and going to a point on the circle

Secant Line that goes through the circle, intersecting with it in two points

Tangent Line that intersects with the circle in one point

Two important formulas associated with circles given below. For this text, the value of 3.14159 will be

used for the constant π.

Formula Description

2C r Circumference (distance around the circle) 2A r Area

Example 1

Find the circumference and area of the circle.

Solution

The radius of this circle is 6 cm.

2

2(3.14159)(6 )

37.7

C r

C cm

C cm

2

2

2

(3.14159)(6 )

113.1

A r

A cm

A cm

So the circumference of the circle is 37.7 cm and the area is 113.1 cm2.

70


Example 2

Find the diameter of a circle that has an area of 48.2m2.

Solution

First the area formula is algebraically manipulated to isolate the radius, and then the values are

substituted. Once the radius is found, it is doubled to find the diameter.

2

2

248.2

3.14159

3.9

A r

Ar

Ar

Ar

mr

r m

Since the radius of the circle is 3.9m, the diameter is 7.8m.

Arcs and associate angles

Arcs represent a portion of the circle and are measured in degrees using the same measurement system

as for angles (i.e. 1/360th of the way around a circle represents 1°).

71


For example, in the following image, the red arc labeled ‘s’ measures 90° since it is ¼ of the circle:

Angles that originate from the center of a circle are called central angles. These angles have the same

measure as the corresponding arc on the circle as demonstrated by the following image:

Angles that originate from a point on a circle are called inscribed angles. These angles also have a

relationship to the corresponding arc on the circle. The measure of an inscribed angle is always half the

measurement of the corresponding arc as demonstrated by the following image:

72


A relationship involving triangles and circles is that a triangle inscribed in circle, and using the diameter

as one of the sides of the triangle, is guaranteed to be a right triangle. Furthermore, the diameter of the

circle will always represent the hypotenuse of the right triangle. Hence the red triangle in the image

below (where C marks the center of the circle) must have a 90° angle as marked in the image.

73


Example 3

If C is the center of the circle, find length of the unknown side ‘x’ in the triangle, and the measure of

the arc ‘s’.

Solution

The red triangle must be a right angle (since it is inscribed in a circle and the diameter represents

one of its sides). This information will be used to find the values.

First the length of ‘x’ can be found using the Pythagorean theorem.

2 24.7 2.4

16.33

4.0

x

x

x m

Next, the measure of the arc ‘s’ can be found using the following two pieces of information:

1. The angle opposite the hypotenuse is 90°

2. The sum of the angles in a triangle is 180°

3. The unknown angle is the inscribed angle associated with the arc ‘s’

The third angle in the triangle is calculated as follows:

Inscribe angle measure = 180 – 90 – 31 = 59°

And the corresponding arc ‘s’ is calculated as follows:

S = 59 x 2 = 118°

Combining what was calculated above, the values are:

x = 4.0 m

s = 118°

74




1. If an angle measures 38°, then:

a. what is the angle’s complement (if it exists)?

b. what is the angle’s supplement (if it exists)?

2. If an angle measures 114°, then:

a. what is the angle’s complement (if it exists)?

b. what is the angle’s supplement (if it exists)?

3. In the image below, what are the two pairs of equal angles?

4. In the image below, find 3 pairs of equal angles and 3 pairs of supplementary angles.

75



1. Fill in the chart for each triangle:

Triangle Acute, right, or obtuse Scalene, Isosceles, or Equilateral

2. In the image below, find the measure of angle A:

3. In the image below, find the value of x and the perimeter of the triangle.

76


4. In the image below, find the value of x and the perimeter of the triangle.

5. Find the area of the triangle in question 3.


7. Find the area of the following triangle:

8. If an isosceles triangle has a perimeter of 176 ft, and the shorter side is 10 feet less than the

measure of the two equal sides, what is the length of the three sides of the triangle?


77


10. When the lengths of the three sides of a triangle are known, but the height is not, Heron’s

formula can be used to calculate the area. Heron’s formula states that if a, b, and c

represent the three sides, then the area is calculated as follows:

( )( )( ) where 2

a b cA s s a s b s c s

Use this formula to find the area of the following triangle:


1. Find the area and perimeter of a rectangle with a length of 30m and a width of 42m.

2. Find the length of a rectangle with an area of 700cm2 and a width of 68cm.

3. Find the area of the trapezoid in the image below:

4. Find the area and perimeter of the trapezoid in the image below:

78


5. A building has a length of 40 feet, a width of 60 feet, and a height of 80 feet. Find the

amount of paint needed (in gallons) and the cost of the paint if:

a. all 4 sides of the building needed to be painted

b. a gallon of paint covers 420 square feet

c. a gallon of paint costs $11


1. Fill in the values in the table below.

Circle radius Circle diameter Circle circumference Circle area

3.4 in

17 ft

3 x 106 km

104 cm

712 mm3

2. Find the measure of the arc s:

79




5. Find the measure of the arc s, and the diameter of the circle:

80


Chapter 7 – Basic Trigonometry

7.1 – The Trigonometric ratios

The six trigonometric ratios are relationships between the lengths of the sides in a right triangle. They

are defined in the following chart (in this chart, ‘A’ represents an angle).

Full ratio name Standard abbreviation Ratio

sine(A) sin(A)

side opposite A commonly referred to as

hypotenuse

O

H

cosine(A) cos(A)

side adjacent to A commonly referred to as

hypotenuse

A

H

tangent(A) tan(A)

side opposite A commonly referred to as

side adjacent to A

O

A

secant(A) sec(A)

hypotenuse commonly referred to as

side adjacent to A

H

A

cosecant(A) csc(A)

hypotenuse commonly referred to as

side opposite A

H

O

cotangent(A) cot(A)

side adjacent to A commonly referred to as

side opposite A

A

O

Secant, cosecant, and cotangent can all be found if the other ratios are known. As a result, they do not

appear on scientific calculators, they are not often used in technical classes, and they will not be covered

in this text.

The acronym SOHCAHTOA (pronounced soh-ca-toe-a) is often used to remember the ratios of the three

major trigonometric functions, since sin , cos , and tanO A O

A A AH H A

.

81


Example 1

Find sinA, cosA, and tanA in the following triangle.

Solution

13sin 0.6190

21

17cos 0.8095

21

13tan 0.7647

17

OA

H

AA

H

OA

A

82


Example 2

Find sinB, cosB, and tanB in the following triangle.

Solution

Since the hypotenuse is needed to find sin and cos, that is calculated first using the Pythagorean

Theorem:

2 24.7 8.2 9.5h m

The three trigonometric ratios can now be calculated:

8.2sin 0.8632

9.5

4.7cos 0.4947

9.5

8.2tan 1.745

4.7

OB

H

AB

H

OB

A

83


7.2 – Finding trigonometric values on a calculator

All scientific calculators can calculate any of the trigonometric values of a particular angle (and can also

calculate an angle given the trigonometric value). Angles can be measured using different measurement

systems. To be consistent with the previous angular discussions, the degree measurement system will

be used. When using a scientific calculator, care must be taken to ensure that the degree measurement

system is used (as opposed to radians or gradians), as this will affect the answers.

In this text, the following two assumptions are made:

1. The calculator used is the standard Windows OS calculator in ‘Scientific’ mode (other calculators

can be used, but they all have their own processes)

2. The degree measurement system is used

This image shows what the calculator looks like. Note the ‘Degrees’ radio button is selected. The ‘sin’,

‘cos’, and ‘tan’ buttons are also visible in this mode.

84


Using the calculator to determine trigonometric values of angles.

Example 1

Find cos(58°).

Solution

The question is asking for the cosine of a 58° angle to be found. This can be done by entering 58 into

the calculator and pressing the ‘cos’ button. The calculator displays the value 0.5299 (trig values are

often rounded to four significant digits), and so:

cos(58°) = 0.5299.

Example 2

Find the sin, cos, and tan of 71.3°.

Solution

Using the calculator, the following values are obtained:

sin(71.3°) = 0.9472

cos(71.3°) = 0.3206

tan(71.3°) = 2.9544

85


Using the calculator to determine angles given their trigonometric values.

Example 3

If cosA = 0.3141, find the measure of angle A.

Solution

The question is the opposite of what is being asked in example 1 above. Rather than being given the

angle and being asked for the trig value, this question gives the trig value and is asking for the angle.

Scientific calculators can calculate this, but the process is slightly different. To calculate this value,

0.3141 is entered in the calculator, and the ‘Inv’ button is pressed (this is often marked as the 2nd

key on Scientific calculators). The image below shows the calculator at this point (the ‘Inv’ button is

in yellow):

The ‘cos-1’ button replaces the ‘cos’ button on the calculator. Pressing this button yields the value

71.69, and so:

If cosA = 0.3141, then the measure of angle A = 71.69°.

86


Example 4

If sinA = 0.2178, find the measure of angle A.

Solution

Using the calculator, the following value is obtained:

If sinA = 0.2178, then the measure of angle A = 12.58°

Example 5

If tanA = 2.668, find sinA and cosA.

Solution

To find sinA and cosA, first the measure of A must be found. Using the calculator (as in examples 3

and 4), the following is obtained:

If tanA = 2.668, then the measure of angle A = 69.45°

Now, cosA and sinA can be found (as in examples 1 and 2):

sin(69.45°) = 0.9364

cos(69.45°) = 0.3510

87


7.3 – Using trigonometry to find unknown components of a triangle

The concepts in sections 7.1 and 7.2 can be used to find the lengths of sides and measures of angles in

right triangles if enough information is provided.

Example 1

Find x in the image below.

Solution:

In this example, the 61° angle and its opposite side are provided. The adjacent side must be

found. Since the sides in question are the opposite and adjacent sides, the appropriate

trigonometric function to use is tangent (since tanO

AA

).

The calculation is performed as follows:

14.1tan 61

(tan 61) 14.1

14.1

tan 61

7.816

x

x

x

x

And so x = 7.816 m.

88


There are many applications of right triangles in technical courses. The following provides one such

example.

Example 2

The relationship between Resistance (R), Inductive reactance (XL), and Impedance (Z) (all three

of these are measured in Ohms), and the phase angle in a circuit is given by the image below.

If R = 11Ω and Z = 15Ω, find XL and the phase angle ( ).

Solution:

XL can be found by using the Pythagorean Theorm:

2 215 11 10.2LX

can be found using R, Z, and cos( ). Cos is used because R and Z represent the adjacent

angle and the hypotenuse.

-1

11cos 0.7333

15

Since cos 0.7333, using cos on the calculator yields the following:

42.84

R

Z

So: XL = 10.2Ω and = 42.84°.

89




1. Find sinA, cosA, and tanA.

2. Find sinA, cosA, and tanA.

3. Find sinB, cosB, and tanB.


1. If A = 67°, find:

a. sinA

b. cosA

c. tanA

2. if B = 23°, find:

a. sinA

b. cosA

c. tanA

90


3. Was there any relationship between the sin and cos values in question 1 and 2? If so, explain

why this happened.

4. If sinA = 0.2513, find A.

5. If cosA = 0.5731, find A.

6. If tanA = 3.614, find A.

7. Explain what happens (and why) if you try to find A if cosA = 2.143.

8. If sinA = 0.3141, find cosA and tanA.


1. Find x and y in the image below.

2. Find x and y in the image below.

91


3. Find the measure of angle A in the image below.



92


For questions 6, 7, and 8, use the image below that represents the relationships between Impedance,

Reactance, Resistance and the phase angle in a circuit:

6. If R and XL are both 14Ω, find Z and the phase angle.

7. If Z = 20Ω and = 38°, find R and XL.

8. If R = 8.5Ω and Z = 12Ω, find and XL.

93


Chapter 8 – Interpreting Charts and Graphs

8.1 – Charts and Graphs

Charts and graphs provide a visual representation of data. The data can be representative of many

different situations, from quality control processes to informational trends over time to the power

output of a device.

While charts and graphs present the data in a concise and more easily-read format, it takes practice to

become comfortable reading and analyzing these charts. Furthermore, there are many different kinds of

charts that can be used. Many software products provide ways to enter data and build charts and

graphs. All of the images in this section were either built from Excel, or by using the TI-83 graphing

calculator. Three samples are given below, and these three types are discussed in more detail on the

following pages.

Chart/Graph type Sample Image

Bar chart

Pie chart

Coordinate system graph

Bar charts

As the name indicates, bar charts use bars to present information. The bars in the chart can be

presented either horizontally or vertically.

0

5000

10000

1 2 3

Series2

1, 23%

2, 30%

3, 47%

1 2 3

94


Example 1

The following chart shows the number of defective parts produced during a manufacturing process.

Quality control made a change to the process in September to reduce the number of defective parts.

Chart analysis

1) Question: How many parts were defective in January and February?

Answer: Looking at the first blue bar, it appears that approximately 720 parts were defective in

these months.

2) Question: How many parts were defective in the months January through August?

Answer: These months comprise the first 4 bars. Adding them together, it appears that there

were approximately 720 + 705 + 800 + 690 = 2915 defective parts in this timeframe.

3) Question: Did the change made to the process in September seem to have a positive affect?

Answer: Additional analysis needs to be done to answer this question. A good approach would

be to calculate the average number of defective parts per month.

For Jan – Aug, the average is: 720 705 800 690

3648

defective parts per month.

For Sep – Dec, the average is: 505 550

2644

defective parts per month.

While more data and analysis should be done to verify the findings, it appears that the process did

have a positive effect since the average number of defective parts per month was lowered by 100.

0

100

200

300

400

500

600

700

800

900

Jan & Feb Mar & Apr May & Jun Jul & Aug Sep & Oct Nov & Dec

Def

ecti

ve p

arts

Timeframe

Quality control report - Defective Parts in Process

95


Example 2

The following chart is a clustered (or stacked) horizontal bar chart showing information regarding

employee salaries and taxes.

Chart analysis

1) Question: Who paid the most in taxes?

Answer: Looking at the legend at the bottom of the chart show that the orange bars represent

taxes. Lori Williams has the longest orange bar, paying approximately $16,000 in taxes.

2) Question: What is the average employee net salary?

Answer: The green bars indicate the net salary. The average is calculated as follows:

88,000 37,000 62,000 45,000 43,00055,500

5

.

3) Question: What is the total company payroll?

Answer: To find the total payroll, the gross salaries needed to be added.

Payroll = 110,00 41,000 72,000 52,000 50,000 325,000 .

0 20,000 40,000 60,000 80,000 100,000 120,000

Jake Smith

Lisa Johnson

John James

Joe Black

Lori Williams

Dollar Value

Emp

loye

e

Payroll Information

Net pay Taxes Gross salary

96


Pie charts

As the name indicates, pie charts use circles with wedges. The wedges are reminiscent of pie-shaped

pieces giving them their name.

Example 3

The following chart shows the same information as in Example 1, only this time a pie-chart is used.

Since this is simply a different way of presenting the data, all of the same questions can be

answered.

Chart analysis

1) Question: How many parts were defective in January and February?

Answer: Looking at the blue area, 4004*.18 = 720 parts were defective in this timeframe.

2) Question: How many parts were defective in the months January through August?

Answer: Looking at the blue, red, green, and purple areas, 4004 * .73 = 2923 parts were

defective in this timeframe.

3) Question: Did the change made to the process in September seem to have a positive affect?

Answer: For Jan – Aug, the average percent is 18 18 20 17

9.1%8

For Sep – Dec, the average is: 13 14

6.8%4

, so again, it seemed to have a positive effect.

Jan & Feb18%

Mar & Apr18%

May & Jun20%

Jul & Aug17%

Sep & Oct13%

Nov & Dec14%

Defective parts (total 4004 defective parts)


97


Example 4

The following chart is a version of a pie chart that focuses on a specific part of the chart and

provides a secondary pie chart with only that data.

Chart analysis

This chart not only shows the defective part totals broken out in their months, but also shows the

timeframe associated with the manufacturing process change. This allows for more detailed analysis

and easier readability so that appropriate decisions can be made regarding the effectiveness of the

change, and if that approach should be continued.

Coordinate system graphs

Coordinate system graphs use a standard x and y axis approach. The x-axis is the horizontal axis and the

y-axis runs vertically. Sometimes these axes will indicate other values (such as time, power, dollar

values, or any value that can be measured).

Jan & Feb18%

Mar & Apr18%

May & Jun20%

Jul & Aug17%

Sep & Oct13%

Nov & Dec14%

Other27%

Defective parts (total 4004 defective parts)


98


Example 5

The following graph displays the relationship from Ohm’s law discussed earlier in the text. Recall

that Ohm’s law states E IR . For a set value of R = 5Ω, the graph appears in the image below.

Note that E is the vertical axis and I is the horizontal axis.

Chart analysis

In interpreting this chart, it can be seen that there is a smooth linear relationship between E and I.

For every unit I goes up, E goes up 5 units. This can be seen in the two images below:

When I = 1, E = 5:

When I = 2, E = 10:

99


Example 6

The following graph displays the relationship Power relationship 2P I R (given earlier in the text).

For a set value of R = 5Ω, the graph appears in the image below. Note that P is the vertical axis and I

is the horizontal axis.

Chart analysis

In interpreting this chart, it can be seen that the relationship is no longer linear. Unlike in the first

example, this time as the value of I increases the value of P increases dramatically. This can be seen

in the two images below:

When I = 1A, P = 5W:

When I = 2A, P jumps all the way to 20W (note that the point is off the graph below). This increase

will get more and more dramatic as I gets larger due to the exponential nature of the formula.

100




1. Use the chart below to answer the questions that follow.

a. In what year did the Cubs win the most games?

b. Approximate the total wins for all even numbered years.

c. If a season consists of 162 games, approximate the number of losses in the year 2000.

d. How many games have the Cubs won since the 2005 season (include 2005 in your

total)?

e. How many games have the Cubs lost since the 2005 season (include 2005 in your total)?

0

20

40

60

80

100

120

2009 2008 2007 2006 2005 2004 2003 2002 2001 2000

Win

s

Season

Chicago Cubs win totals per season

101


2. A small company reported $70,000 of revenue during the calendar year. The chart below

indicates the percentage of that revenue that occurred during each quarter of the year. Use the

chart to answer the questions that follow.

a. What percent of the revenue was generated during the first half of the year?

b. How much combined revenue (in dollars) was generated in the months of July, August,

and September?

c. How much combined revenue (in dollars) was generated in the months of April, May,

June, October, November, and December?

d. A fiscal report provided by the company claims that their sales are greatest in the

months leading up to the holiday season. Explain whether or not the chart supports this

claim.

21%

26%20%

33%

Revenue

Q1 Q2 Q3 Q4

102


3. The chart below uses what is called a line graph to show the number of bolts produced by a

company every 5 years. Use the chart to answer the questions that follow.

a. Approximately how many bolts were produced in 1990?

b. Approximately how many bolts produced in 1990 did not meet specifications?

c. Approximately what percent of bolts produced in 1995 did not meet specifications?

d. A report was published by the company for its stock holders in 1994. The report stated

that a new process would be put into production in 1995 and the expectation was that

while there would be some inefficiencies when the process was first implemented, over

time the process should allow for more bolts to be produced. While the chart only

supplies a small subset of the data, does it corroborate the claims made by the

company?

e. How many bolts were produced in 1985?

86,000

88,000

90,000

92,000

94,000

96,000

98,000

100,000

102,000

1985 1990 1995 2000 2005 2010 2015

Analysing bolt production

Bolts meeting specs Total bolts produced

103


4. The following two graphs show the relationship between Power, Current, and Resistance in 10

situations using the formula 2P I R .

The first graph uses a standard scale:

Because there is an exponential relationship, a standard scale makes it difficult to see the

smaller values of I and R (because P becomes large very quickly). In this case, a logarithmic scale

is often used as in the following chart:

Use these charts to answer the question below:

0

10000

20000

30000

40000

50000

60000

1 2 3 4 5 6 7 8 9 10

P=I2R

Resistance Current (A) Power (W)

1

10

100

1000

10000

100000

1 2 3 4 5 6 7 8 9 10

P=I2R

Resistance Current (A) Power (W)

104


a. In second chart, what is the Resistance in all 10 cases?

b. In the first chart, can the Resistance be identified? If not, why?

c. How does the current change from one scenario to the next? Which chart helps to

answer this question?

d. If the current increases from 10A to 20A, what is the change in the Power?

e. If the current increases from 10A to 100A, what is the change in the Power?

5. Use the graph below to answer the questions that follow.

a. For each value of x, find the corresponding value of y:

i. X = –2

ii. X = 0

iii. X = 3

iv. X = 6

b. For each value of y, find all corresponding values of x:

i. Y = 0

ii. Y = –2

iii. Y = –4

iv. Y = 3

105


Chapter 9 – Basic Statistics

9.1 – Data presentation

Statistics in technical settings often deals with collecting and analyzing data. This process can manifest

itself in many ways including but not limited to the following:

Collecting historical data over time to allow for future predictions that can lead to better

business or processing decision-making

Collecting data to identify trends in industry

Collecting data about a specific process that can lead to better quality control

Collecting data about a specific process that can lead to the determination of whether the

process can meet order specifications in terms of a customer’s tolerance

While there are different mathematical calculations and techniques involved in this process, first and

foremost the data must be presented in a readable format. Consider the following chart of raw data:

Measurement of widgets produced (target measurement: 2.00 mm)

Widgets above target Widgets below target 2.03

1.97 2.01

1.99 2.01

1.99 2.02

1.98 2.01

1.99 2.01

1.99 2.05

1.95 2.01

1.99 2.01

1.99 2.03

1.97 2.02

1.98 2.01

1.99 2.01

1.99 2.02

1.98 2.05

1.97

Looking at this data, it is hard to determine any kind of categorization or trend as it is not grouped

together or in any kind of natural ordering. One common approach is to create what is known as a

frequency table that will group and order the data in a more meaningful way.

106


Measurement of widgets produced (target measurement: 2.00 mm)

Frequency Table

Measurement interval

Frequency 1.93 - 1.95

1 1.96 - 1.98

6 1.99 - 2.01

16 2.02 - 2.04

5 2.05 - 2.07

2

In the frequency table above, the representation helps to form a clearer picture of what is happening. It

can been seen that most of the widgets in this set are close to the target measurement and as the

measurement moves away from the target, fewer and fewer of the widgets fall into that category.

Another way to present data in a more graphical form is to use what is known as a histogram.

Histograms are bar charts that represent the frequency distribution in a way that appeals to many

readers. While histograms can be created by hand, there are many tools that can help to create well-

formatted and visually appealing charts. The chart below was created in Microsoft Excel.

0

2

4

6

8

10

12

14

16

18

1.93 - 1.95 1.96 - 1.98 1.99 - 2.01 2.02 - 2.04 2.05 - 2.07

Freq

uen

cy

Measurement of widget in mm

Analysing widget production

107


Example 1

For the following data set, build the frequency table and the histogram using six 1-unit intervals.

Solution:

The data is tallied and categorized to form the frequency table:

The data is now graphed to form the following histogram:

108


0

1

2

3

4

5

6

7

8

9

6 - 7 7 - 8 8 - 9 9 - 10 10 - 11 11 - 12

Freq

uen

cy

Measurement of acceleration due to gravity

Analysing student experiment results

109


9.2 – Measures of central tendency

When analyzing data sets, there is often a need to have a single number that indicates a ‘middle’ value.

This can give some indication of the overall tendency of the data set. Common examples are the average

score of a standardized test, the average salary of a given population, or the average measurement of a

widget in a production process.

These measurements are referred to as measurements of central tendency. There are three important

such measurements as follows:

Measure of central tendency Notation (if applicable) Description/formula

mean x

The sum of all of the data points divided

by the number of data points

1 2 3 ... nx x x xx

n

median

Middle number of an ordered data set, or

the mean of the two middle numbers if there are an even number of data points

mode

Most frequently occurring data point (there may be none or more than 1

mode)

110


Example 1

Find the mean, median, and mode of the following set of ACT scores:

17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26

Solution:

Finding the mean:

17 17 17 17 18 18 19 19 20 20

17

2

21 21 22 23 25 2

0

5

.2

2

9

6x

x

Finding the median:

Since the data set is an ordered data set with 17 data points, the 9th element is the median

Median = 20

Finding the mode:

Since 17 occurs 4 times, and this is more than any other element, 17 is the mode

Mode = 17

Analyzing these results it can be seen that all 3 of the measurements give a reasonable

representation of the central tendency of this set.

111


Example 2

Find the mean, median, and mode of the following house prices:

70,000 114,000

72,000 117,000

81,000 118,000

90,000 120,000

94,000 8,500,000

Solution:

Finding the mean:

70,000+114,000+72,000+117,000+81,000+118,000+90,000+120,000+94,000+8,500,000

10

937,600

x

x

Finding the median:

Since the data set is an ordered data set with 10 data points, the mean of the 5th and 6th

elements is the median

Median = 94,000 114,000

104,0002

Finding the mode:

Since all of the numbers occur exactly once (i.e. no number occurs more than any other

number), this set does not have a mode

Analyzing these results, the following can be seen:

1. Although the mean was calculated correctly, it is too high to be meaningful in this case.

This is caused by the one house that had a value dramatically higher than all of the other

houses. Data points like this are often called outliers and are sometimes left of

calculations because of the skewing affect they have.

2. Since there is no mode, this calculation is not helpful at all.

3. The median is the only meaningful measure of central tendency for this set of data.

112


9.3 – Measures of dispersion

Whereas measures of central tendency attempt to provide the single number that indicates a ‘middle’

value, sometimes it is important to know how dispersed or ‘spread out’ the data set is. This value will

often be used in many of the same circumstances as above to provide more insight into the data set.

These measurements are referred to as measurements of dispersion. There are three important such

measurements as follows:

Measure of dispersion Notation (if applicable)* Description/formula

range

The largest number in the set minus the smallest

number in the set

variance s2

2

1

( )

1

n

i

i

x x

n

standard deviation s

2

1

( )

1

n

i

i

x x

n

* Sometimes notations other than ‘s’ are used, but for simplicity this text will use ‘s’.

113


Example 1

Find the range, variance, and standard deviation of the set of ACT scores:

17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26

Solution:

Finding the range:

26 17 9range

Finding the variance:

2

2 2 22 1

2

( )(20.29 17) (20.29 17) ... (20.29 26)

1 17 1

9.10

n

i

i

x x

sn

s

Finding the standard deviation:

2

1

2 2 2

( )

1

(20.29 17) (20.29 17) ... (20.29 26)

17 1

9.10

3.02

n

i

i

x x

sn

s

s

s

114


The Windows OS calculator (as well as many other statistical calculators) has built in statistical

functionality which can make these calculations much easier. The image below shows what the

calculator looks like in statistical mode. The important keys for this discussion are the mean and

standard deviation keys (circled in red).

115


Example 2

Use the calculator to find the standard deviation of the following house prices:

70,000 114,000

72,000 117,000

81,000 118,000

90,000 120,000

94,000 8,500,000

Solution:

The data is added to the statistical calculator and the standard deviation is found.

116


9.4 – Normally distributed data sets

Some data sets will have their data naturally categorize into intervals according to a very standard

system. The data sets are called ‘normally distributed data sets’, and occur frequently in both nature

and artificial processes. Normally distributed data sets must meet the following structure (a small

amount of error is allowed in artificial processes):

50% of the data is above the mean, and 50% is below

The data is symmetric when graphed

68.2% of the data falls within one standard deviation of the mean

95.4% of the data falls within two standard deviations of the mean

99.7% of the data falls within three standard deviations of the mean

The following graph encapsulates this structure:

Standardized test scores are typically normally distributed, and analysis can be done based on this fact

as in the following example.

117


Example 1

A standardized exam has the following characteristics:

mean: 57

standard deviation: 9

If 250 students took this exam, how many scored between 48 and 66?

Solution:

48 is one standard deviation below the mean (57 – 9 = 48)

66 is one standard deviation above the mean (57 + 9 = 66)

This question amounts to asking how many scores were within one standard deviation of the

mean. Since this is a standardized exam (meaning it is normally distributed), it is known that

68.2% of the scores will be within this range.

The following calculation generates the number:

250 (.682) = 170.5 (rounded to 171)

Hence, 171 students scored between 48 and 66.

118


Example 2

9000 widgets are produced in a manufacturing process with a weight that follows a normal

distribution according to the following characteristics:

mean: 3.71 g

standard deviation: 0.003 g

a) How many widgets weigh between 3.704 g and 3.716 g?

b) How many widgets weigh over 3.719 g?

Solution:

a) 3.704 = 3.71 – 2(0.003) = 2x s

3.716 = 3.71 + 2(0.003) = 2x s

This question amounts to asking how many widgets’ weights are within two standard

deviations of the mean. Since this is a normally distributed set, it is known that 95.4% of

the weights will be within this range.


9000 (.954) = 8586

Hence, 8586 of the widgets weigh between 3.704 g and 3.716 g.

b) 3.719 = 3.71 + 3(0.003) = 3x s

This question amounts to asking how many widgets’ weights are more than three

standard deviations above the mean. Since this is a normally distributed set, it is known

that 99.7% of the weights will be within three standard deviations, meaning only 0.3%

will not. Of these 0.3%, half (or 0.15%) will be higher than 3x s .


9000 (.0015) = 13.5 (rounded up to 14)

Hence, 14 of the widgets weigh more than 3.719 g.

119


9.5 – Controlled and capable processes

In manufacturing, it is important to have a process work in such a way that all widgets are produced

according to some specific control limits. While common cause variation prevents widgets from all being

produced according to exact specifications, processes can be refined so that the widgets fall within

certain parameters. When a process is refined to this point, it is called a controlled process.

Controlled process – when the measurements of all widgets produced fall within specified control limits

(note this definition is a bit of an oversimplification but is precise enough for the discussion in this text).

When an order is made by a customer, it will often come with tolerance limits (this represents the fact

that the customer understands common cause variation and has identified the limits of measurement

error they can tolerate). If a process is known to have control limits that fall within the tolerance limits,

it is called a capable process.

Capable process – a controlled process where the control limits are within a customer’s specified

tolerance limits.

Example 1

A controlled process produces widgets with the following characteristics (the target

measurement is 7.5 cm):

Upper control limit (UCL): 7.501 cm

Lower control limit (LCL): 7.499 cm

A customer order is entered with the following tolerance limit specification:

7.5 ± 0.003 cm

Is this process capable?

Solution:

The situation can be depicted by the following graphic:

UTL = 7.503: -----------------------------------------------------

UCL = 7.501: -----------------------------------------------------

LCL = 7.499: -----------------------------------------------------

LTL = 7.497: -----------------------------------------------------

Since the control limits are within the tolerance limits (in other words, all widgets will be within

the customer’s tolerance), this is considered a capable process. The order can be accepted and

filled.

120


Example 2

A controlled process produces widgets with the following characteristics (the target

weight is 110 kg):

Control limits: ± 5 kg


110 ± 3.5 kg


Solution:

The situation can be depicted by the following graphic:

UCL = 115: -----------------------------------------------------

UTL = 113.5: -----------------------------------------------------

LTL = 106.5: -----------------------------------------------------

LCL = 105: -----------------------------------------------------

Since the control limits are not within the tolerance limits, it cannot be guaranteed that all

widgets will be within the customer’s tolerance. Hence, this is not a capable process. The order

cannot be accepted.

121




1. In a manufacturing process, a new starter is being installed into a product during assembly.

Quality control personnel are tracking how many of the starters are failing each day over the

course of a month. For the following data set, build the frequency table and the histogram using

four 5-unit intervals. Start the lowest interval with 2 (so the first interval will be 2 – 6).

2. The following test scores were collected over a 5-year period. The grading scale was a straight

90-80-70-60 grading scale to indicate A, B, C, and D grades respectively (anything below a 60 is

considered a failing grade).

Create the frequency table and histogram using 4 intervals – one for each of the passing grades.

122



For questions 1 – 3, find the mean, median, and mode. After finding those measures, discuss whether

any of them are a better reflection of the central tendency for the data set.

1. 4, 7, 9, 9, 9, 14, 14, 16, 23, 31, 32

2. 1,000

700

1800

5300

2300

500

600

400

3. 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 6

4. Is the order of the data set important when calculating the mean? The median? The mode?


1. Manually calculate the range and the standard deviation of the following data set:

3, 5, 7, 7, 9

2. Use a statistical calculator to verify the standard deviation calculated in question 1.

3. Use a statistical calculator to calculate the range and standard deviation of the following data

sets:

a. 17, 18, 18, 22, 44, 45, 45, 45, 65, 71, 73, 78, 80, 80, 91, 92, 93, 99

b. 100, 300, 250, 175, 188, 337, 299, 171, 152, 201

4. Is the order of the data set important when calculating the range? The standard deviation?

123



1. 100,000 widgets are manufactured with a target measurement of 11.5 cm. After the widgets are

measured and the data is collected, the distribution characteristics are:

11.499 cm

0.001 cm

x

s

a. How many of the widgets measure between 11.498 cm and 11.5 cm?

b. How many of the widgets measure between 11.497 cm and 11.501 cm?

c. How many of the widgets measure between 11.496 and 11.499 cm?

d. How many of the widgets measure between 11.496 and 11.501 cm?

e. How many of the widgets measure at least 11.499 cm?

f. If an order is placed for 15,000 widgets, but only under the condition that at least 90%

of them are between 11.496 cm and 11.502 cm, can this order be accepted?

2. Professional baseball batting averages that fell between .280 and .289 were tracked over a

season. The distribution of a sampling of 10 players is:

0.280, 0.282, 0.282, 0.284, 0.285, 0.287, 0.287, 0.287, 0.289, 0.289

Assuming this data is normally distributed, use the sampling to determine the pertinent

characteristics of the data set, and then determine how many of the 73 players who had a

percentage between .280 and .289 were either above 0.28832 or below 0.28208.

124



1. A controlled process produces widgets with the following characteristics (the target

measurement is 33.2 mm):

Upper control limit (UCL): 33.204 cm

Lower control limit (LCL): 33.196. cm


33.2 ± 0.01 cm.


2. Another customer order for the widgets in question 1 is entered with the following tolerance

limit specification:

33.2 ± 0.001 cm.


3. A controlled process produces widgets with the following characteristics (the target weight is

100 kg):

Control limits: ±2.5 kg


220.5 ± 3.3 (units are measured in lb)


125


Chapter 10 – Basic Logic

10.1 – Logical expressions

Logical expressions revolve around situations that have only one of two ways to be resolved. These

situations are often referred to as Boolean scenarios, and are not entirely unrelated to the binary

discussion in the Number systems chapter.

Situations in technical settings often revolve around circuitry or computer programming. They also occur

in many other areas. A small sample of situations in technology are given below.

In circuitry the Boolean values occur in situations such as:

o Determining if the voltage is high or low

o Determining whether a given scenario occurs, and taking action based on that scenario

(for example, if a car is waiting in a left turn lane the logic controller will detect that and

give a green turn arrow when the time comes)

In computer programming the Boolean values occur and dictate the logic flow:

o The value may determine what module is executed (for example, if the employee is a

manager one routine might be executed, otherwise some other routine might be

executed)

o The value may determine how many times a module is executed (for example, as long as

there is data in a file the records are processed and as soon as there is no more data the

processing ends)

Evaluating logical expressions is very similar to evaluating algebraic expressions. Recall that in algebraic

expressions, we use the operations +, –, x, and .

Example 1 (Reviewing algebraic expressions)

If x = 3, y = 2, and z = 11, evaluate the algebraic expression: 2

2 3 4xy y z .

Solution:

2

2

2 3 4 when 3, 2, and 11

2(3)(2) 3(2) 4(11)

12 12 44

20

xy y z x y z

126


A similar process is used to evaluate logical expressions. However, the operations are no longer the four

familiar arithmetic operations. The logical operators are:

Boolean operator Notation Description

And ˄ Evaluates to true if both operands are true

Or ˅ Evaluates to true if either operand (or both) is true

Not ‘ (the single apostrophe) Inverts the value of the operand

These three operators are more formally defined by their truth tables. A truth table provides all of the

necessary information for a logical operator.

The ‘And’ logical operator

When logical variables are needed, p, q, and r are often used (as opposed to x, y, and z in algebra).

Additional, the two values used are typically 0 and 1, and these stand for the two states in the system

(the states might be 1 = high voltage and 0 = low voltage in an electronics context, 1 = true and 0 = false

in a programming context, etc.)

The ‘And’ logical operator is defined by the following truth table (where p and q are Boolean variables):

p q p ^ q

0 0 0

0 1 0

1 0 0

1 1 1

This allows for expressions to be evaluated.

Example 2

If p = 1 and q = 0, evaluate the logical expression: ( )p q p .

Solution:

In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in

the appropriate data, performing the operations, and obtaining the final value.

( ) when 1 and 0

(1 0) 1

0 1

0

p q p p q

127


The ‘Or’ logical operator

The ‘Or’ logical operator is defined by the following truth table:

p q p ˅ q

0 0 0

0 1 1

1 0 1

1 1 1

Example 3

If p = 1 and q = 0, evaluate the logical expression: ( )p q p .

Solution:

In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in

the appropriate data, performing the operations, and obtaining the final value.

( ) when 1 and 0

(1 0) 1

1 1

1

p q p p q

The ‘Not’ logical operator

The ‘Not’ logical operator is defined by the following truth table (since the ‘Not’ operator simply inverts

the operand, only one variable is needed):

p p’

0 1

1 0

128


10.2 – Logic truth tables

Now that the three fundamental operations have been defined, larger truth tables that encompass

more complicated situations can be formed. Again, there are many scenarios in technology where this

process is needed. A couple of examples would be:

In the stop light example above, the controller will have to sense many possible situations. For

example, if there is a car in one left turn lane and a car in the opposite turn lane, then provide

both of them with left turn arrow, if there is only a car in one turn lane then provide the cars on

that side of the street with both a green arrow and a green light, if there is not a car in either

turn lane, then just provide a green light, etc.

In the programming example above, it may be a case where one module is executed if the

employee is a manager and has a salary above a certain value or if they are an administrator,

another module might be executed if they are a programmer, analyst, or DBA, and yet a third

module is executed in all other cases.

In these situations, the logic device will have to evaluate the particular expression and execute the

appropriate commands.

Truth tables allow designers to lay out all of the possibilities so all scenarios can be expressed in one

structure. This allows for analysis and problem-solving of the given situation.

Truth tables for logical expressions will be tables with a certain number of rows and columns. They are

created by following this algorithm (an algorithm is a step-by-step step set of instructions for

accomplishing some task).

1. Create columns for each of the variables in the expression, in alphabetical order

2. Use additional columns to build up to the expression, one operation at a time

3. Create 2n rows, where n is the number of unique variables in the expression

4. Enter all possible combinations of initial conditions in the columns for the variables

5. For all of the remaining columns, fill them in one at the time applying the appropriate rules for

the current operation

The following example shows every step in this process:

Example 4

Create a truth table for the logical expression: ( ) ' 'p q p .

Solution:

Steps 1, 2, and 3 form the structure of the truth table

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

129


Step 4 provides all combinations of initial conditions (these are the first 2 columns)

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

0 0

0 1

1 0

1 1

Step 5 is done one column at a time, focusing on the current operands and operator.

Column 3:

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

0 0 0

0 1 1

1 0 1

1 1 1

Column 4:

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

0 0 0 1

0 1 1 0

1 0 1 0

1 1 1 0

Column 5:

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

0 0 0 1 1

0 1 1 0 1

1 0 1 0 0

1 1 1 0 0

And finally column 6 to complete the table:

p q (p ˅ q) (p ˅ q)’ p’ (p ˅ q)’ ˄ p’

130


0 0 0 1 1 1

0 1 1 0 1 0

1 0 1 0 0 0

1 1 1 0 0 0

Analyzing the results:

While all of the columns are an important part of the process, the most important column is the

last one. This provides the ultimate result for this expression given a set of initial conditions. For

example if p = 0 and q = 1, then the expression evaluates to 0. This can be seen by identifying

row 2 as the row associated with the initial conditions p = 0 and q = 1, and then looking to the

far right column to see the result of 0.

Typically the table is not replicated for each new column. The steps above are followed, but only one

table is needed as demonstrated in the next example.

Example 5

Create a truth table for the logical expression: ( ' ) ( ')p r q p .

Solution:

p q r p’ (p’ ˄ r) (q ˅ p’) (p’ ˄ r) ˅ (q ˅ p’)

0 0 0 1 0 1 1

0 0 1 1 1 1 1

0 1 0 1 0 1 1

0 1 1 1 1 1 1

1 0 0 0 0 0 0

1 0 1 0 0 0 0

1 1 0 0 0 1 1

1 1 1 0 0 1 1

Analyzing the results:

Since there are 3 variables, 23 = 8 rows are needed to handle all possible combinations of initial

conditions. Other than that, the table works in the same way as any other table.

131


10.3 – Schematics

Logical expressions can be expressed as in sections 10.1 and 10.2, or then can be expressed using

technical schematics. When schematics are used, the three fundamental operations are defined and

notated as in the following table:

Operator Image Image name

And

And gate

Two inputs, one output

OR

Or gate

Two inputs, one output

Not

Inverter

One input, one output

Building the actual schematic follows a similar process as choosing the columns in the corresponding

truth table. The following 2-step algorithm indicates what must be done:

1. Write each of the variables in the expression vertically along the left-hand side of the schematic

2. Build the schematic one operation at a time, thinking through and indicating the inputs to each

gate/inverter

132


Example 1

Create the logic schematic for the logical expression from the example above:

( ) ' 'p q p .

Solution:

Following the two steps above, the following schematic is created:

p

q

Schematics can also be used to generate a logical expression. It is important to read the schematics from

left to right to obtain the correct expression.

Example 2

Determine the logical expression associated with the following schematic:

Solution:

Reading the schematic from left to right, the following expression is determined:

( ) ' (( ) r)p q q r

133




1. If x, y, and z all have the value 1, evaluate the logical expression: ( ')x y z .

2. If x is 1 and y is 0, evaluate the expression: ( ') ( ( ' ))x y y x y


3. Construct the truth table for the expression in question 1.

4. Construct the truth table for the expression in question 2.


5. Construct the logic schematic for the expression in question 1.



7. Determine the logical expression associated with the following schematic:

p

q

134


8. Determine the logical expression associated with the following schematic:

135


Supplemental Chapter – Systems of Linear Equations

S.1 – Solving systems of 2 linear equations using a graphing tool

There are often times in technological settings when two linear equations (which would graph as lines)

are part of a system. When this happens, certain applications require knowing when the two lines

intersect, as this point of intersection would provide the (often unique) solution to the system. For

example, consider the following two lines:

12

3 2

4

y x

y x

When graphed on the TI-83, these two lines look as follows:

If the point of intersection needs to be determined, it can be done on the calculator using the intersect

functionality resulting in the following image:

The intersection point calculated by the TI-83 is (2.4, 5.2). What this means is that while there are an

infinite number of points that satisfy the line 3 2y x , and an infinite number of points that satisfy

the line 12

4y x , there is ONE AND ONLY ONE point that satisfies both of them – specifically the point

(2.4, 5.2) referenced above.

136


The following example illustrates the use of a TI-83 to solve a system in more detail.

Example 1 – Solve the following system of equations using a TI-83

1.5 1

2 3.2

y x

y x

Solution:

1. Enter the equations into the calculator (press the ‘Y=’ button to do this)

2. Graph the equations (press the ‘GRAPH’ button to do this)

3. Use the TI-83 to calculate the point of intersection by doing the following:

Press ‘2nd' and then ‘TRACE’

Press ‘5’ to choose ‘intersect’

Press ‘ENTER’ to choose the first graph

Press ‘ENTER’ to choose the second graph

Move the cursor close to the point of intersection and press ‘ENTER’

137


4. The calculator will now display the point of intersection:

Hence, these two lines intersect at the point (1.2, –0.8)

138


S.2 – Solving systems of 2 equations using the Addition-Subtraction method

When a graphing tool is not available, systems of equations can be solved algebraically. Two techniques

for doing this are discussed here:

1. Algebraically solving the system using the Addition-Subtraction method

2. Algebraically solving the system using the Substitution method

The Addition-Subtraction method is discussed in this section. The following algorithm is used when

implementing this method:

1. If necessary, algebraically manipulate both equations so they are in the form ax by c

2. If necessary, multiply one or both of the equations so that either the coefficients on the x terms

or the coefficients on the y terms are the same (without regard to the sign).

3. Add or subtract (as appropriate) one equation from the other in order to remove one of the

variables from the equations

4. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary

value

5. Substitute the solution from step 4 into either of the original equations to find the second

necessary value

6. Check the solution by substituting the values into both of the original equations

This algorithm is demonstrated in the following example which uses the equations graphed previously:

Example 1 – Solve the following system of equations using the Addition-Subtraction method:

12

3 2

4

y x

y x

Solution:

1. Algebraically manipulate the equations (step 1 from algorithm)

12

3 2

4

y x

y x

2. Multiply the second equation by 6 so the coefficients on x are the same (step 2 from algorithm)

(note that this step is technically not necessary for this problem since the y variables already

have a common coefficient, but is done to demonstrate the process)

3 2

6 3 24

y x

y x

139


3. Subtract equation 2 from equation 1 to eliminate the terms with x (step 3 from algorithm)

3 2

6 3 24

y x

y x

----------------------

5 26y

4. Solve 5 26y for y (step 4 from the algorithm)

5 26

5.2

y

y

5. Substitute the value 5.2 in for y in either of the original equations (step 5 from algorithm)

3 2

5.2 3 2

7.2 3

2.4

y x

x

x

x

6. Check (step 6 from algorithm)

3 2

5.2 3(2.4) 2

5.2 7.2 2

5.2 5.2

y x

12

12

4

5.2 (2.4) 4

5.2 1.2 4

5.2 5.2

y x

The solutions check out, and hence the solution is: 2.4 and 5.2x y which is another way to

notate the point (2.4, 5.2). Note that this is the same solution previously found using the graphing

technique.

140


In the next example, both equations must be multiplied by a value in step 2:

Example 2 – Solve the following system of equations using the Addition-Subtraction method:

2 7 10

3 4 12

x y

x y

Solution:

1. Algebraically manipulate the equations – this step is unnecessary as the equations are already in

the appropriate form.

2. Multiply the first equation by 3 and the second equation by 2 so the coefficients on x are the

same

6 21 30

6 8 24

x y

x y

3. Subtract equation 2 from equation 1 to eliminate the terms with x

6 21 30

6 8 24

x y

x y

----------------------

13 6y

4. Solve 13 6y for y (step 4 from the algorithm)

613

13 6y

y

5. Substitute the value 6

13 in for y in either of the original equations

613

4213

8813

4413

2 7 10

2 7( ) 10

2 10

2

x y

x

x

x

x

141



64413 13

88 4213 13

13013

2 7 10

2( ) 7( ) 10

10

10

10 10

x y

64413 13

132 2413 13

15613

3 4 12

3( ) 4( ) 12

12

12

12 12

x y

The solutions check out, and hence the solution is: 64413 13

and x y which is another way to notate

the point (4413 ,

613 ).

142


S.3 – Solving systems of 2 equations using the Substitution method

As discussed above, the two techniques for solving a system of equations are:

1. Algebraically solving the system using the Addition-Subtraction method

2. Algebraically solving the system using the Substitution method

In this section, the Substitution method is discussed. The following algorithm is used when

implementing this method:

1. If necessary, algebraically manipulate one of the equations so that one of the variables is

isolated

2. Substitute the corresponding expression for the isolated variable into the second equation

3. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary

value

4. Substitute the solution from step 3 into either of the original equations to find the second

necessary value

5. Check the solution by substituting the values into both equations

This algorithm is demonstrated in the following example which uses the system of equations solved in

the previous sections:

Example 1 – Solve the following system of equations using the Substitution method:

12

3 2

4

y x

y x

Solution:

1. Algebraically manipulate one of the equations to isolate a variable (step 1 from algorithm) – this

step is unnecessary since the equations already have the y variable isolated

2. Substitute the corresponding value in for y (step 2 from algorithm)

12

3 2

4 3 2

y x

x x

143


3. Solve 12

4 3 2x x for x (step 3 from the algorithm)

12

12

52

4 3 2

3 2 4

6

5 12

2.4

x x

x x

x

x

x

4. Substitute the value 2.4 in for x in either of the original equations (step 4 from algorithm)

3 2

3(2.4) 2

7.2 2

5.2

y x

y

y

y


3 2

5.2 3(2.4) 2

5.2 7.2 2

5.2 5.2

y x

12

12

4

5.2 (2.4) 4

5.2 1.2 4

5.2 5.2

y x

The solutions check out, and hence the solution is: 2.4 and 5.2x y which is another way to

notate the point (2.4, 5.2). Note that this is the same solution found using the graphing technique

and the addition-subtraction technique used previously.

144


In the next example, step 1 is required as neither equation is isolated for a variable. In this case, the

addition-subtraction method is typically the preferred method but any method can be used.

Example 2 – Solve the following system of equations using the Substitution method:

2 5 27

11 3 41

x y

x y

Solution:

1. Algebraically manipulate one of the equation to isolate a variable

5 272 2

2 5 27

2 5 27

x y

x y

x y

2. Substitute the corresponding value in for x into the other equation (step 2 from algorithm)

5 272 2

11 3 41

11( ) 3 41

x y

y y

3. Solve 5 272 2

11( ) 3 41y y for y

5 272 2

55 297 6 822 2 2 2

49 3792 2

37949

11( ) 3 41

49 379

y y

y y

y

y

y

4. Substitute the value 37949 in for y in either of the original equations

379 132349 49

1895 132349 49

57249

28649

2 5 27

2 5( )

2

2

x y

x

x

x

x

145



286 37949 49

572 189549 49

132349

2 5 27

2( ) 5( ) 27

27

27

27 27

x y

286 37949 49

3146 113749 49

200949

11 3 41

11( ) 3( ) 41

41

41

41 41

x y

The solutions check out, and hence the solution is: 286 37949 49

and x y which is another way to

notate the point 286 37949 49

( , ) .

146


S.4 – Applications associated with systems of equations

There are many technological situations where a system of equations is needed to find solutions. The

following provide examples of this situation.

Example 1

A chemist needs 50 mL of an 8% acid solution. She only has a 5% and a 13% solution on hand. How

much of each needs to be combined to obtain the desired 8% solution?

Solution:

1. The first step is to determine the necessary relationships in order to obtain the relevant system

of equations. Variables are used to symbolize the needed quantities:

= the mL of the 5% solution needed

= the mL of the 13% solution needed

Considering the problem, it can be determined that:

50 (since the amount of the two solutions must

A

B

A B add to 50mL)

0.05 0.13 0.08(50) (since the amount of acid will be constant)A B

2. Now that the system has been found, it is solved using any technique above. Here the Addition-

Subtraction technique is used.

50

0.05 0.13 0.08(50)

50

2.6 80 (this equation was multiplied by 20)

1.6 30 (the bottom equation was subtracted from the top equation)

18.75

18.75 50 (the value of B is subst

A B

A B

A B

A B

B

B

A

itued in to one of the equations)

31.25A

Since A represented the amount of the 5% solution, and B represented the amount of the 13%

solution, 31.25 mL of the 5% solution must be mixed with 18.75 mL of the 13% solution.

147


Example 2

If a concrete mix must have 5 times as much gravel as cement, and enough product must be mixed

to fill a volume of 15 m3, how much gravel and how much cement is needed to make this mix?

Solution:

1. Variables are used to symbolize the needed quantities:

3

3

3

C = m of cement needed

G = m of gravel needed

Considering the problem, it can be determined that:

15 (since the amount of the two quantities must add to 15m )

5

C G

G C

(since there is 5 times more gravel)

2. Now that the system has been found, it is solved using any technique. Here the substitution

method is used since the second equation is already solved for G.

15

5 15 (5C is substituted in place of G)

6 15

2.5

2.5 15 (The value of C determined above is substituted into one of the equations)

12.5

C G

C C

C

C

G

G

Since C represented the amount of cement and G represented the amount of gravel, 2.5 m3 of

cement is needed, and 12.5 m3 of gravel is needed.

148


S.5 – Systems of 3 linear equations

In certain instances there are three unknown quantities which need to be determined. In this case, three

equations are needed. While the process is more complicated than that for systems of two equations,

the actual steps use similar techniques. The algorithm is as follows:

1. If necessary, algebraically manipulate all equations so they are in the form ax by cz d .

2. Eliminate one of the variables from any pair of the three equations.

3. Eliminate the same variables from any other pair of the three equations.

4. Solve the resulting system of two linear equations using any of the three earlier techniques.

5. Substitute the two values obtained in step 4 to obtain the last unknown.

6. Check the solution by substituting the values into all three equations.

Example 1 – Solve the following system of equations:

10

2 2 5 4

3 2 5 20

x y z

x y z

x y z

Solution:

1. All equations are already in the form ax by cz d (step 1 from algorithm)

2. Eliminate y from the first two equations (step 2 from algorithm)

10

2 2 5 4

2 2 2 20

2 2 5 4

4 7 24

x y z

x y z

x y z

x y z

x z

149


3. Eliminate y from the first and third equations (step 3 from algorithm)

10

3 2 5 20

2 2 2 20

3 2 5 20

5 7 40

x y z

x y z

x y z

x y z

x z

4. Solve the system of two equations formed by combining the equations that resulted from the

previous two steps (step 4 from algorithm)

407

4 7 24

5 7 40

16

16

4(16) 7 24

64 7 24

7 40

x z

x z

x

x

z

z

z

z

5. Substitute the values to find y (step 5 from algorithm)

407

40 701127 7 7

72 707 7

27

10

16 10

x y z

y

y

y

y

150



4027 7

10

16 10

16 6 10

10 10

x y z

4027 7

20047 7

287

2 2 5 4

2(16) 2( ) 5( ) 4

32 4

4

4 4

x y z

4027 7

20047 7

336 20047 7 7

3 2 5 20

3(16) 2( ) 5( ) 20

48 20

20

20 20

x y z

The solutions check out, and hence the solution is: 4027 7

16, , and x y z .

151


Chapter S Homework problems

Section S.1 Homework

Use the Virtual TI-83 (or any graphing calculator) to find the point of intersection of the following lines.

1. 4 5

3 9

y x

y x

2. Note: For this problem you will need to manipulate the equation to isolate y, and you will need

to change the Window to:

Xmin = 70

Xmax = 80

Ymin = –30

Ymax = –40

2 168

2 128

y x

y x

3. Note: For this problem you will need to manipulate the equation to isolate y, and you will need

to change the Window back to the default values:

Xmin = –10

Xmax = 10

Ymin = –10

Ymax = 10

4 27.8 2

24 153.3 3

y x

y x

4. Enter the following two equations as written below, and graph them to try to determine the

point of intersection.

3 5

3 2

y x

y x

Can you explain what happened in this case?

152



Use the Addition-Subtraction method to solve the following system of equations (note that the first 3

problems are the same as the preceding section and will have the same answers).

1. 4 5

3 9

y x

y x

2. 2 168

2 128

y x

y x

3. 4 27.8 2

24 153.3 3

y x

y x

4. 3 11 172

7 4 57

a b

a b


Use the Substitution method to solve the following system of equations (note that the first 3 problems

are the same as the preceding sections and will have the same answers).

1. 4 5

3 9

y x

y x

2. 2 168

2 128

y x

y x

3. 4 27.8 2

24 153.3 3

y x

y x

4. 4 30 12

7 3 17

m n

m n

153


Section S.4 Homework – use any of the three methods when solving the following problems.

1. E10 is a common gasohol blend, as typically no modifications need to be made to a vehicle’s

engine in order to use this mixture. E10 means the mixture would have 10% ethanol mixed with

90% gasoline.

If 30 liters of E10 are needed, but only E12 and E5 are on hand (these are common mixtures in

other countries), how much of the E12 must be mixed with the E5 in order to obtain the 30 liters

of E10?

Note: E12 would contain 12% ethanol and E5 would contain 5% ethanol.

2. A rectangular yard has been fenced using a total of 344m of fencing. Find the dimensions of the

yard if the length is 14m more than the width.

3. A plane travels 1400 miles with the wind in 5 hours. It makes the return trip in 6 hours.

Assuming the wind speed is constant, find the speed of the wind and the speed of the plane.

4. A circuit has the following relationships between two currents 1 2 and I I (currents are measured

in mA):

1 2

1 2

17.6 8 11.2

18 4 22

I I

I I

Find 1 2 and I I .

154



Solve the following systems of equations.

1.

2 4 6 14

4 2 2 8

9 6 6 30

x y z

x y z

x y z

2.

3 2 20

2 6 24

3 4 4 28

x y

x y z

x y z

3.

3 2 4 14

10 5 4

5 4 5 2

x y z

x y z

x y z

155


Answers to Homework Problems

Chapter 1 Answers


1. 8/7

2. 8/11

3. –23/14

4. –6

5. 10/27

6. 2/7

7. –27/143

8. 34/12

9. –24/13

10. 1


Fill in the following chart.

Mixed number Improper Fraction Decimal

74/7 53/7 7.571

–1113/41 –464/41 –11.317

22/3 24/9 2.667

113 1582/14 113

–662/100 –662/100 –6.62

746/10 373/5 74.6

156




1. 25

2. 95

3. 13.5

4. 1369

5. 145

6. 6.09


In each of the equations, solve for the variable.

1. x = 39

2. x = 32

3. x = 10

4. x = -5/4

5. x = 2474/7 = 353.43


In each of the following, find the value of x which solves the proportion.

1. x = 11

2. x = –12/5

3. x = 11

4. x = 135

5. x = 5018300/21689 = 231.38

157



Fill in the following chart:

Scientific Notation Decimal Notation

9.21 x 108 921,000,000

8.7 x 10-5 0.000087

5 x 1014 500,000,000,000,000

5.09 x 1011 509,000,000,000

7.0 x 10-5 0.00007

3.14 x 10-8 0.0000000314

Calculate the following and express the answer in scientific notation.

1. 8 x 1012

2. 7.48 x 104

3. 9 x 1070

4. 1.632 x 10-4

5. 9.02 x 108

6. 1.33 x 1015

158


Chapter 2 Answers


1.

a. 461,200 mm

b. 461.20 m

c. 46.120 dam

d. 0.46120 km

2.

a. 1.7 x 1010 mW

b. 1.7 x 107 W

c. 17,000 kW

d. 0.017 GW

3.

a. 0.005 dL

b. 0.05 cL

c. 5 x 10-7 kL

d. 0.00005 daL


1.

a. 1.0 x 1015 µm2

b. 1.0 m2

c. 1.0x10-6 km2

2.

a. 321 cm3

b. 3.21 x 10-4 m3

c. 3.21 x 10-40 Tm3

159



1.

Degrees Fahrenheit Degrees Celcius

-48º -44.444º

111º 43.888º

37.76º 3.2º

80.6º 27º

2. -273 ºC and -459.4 ºF are absolute zero.


1. 27.255 liters

2. 3048 meters

3. 1.60276 pounds

4. 24.7105 acres

5. 170.8665 in3

160


Chapter 3 Answers


Fractional representation Decimal representation Percentage representation 1_ 10

0.1 10%

13 74

0.17568 17.568%

3.25 325%

5_ 100

0.05 5%

_23_ 10,000

0.0023 0.23%

1744 100

17.44 1,744%

71 100

0.71 71%

1733 10,000

0.1733 17.33%

__56__ 1,000,000

.000056 0.0056%


1. 21

2. 81.72

3. 0.0441

4. 68.182%

5. 400%

6. 0.646%

7. 90.625

8. 364.86486

9. 6

10. $31,364.49

161



Measurement Precision Greatest Possible Error

Relative Error Percent of Error

30 V 10 V 5 V 0.16667 V 16.667%

21.5 m 0.1 m 0.05 m 0.00233 m 0.233%

0.0003 g 0.0001 g 0.00005 g 0.16667 g 16.667%

105,000,000 mi 1,000,000 mi 500,000 mi 0.00476 mi 0.476%


Specification Tolerance Upper Limit Lower Limit Tolerance Interval

17 m 1 m 18 m 16 m 2 m

60.01 mL 0.005 mL 60.015 mL 60.005 mL 0.01 mL

1.0 oz .5 oz 1.5 oz 0.5 oz 1.0 oz

60.0001 W 0.0002 W 60.0003 W 59.9999 W 0.0004 W

2.5 cm 0.00015 cm 2.50015 cm 2.49985 cm .0003 cm

4 m 0.5 m 4.5 m 3.5 m 1 m

162


Chapter 4 Answers


1. 6(1000) + 1(100) + 9(10) + 2(1)

2. 1(22) + 0(21) + 1(20)

3. 1(28) + 0(27) + 1(26) + 0(25) + 1(24) + 1(23) + 0(22) + 0(21) + 1(20)

4. 7(162) + 2(161) + 9(160)

5. 10(162) + 14(161) + 3(160)

6. 1(165) + 3(164) + 12(163) + 2(162) + 13(161) + 0(160)


1. 5

2. 345

3. 91

4. 1010012

5. 10001000112

6. 100000000002


1. 2,787

2. 1,295,056

3. 2C816

4. 7A64A16

5. 4B16

6. 334B16

7. 101000112

8. 11001110011012

163


Chapter 5 Answers


1. 450 N

2. 48W

3. 18W

4. 18W

5. 183.2 cm3


1. 4.17A

2. 5.56V

3. 0.617Ω

4. 3.16A

5. 7.50 cm

164


Chapter 6 Answers


1.

a. The angle’s complement is 52⁰.

b. The angle’s supplement is 142⁰.

2.

a. The angle’s complement does not exist.

b. The angle’s supplement is 66⁰.

3. Two pairs of equal angles are W & Y and X & Z.

4. Three pairs of equal angles are E & B, B & G, and E & G. Three pairs of supplementary angles are

E & F, F & G, and G & H.


1.

Acute, Right, or Obtuse Scalene, Isosceles, or Equilateral

Acute Equilateral

Obtuse Isosceles

Right Scalene

2. A = 62⁰

3. x = 20.24846. The perimeter is 48.24846.

4. x = 20.19332. The perimeter is 47.69332.

5. The area is 93.5 m.

6. The area is 99.95693 in.

7. The area is 37,960.12 m.

8. The three sides are 83 ft, 83 ft, and 10 ft.

9. The area is 415.75 ft.

10. The area is 2.24 cm.

165



1. The area is 1,260 m2 and the perimeter is 144 m.

2. The width is 10.29412 cm.

3. The area is 592.5 mm2.

4. The area is 25,750 m2 and the perimeter is 746.072 m.

5. 39 gallons of paint are needed and the cost is $429.


1.

Circle radius Circle diameter Circle circumference Circle area

3.4 in 6.8 in 6.8 π 11.56 π

17 ft 34 ft 34 π 289 π

1.5 x 106 km 3 x 106 km 3 π x 106 km 1.125 π x 1012 km

52 π

104 π

104 cm 2704

π

15.0545 mm 30.1089 mm 30.1089 π mm 712 mm3

2. 137o

3. 92o

4. 186o

5. Arc s = 76o and the diameter of the circle is 3.56 in.

166


Chapter 7 Answers


1. sinA = 0.8, cosA = 0.6, tanA = 1.333

2. sinA = 0.5519, cosA = 0.8340, tanA = 0.6617

3. sinB = 0.7273, cosA = 0.6864, tanB = 1.0596


1. If A = 67°, find:

a. sinA 0.9205

b. cosA 0.3907

c. tanA 2.3559

2. If B = 23°, find:

a. sinA 0.3907

b. cosA 0.9205

c. tanA 0.4245

3. Yes there was a relationship between the two. This occurs because if a right triangle has an

angle of 23°, the other must be 67º (the values used in both questions 1 and 2). If you look at

this triangle more in depth, you will quickly see that the cosine of one angle is equal to the sine

of the other and vice versa.

4. 14.55°

5. 55.03°

6. 74.53°

7. The value you get back is ‘not a number.’ This occurs because two possible values for the

adjacent and hypotenuse lengths cannot exist to satisfy a value of 2.143 in a right triangle. Since

the length of the hypotenuse will always be greater than the length of one of the legs, and the

hypotenuse is in the denominator, the cos of an angle will always be less than 1.

8. cosA = 0.9494, tanA = 0.3309


1. x = 12.72 cm, y = 11.87 cm

2. x = 82.89 ft, y = 137.73 ft

3. A = 50.91°

4. A = 61.95°

5. A = 52.61°

6. Z = 19.8, phase angle = 45°

7. R = 15.76, XL = 12.31

8. phase angle = 44.9°, XL = 8.47

167


Chapter 8 Answers


1.

a. 2008

b. ~388

c. ~96

d. ~479

e. ~399

2.

a. 47%

b. $14,000

c. $41,300

d. This does support the claim. In the fourth quarter (the months leading to the holidays)

revenue was approximately $23,100 dollars which is appreciably higher than the other

quarters.

3.

a. ~95,000 bolts

b. ~500 bolts

c. ~4.74%

d. Yes. While 1995 produced a larger number of bolts not meeting specifications, the

average number of bolts produced, as well as bolts meeting specifications, has been

greater since 2000.

e. The chart does not specify the bolts produced in 1985. Either no data was collected and

recorded at that time, or the chart doesn’t provide the y-axis numbers necessary for

representing the bolts produced in 1985.

4.

a. ~7Ω

b. No, the y-axis (Power) scale is too large to represent the resistance values.

c. The current increases by 10 amps in each subsequent situation. The second chart.

d. From 700W to 2,800W => increases by ~2,100W.

e. From 700W to 700,000W => increases by ~699,300W.

168


5.

a. For each value of x, find the corresponding value of y:

i. Y = 8

ii. Y = -1

iii. Y = -4

iv. Y = 11

b. For each value of y, find the corresponding value of x:

i. X = -0.25

ii. X = 0.25

iii. X = 1

iv. X = -1

169


Chapter 9 Answers


1. Frequency table:

Histogram:

2. Frequency table:

0

2

4

6

8

10

12

2 - 6 7 - 11 12 - 16 17 - 21

Freq

uen

cy

Number of failures

Analyzing new starter product

170


Histogram:


1. Mean: 15.27

Median: 14

Mode: 9

All values seem to provide a reasonable measure of central tendency.

2. Mean: 1950

Median: 850

Mode: none

The median provides the most reasonable measurement of central tendency. The mean is not

terrible, but it is skewed somewhat as indicated by the fact that it is greater than all but 2 of the

data elements. The mode provides nothing meaningful.

3. The mean, median, and mode are all 3. All provide a reasonable measure of central tendency.

4. Order only matters when calculating the median.

0

100

200

300

400

500

600

700

800

900

60 - 69 70 - 79 80 - 89 90 - 99

Freq

uen

cy

Grades

Analyzing student grades

171



1. Range: 6, Standard deviation: 2.28

2. Range: 6, Standard deviation: 2.28

3.

a. Range: 82, Standard deviation: 28.48

b. Range: 237, Standard deviation: 76.01

4. The order does not matter when calculating the standard deviation, and only effects the range

to the extent that the greatest and least element need to be identified.


1.

a. 68,200 widgets

b. 95,400 widgets

c. 49,850 widgets

d. 97,550 widgets

e. 50,000 widgets

f. Yes, the order can be accepted because the statistics show that 99.7% are within the

requested range

2. Mean: 0.2852, Standard deviation: 0.00312

49 of the players are within the stated range


1. Yes because control limits are within tolerance

2. No, because control limits are not within tolerance

3. Yes because control limits are within tolerance

172


Chapter 10 Answers


1. 1

2. 1


3.

x y z (y ∧ z’) x ∨(y ∧ z’)

0 0 0 0 0

0 0 1 0 0

0 1 0 1 1

1 0 0 0 1

1 0 1 0 1

1 1 0 1 1

1 1 1 0 1

4.

x y (x ∧ y’) (x’ ∨ y) (y ∧ (x’ ∨ y)) (x ∧ y’) ∨ (y ∧ (x’ ∨ y))

0 0 0 1 0 0

0 1 0 1 1 1

1 0 1 0 0 1

1 1 0 1 1 1

173



5.

x

y

z


x

y


7. ((p ∧ q) ∧ q’)’

8. ((p ∧ q) ∧ ((q ∨ r) ∧ r))’

174


Supplemental Chapter Answers


1. 2 and 3x y

2. 74 and 20x y

3. 1.5 and 6.2x y

4. The two lines are parallel and hence there is no point of intersection


1. 2 and 3x y

2. 74 and 20x y

3. 1.5 and 6.2x y

4. 0.937 and 15.892a b


1. 2 and 3x y

2. 74 and 20x y

3. 1.5 and 6.2x y

4. 2.135 and 0.685m n


1. 21.4 L of E12 and 8.6 L of E10 are needed

2. Length = 93 m and width = 79 m

3. Plane speed = 256.7 mph and wind speed = 23.3 mph

4. 1 23mA and 8mAI I

175



1. 2, 1, and 1x y z

2. 28 128 3017 17 17

, , and x y z

3. 452 139 35023 23 23

, , and x y z

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