Step by step business math and statisticsby jin w choi

Step-by-Step Business
Math and Statistics
By Jin W. Choi

Included in this preview:
• Copyright Page
• Table of Contents
• Excerpt of Chapter 1

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Step-by-Step Business Math
and Statistics
Jin W. Choi
DePaul University


Copyright © 2011 by Jin W. Choi. All rights reserved. No part of this publication may be reprinted,
reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now
known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of University Readers, Inc.
First published in the United States of America in 2011 by University Readers, Inc.
Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are
used only for identification and explanation without intent to infringe.

15 14 13 12 11

12345

Printed in the United States of America

ISBN: 978-1-60927-872-4


Contents

Acknowledgments

Part 1.

v

Business Mathematics

Chapter 1.

Algebra Review

1

Chapter 2.

Calculus Review

42

Chapter 3.

Optimization Methods

67

Chapter 4.

Applications to Economics

85

Part 2.

Business Statistics

Chapter 1.

Introduction

108

Chapter 2.

Data Collection Methods

115

Chapter 3.

Data Presentation Methods

122

Chapter 4.

Statistical Descriptive Measures

133

Chapter 5.

Probability Theory

157

Chapter 6.

Discrete Probability Distributions

179

Chapter 7.

The Normal Probability Distribution

195

Chapter 8.

The t-Probability Distribution

218

Chapter 9.

Sampling Distributions

228

Chapter 10.

Confidence Interval Construction

249

Chapter 11.

One-Sample Hypothesis Testing

264

Chapter 12.

Two-Sample Hypothesis Testing

312

Chapter 13.

Simple Regression Analysis

334

Chapter 14.

Multiple Regression Analysis

382

Chapter 15.

The Chi-Square Test

412

Appendix:

Statistical Tables

428

Subject Index

437


Acknowledgments

would like to thank many professors who had used this book in their classes. Especially, Professors
Bala Batavia, Burhan Biner, Seth Epstein, Teresa Klier, Jin Man Lee, Norman Rosenstein, and Cemel
Selcuk had used previous editions of this book in teaching GSB420 Applied Quantitative Analysis
at DePaul University. Their comments and feedbacks were very useful in making this edition more

user-friendly.
Also, I would like to thank many current and past DePaul University’s Kellstadt Graduate School of
Business MBA students who studied business mathematics and statistics using the framework laid out
in this book. Their comments and feedbacks were equally important and useful in making this book an
excellent guide into the often-challenging fields of mathematics and statistics. I hope and wish that the
knowledge gained via this book would help them succeed in their business endeavors.
As is often the case with equations and numbers, I am sure this book still has some errors. If you find
some, please let me know at
Best wishes to those who use this book.

I

Jin W. Choi, Ph.D.
Kellstadt Graduate School of Business
DePaul University
Chicago, IL 60604


Acknowledgments

v


Math. Chapter 1. Algebra Review

Part 1. Business Mathematics
There are 4 chapters in this part of business mathematics: Algebra review, calculus
review, optimization techniques, and economic applications of algebra and calculus.

Chapter 1. Algebra Review

A.

The Number System

The number system is comprised of real numbers and imaginary numbers. Real numbers
are, in turn, grouped into natural numbers, integers, rational numbers, and irrational
numbers.
1.

Real Numbers = numbers that we encounter everyday during a normal course
of life Æ the numbers that are real to us.
i.

ii.

iii.

iv.

Natural numbers = the numbers that we often use to count items Æ
counting trees, apples, bananas, etc.: 1, 2, 3, 4, …
a.

odd numbers: 1, 3, 5, …

b.

even numbers: 2, 4, 6, …

Integers = whole numbers without a decimal point: 0, +1, +2, +3, +4, ….

a.

positive integers: 1, 2, 3, 4, …

b.

negative integers: –1, –2, –3, –4, …

Rational numbers = numbers that can be expressed as a fraction of integers
such as a/b (= a÷b) where both a and b are integers
a.

finite decimal fractions: 1/2, 2/5, etc.

b.

(recurring or periodic) infinite decimal fractions: 1/3, 2/9, etc.

Irrational Numbers = numbers that can NOT be expressed as a fraction of
integers = nonrecurring infinite decimal fractions:
a.

n-th roots such as 2 , 3 5 , 7 3 , etc.

b.

special values such as ʌ (=pi), or e (=exponential), etc.

Chapter 1: Algebra Review

1


Math. Chapter 1. Algebra Review

v.

vi.

Undefined fractions:
a.

any number that is divided by a zero such as k/0 where k is any
number

b.

a zero divided by a zero = 0/0

c.

an infinity divided by an infinity =

d.

a zero divided by an infinity =

f
f

0
f

Defined fractions:

a.

a one divided by a very small number Æ
1
1
1010 10,000,000,000 | a very large
10
0.0000000001 10
number such as a number that can approach ’

b.

a one divided by a very large number Æ
1/(a large number) = a small number Æ

c.

1
|0
f

a scientific notion Æ the use of exponent
2.345E+2 = 2.345 x 102 = 234.5
2.345E+6 = 2.345 x 106 = 2,345,000
2.345E–2 = 2.345 x 10-2 = 2.345 ˜

1
10 2

2.345 ˜

2.345E–6 = 2.345 x 10-6 =
1
1
2.345 ˜ 6 2.345 ˜
1,000,000
10

1
100

0.02345

0.000002345

Similarly, a caret (^) can be used as a sign for an exponent:
Xn = X^n

Æ

X10 = X^10

Note: For example, E+6 means move the decimal point 6 digits to the
right of the original decimal point whereas E-6 means move the
decimal point 6 digits to the left of the original decimal point.

2

Step by Step Business Math and Statistics


Math. Chapter 1. Algebra Review

2. Imaginary Numbers = numbers that are not easily encountered and recognized on
a normal course of life and thus, not real enough (or imaginary) to an individual.
Æ Often exists as a mathematical conception.

1

i

4
B.

2

2i

i 2

(5i)2 = –25

2i

Rules of Algebra
1.

ab

ba

2.

ab

3.

aa 1

4.

a (b  c)

5.

a  ( a)

6.

(a)b

7.

(a )(b)

8.

( a  b) 2

9.

( a  b) 2

10.

(a  b)(a  b)

11.

a
b

(a) /(b)

12.

a
b

a
b

13.

a

14.

a c

b d

Æ

2+3=3+2

Æ

5

Æ

2x3=3x2

Æ

6

Æ

2 x 2-1 = 20 = 1

ab  ac

Æ

2 x (3 + 4) = 2 x 3 + 2 x 4

Æ

14

a  ( a)

0Æ

2 + (–2) = 2 – (+2) = 2 – 2 = 0

ba

b
c

1 for a z 0

a(b)

ab Æ

(–2) x 3 = 2 x (–3)

Æ

–6

Æ

(–2) x (–3) = 2 x 3

Æ

6

a 2  2ab  b 2 Æ (2 + 3)2 = 22 + 2(2)(3) + 32

Æ

25

a 2  2ab  b 2 Æ (2 – 3)2 = 22 – 2(2)(3) + 32

Æ

1

(2 + 3)(2 – 3) = 22 – 32

Æ

–5

2
3

Æ

2
3

Æ



2˜4  3
4

Æ

11
4

2˜5  3˜ 4
3˜5

Æ

22
15

ab

a2  b2 Æ

1 ˜

a/b Æ

a
b



(2) /(3)

a
2
Æ
b
3

(2) /(3)

ac  b
c

Æ

2

ad  bc
bd

Æ

2 4


3 5

3
4

2/3
2
3

2
3

Chapter 1: Algebra Review

3


Math. Chapter 1. Algebra Review

C.

b
c

15.

au

a˜

16.

a
b
c
d

a c
y
b d

17.

a1 / 2

a 0.5

18.

a 1/n =

n

a

b
c

ab
c

Æ

a d
u
b c

ad
bc

2u

3
4

2˜

2
3
4
5

Æ

2 4
y
3 5

a where a • 0 Æ
Æ

where a • 0

19.

ab =

a* b

20.

a
b

21.

ab z a  b

22.

a
z
b

a
b

a
b

2˜3
4

3
4

Æ

2˜5
Æ
3˜ 4

2 5
u
3 4

21 / 2

2 0.5

2 1/3 =

3

2

2

Æ

2˜3 =

Æ

2
3

Æ

23 z 2  3

Æ

2
z
3

2* 3

6

2
3

1.4142

Æ

1.2599

Æ

2.4495

Æ

0.8165

2
3

It is very important that we know the following properties of exponents:
Note that 00 = undefined

Æ

X0

2.

1
Xb

3.

Xa *Xb

1

X b = X ^ (b)

Xa ˜Xb

Æ 23 ˜ 2 4

X2˜X3

4.

( X a )b

X a˜b

X a*b
Æ (X )

2 3

4

2 3 4

X

Step by Step Business Math and Statistics

27

X

X ^ ( a  b)

Æ

X 2X 3

X 23

X5

= 128
Æ

X ab
2*3

X 10 = X ^ (10)

X a b

X aXb

Æ X2*X3

1
X 10

Æ

2˜3

X

6

X ^ ab

10
12

Æ

Properties of Exponents Æ Pay attention to equivalent notations

1.

6
4


Math. Chapter 1. Algebra Review

5.

6.

7.

Xa
Xb

Æ (2 3 ) 4

2 3˜4

X a ˜ X b

X a b

X2
Æ 3
X

X 2 ˜ X 3

X

X aY a

Æ ( XY ) 2

X 2 *Y 2

X 2 ˜Y 2

X

1
n

D.

X p/q

1
X

X 1

X a ˜Y a

Æ

8.

X 2 3

X a *Y a

( XY ) a

n

212 = 4096

X 2Y 2

X 1/ n

4

4

( X 1/ q ) p

1
2

41 / 2

4 0.5
q

( X p )1 / q

Æ 210/5

22

Æ 82 / 3

(2 3 ) 2 / 3

(21/5 )10
22

2 2˜0.5

( 2 2 ) 0. 5

21

2

Xp
5

(210 )1/5
3

82

3

210 = 22 = 4

64

4

Linear and Nonlinear Functions
1.

Linear Functions
Linear Functions have the general form of:
Y=a+bX
where Y and X are variables and a and b are constants. More specifically,
a is called an intercept and b, a slope coefficient. The most visually
distinguishable character of a linear function is that it is a straight line.
Note that +b means a positive slope and –b means a negative slope.

2.

Nonlinear Functions
There are many different types of nonlinear functions such as polynomial,
exponential, logarithmic, trigonometric functions, etc. Only polynomial,
exponential and logarithmic functions will be briefly explained below.
i) The n-th degree polynomial functions have the following general form:

Chapter 1: Algebra Review

5


Math. Chapter 1. Algebra Review

Y a  bX  cX 2  dX 3  ......  pX n 1  qX n
Or alternatively expressed as:
Y

qX n  pX n 1  ......  dX 3  cX 2  bX  a

where a, b, c, d, …, p and q are all constant numbers called coefficients
and n is the largest exponent value.
Note that the n-th degree polynomial function is named after the highest
value of n. For example, when n = 2, it is most often called a quadratic
function, instead of a second-degree polynomial function, and has the
following form:

Y

a  bX  cX 2

When n = 3, it is called a third-degree polynomial function or a cubic
function and has the following form:

Y

a  bX  cX 2  dX 3

ii) Finding the Roots of a Polynomial Function
Often, it is important and necessary to find roots of a polynomial function,
which can be a challenging task. An n-th degree polynomial function will
have n roots. Thus, a third degree polynomial function will have 3 roots
and a quadratic function, two roots. These roots need not be always
different and in fact, can have the same value. Even though finding roots
to higher-degree polynomial functions is difficult, the task of finding the
roots of a quadratic equation is manageable if one relies on either the
factoring method or the quadratic formula.
If we are to find the roots to a quadratic function of:

aX 2  bX  c

0

we can find their two roots by using the following quadratic formula:

X1, X 2

 b r b 2  4ac
2a

iii) Examples:
Find the roots, X1 and X2, of the following quadratic equations:
(a)
6

X 2  3X  2

Step by Step Business Math and Statistics

0


Math. Chapter 1. Algebra Review

Factoring Method1:

X 2  3 X  2 ( X  1) ˜ ( X  2)

0

Therefore, we find two roots as: X1 = 1 and X2 = 2.
Quadratic Formula2:
Note: a 1 , b

 b r b 2  4ac
2a

X1, X 2

=

(b)

3 , and c

3r 98
2

4 X 2  24 X  36

2

 (3) r (3) 2  4(1)(2)
2 ˜1

3 r1
1, 2
2

0

Factoring Method:

4 X 2  24 X  36 (2 X  6) ˜ (2 X  6) (2 X  6) 2

4( X  3)2

0

Therefore, we find two identical roots (or double roots) as:

X1

X2

3

Quadratic Formula:
Note: a 4 , b

X1, X 2

24 , and c

 b r b 2  4ac
2a

36

 (24) r (24) 2  4(4)(36)
2˜4

1

The factoring method often seems more convenient for people with great experience with algebra. That is,
the easiness comes with experience. Those who lack algebraic skill may be better off using the quadratic
formula.
2
In order to use the quadratic formula successfully, one must match up the values for a, b, and c correctly.

Chapter 1: Algebra Review

7


Math. Chapter 1. Algebra Review

=
(c)

 24 r 576  576
8

4 X 2  9Y 2

 24 r 0
8



24
8

3

0

Factoring Method:

4 X 2  9Y 2

(2 X  3Y ) ˜ (2 X  3Y )

0

Therefore, we find two roots as:
X1

3Y
2

1.5Y and X 2



3Y
2

1.5Y

Quadratic Formula3:

9Y 2

4, b

X1, X 2

b r b 2  4ac

2a

(0) r (0) 2  4(4)(9Y 2 )
2˜4

0 r 0  144Y 2
8

r 12Y
8

=

E.

0 , and c

Note: a

3
r Y
2

1.5Y ,  1.5Y

Exponential and Logarithmic Functions
1.

Exponential Functions
An exponential function has the form of Y a ˜ b X where a and b are

constant numbers. The simplest form of an exponential function is Y b X
where b is called the base and X is called an exponent or a growth factor.
A unique case of an exponential function is observed when the base of e is
used. That is, Y e X where e | 2.718281828 . Because this value of e is
often identified with natural phenomena, it is called the “natural” base4.

3

One must be very cognizant of the construct of this quadratic equation. Because we are to find the roots
associated with X, –9Y2 should be considered as a constant term, like c in the quadratic equation.
n

§ 1·
Technically, the expression ¨1  ¸ approaches e as n increases. That is, as n approaches  f ,
© n¹
e | 2.718281828 .

4

8

Step by Step Business Math and Statistics


Math. Chapter 1. Algebra Review

Examples>
In order to be familiar with how exponential functions work, please verify the
following equalities by using a calculator.

5e 2 4

a.

5e 2 ˜ e 4

b.

(5e 3 ) ˜ (3e 4 ) 15e 7

c.

10e 3 y 2e 4

5e 6

10 34
˜e
2

5 ˜ 403.4287935

2017.143967

15 ˜ 1096.633158 16449.49738
5e 1

5
e

5
1.839397206
2.718281828

2. Logarithmic Functions
The logarithm of Y with base b is denoted as “ log b Y ” and is defined as:

log b Y

X if and only if b X

Y

provided that b and Y are positive numbers with b z 1 . The logarithm
enables one to find the value of X given 2 X 4 or 5 X 25 . In both of
these cases, we can easily find X=2 due to the simple squaring process
involved. However, finding X in 2 X 5 is not easy. This is when
knowing a logarithm comes in handy.
Examples>
Convert the following logarithmic functions into exponential functions:

log 2 8 X
log 5 1 0

Æ
Æ

2X 8
50 1

log 4 4 1

Æ

41

log1 / 2 4

2

Æ

§1·
¨ ¸
©2¹

Æ

X=3

4
2

2


1  2

2 2

22

4

a. Special Logarithms: A common logarithm and a natural logarithm.
i)

A Common Logarithm = a logarithm with base 10 and often
denoted without the base value.
That is, log10 X

ii)

log X Æ read as "a (common) logarithm of X."

A Natural Logarithm = a logarithm with base e and often denoted
as ‘ln”.
Chapter 1: Algebra Review

9


Math. Chapter 1. Algebra Review

That is, log e X

ln X Æ read as "a natural logarithm of X."

b. Properties of Logarithms
i)

Product Property:

log b mn

ii)

Quotient Property:

log b

iii)

Power Property:

log b m n

m
n

log b m  log b n
log b m  log b n

n ˜ log b m

Example 1> Using the above 3 properties of logarithm, verify the following
equality or inequality by using a calculator.

ln(5 ˜ 6)

i)

ln 30

ii)

ln

iii)

ln 20
20
z ln
ln 40
40

iv)

ln 10 3

20
40

ln 5  ln 6

ln 20  ln 40

3 ˜ ln 10

Example 2> Find X in 2 X

Æ

ln 0.5 Æ

ln 1000

Æ

3.401197
–0.693147

6.907755

5 . (This solution method is a bit advanced.)

In order to find X,
(1) we can take a natural (or common) logarithm of both sides as:
ln 2 X ln 5
(2) rewrite the above as: X ˜ ln 2 ln 5 by using the Power Property
ln 5
(3) solve for X as: X
ln 2
(4) use the calculator to find the value of X as:
ln 5 1.6094379
X
2.321928095
ln 2 0.6931471

10

Step by Step Business Math and Statistics


Math. Chapter 1. Algebra Review

Additional topics of exponential and logarithmic functions are complicated and
require many additional hours of study. Because it is beyond our realm, no
additional attempt to explore this topic is made herein5.
F.

Useful Mathematical Operators
n

1.

Summation Operator = Sigma = Ȉ Æ ¦ X i

¦

i 1

n

¦X

n

Xi

i 1

¦X

i

X 1  X 2  .....  X n 1  X n = Sum Xi’s where i goes from 1 to n.

i

i 1

Examples: Given the following X data, verify the summation operation.
i =

1

2

3

4

5

Xi =

25

19

6

27

23

3

¦X

a.

i

X1  X 2  X 3

25  19  6

i

X1  X 2  X 3  X 4  X 5

i

X3  X4  X5

50

i 1

5

¦X

b.

25  19  6  27  23 100

i 1

5

¦X

c.

6  27  23 56

i 3

3

5

i 1

i 3

3

5

i 1

i 3

¦ X i  ¦ X i (X1  X 2  X 3 )  (X 3  X 4  X 5 )

d.

(25  19  6)  (6  27  23)

¦ X i  ¦ X i (X1  X 2  X 3 )  (X 3  X 4  X 5 )

e.

(25  19  6)  (6  27  23)
n

2.

Multiplication Operator = pi = Ȇ Æ – X i
i 1

n

–X

50  56 106

i

50  56

–X

6

i

X 1 ˜ X 2 ˜ .... ˜ X n 1 ˜ X n = Multiply Xi’s where i goes from 1 to n.

i 1

5

For detailed discussions and examples on this topic, please consult high school algebra books such as
Algebra 2, by Larson, Boswell, Kanold, and Stiff. ISBN=13:978-0-618-59541-9.

Chapter 1: Algebra Review

11


Math. Chapter 1. Algebra Review

Examples: Given the following X data, verify the multiplication operation.
i =

1

2

3

4

5

Xi =

3

5

6

2

4

a.
3

–X

X1 ˜ X 2 ˜ X 3

i

3˜5˜ 6

90

i 1

5

b.

–X

i

X1 ˜ X 2 ˜ X 3 ˜ X 4 ˜ X 5

3˜5˜6˜ 2˜ 4

i

X3 ˜ X4 ˜ X5

48

720

i 1
5

c.

–X

6˜2˜4

i 3

d.

e.

f.

G.

3

5

i 1

i 3

3

5

i 1

i 4

– X i  – X i (X1 ˜ X 2 ˜ X 3 )  (X 3 ˜ X 4 ˜ X 5 )
(3 ˜ 5 ˜ 6)  (6 ˜ 2 ˜ 4)

– Xi – Xi

(X1 ˜ X 2 ˜ X 3 )  (X 4 ˜ X 5 )

(3 ˜ 5 ˜ 6)  (2 ˜ 4)

2

5

i 1

3

¦ Xi  – Xi

90  48 138

90  8

72

(X1  X 2 )  (X 3 ˜ X 4 ˜ X 5 )

(3  5)  (6 ˜ 2 ˜ 4)

8  48

40

Multiple-Choice Problems for Exponents, Logarithms, and Mathematical
Operators:
Identify all equivalent mathematical expressions as correct answers.

1.

12

(X + Y)2 =
a.

X2 + 2XY + Y2

b.

X2 – 2XY + Y2

c.

X2 + XY + Y2

d.

X2 + 2XY + 2Y2

e.

none of the above

Step by Step Business Math and Statistics


Math. Chapter 1. Algebra Review

2.

3.

4.

5.

6.

7.

(X – Y)2 =
a.

X2 + 2XY + Y2

b.

(X – Y) (X – Y)

c.

X2 – 2XY + Y2

d.

X2 – XY + Y2

e.

only (b) and (c) of the above

(2X + 3Y)2 =
a.

4X2 + 6YX + 9Y2

b.

4X2 + 12XY + 9Y2

c.

2X2 + 6XY + 3Y2

d.

4X2 + 9Y2

e.

none of the above

(2X – 3Y)2 =
a.

4X2 – 9Y2

b.

2X2 + 6XY + 3Y2

c.

4X2 – 12XY + 9Y2

d.

4X2 + 9Y2

e.

none of the above

(2X3)(6X10) =
a.

2X3+10

b.

12X30

c.

48X3/10

d.

12X13

e.

none of the above

(12X6Y2)(2Y3X2)(3X3Y4) =
a.

72X11Y9

b.

72X12Y8

c.

17X10Y10

d.

72Y8 X12

e.

only (b) and (d) of the above

X2(X + Y)2 =

Chapter 1: Algebra Review

13


Math. Chapter 1. Algebra Review

8.

9.

10.

11.

a.

X2(X2 + 2XY + Y2)

b.

X2+2 + 2X1+2Y + X2Y2

c.

X4 + 2X3Y + X2Y2

d.

all of the above

e.

none of the above

X3 6
˜
=
2 X2
a.

3X 5

b.

3X

c.

3 X 1

d.

12 X

e.

none of the above

(2X3)/(6X10) =
a.

0.33333333X 7

b.

1
3X 7

c.

1 7
X
3

d.

only (a) and (c) of the above

e.

all of the above

10
˜ X 5Y 3 =
9 5
X Y
a.

10 X 4Y 2

b.

c.

10 X 9Y 5 X 5Y 3

d.

e.

all of the above

only (a) and (b) of the above

24 X 0.5Y 1.5 y 12 X 1.5Y 0.5 =
a.
d.

14

10

X 4Y 2

2Y
X
Y
X

b.
e.

Step by Step Business Math and Statistics

2X
Y
X
Y

c.

Y
2X


Math. Chapter 1. Algebra Review

12.

15.

2. 5

2

b.

2

X Y
2Y
X2

d.

14.

4
3

(64 X ) (8Y ) y 8 X Y =

a.

13.

1
3

1
2

2
XY

2
Y X

c.

e.

none of the above

2

(2X2)3 =
a.

2X6

b.

8X6

d.

16X6

e.

(2X3)2

c.

8X5

c.

81X16Y10

c.

1/Y2

[(3X4Y3)2]2 =
a.

9X8Y7

b.

9X16Y12

d.

81X16Y12

e.

81X8Y7

(4X4Y3)2/(2X2Y2)4 =
a.

Y-3

b.

Y2

d.

X2

e.

1/X2

Using the following data, answer Problems 16 – 20.
i =

1

2

3

4

5

6

7

Xi =

30

52

67

22

16

42

34

3

16.

¦X

i

i 1

a.
d.

6
104520

b.
e.

140
c.
none of the above

149

Chapter 1: Algebra Review

15


Math. Chapter 1. Algebra Review

6

17.

–X

i

i 4

a.
d.
18.

6
14784

2

5

i 1

3

– 23
2350

7

5

i 5

3

92
46432

7

i

a.
d.

0
c.
none of the above

2352

b.
e.

23584
c.
none of the above

23676

i

i 3

0
672

b.
e.

1428
84

89

i

5

Find the value of X in 3 X
a.
d.

22.

b.
e.

6

4

– X ¦ X  – X
i 6

21.

120

¦ Xi  – Xi
a.
d.

20.

80
c.
none of the above

¦ Xi – Xi
a.
d.

19.

b.
e.

20
5

b.
e.

c.

59049 .

15
c.
none of the above

10

Identify the correct relationship(s) shown below:
a.
c.
e.

X log 20 log 20 X
b.
2
5 ln
ln 2
d.
5
none of the above is correct.

15 X z 3 X ˜ 5 X
ln X
ln X  ln Y
ln Y

Answers to Exercise Problems for Exponents and Mathematical
Operators
1.

(X + Y)2 =

a.*

16

X2 + 2XY + Y2 because
(X + Y) (X + Y) = X2 + XY + YX + Y2 = X2 + 2XY + Y2

Step by Step Business Math and Statistics


Math. Chapter 1. Algebra Review

2.

(X – Y)2 =
e.*

3.

(2X + 3Y)2 =
b.*

4.

3 X because

6X 3
2X 2

3 X 3 2

3X

(2X3)/(6X10) =
e.*

10.

all of the above because
X2(X2 + 2XY + Y2) = X2+2 + 2X1+2Y + X2Y2 = X4 + 2X3Y + X2Y2

X3 6
˜
=
2 X2
b.*

9.

72X11Y9 because (12)(2)(3)X6+2+3Y2+3+4 = 72X11Y9

X2(X + Y)2 =
d.*

8.

12X13 because (2)(6)X3+10 = 12X3+10 = 12X13

(12X6Y2)(2Y3X2)(3X3Y4) =
a.*

7.

4X2 – 12XY + 9Y2 because
(2X – 3Y)2 = (2X – 3Y) (2X – 3Y) = 4X2 – 6XY – 6YX + 9Y2
= 4X2 – 12XY + 9Y2

(2X3)(6X10) =
d.*

6.

4X2 + 12XY + 9Y2 because
(2X + 3Y) (2X + 3Y) = 4X2 + 6XY + 6YX + 9Y2
= 4X2 + 12XY + 9Y2

(2X – 3Y)2 =
c.*

5.

only (b) and (c) of the above because
(X – Y) (X – Y) = X2 – XY – YX + Y2 = X2 – 2XY + Y2

all of the above because (2/6)X3-10 = (1/3)X-7 = 0.3333 X 7

1
3X 7

10

˜ X 5Y 3 =
X 9Y 5
d.*

only (a) and (b) of the above because

Chapter 1: Algebra Review

17


Math. Chapter 1. Algebra Review

10 X 9Y 5 X 5Y 3
11.

24 X 0.5Y 1.5
2Y
because
X
12 X 1.5Y 0.5
1

1

2 X 0.51.5Y 1.50.5

2

a.*

X 2Y

=

because

1

8X Y

1

1

1

(64) 2 X 2 (8) 3 Y 3

4
3

2 .5

8X Y

4
3

1

(8) X 2 ( 2)Y 3
8X Y

1

(64 X ) 2 (8Y ) 3
2 .5

2 .5

X

4
3

1
 2 .5
2

1 4

3

( 2)Y 3

2 X  2Y 1

2
X 2Y

(2X2)3 =
8X6 because (2X2) (2X2) (2X2) = (2)3X2+2+2 = 23X2x3 = 8X6

b.*

[(3X4Y3)2]2 =
81X16Y12 because [32X4x2Y3x2]2 = 32x2X8x2Y6x2 = 81X16Y12

d.*
15.

2Y
X

4

1

14.

2 X 1Y 1

(64 X ) 2 (8Y ) 3 y 8 X 2.5Y 3 =
1

13.

10
X 4Y 2

10 X  4Y  2

24 X 0.5Y 1.5 y 12 X 1.5Y 0.5 =
a.*

12.

10 X 95Y 53

(4X4Y3)2/(2X2Y2)4 =
1/Y2

c.*

because

(4X4Y3)2(2X2Y2)-4 = [(22)2X8Y6][(2)-4 X-8Y-8]= Y-2 = 1/Y2
3

16.

¦X

i

i 1

3

c.*

149

because

¦X

i

X1  X 2  X 3

30  52  67 149

i 1

6

17.

–X

i

i 4

6

d.*

14784 because – X i
i 4

18

Step by Step Business Math and Statistics

X4 ˜ X5 ˜ X6

22 ˜ 16 ˜ 42 14784


Math. Chapter 1. Algebra Review

5

2

18.

¦ X – X
i

i 1

i

i 3

e.*

none of the above
2

5

i 1

i 3

because ¦ X i  – X i

( X1  X 2 )  ( X 3 ˜ X 4 ˜ X 5 )

(30  52)  (67 ˜ 22 ˜ 16)

19.

¦ X – X
i

i 5

c.*

82  23584

i

i 3

23676
7

5

i 5

i 3

because ¦ X i  – X i

(X5  X6  X7 )  ( X3 ˜ X4 ˜ X5)

(16  42  34)  (67 ˜ 22 ˜ 16)

20

7

4

6

i 6

i 3

i 5

e.*

845

– X i ¦ X i  – X i

7

4

6

i 6

i 3

i 5

23676

( X 6  X 7 ) ( X 3  X 4 )  ( X 5 ˜ X 6 )

(42 ˜ 34)  (67  22)  (16 ˜ 42) 1428  89  672 845

Find the value of X in 3 X
c.*

92  23584

20.

because – X i  ¦ X i  – X i

21.

23502

5

7

59049 .

10

In order to find X,
(1) we can take a natural (or common) logarithm of both sides as:
ln 3 X ln 59049
(2) rewrite the above as: X ˜ ln 3 ln 59049 by using the Power Property
ln 59049
(3) solve for X as: X
ln 3
(4) use the calculator to find the value of X as:
ln 59049 10.9861
X
10
ln 3
1.09861
22.

Identify the correct relationship(s) shown below:
e.*

none of the above is correct.

Note that

Chapter 1: Algebra Review

19