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8.5: Putting It Together- Roots and Rational Exponents


In the beginning of this module, we proposed that one of the most well known uses of roots and rational exponents is found in quadratic equations.

Quadratic equations, when graphed on the Cartesian coordinate plane, form the shape of a parabola. Parabolas are unique and interesting shapes because no matter how you stretch, shrink, or enlarge them, they are all identical to each other. What does that mean?

A parabola is the set of all points (left(x,y ight)) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.

This feature of parabolas is what makes them so useful in so many ways, and what makes them identical to each other – this relationship holds no matter how you stretch or shrink them. Ellipses and circles do not share this quality. You can stretch and shrink them and they are not identical to each other. This unique fact about parabolas has some interesting impacts in our real lives.

For example, did you know that stage lighting and car headlamps are often parabolic in shape? The reason for this stems from the shape of the paraboloid.

According to Wikipedia, a parabolic (or paraboloid or paraboloidal) reflector (or dish or mirror) is a reflective surface used to collect or project energy such as light, sound, or radio waves.

Parabolic reflectors are used to collect energy from a distant source (for example sound waves or incoming star light) and bring it to a common focal point, as in the image below:

Parallel rays coming in to a parabolic mirror are focused at a point F. The vertex is V, and the axis of symmetry passes through V and F. For off-axis reflectors (with just the part of the paraboloid between the points P1 and P3), the receiver is still placed at the focus of the paraboloid, but it does not cast a shadow onto the reflector.

Since the principles of reflection are reversible, parabolic reflectors can also be used to project energy of a source at its focus outward in a parallel beam, used in devices such as spotlights and car headlights.


Share All sharing options for: Let’s Expand the Playoff! Part 5 - Putting It Into Practice

Photo by David J. Griffin/Icon Sportswire/Corbis/Icon Sportswire via Getty Images

I’ve laid out my reasoning for why the Playoff should expand. But what would that even look like?

This past week the College Football Playoff Management Committee met to discuss, among other things, the possibility of expanding the Playoff. Those discussion apparently included 63 different models of expansion, including 6-, 8-, 10-, 12- and 16-team options, each with a variety of different scenarios. While no formal recommendation for change was reached, there is enough chatter among the the college football culture - and from important, influential college football characters as well - that an expanded Playoff is inevitable.

So, given that the CFP Management Committee apparently heard my plea and started talking about it, let’s take them at their word and explore the options that we are aware of. I have no idea what those 63 models looked like but I do know how to whip up a bracket with the numbers they’re working with. Using the 2020 season as our model, I’ll project a sample of each bracket size, starting with the smallest, to give you an idea of what that would look like. Also, because I’m a monster and have been doing mock brackets for years, I’ll tell you if any Missouri team from the past 20 years would have made it into the scenario given.

Ok! Let’s start with a 6-team Playoff!


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ISBN10: 0073512915 | ISBN13: 9780073512914

The estimated amount of time this product will be on the market is based on a number of factors, including faculty input to instructional design and the prior revision cycle and updates to academic research-which typically results in a revision cycle ranging from every two to four years for this product. Pricing subject to change at any time.

The estimated amount of time this product will be on the market is based on a number of factors, including faculty input to instructional design and the prior revision cycle and updates to academic research-which typically results in a revision cycle ranging from every two to four years for this product. Pricing subject to change at any time.

Program Details

Beginning & Intermediate Algebra with POWER Learning

Chapter 1 The Real Number System and Geometry

1.2 Exponents and Order of Operations

1.4 Sets of Numbers and Absolute Value

1.5 Addition and Subtraction of Real Numbers

1.6 Multiplication and Division of Real Numbers

1.7 Algebraic Expressions and Properties of Real Numbers

Chapter 2 Rules of Exponents

2.1A Basic Rules of Exponents: The Product Rule and Power Rules

2.1B Basic Rules of Exponents: Combining the Rules

2.2A Integer Exponents Bases: Real-Number Bases

2.2B Integer Exponents Bases: Variable Bases

Chapter 3 Linear Equations and Inequalities in One Variable

3.1 Solving Linear Equations Part I

3.2 Solving Linear Equations Part II

3.3 Applications of Linear Equations

3.4 Applications Involving Percents

3.5 Geometry Applications and Solving Formulas

3.6 Applications of Linear Equations to Proportions, Money Problems, and d = rt

3.7 Linear Inequalities in One Variable

3.8 Compound Inequalities in One Variable

Chapter 4 Linear Equations in Two Variables and Functions

4.1 Introduction to Linear Equations in Two Variables

4.2 Graphing by Plotting Points and Finding Intercepts

4.4 The Slope-Intercept Form of a Line

4.5 Writing an Equation of a Line

4.6 Introduction to Functions

Chapter 5 Solving Systems of Linear Equations

5.1 Solving Systems by Graphing

5.2 Solving Systems by the Substitution Method

5.3 Solving Systems by the Elimination Method

5.4 Applications of Systems of Two Equations

5.5 Solving Systems of Three Equations and Applications

5.6 Solving Systems of Linear Equations Using Matrices

6.1 Review of the Rules of Exponents

6.2 Addition and Subtraction of Polynomials

6.3 Multiplication of Polynomials

6.4 Division of Polynomials

Chapter 7 Factoring Polynomials

7.1 The Greatest Common Factor and Factoring by Grouping

7.2 Factoring Trinomials of the Form x 2 + bx + c

7.3 Factoring Trinomials of the Form ax 2 + bx + c ( a ≠1)

7.4 Factoring Special Trinomials and Binomials

7.5 Solving Quadratic Equations by Factoring

7.6 Applications of Quadratic Equations

Chapter 8 Rational Expressions

8.1 Rational Expressions and Functions

8.2 Multiplying and Dividing Rational Expressions

8.3 Finding the Least Common Denominator

8.4 Adding and Subtracting Rational Expressions

8.5 Simplifying Complex Fractions

8.6 Solving Rational Equations

8.7 Applications of Rational Equations

Chapter 9 More Equations and Inequalities

9.1 Absolute Value Equations

9.2 Absolute Value Inequalities

9.3 Linear and Compound Linear Inequalities in Two Variables

Chapter 10 Radicals and Rational Exponents

10.3 Simplifying Expressions Containing Square Roots

10.4 Simplifying Expressions Containing Higher Roots

10.5 Adding, Subtracting, and Multiplying Radicals

10.7 Solving Radical Equations

Chapter 11 Quadratic Equations

11.1 Review of Solving Equations by Factoring

11.2 The Square Root Property and Completing the Square

11.3 The Quadratic Formula

11.4 Equations in Quadratic Form

11.5 Formulas and Applications

Chapter 12 Functions and Their Graphs

12.1 Relations and Functions

12.2 Graphs of Functions and Transformations

12.3 Quadratic Functions and Their Graphs

12.4 Applications of Quadratic Functions and Graphing Other Parabolas

12.5 Quadratic and Rational Inequalities

12.6 The Algebra of Functions

Chapter 13 Exponential and Logarithmic Functions

13.2 Exponential Functions

13.3 Logarithmic Functions

13.4 Properties of Logarithms

13.5 Common and Natural Logarithms and Change of Base

13.6 Solving Exponential and Logarithmic Equations

Chapter 14 Conic Sections, Nonlinear Systems, and Nonlinear Inequalities

14.2 The Ellipse and the Hyperbola

14. PIAT Putting It All Together

14.4 Second-Degree Inequalities and Systems of Inequalities

Chapter 15 Sequences and Series (online only)

15.2 Arithmetic Sequences and Series

15.3 Geometric Sequences and Series


Table of specific heat capacities

The following table of specific heat capacities gives the volumetric heat capacity, as well as the specific heat capacity of some substances and engineering materials, and (when applicable) the molar heat capacity.

Generally, the most constant parameter is notably the volumetric heat capacity (at least for solids), which is notably around the value of 3 megajoule per cubic meter per kelvin: [1]

Note that the especially high molar values, as for paraffin, gasoline, water and ammonia, result from calculating specific heats in terms of moles of molecules. If specific heat is expressed per mole of atoms for these substances, none of the constant-volume values exceed, to any large extent, the theoretical Dulong–Petit limit of 25 J⋅mol −1 ⋅K −1 = 3 R per mole of atoms (see the last column of this table). Paraffin, for example, has very large molecules and thus a high heat capacity per mole, but as a substance it does not have remarkable heat capacity in terms of volume, mass, or atom-mol (which is just 1.41 R per mole of atoms, or less than half of most solids, in terms of heat capacity per atom).

In the last column, major departures of solids at standard temperatures from the Dulong–Petit law value of 3 R, are usually due to low atomic weight plus high bond strength (as in diamond) causing some vibration modes to have too much energy to be available to store thermal energy at the measured temperature. For gases, departure from 3 R per mole of atoms in this table is generally due to two factors: (1) failure of the higher quantum-energy-spaced vibration modes in gas molecules to be excited at room temperature, and (2) loss of potential energy degree of freedom for small gas molecules, simply because most of their atoms are not bonded maximally in space to other atoms, as happens in many solids.

A Assuming an altitude of 194 metres above mean sea level (the worldwide median altitude of human habitation), an indoor temperature of 23 °C, a dewpoint of 9 °C (40.85% relative humidity), and 760 mmHg sea level–corrected barometric pressure (molar water vapor content = 1.16%).

B Calculated values
*Derived data by calculation. This is for water-rich tissues such as brain. The whole-body average figure for mammals is approximately 2.9 J⋅cm −3 ⋅K −1 [11]


Rules of Logarithms

Study the description of each rule to get an intuitive understanding of it.

Description of Each Logarithm Rule

Rule1: Product Rule

The logarithm of the product of numbers is the sum of logarithms of individual numbers.

Rule 2: Quotient Rule

The logarithm of the quotient of numbers is the difference of the logarithm of individual numbers.

Rule 3: Power Rule

The logarithm of an exponential number is the exponent times the logarithm of the base.

Rule 4: Zero Rule

The logarithm of 1 with b > 0 but b e 1 equals zero.

Rule 5: Identity Rule

The logarithm of a number that is equal to its base is just 1 .

Rule 6: Log of Exponent Rule

The logarithm of an exponential number where its base is the same as the base of the log equals the exponent.

Rule 7: Exponent of Log Rule

Raising the logarithm of a number to its base equals the number.

Examples of How to Combine or Condense Logarithms

Example 1: Combine or condense the following log expressions into a single logarithm:

This is the Product Rule in reverse because they are the sum of log expressions. That means we can convert those addition operations (plus symbols) outside into multiplication inside.

Since we have “condensed” or “compressed” three logarithmic expressions into one log expression, then that should be our final answer.

Example 2: Combine or condense the following log expressions into a single logarithm:

The difference between logarithmic expressions implies the Quotient Rule. I can put together that variable x and constant 2 inside a single parenthesis using division operation.

Example 3: Combine or condense the following log expressions into a single logarithm:

Start by applying Rule 2 (Power Rule) in reverse to take care of the constants or numbers on the left of the logs. Remember that Power Rule brings down the exponent, so the opposite direction is to put it up.

The next step is to use the Product and Quotient rules from left to right. This is how it looks when you solve it.

Example 4: Combine or condense the following log expressions into a single logarithm:

I can apply the reverse of Power rule to place the exponents on variable x for the two expressions and leave the third one for now because it is already fine. Next, utilize the Product Rule to deal with the plus symbol followed by the Quotient Rule to address the subtraction part.

In this problem, watch out for the opportunity where you will multiply and divide exponential expressions. Just a reminder, you add the exponents during multiplication and subtract during division.

Example 5: Combine or condense the following log expressions into a single logarithm:

I suggest that you don’t skip any steps. Unnecessary errors can be prevented by being careful and methodical in every step. Check and recheck your work to make sure that you don’t miss any important opportunity to simplify the expressions further such as combining exponential expressions with the same base.

So for this one, start with the first log expression by applying the Power Rule to address that coefficient of large <1 over 2>. Next, think of the power large <1 over 2>as a square root operation. The square root can definitely simplify the perfect square 81 and the > because it has an even power.

Example 6: Combine or condense the following log expressions into a single logarithm:

The steps involved are very similar to previous problems but there’s a “trick” that you need to pay attention to. This is an interesting problem because of the constant 3 . We have to rewrite 3 in the logarithmic form such that it has a base of 4 . To construct it, use Rule 5 (Identity Rule) in reverse because it makes sense that 3 = left( <<4^3>> ight) .


Multiple Choice Questions
Question. 1 In 2 n , n is known as
(a) base (b) constant
(c) exponent (d) variable
Solution.
(c) We know that an is called the nth power of a and is also read as a raised to the power n.
The rational number a is called the base and n is called the exponent (power or index). In the same way in 2 n ,n is known as exponent.

Question. 2 For a fixed base, if the exponent decreases by 1, the number becomes
(a) one-tenth of the previous number
(b) ten times of the previous number
(c) hundredth of the previous number
(d) hundred times of the previous number
Solution.
(a) For a fixed base, if the exponent decreases by 1, the number becomes one-tenth of the previous number.

Question. 3

Solution.

Question. 4 The value of 1/4 -2 is
Solution.

Question. 5 The value of 3 5 ÷ 3 -6 is
(a) 3 5 (b) 3 -6 (c) 3 11 (d) 3 -11
Solution.

Question. 6

Solution.

Question. 7

Solution.

Question. 8 The multiplicative inverse of 10 -100 is
(a) 10 (b) 100 (c) 10 100 (d)10 -100
Solution.

Question.9 The value of (-2) 2 x 3-1 is
(a) 32 (b) 64 (c) -32 (d) -64
Solution.

Question.10

Solution.

Question. 11

Solution.

Question. 12 If x be any non-zero integer and w, n be negative integers, then x m x x n is equal to
(a) x m (b)x (m+n) (c) x n (d) x (m-n)
Solution.

Question. 13 If y be any non-zero integer, then y 0 is equal to
(a) 1 (b) 0 (c) – 1 (d) not defined
Solution.

Question.14 If x be any non-zero integer, then X -1 is equal to
(a) x (b) 1/x (c) – x (d) -1/x
Solution.

Question. 15

Solution.

Question. 16

Solution.

Question. 17

Solution.

Question. 18

Solution.

Question. 19

Solution.

Question. 20

Solution.

Question. 21

Solution.

Question. 22

Solution.

Question. 23

Solution.

Question.24 The standard form for 0.000064 is
(a) 64 x 10 4 (b) 64 x 10 -4 (c) 6.4 x 10 5 (d) 6.4 x 10 -5
Solution.
(d) Given, 0.000064 = 0. 64 x 10 -4 =6.4 x 10 -5
Hence, standard form of 0.000064 is 6.4 x 10 -5 .

Question. 25 The standard form for 234000000 is
(a) 2.34 x 10 8 (b) 0.234 x 10 9
(c) 2.34 x 10 -8 (d) 0.234 x 10 – 9
Solution.
(a) Given, 234000000 = 234 x 10 6 = 2.34 x 10 +6 = 2.34 x 10 8
Hence, standard form of 234000000 is 2.34 x 10 8 .

Question.26 The usual form for 2.03 x 10 -5 is
(a) 0.203 (b) 0.00203 (c) 203000 (d) 0.0000203
Solution.

Question. 27

Solution.

Question. 28

Solution.

Question. 29

Solution.

Question. 30

Solution.

Question. 31

Solution.

Question. 32

Solution.

Question. 33

Solution.

Fill in the Blanks
In questions 34 to 65, fill in the blanks to make the statements true.

Question. 34 The multiplicative inverse of 10 10 is_________
Solution.

Question.35 a 3 x a -10 = _________
Solution.

Question.36 5 0 = _________
Solution.

Question.37 5 5 x 5 -5 = _________
Solution.

Question.38

Solution.

Question. 39 The expression for 8 -2 as a power with the base 2 is_________
Solution.

Question. 40 Very small numbers can be expressed in standard form by using_________
exponents
Solution.
Very small numbers can be expressed in standard form by using negative exponents, i.e. 0.000023 = 2.3 x 10 -3

Question. 41 Very large numbers can be expressed in standard form by using
exponents.
Solution.
Very large numbers can be expressed in standard form by using positive exponents,
i.e. 23000 = 23 x 10 3 =2.3 x 10 3 x 10 1 =2.3 x 10 4

Question. 42 By multiplying (10) 5 by (10) -10 , we get
Solution.

Question.43

Solution.

Question.44

Solution.

Question.45

Solution.

Question.46

Solution.

Question.47

Solution.

Question.48

Solution.

Question.49

Solution.

Question.50

Solution.

Question.51

Solution.

Question.52

Solution.

Question.53 The value of 3 x 10 -7 is equal to_______
Solution
Given, 3 x 10 -7 = 3.0 x 10 -7
Now, placing decimal seven place towards left of original position, we get 0.0000003. Hence, the value of 3 x 10 -7 is equal to 0.0000003.

Question.54 To add the numbers given in standard form, we first convert them into number with_______exponents.
Solution.
To add the numbers given in standard form, we first convert them into numbers with equal exponents.
e.g. 2.46 x 10 6 + 24.6 x 105 = 2.46 x 10 5 + 2.46 x 10 6 = 4.92 x 10 6

Question.55 The standard form for 32500000000 is_______.
Solution.
For standard form, 32500000000 = 3250 x 10 2 x 10 2 x 10 3
= 3250 x 10 7 = 3.250 x 10 10 or 3.25 x 10 10
Hence, the standard form for 32500000000 is 3.25 x 10 10 .

Question. 56 The standard form for 0.000000008 is_______.
Solution.
For standard form, 0.000000008 = 0.8 x 10 -8 = 8 x 10 -9 =8.0 x 10 -9
Hence, the standard form for 0.000000008 is 8.0 x 10-9

Question.57 The usual form for 2.3 x 10 -10 is_______.
Solution. For usual form, 2.3 x 10 -10 = 0.23 x 10 -11
= 0.00000000023
Hence, the usual form for 2.3 x 10 -10 is 0.00000000023.

Question. 58 On dividing 8 5 by_______. we get 8.
Solution.

Question. 59

Solution.

Question. 60

Solution.

Question.61

Solution.

Question.62

Solution.

Question.63

Solution.

Question.64

Solution.

Question.65

Solution.

True / False
In questions 66 to 90, state whether the given statements are True or False.

Question.66

Solution.

Question.67

Solution.

Question.68

Solution.

Question.69 24.58 = 2 x 10 + 4 x 1+5 x 10 + 8 x 100
Solution. False
R H S = 2 x 10+ 4 x 1+ 5 x 10+ 8 x 100=20+ 4 + 50+ 800=874 L H S ≠ R H S

Question.70 329.25 = 3 x 10 2 + 2 x 10 1 + 9 x 10 0 + 2 x 10 -1 + 5 x 10 -2
Solution.

Question.71

Solution.

Question.72

Solution.

Question.73

Solution.

Question. 74 5 0 = 5
Solution.

Question. 75 (-2) 0 = 2
Solution.

Question.76

Solution.

Question. 77 (-6)° = – 1
Solution.

Question. 78

Solution.

Question. 79

Solution.

Question. 80

Solution.

Question. 81

Solution.

Question. 82

Solution.

Question. 83

Solution.

Question. 84

Solution.

Question.85 The standard form for 0.000037 is 3.7 x 10 -5
Solution. True
For standard form, 0.000037 = 0.37 x 10 -4 = 3.7 x 10 -5

Question. 86 The standard form for 203000 is 2.03 x 105.
Solution. True
For standard form, 203000 = 203 x 10 x 10 x 10 = 203 x 10 3
= 2.03 x 10 2 x 10 3 = 2.03 x 10 5

Question. 87 The usual form for 2 x 10 -2 is not equal to 0.02.
Solution.

Question. 88 The value of 5 -2 is equal to 25.
Solution. False

Question. 89 Large numbers can be expressed in the standard form by using positive exponents.
Solution.True
e.g. 2360000 = 236 x 10 x 10 x 10 x 10= 236 x 10 4
‘ = 2.36 x 10 4 x 10 2 =2.36 x 10 6

Question. 90

Solution.

Question. 91 Solve the following,

Solution.

Question. 92 Express 3 -5 x 3 -4 as a power of 3 with positive exponent.
Solution.

Question. 93 Express 16 -2 as a power with the base 2.
Solution.

Question. 94

Solution.

Question. 95

Solution.

Question. 96 Express as a power of a rational number with negative exponent.

Solution.

Question. 97 Find the product of the cube of (-2) and the square of (+4).
Solution.

Question.98 Simplify

Solution.

Question. 99 Find the value of x, so that

Solution.


Question. 100 Divide 293 by 1000000 and express the result in standard form.
Solution.

Question. 101

Solution.

Q. 102 By what number should we multiply (-29)°, so that the product becomes (+29)°.
Solution.

Question. 103 By what number should (-15) -1 be divided so that quotient may be equal to (-15) -1 ?
Solution.

Question.104 Find the multiplicative inverse of (-7) 2 ÷ (90) -1
Solution.

Question.105

Solution.

Question.106 Write 390000000 in the standard form.
Solution.

Question. 107 Write 0.000005678 in the standard form.
Solution.
For standard form, 0.000005678 = 0.5678 x 10 -5 = 5.678 x 10 -5 x 10 -1 = 5.678 x 10 -6 Hence, 5.678 x 10 -6 is the standard form of 0.000005678.

Question.108 Express the product of 3.2 x 10 6 and 4.1 x 10 1 in the standard form.
Solution.

Question.109

Solution.

Question. 110 Some migratory birds travel as much as 15000 km to escape the extreme climatic conditions at home. Write the distance in metres using scientific notation.
Solution.

Question. 111 Pluto is 5913000000 m from the Sun. Express this in the standard form.
Solution.

Question. 112 Special balances can weigh something as 0.00000001 gram. Express this number in the standard form.
Solution.

Question. 113 A sugar factory has annual sales of 3 billion 720 million kilograms of sugar. Express this number in the standard form.
Solution.

Question. 114 The number of red blood cells per cubic millimetre of blood is approximately mm 3 )
Solution. The average body contain 5 L of blood.
Also, the number of red blood cells per cubic millimetre of blood is approximately 5.5 million.
Blood contained by body = 5 L = 5 x 100000 mm 3
Red blood cells = 5 x 100000 mm 3
Blood = 5.5 x 1000000 x 5 x 100000= 55 x 5 x 10 5 + 5
= 275 x 10 10 = 2.75 x 10 10 x 10 2 = 2.75 x 10 12

Question. 115 Express each of the following in standard form:

Solution.


Question.116

Solution.

Question.117

Solution.

In questions 118 and 119, find the value of n.
Question.118

Solution.

Question.119

Solution.

Question.120

Solution.

Question.121

Solution.

Question.122

Solution.

Question.123 A new born bear weights 4 kg. How many kilograms might a five year old bear weight if its weight increases by the power of 2 in 5 yr?
Solution.
Weight of new born bear = 4 kg
Weight increases by the power of 2 in 5 yr.
Weight of bear in 5 yr = (4) 2 = 16 kg

Question.124 The cell of a bacteria doubles in every 30 min. A scientist begins with a single cell. How many cells will be thereafter (a) 12 h (b) 24 h ?
Solution.

Question.125 Planet A is at a distance of 9.35 x 10 6 km from Earth and planet B is 6.27 x 107 km from Earth. Which planet is nearer to Earth?
Solution.
Distance between planet A and Earth = 9.35 x 10 6 km Distance between planet B and Earth = 6.27 x 10 7 km
For finding difference between above two distances, we have to change both in same exponent of 10, i.e. 9.35 x.10 6 = 0.935 x 10 7 , clearly 6.27 x 10 7 is greater.
So, planet A is nearer to Earth.

Question.126 The cells of a bacteria double itself every hour. How many cells will be there after 8 h, if initially we start with 1 cell. Express the answer in powers.
Solution.

Question. 127 An insect is on the 0 point of a number line, hopping towards 1. She covers half the distance from her current location to 1 with each hop.
So, she will be at 1/2 after one hop, 3/4 after two hops and so on.

(a) Make a table showing the insect’s Location for the first 10 hops.
(b) Where will the insect be after n hops?
(c) Will the insect ever get to 1? Explain.
Solution.
(a) On the basis of given information in the question, we can arrange the following table which shows the insect’s location for the first 10 hops.

Question. 128 Predicting the ones digit, copy and complete this table and answer the questions that follow.

Solution.
(a) On the basis of given pattern in 1 x and 2 x , we can make more patterns for 3 x 4 x , 5 x ,6 x , 7 x , 8 x , 9 x , 10 x .
Thus, we have following table which shows all details about the patterns.

Question. 129 Astronomy The table shows the mass of the planets, the Sun and the Moon in our solar system.

Solution.

Question. 130 Investigating Solar System The table shows the average distance from each planet in our solar system to the Sun.

Solution.

Question. 131 This table shows the mass of one atom for five chemical elements.
Use it to answer the question given.

(a) Which is the heaviest element?
(b) Which element is lighter, Silver or Titanium?
(c) List all the five elements in order from lightest to heaviest.
Solution.

Question. 132 The planet Uranus is approximately 2,896,819,200,000 metres away from the Sun. What is this distance in standard form?
Solution.
Distance between the planet Uranus and the Sun is 2896819200000 m.
Standard form of 2896819200000 = 28968192 x 10 x 10 x 10 x 10 x 10
= 28968192 x 10 5 = 2.8968192 x 10 12 m

Question. 133 An inch is approximately equal to 0.02543 metres. Write this distance in standard form.
Solution. Standard form of 0.02543 m = 0.2543 x 10 -1 m = 2.543 x 10 -2 m Hence,’ standard form of 0.025434s 2.543 x 10 -2 m.

Question.134 The volume of the Earth is approximately 7.67 x 10 -7 times the volume
of the Sun. Express this figure in usual form.
Solution.

Question.135 An electron’s mass is approximately 9.1093826 x 10 -31 kilograms. What is its mass in grams?
Solution.

Question. 136 At the end of the 20th century, the world population was approximately 6.1 x 10 9 people. Express this population in usual form. How would you say this number in words?
Solution.
Given, at the end of the 20th century, the world population was 6,1 x 10 9 (approx). People population in usual form = 6.1 x 10 9 = 6100000000 Hence, population in usual form was six thousand one hundred million.

Question.137 While studying her family’s history, Shikha discovers records of ancestors 12 generations back. She wonders how many ancestors she had in the past 12 generations. She starts to make a diagram to help her figure this out. The diagram soon becomes very complex

Solution.
(a) On the basis of given diagram, we can make a table that shows the number of ancestors in each of the 12 generations.

Question. 138 About 230 billion litres of water flows through a river each day, how many litres of water flows through that river in a week? How many litres of water flows through the river in an year? Write your answer in standard notation.
Solution.

Question. 139 A half-life is the amount of time that it takes for a radioactive substance to decay one-half of its original quantity.
Suppose radioactive decay causes 300 grams of a substance to decrease 300 x 2 -3 grams after 3 half-lives. Evaluate 300 x 2 -3 to determine how many grams of the substance is left.
Explain why the expression 300 x 2 -n can be used to find the amount of the substance that remains after n half-lives.
Solution.

Question. 140 Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression 3 -t .
(a) What fraction of the substance remains after 7 half-lives?
(b) After how many half-lives will the fraction be 1/243 of the original?
Solution.

Question. 141 One fermi is equal to 10 -15 metre. The radius of a proton is 1.3 fermi. Write the radius of a proton (in metres) in standard form.
Solution. The radius of a proton is 1.3 fermi.
One fermi is equal to 10 -15 m.
So, the radius of the proton is 1.3 x 10 -15 m.
Hence, standard form of radius of the proton is 1.3 x 10 -15 m.

Question. 142 The paper clip below has the indicated length. What is the length in Standard form.

Solution.

Length of the paper clip = 0.05 m
In standard form, 0.05 m = 0.5 x 10 -1 = 5.0 x 10 -2 m
Hence, the length of the paper clip in standard form is 5.0 x 10 -2 m

Question.143 Use the properties of exponents to verify that each statement is true.

Solution.

Question. 144 Fill in the blanks.

Solution.

Question. 145 There are 86400 sec in a day. How many days long is a second? Express your answer in scientific notation.
Solution. Total seconds in a day = 86400
So, a second is long as 1/86400 = 0.000011574
Scientific notation of 0.000011574= 1.1574 x 10 -5 days

Question. 146 The given table shows the crop production of a state in the year 2008 and 2009. Observe the table given below and answer the given questions.

(a) For which crop(s) did the production decrease?
(b) Write the production of all the crops in 2009 in their standard form.
(c) Assuming the same decrease in rice production each year as in 2009, how many acres will be harvested in 2015? Write in standard form.
Solution.

Question. 147 Stretching Machine
Suppose you have a stretching machine which could stretch almost anything, e.g. If you put a 5 m stick into a (x 4) stretching machine (as shown below), you get a 20 m stick.
Now, if you put 10 cm carrot into a (x 4) machine, how long will it be when it comes out?

Solution.
According to the question, if we put a 5 m stick into a (x 4) stretching machine, then machine produces 20 m stick.
Similarly, if we put 10 cm carrot into a (x 4) stretching machine, then machine produce 10 x 4= 40 cm stick.

Question. 148 Two machines can be hooked together. When something is sent through this hook up, the output from the first machine becomes the input for the second.

Solution.

Question. 149 Repeater Machine
Similarly, repeater machine is a hypothetical machine which automatically enlarges items several times, e.g. Sending a piece of wire through a (x 2 4 ) machine is the same as putting it through a (x 2) machine four times. ‘
So, if you send a 3 cm piece of wire thorugh a (x 2)4 machine, its length becomes 3 x 2 x 2 x 2 x 2 = 48 cm. It can also be written that a base (2) machine is being applied 4 times.

What will be the new length of a 4 cm strip inserted in the machine?
Solution.
According to the question, if we put a 3 cm piece of wire through a (x 2 4 ) machine, its length becomes 3 x 2 x 2 x 2 x 2 = 48 cm.
Similarly, 4 cm long strip becomes 4 x 2 x 2 x 2 x 2 = 64 cm.

Question. 150 For the following repeater machines, how many times the base machine is applied and how much the total stretch is?

Solution.
In machine (a), (x 100 2 ) = 10000stretch. Since, it is two times the base machine.
In machine (b), (x 7 5 ) = 16807 stretch.
Since, it is fair times the base machine.
In machine (c), (x 5 7 ) = 78125stretch.
Since, it is 7 times the base machine.

Question. 151 Find three repeater machines that will do the same work as a (x 64) machine. Draw them, or describe them using exponents.’
Solution.
We know that, the possible factors of 64 are 2, 4, 8. :
If 2 6 =64, 4 3 =64 and 8 2 =64
Hence, three repeater machines that would work as a (x 64) will be (x 2 6 ), (x 4 3 ) and (x 8 2 ). The diagram of (x 2 6 ), (x 4 3 )and (x 8 2 )is given below.

Question. 152 What will the following machine do to a 2 cm long piece of chalk?

Solution.
The machine produce x 1 100 =1
So, if we insert 2 cm long piece of chalk in that machine, the piece of chalk remains same.

Question. 153 In a repeater machine with 0 as an exponent, the base machine is applied 0 times.
(a) What do these machines do to a piece of chalk?

(b) What do you think the value of 6° is?
You have seen that a hookup of repeater machines with the same base can be replaced by a single repeater machine. Similarly, when you multiply exponential expressions with the same base, you can replace them with a single expression.
Asif Raza thought about how he could rewrite the expression 220 x 2 5 .

Asif Raza’s idea is one of the product laws of exponents, which can be expressed like this
Multiplying Expressions with the Same Base ab x ac = ab+ c
Actually, this law can be used with more than two expressions. As long as the bases are the same, to find the product you can add the exponents and use the same base.
Solution.

Question. 154 Shrinking Machine In a shrinking machine, a piece of stick is compressed to reduce its length. If 9 cm long sandwich is put into the shrinking machine below, how long will it be when it emerges?

Solution.
According to the question, in a shrinking machine, a piece of stick is compressed to reduce its length. If 9 cm long sandwich is put into the shrinking machine, then the length
of sandwich will be 9 x 1/ 3 -1 = 9 x 3 = 27 cm.

Question. 155 What happens when 1 cm worms are sent through these hook-ups?

Solution.

Question. 156 Sanchay put a 1 cm stick of gum through a (1 x 3 -2 ) machine. How
long was the stick when it came out?
Solution.

Question. 157 Ajay had a 1 cm piece of gum. He put it through repeater machine
given below and it came out 1/100000 cm long. What is the missing
value?

Solution.

Question. 158 Find a single machine that will do the same job as the given hook-up.


Solution.

Question. 159 Find a single repeater machine that will do the same work as each hook-up.


Solution.

Question. 160 For each hook-up, determine whether there is a single repeater
machine that will do the same work. If so, describe or draw it.

Solution.

Question. 161 Shikha has an order from a golf course designer to put palm trees through a (x 2 3 ) machine and then through a (x 3 3 ) machine. She thinks that she can do the job with a single repeater machine. What single repeater machine she should use?

Solution.
Sol. The work done by hook-up machine is equal to 2 x 2 x 2 x 3 x 3 x 3 = 216 = 6 3 So, she should use (x 6 3 ) single machine for the purpose.

Question. 162 Neha needs to stretch some sticks to 25 2 times of their original lengths, but her (x 25) machine is broken. Find a hook-up of two repeater machines that will do the same work as a (x 25 2 ) machine. To get started, think about the hook-up you could use to replace the (x 25) machine.

Solution.
Work done by single machine (x 25 2 ) = 25 x 25 = 625 or 5 x 5 x 5 x 5 or 52 x 52
Hence, (x 5 2 ) and (x 5 2 ) hook-up machine can replace the (x 25) machine.

Question.163 Supply the missing information for each diagram.

solution.

Question. 164 If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use (x 1) machines.

solution.
(a) Single machine work = 100
Hook-up machine of prime base number that do the same work down by x 100
= 2 2 x 5 2
=4吕
= 100
(b) x 99 = 3 2 x 111 hook-up machine.
(c) x 37 machine cannot do the same work.
(d) x 1111 = 101 x 11 hook-up machine.

Question. 165 Find two repeater machines that will do the same work as a (x 81) machine.
Solution. Two repeater machines that do the same work as (x 81) are (x 3 4 ) and (x 9 2 ).
Since, factor of 81 are.3 and 9.

Question. 166

Solution.

Question. 167 Find three machines that can be replaced with hook-up of (x 5) machines.
Solution.
Since, 5 2 = 25, 5 3 = 125, 5 4 = 625
Hence, (x 5 2 ), (x 5 3 )and (x 5 4 ) machine can replace (x 5) hook-up machine.

Question. 168 The left column of the chart lists is the length of input pieces of ribbon. Stretching machines are listed across the top.
The other entries are the outputs for sending the input ribbon from that row through the machine from that column. Copy and complete the chart.

Solution.
In the given table, the left column of chart list is the length of input piece of ribbon. Thus, the outputs for sending the input ribbon are given in the following table.

Question. 169 The left column of the chart lists is the length of input chains of gold. Repeater machines are listed across the top. The other entries are the outputs you get when you send the input chain from that row through the repeater machine from that column. Copy and complete the chart.

Solution.
In the given table, the left column of the chart lists is the length of input chains of gold. Thus, the output we get when we send the input chain from the row through the repeater machine are detailed in the following table.

Question.170 Long back in ancient times, a farmer saved the life of a king’s daughter. The king decided to reward the farmer with whatever he wished. The farmer, who was a chess champion, made an unusal request
“I would like you to place 1 rupee on the first square of my chessboard. 2 rupees on the second square, 4 on the third square, 8 on the fourth square and so on, until you have covered all 64 squares. Each square should have twice as many rupees as the previous square.” The king thought this to be too less and asked the farmer to think of some , better reward, but the farmer didn’t agree.
How much money has the farmer earned?
[Hint The following table may help you. What is the first square on which the king will place atleast Rs. 10 lakh?]

Solution.

Question. 171 The diameter of the Sun is 1.4 x 10 9 m and the diameter of the Earth is 1.2756 x 10 7 m. Compare their diameters by division.
Solution.

Question. 172 Mass of Mars is 6.42 x 10 29 kg and mass of the Sun is 1.99 x 10 30 kg. What is the total mass?
Solution.
Mass of Mars = 6.42 x 10 29 kg
Mass of the Sun = 1.99 x 10 30 kg
Total mass of Mars and Sun together = 6.42 x 10 29 + 1.99 x 10 30
= 6.42 x 10 29 + 19.9 x 10 29 = 26.32 x 10 29 kg

Question. 173 The distance between the Sun and the Earth is 1.496 x 10 8 km and : distance between the Earth and the Moon is 3.84 x 10 8 m. During
solar eclipse, the Moon comes in between the Earth and the Sun. What is the distance between the Moon and the Sun at that particular time?
Solution.
The distance between the Sun and the Earth is 1.496 x 10s km
= 1.496 x 10 8 x 10 3 m = 1496 x 10 8 m
The distance between the Earth and the Moon is 3.84 x10 8 m.
The distance between the Moon and the Sun at particular time (solar eclipse) = (1496 x 10 8 -3.84 x 10 8 )m = 1492. 16 x 10 8 m

Question. 174 A particular star is at a distance of about 8.1 x 10 13 km from the Earth. Assuring that light travels at 3 x 10 8 m per second, find how long does light takes from that star to reach the Earth?
Solution.

Question. 175 By what number should (-5) -1 be divided so that the quotient may be equal to (-5) -1 ?
Solution.

Question. 176 By what number should (-8) -3 .be multiplied so that the product may be equal to (-6) -3 ?
Solution.

Question. 177 Find x.

Solution.


Question. 178 If a = – 1, b = 2,-then find the value of the following,
(i) a b + b a (ii) a b – b a
(iii) a b x b a (iv) a b ÷ b a
Solution.

Question.179 Express each of the following in exponential form.

Solution.

Question. 180 Simplify

Solution.



Lesson 2-7: Adding and Subtracting Rational Numbers

Students will be adding and subtracting with negative numbers. We can represent this on a number line using arrows. The arrow for a positive number points right, and the arrow for a negative number points left. We add numbers by putting the arrows tail to tip.

For example, here is a number line that shows - 5 + 12 = 7 .

The first number is represented by an arrow that starts at 0 and points 5 units to the left. The next number is represented by an arrow that starts directly above the tip of the first arrow and points 12 units to the right. The answer is 7 because the tip of this arrow ends above the 7 on the number line.

In elementary school, students learned that every addition equation has two related subtraction equations. For example, if we know 3 + 5 = 8 , then we also know 8 &minus 5 = 3 and 8 &minus 3 = 5 .

The same thing works when there are negative numbers in the equation. From the previous example, - 5 + 12 = 7 , we also know 7 &minus 12 = - 5 and 7 &minus - 5 = 12 .

Here is a task to try with your student:

2. What does your answer tell you about the value of:

  1. The first arrow starts at 0 and points 3 units to the right. The next arrow starts at the tip of the first arrow and points 5 units to the left. This arrow ends above the -2, so 3 + - 5 = - 2 .

2. From the addition equation 3 + - 5 = - 2 , we get the related subtraction equations:


5 Putting It Together: Which Method Do I Use?

approach to take in testing hypotheses for the three parameters discussed in this chapter.

is greater than 30 or the

population that is normally

from a population that is

10.5 Assess Your Understanding

1. A simple random sample of size n = 19 is drawn from a

population that is normally distributed. The sample mean is

found to be 0.8, and the sample standard deviation is found to

be 0.4. Test whether the population mean is less than 1.0 at the

a = 0.01 level of significance.

2. A simple random sample of size n = 200 drivers were asked

if they drive a car manufactured in a certain country. Of the

200 drivers surveyed, 109 responded that they did. Determine if

more than half of all drivers drive a car made in this country at

the a = 0.05 level of significance.

3. A simple random sample of size n = 20 is drawn from a

population that is normally distributed. The sample variance is

found to be 49.3. Determine whether the population variance is

less than 95 at the a = 0.1 level of significance.

4. A simple random sample of size n = 15 is drawn from a

population that is normally distributed. The sample mean is

found to be 23.8, and the sample standard deviation is found to

be 6.3. Is the population mean different from 25 at the a = 0.01

5. A simple random sample of size n = 16 is drawn from a

population that is normally distributed. The sample variance is

found to be 13.7. Test whether the population variance is greater

than 10 at the a = 0.05 level of significance.

6. A simple random sample of size n = 65 is drawn from a

population. The sample mean is found to be 583.1, and the

sample standard deviation is found to be 114.9. Is the population

mean different from 600 at the a = 0.1 level of significance?

7. A simple random sample of size n = 40 is drawn from a

population. The sample mean is found to be 108.5, and the

sample standard deviation is found to be 17.9. Is the population

mean greater than 100 at the a = 0.05 level of significance?

8. A simple random sample of size n = 320 adults was asked

their favorite ice cream flavor. Of the 320 individuals surveyed,

58 responded that they preferred mint chocolate chip. Do less

than 25% of adults prefer mint chocolate chip ice cream? Use

the a = 0.01 level of significance.

9. Smarter Kids? A psychologist obtains a random sample of

20 mothers in the first trimester of their pregnancy. The mothers

are asked to play Mozart in the house at least 30 minutes each

day until they give birth. After 5 years, the child is administered

an IQ test. We know that IQs are normally distributed with a

mean of 100. If the IQs of the 20 children in the study result in

a sample mean of 104.2, and a sample standard deviation of 14.7,

is there evidence that the children have higher IQs? Use the

a = 0.05 level of significance.

10. The Atomic Bomb In October 1945, the Gallup

organization asked 1487 randomly sampled Americans, “Do you

think we can develop a way to protect ourselves from atomic

bombs in case other countries tried to use them against us?”

with 788 responding yes. Did a majority of Americans feel the

United States could develop a way to protect itself from atomic

bombs in 1945? Use the a = 0.05 level of significance.

11. Yield Strength A manufacturer of high-strength, lowalloy steel beams requires that the standard deviation of yield

strength not exceed 7000 pounds per square inch (psi). The

quality-control manager selected a sample of 20 steel beams

and measured their yield strength. The standard deviation of the

sample was 7500 psi. Assume that yield strengths are normally

distributed. Does the evidence suggest that the standard

deviation of yield strength exceeds 7000 psi at the a = 0.01 level

12. Pharmaceuticals A pharmaceutical company manufactures

a 200-milligram (mg) pain reliever. Company specifications

require that the standard deviation of the amount of the active

ingredient must not exceed 5 mg. The quality-control manager

selects a random sample of 30 tablets from a certain batch and

finds that the sample standard deviation is 7.3 mg. Assume that

the amount of the active ingredient is normally distributed.

Determine whether the standard deviation of the amount of the

active ingredient is greater than 5 mg at the a = 0.05 level of

13. Course Redesign Pass rates for Intermediate Algebra at a

community college are 52.6%. In an effort to improve pass rates

in the course, faculty of a community college develop a masterybased learning model where course content is delivered in a lab

through a computer program. The instructor serves as a learning

mentor for the students. Of the 480 students who enroll in the

mastery-based course, 267 pass.

(a) What is the variable of interest in this study? What type of

(b)At the 0.01 level of significance, decide whether the sample

evidence suggests the mastery-based learning model

(c) Explain why a 0.01 level of significance might be used to test

14. Number of Credit Cards According to the Federal Reserve

Bank of Boston, among individuals who had credit cards in

2014, the mean number of cards was 3.5. Treat the individuals

who have credit cards in the SullivanStatsSurveyI as a random

sample of credit card holders. Go to the book’s website to obtain

the data file SullivanStatsSurveyI using the file format of your

choice for the version of the text you are using. The results of the

survey are under the column “Number of cards.”

(a) What is the variable of interest in this study? Is it qualitative

Section 10.5 Putting It Together: Which Method Do I Use?

(b)Do the results of the survey imply that the mean number of

cards per individual is less than 3.5? Use the a = 0.05 level

15. Gas Mileage The Environmental Protection Agency (EPA)

states that a 2013 Kia Optima should get 28 miles per gallon,

on average. The website www.fueleconomy.gov allows users to

report the miles per gallon that they get on their vehicle. Treat

the following data as a random sample of ten 2013 Kia Optima

owners. The data represent the miles per gallon on their vehicle.

Is there reason to believe that individuals are getting different

gas mileage than the EPA states should be attained? Be sure to

verify all conditions necessary to conduct the appropriate test.

16. Sleepy? According to the National Sleep Foundation, children

between the ages of 6 and 11 years should get 10 hours of sleep each

night. In a survey of 56 parents of 6 to 11 year olds, it was found that

the mean number of hours the children slept was 8.9 with a standard

deviation of 3.2. Does the sample data suggest that 6 to 11 year olds

are sleeping less than the required amount of time each night? Use

the 0.01 level of significance.

17. Text while Driving According to the research firm Toluna,

the proportion of individuals who text while driving is 0.26.

Suppose a random sample of 60 individuals are asked to disclose

if they text while driving. Results of the survey are shown next,

where 0 indicates no and 1 indicates yes. Do the data contradict

the results of Toluna? Use the a = 0.05 level of significance.

18. Student Loan Balances Student loan debt has reached

record levels in the United States. In a random sample of 100

individuals who have student loan debt, it was found the mean

debt was $23,979 with a standard deviation of $31,400. Data

based on results from the Federal Reserve Bank of New York.

(a) What do you believe is the shape of the distribution of

student loan debt? Explain.

(b)Use this information to estimate the mean student loan debt

among all with such debt at the 95% level of confidence.

(c) What could be done to increase the precision of the

19. Political Decision Politicians often form their positions on

various policies through polling. Suppose the U.S. Congress is

considering passage of a tax increase to pay down the national

debt and national polls suggest the general population is

equally split on the matter. A congresswoman wants to poll her

constituency on this controversial tax increase. To get a good sense

as to how the citizens of her very populous district feel (well over

1 million registered voters), she decides to poll 8250 individuals

within the district. Of those surveyed, 4205 indicated they are in

favor of the tax increase. Given that politicians are generally leery

of voting for tax increases, what level of significance would you

Chapter 10 Hypothesis Tests Regarding a Parameter

recommend the congresswoman use in conducting this hypothesis

test? Do the results of the survey represent statistically significant

evidence a majority of the district favor the tax increase? What

would you recommend to the congresswoman?

20. Quality Control Suppose the mean wait-time for a telephone

reservation agent at a large airline is 43 seconds. A manager

with the airline is concerned that business may be lost due to

customers having to wait too long for an agent. To address this

concern, the manager develops new airline reservation policies

that are intended to reduce the amount of time an agent needs

to spend with each customer. A random sample of 250 customers

results in a sample mean wait-time of 42.3 seconds with a standard

deviation of 4.2 seconds. Using an a = 0.05 level of significance,

do you believe the new policies were effective? Do you think the

results have any practical significance?

21. Ideal Number of Children A survey from the Gallup

organization asked, “What do you think is the ideal number

of children for a family to have?” Go to the book’s website to

obtain the data file 10_5_21 using the file format of your

choice for the version of the text you are using to get the

(a) Draw a dot plot of the data. Comment on the shape of the

(b)What is the mode ideal number of children?

(c) Determine the mean, median, standard deviation, and

interquartile range ideal number of children. Round your

answers to the nearest thousandth.

(d)Explain why a large sample size is needed to perform any

inference regarding this population.

(e) In May 1997, the ideal number of children was considered

to be 2.64. Do the results of this poll indicate that people’s

beliefs as to the ideal number of children have changed?

Use the 0.05 level of significance.

22. Confidence Intervals Suppose you wish to determine if the

mean IQ of students on your campus is different from the mean IQ

in the general population, 100. To conduct this study, you obtain a

simple random sample of 50 students on your campus, administer

an IQ test, and record the results. The mean IQ of the sample of 50

students is found to be 107.3 with a standard deviation of 13.6.

(a) Conduct a hypothesis test (preferably using

technology) H0: m = m0 versus H1: m ≠ m0 for

m0 = 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 at the

a = 0.05 level of significance. For which values of m0 do you

not reject the null hypothesis?

(b) Construct a 95% confidence interval for the mean IQ of

students on your campus. What might you conclude about how

the lower and upper bounds of a confidence interval relate to

the values for which the null hypothesis is rejected?

(c) Suppose you changed the level of significance in conducting

the hypothesis test to a = 0.01. What would happen to the

range of values of m0 for which the null hypothesis is not

rejected? Why does this make sense?

In Problems 23–29, decide whether the problem requires a

confidence interval or hypothesis test, and determine the variable

of interest. For any problem requiring a confidence interval,

state whether the confidence interval will be for a population

proportion or population mean. For any problem requiring a

hypothesis test, write the null and alternative hypothesis.

23. An investigator with the Food and Drug Administration

wanted to determine whether a typical bag of potato chips

contained less than the 16 ounces claimed by the manufacturer.

24. A researcher wanted to estimate the average length of time

mothers who gave birth via Caesarean section spent in a hospital

after delivery of the baby.

25. An official with the Internal Revenue Service wished to

estimate the proportion of high-income (greater than $100,000

annually) earners who under-reported their net income (and,

therefore, their tax liability).

26. According to the Pew Research Center, 55% of adult

Americans support the death penalty for those convicted of

murder. A social scientist wondered whether a higher proportion of

adult Americans with at least a bachelor’s degree support the death

penalty for those convicted of murder.

27. In 2014, of the 37 million borrowers who have outstanding

student loan balances, 14% have at least one past due student

loan account. A researcher with the United States Department

of Education believes this proportion has increased since then.

Source: American Student Assistance

28. Researchers measured regular testosterone levels in a random

sample of athletes and then measured testosterone levels prior

to an athletic event. They wanted to know whether testosterone

levels increase prior to athletic events.

29. Is the dispersion of IQ scores among undergraduate students

lower than that of the general population, 15?

10.6The Probability of a Type II Error

and the Power of the Test

Preparing for This Section Before getting started, review the following:

• Computing normal probabilities (Section 7.2, pp. 394–398)

❶Determine the probability of making a Type II error

❷Compute the power of the test

❶ Determine the Probability of Making a Type II Error

The probability of making a Type I error, rejecting a true null hypothesis, is denoted a,

and is also called the level of significance. So, if the null hypothesis is true and a = 0.05,

the probability of rejecting H0 is 0.05.

SECTION 10.6 The Probability of a Type II Error and the Power of the Test

If the null hypothesis is false, but we fail to reject it, we have made a Type II error.

We denote the probability of making a Type II error b. We know there is an inverse

relation between a and b, but how can we determine the probability of b? Because the

alternative hypothesis contains a range of values (such as p 7 0.3), the probability of

making a Type II error has multiple values, each corresponding to a specific value of the

parameter from the alternative hypothesis.

EXAMPLE 1 Computing the Probability of a Type II Error

Problem In Example 1 from Section 10.2 on page 512, we tested H0: p = 0.5 versus

H1: p 6 0.5, where p is the proportion of Illinois high school students who have taken

the ACT and are prepared for college-level mathematics. To conduct this test, a random

sample of n = 500 high school students who took the ACT was obtained and the number

of students who were prepared for college mathematics (ACT score at least 22) was

determined. The test was conducted with a = 0.05. If the true population proportion of

Illinois high school students with a score of 22 or higher on the ACT is 0.48, which means

the alternative hypothesis is true, what is the probability of making a Type II error, b? That

is, what is the probability of failing to reject the null hypothesis when, in fact, p 6 0.5?

Approach To compute b, we need to determine the likelihood of obtaining a sample

proportion that would lead to failing to reject the null hypothesis assuming the true

population proportion is 0.48. This can be done in two steps.

Step 1 Determine the values of the sample proportion that lead to rejecting H0.

Step 2 Determine the probability that we do not reject H0 assuming the true

population proportion is 0.48.

Step 1 Because we are testing H0 : p = 0.5 and n = 500, the distribution for the

sample proportion is approximately normally distributed with mpn = 0.5 and standard

. Any test statistic less than z = -1.645 will lead to

rejecting the statement in the null hypothesis (since this results in a [email protected] 6 0.052.

Put another way, if pn is more than 1.645 standard errors below p0 = 0.5, the null

hypothesis will be rejected. So we reject H0 if

Figure 25 shows the reject and do not reject regions.

Note: We agree to round the sample proportion that makes the “cut-off” point to three

Step 2 If the true population proportion is 0.48, then the mean of the sampling

distribution is mpn = 0.48 and standard deviation spn =

the probability of a Type II error is the probability of not rejecting H0 when H1 is

true. So we need to determine the probability of obtaining a sample proportion of

pn = 0.463 or higher from a population whose proportion is p = 0.48.

= P1do not reject H0 when H1 is true2

= P1pn 7 0.463 given that p = 0.482

The probability of not rejecting H0: p = 0.50 when, in fact, H1: p 6 0.5 is true is 0.7764. •

CHAPTER 10 Hypothesis Tests Regarding a Parameter

Figure 26 shows the distributions of pn for both the assumption that the null hypothesis

is true, p = 0.5 (red), and if a specific alternative hypothesis is true, p = 0.48 (blue).

For each possible value of p less than 0.5, there is a unique value of b. For example,

if the true population proportion for the scenario in Example 1 was p = 0.45, then

b = 0.2810. You should confirm this for yourself. Notice the probability of making a

Type II error decreases as the specific value in the alternative moves farther away from

the statement in the null 1p = 0.52.

The procedure for computing the probability of making a Type II error is shown

Determining the Probability of a Type II Error

Step 1 Determine the sample proportion that separates the rejection region from the non-rejection region.

Left-Tailed Test 1 H1 : p * p 0 2

Right-Tailed Test 1H1 : p + p 0 2

Note: za is the critical z-value corresponding to the level of significance, a.

Two-Tailed Test 1H1 : p 3 p 0 2

Left-Tailed Test 1 H1 : p * p 0 2

Find the area under the normal curve to the

right of the sample proportion found in Step 1

assuming a particular value of the population

proportion in the alternative hypothesis.

Right-Tailed Test 1H1 : p + p 0 2

Find the area under the normal curve to the

left of the sample proportion found in Step 1

assuming a particular value of the population

proportion in the alternative hypothesis.

Two-Tailed Test 1H1 : p 3 p 0 2

Find the area under the normal curve between

pn L and pn U found in Step 1 assuming a

particular value of the population proportion

in the alternative hypothesis.

Note two important points about the probability of making a Type II error. For a given

value of the level of significance, a:

• As the value of the parameter moves farther away from the value stated in the null

hypothesis, the likelihood of making a Type II error decreases.

• As the sample size n increases, the probability of making a Type II error decreases.

The second point should not shock you. We know as the sample size increases the

standard error decreases, thereby reducing the chance of making an error. The second

point also emphasizes the need to determine a sample size appropriate for your level

of comfort in making an error. Whenever designing an experiment or survey, it is good

practice to determine b for the level of a chosen. The value of the parameter that

should be used to represent a possible value in the alternative hypothesis should be

one that represents the effect size you are looking for. For example, if the researchers

from Example 1 were really concerned if p actually was 0.45, this is a good choice for

SECTION 10.6 The Probability of a Type II Error and the Power of the Test

determining the probability of making a Type II error for the sample size chosen. If this

value of b is too large, the researchers should increase the sample size.

❷ Compute the Power of the Test

The power of the test is the

probability that you will

The probability of rejecting the null hypothesis when the alternative hypothesis is true

is 1 - b. The value of 1 - b is called the power of the test. The higher the power of the

test, the more likely the test will reject the null when the alternative hypothesis is true.

EXAMPLE 2 Computing the Power of the Test

Problem Compute the power of the test for the situation in Example 1.

Approach The power of the test is 1 - b. In Example 1, we found that b is 0.7764

when the true population proportion is 0.48.

Solution The power of the test is 1 - 0.7764 = 0.2236. There is a 0.2236 probability of

rejecting the null hypothesis if the true population proportion is 0.48.•

As the true proportion gets

closer to the value of the

proportion stated in the null

hypothesis, it becomes more

difficult to correctly reject the

Table 3 shows both the probability of making a Type II error and the power of the

test for a number of different values of the population proportion that are less than 0.5.

Notice the closer the population proportion gets to 0.5, the larger the probability of a

Type II error becomes. This is because when the true population proportion is close to

the population proportion stated in the null hypothesis, there is a high probability that

we will not reject the null hypothesis and will make a Type II error.

We can use the values of p and their corresponding powers to construct a power curve.

A power curve is a graph that shows the power of the test against values of the parameter

that make the null hypothesis false. Figure 27 shows the power curve for the values in

Table 3. The power curve shows that as the true population proportion approaches the

proportion stated in the null hypothesis, p = 0.5, the power of the test decreases.

Technology Step-by-Step Probability of a Type II Error and Power

1. Select Stat, highlight T Stats or Proportion Stats

(depending on whether finding power for a mean or

proportion), select One Sample, then Power/Sample

2. For means: Enter Alpha (the level of significance), the

standard deviation, whether the alternative hypothesis is

one-sided or two-sided, the difference between the mean

stated in the null hypothesis and the true mean, and the

sample size. Delete any entry in the Power cell. Leave

the cursor in the Power cell and click Compute.

For proportions: Enter Alpha (the level of

significance), the target proportion (proportion

stated in the null hypothesis), whether the alternative

hypothesis is one-sided or two-sided, the true population

proportion, and the sample size. Delete any entry in the

Power cell. Leave the cursor in the Power cell and click

CHAPTER 10 Hypothesis Tests Regarding a Parameter

10.6 Assess Your Understanding

Vocabulary and Skill Building

1. Explain what it means to make a Type II error.

2. Explain the term power of the test.

3. To test H0: p = 0.30 versus H1: p 6 0.30, a simple random

sample of n = 300 individuals is obtained and x = 86 successes

(a)What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.28. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.25.

4. To test H0: p = 0.40 versus H1: p 7 0.40, a simple random

sample of n = 200 individuals is obtained and x = 84 successes

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.44. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.47.

5. To test H0: p = 0.65 versus H1: p 7 0.65, a simple random

sample of n = 100 individuals is obtained and x = 69 successes

(a)What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.01 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.70. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.72.

6. To test H0: p = 0.75 versus H1: p 6 0.75, a simple random

sample of n = 400 individuals is obtained and x = 280 successes

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.01 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.71. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.68.

7. To test H0: p = 0.45 versus H1: p ≠ 0.45, a simple random

sample of n = 500 individuals is obtained and x = 245 successes

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.1 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.49. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.47.

8. To test H0: p = 0.25 versus H1: p ≠ 0.25, a simple random

sample of n = 350 individuals is obtained and x = 74 successes

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, compute the probability of

making a Type II error if the true population proportion is

0.23. What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.28.

9. Worried about Retirement? In April 2009, the Gallup

organization surveyed 676 adults aged 18 and older and found

that 352 believed they would not have enough money to live

comfortably in retirement. The folks at Gallup want to know if

this represents sufficient evidence to conclude a majority (more

than 50%) of adults in the United States believe they will not

have enough money in retirement.

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.53. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.55.

10. Ready for College? ACT, a college entrance exam used for

admission, looked at historical records and established 21 as the

minimum score on the ACT reading portion of the exam for a

student to be considered prepared for social science in college.

(Note: “Being prepared” means there is a 75% probability of

successfully completing a social science course in college.) An

official with the Illinois State Department of Education wonders

whether a majority of the students in her state who took the ACT

are prepared to take social science. She obtains a simple random

sample of 500 records of students who have taken the ACT and

finds that 269 are prepared. Does this represent significant evidence

that a majority (more than 50%) of the students in the state of

Illinois are prepared for social science in college upon graduation?

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.52. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.55.

11. Quality of Education In August 2002, 47% of parents who had

children in grades K–12 were satisfied with the quality of education

the students receive. A recent Gallup poll found that 437 of 1013

parents of children in grades K–12 were satisfied with the quality of

education the students receive. Does this suggest the proportion of

parents satisfied with the quality of education has decreased?

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.10 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.42. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.46.

12. Eating Together In December 2001, 38% of adults with

children under the age of 18 reported that their family ate dinner

together seven nights a week. In a recent poll, 403 of 1122 adults

with children under the age of 18 reported that their family

ate dinner together seven nights a week. Has the proportion

of families with children under the age of 18 who eat dinner

together seven nights a week decreased?

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.10 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.35. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.32.

13. Taught Enough Math In 1994, 52% of parents of children

in high school felt it was a serious problem that high school

students were not being taught enough math and science. A

recent survey found that 256 of 800 parents of children in high

school felt it was a serious problem that high school students

were not being taught enough math and science. Do parents feel

differently today than they did in 1994?

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.50. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.48.

14. Infidelity According to menstuff.org, 22% of married men

have “strayed” at least once during their married lives. A survey

of 500 married men indicated that 122 have strayed at least once

during their married life. Does this survey result contradict the

(a) What does it mean to make a Type II error for this test?

(b)If the researcher decides to test this hypothesis at the

a = 0.05 level of significance, determine the probability of

making a Type II error if the true population proportion is

0.25. What is the power of the test?

(c) Redo part (b) if the true proportion is 0.20.

15. Effect of A Redo Problem 3(b) with a = 0.01. What effect

does lowering the level of significance have on the power of the

test? Why does this make sense?

16. Effect of A Redo Problem 4(b) with a = 0.01. What effect

does lowering the level of significance have on the power of the

test? Why does this make sense?

17. Power Curve Draw a power curve for the scenario

in Problem 3 by finding the power of the test for

p = 0.2, 0.22, 0.24, 0.26, 0.28 [done in part (b)], and 0.29.

18. Power Curve Draw a power curve for the scenario

in Problem 4 by finding the power of the test for

p = 0.41, 0.42, 0.44 [done in part (b)], 0.46, 0.48, and 0.50.

19. Power in Tests on Means To test H0: m = 50 versus

H1: m 6 50, a simple random sample of size n = 24 is obtained

from a population that is known to be normally distributed, and

the sample standard deviation is found to be 6.

(a) A researcher decides to test the hypothesis at the a = 0.05

level of significance. Determine the sample mean that

separates the rejection region from the nonrejection region.

[Hint: Follow the same approach as that laid out on

page 542, but use Student’s t-distribution to find the critical

(b)Suppose the true population mean is m = 48.9. Use

technology to find the area under the t-distribution to

the right of the sample mean found in part (a) assuming

m = 48.9. [Hint: This can be accomplished by performing a

one-sample t-test.] This represents the probability of making

a Type II error, b. What is the power of the test?

20. What happens to the power of the test as the true value of

the parameter gets closer to the value of the parameter stated in

the null hypothesis? Why is this result reasonable?

21. What effect does increasing the sample size have on the

power of the test, assuming all else remains unchanged?

22. How do the probability of making a Type II error and effect

size play a role in determining an appropriate sample size when

performing a hypothesis test?

In this chapter, we discussed hypothesis testing. Hypothesis

testing is the second type of inferential statistics (recall

the other type of inference is estimation via confidence

In hypothesis testing, a statement is made regarding

a population parameter, which leads to a null, H 0, and

alternative hypothesis, H 1 . The null hypothesis is a

statement of “no change” or “no difference”. We build a

probability model under the assumption the statement

in the null hypothesis is true, and we use sample data to

decide whether to reject or not reject the statement in the

null hypothesis. There are three options for structuring the

null and alternative hypotheses on a single parameter.

Determining the Null and Alternative Hypotheses

E qual hypothesis versus not equal hypothesis

E qual hypothesis versus less than hypothesis (lefttailed test)

H1: parameter 6 some value

Equal hypothesis versus greater than hypothesis (righttailed test)

H1: parameter 7 some value

In performing a hypothesis test, there is always the

possibility of making a Type I error (rejecting the null

hypothesis when it is true) or making a Type II error

(not rejecting the null hypothesis when it is false). The

probability of making a Type I error is equal to the level of

significance, a, of the test.

The first test introduced was hypothesis tests about a

population proportion p. Provided certain requirements

were satisfied (simple random sample or randomized

CHAPTER 10 Hypothesis Tests Regarding a Parameter

experiment, independence, and a large sample size), we

were able to use the normal model to assess statements

made in the null hypothesis.

Then we discussed hypothesis testing on the mean.

Here we also required a simple random sample (or

randomized experiment) and independence, but we also

required that either the sample come from a population

that is normally distributed with no outliers or a large

1n Ú 302 sample size. When dealing with small sample

sizes, we tested the normality requirement with a normal

probability plot and the outlier requirement with a

boxplot. These requirements allow us to use Student’s

t-distribution to test the hypothesis.

Finally, we performed tests regarding a population

standard deviation (or variance). This test requires that the

population from which the sample is drawn be normally

distributed. This test is not robust, so deviation from the

normality requirement is not allowed.

All three hypothesis tests were performed using

both the classical method and the P-value approach. The

P-value approach to testing hypotheses has appeal because

the rejection rule is always to reject the null hypothesis if

the P-value is less than the level of significance, a.

In testing hypotheses, it is important to remember

that we never accept the null hypothesis, because, without

having access to the entire population, we don’t know the

exact value of the parameter stated in the null hypothesis.

Rather, we say that we do not reject the null hypothesis.

Remember, statistical significance refers to the

idea that the observed results are unlikely to occur if

the statement in the null hypothesis is true. Practical

significance, on the other hand, refers to the idea that,

while small differences between the statistic and parameter

stated in the null hypothesis are statistically significant, the

difference may not be large enough to cause concern or be

We concluded the section by learning how to compute

the probability of making a Type II error and the power

of the test. The power of the test is the probability of rejecting

the null hypothesis when the alternative hypothesis is true.

Alternative hypothesis (p. 502)

Level of significance (p. 505)

Statistically significant (p. 509)

Test statistic (pp. 512, 523, 534)

Practical significance (p. 527)

follows the standard normal

distribution if np0 11 - p0 2 Ú 10 and n … 0.05N

follows Student’s t-distribution with n - 1

degrees of freedom if the population from which the

sample was drawn is normal or if the sample size is

follows the x2-distribution with n - 1

degrees of freedom if the population from which the

sample was drawn is normal.

Type I and Type II Errors

= P1rejecting H0 when H0 is true2

= P(not rejecting H0 when H1 is true)

Example(s) Review Exercises

1 Determine the null and alternative hypotheses (p. 501)

2 Explain Type I and Type II errors (p. 504)

3 State conclusions to hypothesis tests (p. 506)

1 Explain the logic of hypothesis testing (p. 509)

2 Test hypotheses about a population proportion (p. 511)

3 Test hypotheses about a population proportion using the

binomial probability distribution (p. 516)

Example(s) Review Exercises

1 Test hypotheses about a mean (p. 522)

2 Understand the difference between statistical significance and

practical significance (p. 526)

1 Test hypotheses about a population standard deviation (p. 533)

1 Determine the appropriate hypothesis test to perform (p. 538)

1 Determine the probability of making a Type II error (p. 540)

2 Compute the power of the test (p. 543)

For Problems 1 and 2, (a) determine the null and alternative

hypotheses, (b) explain what it would mean to make a Type I

error, (c) explain what it would mean to make a Type II error,

(d) state the conclusion that would be reached if the null

hypothesis is not rejected, and (e) state the conclusion that would

be reached if the null hypothesis is rejected.

1. Credit-Card Debt According to creditcard.com, the mean

outstanding credit-card debt of college undergraduates was

$3173 in 2010. A researcher believes that this amount has

2. More Credit-Card Debt Among all credit cards issued, the

proportion of cards that result in default was 0.13 in 2010. A

credit analyst with Visa believes this proportion is different

3. A test is conducted at the a = 0.05 level of significance. What

is the probability of a Type I error?

4. b is computed to be 0.113. What is the probability of a Type II

error? What is the power of the test? How would you interpret

5. To test H0: m = 100 versus H1: m 7 100, a simple random

sample of size n = 35 is obtained from an unknown distribution.

The sample mean is 104.3 and the sample standard deviation

(a) To use the t-distribution, why must the sample size be large?

(b)Use the classical or P-value approach to decide whether to

reject the statement in the null hypothesis at the a = 0.05

6. To test H0: m = 50 versus H1: m ≠ 50, a simple random

sample of size n = 15 is obtained from a population that is

normally distributed. The sample mean is 48.1 and the sample

(a) Why must it be the case that the population from which the

sample was drawn is normally distributed?

(b)Use the classical or P-value approach to decide whether to

reject the statement in the null hypothesis at the a = 0.05

In Problems 7 and 8, test the hypothesis at the a = 0.05 level of

significance, using (a) the classical approach and (b) the P-value

approach. Be sure to verify the requirements of the test.

7. H0: p = 0.6 versus H1: p 7 0.6

8. H0: p = 0.35 versus H1: p ≠ 0.35

9. To test H0: s = 5.2 versus H1: s ≠ 5.2, a simple random

sample of size n = 18 is obtained from a population that is

known to be normally distributed.

(a) If the sample standard deviation is determined to be s = 4.9,

compute the test statistic.

(b)Test this hypothesis at the a = 0.05 level of significance.

10. To test H0: s = 15.7 versus H1: s 7 15.7, a simple random

sample of size n = 25 is obtained from a population that is

known to be normally distributed.

(a) If the sample standard deviation is determined to be

s = 16.5, compute the test statistic.

(b)Test this hypothesis at the a = 0.1 level of significance.

11. Sneeze According to work done by Nick Wilson of Otago

University Wellington, the proportion of individuals who cover

their mouth when sneezing is 0.733. As part of a school project,

Mary decides to confirm this by observing 100 randomly

selected individuals sneeze and finds that 78 covered their

(a) What are the null and alternative hypotheses for Mary’s

(b)Verify the requirements that allow use of the normal model

to test the hypothesis are satisfied.

(c) Does the sample evidence contradict Professor Wilson’s

12. Emergency Room The proportion of patients who visit

the emergency room (ER) and die within the year is 0.05.

Source: SuperFreakonomics. Suppose a hospital administrator is

concerned that his ER has a higher proportion of patients who

die within the year. In a random sample of 250 patients who

have visited the ER in the past year, 17 have died. Should the

administrator be concerned?

13. Linear Rotary Bearing A linear rotary bearing is designed

so that the distance between the retaining rings is 0.875 inch. The

quality-control manager suspects that the manufacturing process

needs to be recalibrated because the mean distance between the

retaining rings is greater than 0.875 inch. In a random sample

of 36 bearings, he finds the sample mean distance between the

retaining rings is 0.876 inch with standard deviation 0.005 inch.

(a) Are the requirements for conducting a hypothesis test

(b)State the null and alternative hypotheses.

(c) The quality-control manager decides to use an a = 0.01 level

of significance. Why do you think this level of significance was

(d)Does the evidence suggest the machine be recalibrated?

(e) What does it mean for the quality-control engineer to make

a Type I error? A Type II error?

CHAPTER 10 Hypothesis Tests Regarding a Parameter

14. Normal Temperature Carl Reinhold August Wunderlich

said that the mean temperature of humans is 98.6°F. Researchers

Philip Mackowiak, Steven Wasserman, and Myron Levine

[JAMA, Sept. 23–30 1992 268(12):1578–80] thought that the mean

temperature of humans is less than 98.6°F. They measured the

temperature of 148 healthy adults one to four times daily for 3

days, obtaining 700 measurements. The sample data resulted in a

sample mean of 98.2°F and a sample standard deviation of 0.7°F.

Test whether the mean temperature of humans is less than 98.6°F

at the a = 0.01 level of significance.

15. Conforming Golf Balls The U.S. Golf Association (USGA)

requires that golf balls have a diameter that is 1.68 inches. To

determine if Maxfli XS golf balls conform to USGA standards,

a random sample of Maxfli XS golf balls was selected. Their

diameters are shown in the table.

Source: Michael McCraith, Joliet Junior College

(a) Because the sample size is small, the engineer must verify

that the diameter is normally distributed and the sample

does not contain any outliers. The normal probability

plot and boxplot are shown. The correlation between ball

diameter and expected z-scores is 0.951. Are the conditions

for testing the hypothesis satisfied?

(b)Construct a 95% confidence interval to judge whether the

golf balls conform to USGA standards. Be sure to state the

null and alternative hypotheses and write a conclusion.

16. Studying Enough? College mathematics instructors suggest

that students spend 2 hours outside class studying for every hour

in class. So, for a 4-credit-hour math class, students should spend

at least 8 hours (480 minutes) studying each week. The given

data, from Michael Sullivan’s College Algebra class, represent

the time spent on task recorded in MyMathLab (in minutes) for

randomly selected students during the third week of the semester.

Determine if the evidence suggests students may not, in fact, be

following the advice. That is, does the evidence suggest students

are studying less than 480 minutes each week? Use a = 0.05

level of significance. Note: A normal probability plot and boxplot

indicate that the data come from a population that is normally

distributed with no outliers.

17. Sleeping Patterns of Pregnant Women A random sample

of 150 pregnant women indicated that 81 napped at least twice

per week. Do a majority of pregnant women nap at least twice a

week? Use the a = 0.05 level of significance.

Source: National Sleep Foundation

(a) Explain what it would mean to make a Type II error for this

(b)Determine the probability of making a Type II error if the

true population proportion of pregnant women who nap is

0.53. What is the power of the test?

19. Birth Weights An obstetrician maintains that preterm babies

(gestation period less than 37 weeks) have a higher variability

in birth weight than do full-term babies (gestation period 37 to

41 weeks). According to the National Vital Statistics Report, the

birth weights of full-term babies are normally distributed, with

standard deviation 505.6 grams. A random sample of 41 preterm

babies results in a standard deviation equal to 840 grams. Test the

researcher’s hypothesis that the variability in the birth weight of

preterm babies is more than the variability in birth weight of fullterm babies, at the a = 0.01 level of significance.

20. Grim Report Throughout the country, the proportion of firsttime, first-year community college students who return for their


Table of Contents

Chapter 1 - Operations on Real Numbers and Algebraic Expressions

1.1 Success in Mathematics

1.2 The Number Systems and the Real Number Line

1.3 Adding, Subtracting, Multiplying, and Dividing Integers

1.4 Adding, Subtracting, Multiplying, and Dividing Rational Numbers Expressed as Fractions and Decimals

Putting the Concepts Together (Sections 1.1 - 1.4)

1.5 Properties of Real Numbers

1.6 Exponents and the Order of Operations

1.7 Simplifying Algebraic Expressions

Chapter 2-Equations and Inequalities in One Variable

2.1 Linear Equations: The Addition and Multiplication Properties of Equality

2.2 Linear Equations: Using the Properties Together

2.3 Solving Linear Equations Involving Fractions and Decimals Classifying Equations

2.4 Evaluating Formulas and Solving Formulas for a Variable

Putting the Concepts Together (Sections 2.1 - 2.4)

2.5 Introduction to Problem Solving: Direct Translation Problems

2.6 Problem Solving: Direct Translation Problems Involving Percent

2.7 Problem Solving: Geometry and Uniform Motion

2.8 Solving Inequalities in One Variable

Chapter 3 - Introduction to Graphing and Equations of Lines

3.1 The Rectangular Coordinate System and Equations in Two Variables

3.2 Graphing Equations in Two Variables

3.4 Slope-Intercept Form of a Line

3.5 Point-Slope Form of a Line

3.6 Parallel and Perpendicular Lines

Putting the Concepts Together (Sections 3.1 - 3.6)

3.8 Linear Inequalities in Two Variables

Cumulative Review Chapters 1-3

Chapter 4 - Systems of Linear Equations and Inequalities

4.1 Solving Systems of Linear Equations in Two Variables by Graphing

4.2 Solving Systems of Linear Equations in Two Variables Using Substitution

4.3 Solving Systems of Linear Equations in Two Variables Using Elimination

Putting the Concepts Together (Sections 4.1 - 4.3)

4.4 Solving Direct Translation, Geometry, and Uniform Motion Problems Using Systems of Linear Equations

4.5 Solving Mixture Problems Using Systems of Linear Equations

4.6 Systems of Linear Inequalities

Chapter 5 - Exponents and Polynomials

5.1 Adding and Subtracting Polynomials

5.2 Multiplying Monomials: The Product and Power Rules

5.3 Multiplying Polynomials

5.4 Dividing Monomials: The Quotient Rule and Integer Exponents

Putting the Concepts Together (Sections 5.1 - 5.4)

5.6 Applying Exponent Rules: Scientific Notation

Cumulative Review Chapters 1-5

Chapter 6 - Factoring Polynomials

6.1 Greatest Common Factor and Factoring by Grouping

6.2 Factoring Trinomials of the Form x2 + bx + c

6.3 Factoring Trinomials of the Form ax2 + bx + c , a ¿ 1

6.4 Factoring Special Products

6.5 Summary of Factoring Techniques

Putting the Concepts Together (Sections 6.1 - 6.5)

6.6 Solving Polynomial Equations by Factoring

6.7 Modeling and Solving Problems with Quadratic Equations

Getting Ready for Intermediate Algebra: A Cumulative Review of Chapters 1-6

Chapter 7 - Rational Expressions and Equations

7.1 Simplifying Rational Expressions

7.2 Multiplying and Dividing Rational Expressions

7.3 Adding and Subtracting Rational Expressions with a Common Denominators

7.4 Finding the Least Common Denominator and Forming Equivalent Rational Expressions

7.5 Adding and Subtracting Rational Expressions with Unlike Denominators

7.6 Complex Rational Expressions

Putting the Concepts Together (Sections 7.1 - 7.6)

7.8 Models Involving Rational Expressions

Getting Ready for Intermediate Algebra: A Cumulative Review of Chapters 1-7

Chapter 8- Graphs, Relations, and Functions

Putting the Concepts Together (Section 8.1 - 8.2)

8.3 An Introduction to Functions

8.4 Functions and Their Graphs

8.7 Absolute Value Equations and Inequalities

Chapter 9- Radicals and Rational Exponents

9.2 n th Roots andRational Exponents

9.3 Simplify Expressions Using the Laws of Exponents

9.4 Simplifying Radical Expressions

9.5 Adding, Subtracting, and Multiplying Radical Expressions

9.6 Rationalizing Radical Expressions

Putting the Concepts Together (Sections 9.1 - 9.6)

9.7 Functions Involving Radicals

9.8 Radical Equations and Their Applications

9.9 The Complex Number System

Chapter 10- Quadratic Equations and Functions

10.1 Solving Quadratic Equations by Completing the Square

10.2 Solving Quadratic Equations by the Quadratic Formula

10.3 Solving Equations Quadratic in Form

Putting the Concepts Together (Sections 10.1 - 10.3)

10.4 Graphing Quadratic Functions Using Transformations

10.5 Graphing Quadratic Functions Using Properties

10.6 Quadratic Inequalities

10.7 Rational Inequalities

Cumulative Review Chapters 1-10

Chapter 11- Exponential and Logarithmic Functions

11.1 Composite Functions and Inverse Functions

11.2 Exponential Functions

11.3 Logarithmic Functions

Putting the Concepts Together (Sections 11.1 - 11.3)

11.4 Properties of Logarithms

11.5 Exponential and Logarithmic Equations

12.1 Distance and Midpoint Formulas

Putting the Concepts Together (Sections 12.1 - 12.3)

12.6 Nonlinear Systems of Equations

Cumulative Review Chapters 1-12

Chapter 13- Sequences, Series, and The Binomial Theorem

13.3 Geometric Sequences and Series

Putting the Concepts Together (Section 13.1 - 13.3)

Appendix A: Review of Fractions, Decimals, and Percents

Appendix B: Division of Polynomials Synthetic Division

Appendix C: The Library of Functions

Appendix E: More on Systems of Equations

E.1 Systems of Linear Equations in Three Variables

E.2 Using Matrices to Solve Systems

E.3 Determinants and Cramer's Rule


Myths are a way of understanding the world. This lesson has been about Greek mythology, but every culture has myths. Myths define social customs and beliefs, explain natural and psychological phenomena, and provide a way for people to discuss things that cause anxiety.

Mythology is all around us. Here are just a few examples of places we find myths today:

  • Days of the week &mdash Wednesday (Woden or Odin &mdash Norse god) Thursday (Thor &mdash Norse god) Friday (Freya &mdash Norse goddess) Saturday (Saturn &mdash Roman god who ruled before Jove)
  • Cars &mdash Toyota Avalon and Cressida, Cadillac El Dorado, Honda Odyssey, Mercury
  • Shoes &mdash Nike
  • Tires &mdash Midas
  • Astronomy &mdash Constellations like Orion, the Argo, all of the planets and the Pleiades. The Milky Way itself was supposedly the road over which the stars traveled to Jupiter&rsquos palace.
  • NASA &mdash The first part of U.S. space program was Project Mercury, named after the messenger of the gods because the project&rsquos purpose was to send a message to the Soviets that America was in the space race. The Gemini Project was next Gemini is Latin for "twins," and the project was called this because the capsule held two astronauts. Apollo astronauts rode on Saturn rockets.

Myths also make great stories. They come up in literature all over the place, from really serious stuff like Dante to comic strips. Myths inspire music actually the word music comes from the mythological muses who inspired art of all kinds. Painters such as Michelangelo and Botticelli were inspired by myths. Even children&rsquos movies are a good place to look for myths you will find them everywhere, including Snow White, Star Wars and The Lord of the Rings. Can you find at least three examples of mythology connections in the world?


Watch the video: Simplifying Radicals With Variables, Exponents, Fractions, Cube Roots - Algebra (October 2021).