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Test Bank for Business Mathematics In Canada 11th Edition By F. Ernest Jerome, Tracy Worswick

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Test Bank for Business Mathematics In Canada 11th Edition By F. Ernest Jerome, Tracy Worswick. CHAPTER 1 Review and Applications of Basic Mathematics Appendix 1A: The Texas Instruments BA II PLUS CHAPTER 2 Review and Applications of Algebra CHAPTER 3 Percent and Percent Change CHAPTER 4 Ratios and ...

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  • Business Mathematics In Canada 11th Edition
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TEST BANK
Business Mathematics in Canada 11th Edition by
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Tracy Worswick F. Ernest Jerome
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CHAPTER 1 Review and Applications of Basic Mathematics
Appendix 1A: The Texas Instruments BA II PLUS
CHAPTER 2 Review and Applications of Algebra
CHAPTER 3 Percent and Percent Change
CHAPTER 4 Ratios and Proportions
CHAPTER 5 Mathematics of Merchandising
5.2 Supplement: Other Notations for Terms of Payment (on Connect)
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5.3 Supplement: Diagram Model for Markup Problems (on Connect)
CHAPTER 6 Applications of Linear Equations
Appendix 6A: The Texas Instruments BA II PLUS Break-Even Worksheet
CHAPTER 7 Simple Interest
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Appendix 7A: An Aid for Determining the Number of Days in Each Month
Appendix 7B: The Texas Instruments BA II PLUS Date Worksheet
CHAPTER 8 Applications of Simple Interest
Appendix 8A: Promissory Notes
CHAPTER 9 Compound Interest: Future Value and Present Value
Appendix 9A: Instructions for Specific Models of Financial Calculators
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CHAPTER 10 Compound Interest: Further Topics and Applications
Appendix 10A: The Texas Instruments BA II PLUS Interest Conversion Worksheet
Appendix 10B: Logarithms
Appendix 10C: Annualized Rates of Return and Growth (on Connect)
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CHAPTER 11 Ordinary Annuities: Future Value and Present Value
CHAPTER 12 Ordinary Annuities: Periodic Payment, Number of Payments, and Interest Rate
Appendix 12A: Derivation of the Formula for n from the Formula for FV_(on Connect)
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Appendix 12B: The Trial-and-Error Method for Calculating the Interest Rate per Payment Interval_(on Connect)
CHAPTER 13 Annuities Due
Appendix 13A: Setting Your Calculator in the Annuity Due Mode
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CHAPTER 14 Annuities: Special Situations
CHAPTER 15 Loan Amortization: Mortgages
15.4: Mortgage Loans: Additional Topics (on Connect)
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Appendix 15A: Instructions for the Texas Instruments BA II PLUS Amortization Worksheet
Appendix 15B: Amortization Functions on the Sharp EL-738 Calculator
CHAPTER 16 Bonds and Sinking Funds
Appendix 16A: Instructions for the Texas Instruments BA II PLUS Bond Worksheet
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CHAPTER 17 Business Investment Decisions
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1
Medconnoisseurlibraries.com
Review
Reviewand
and Applications
Applications of
of
BasicBasic
Mathematics
Mathematics
Exercise 1.1
a. 10 + 10 x 0 = 10 + 0 = 10
b. 2x2+4–8=4+4–8=0
c. (10 + 10) x 0 = 20 x 0 = 0
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d. 2 x (2 + 4) – 8 = 2 x 6 – 8 = 12 – 8 = 4
e. 0 + 3 x 3 – 32 + 10 = 0 + 9 – 9 + 10 = 10
f. 12 – 2 x 5 + 22 x 0 = 12 – 10 + 4 x 0 = 12 – 10 + 0 = 2
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g. 0 + 3 x 3 – (32 + 10) = 0 + 9 – 19 = -10
h. (12 – 2) x (5 + 22) x 0 = 10 x 9 x 0 = 0
22 −4 4 −4 0
i. = = =0
(4−2)2 22 4

(2 −4)2 (−2)2 4
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= = =4
5−22 5 −4 1
j.
1. 20 − 4  2 − 8 = 20 − 8 − 8 = 4
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2. 18  3 + 6  2 = 6 + 12 = 18
3. (20 − 4)  2 − 8 = 16  2 − 8 = 32 − 8 = 24
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4. 18  (3 + 6)  2 = 18  9  2 = 2  2 = 4

5. 20 − (4  2 − 8) = 20 − (8 − 8) = 20
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6. (18  3 + 6)  2 = (6 + 6)  2 = 24
7. 54 − 36  4 + 22 = 54 − 9 + 4 = 49
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8. (5 + 3)2 − 32  9 + 3 = 82 − 9  9 + 3 = 64 − 1+ 3 = 66
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9. (54 − 36)  (4 + 2)2 = 18  62 = 18  36 = 0.5
(
10. 5 + 32 − 3 )2  (9 + 3) = 5 + (9 − 3)2  12 = 5 + 36  12 = 5 + 3 = 8
82 − 42 64 − 16 = 48 = 6
11. =
(4 − 2)3 23 8
2
(8 − 4)2 4 16
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12. = = =−4
4 − 23 4−8 −4
13. 3(6 + 4)2 − 5(17 − 20)2 = 3 10 2 − 5(− 3)2 = 3 100 − 5  9 = 300 − 45 = 255
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14. (4  3 − 2)2  (4 − 3  22 ) = (12 − 2)2  (4 − 3  4) = 102  (4 − 12) = 100  (− 8) = − 12.5
15. (20 + 8  5) − 7  (− 3) 9 = (20 + 40 + 21)  9 = 81 9 = 9

( )  
519 + 52 − 16 2  = 5 19 + (25 − 16 )2 2 = 5(19 + 81)2 = 5  100 2 = 50,000
2
16.
 
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17. $100(1+ 0.06  365
45
) = $100(1+ 0.00739726) = $100.74
$200 $200 $200
= = = $194.17
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18.
1+ 0.09  12 1+ 0.03 1.03
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$500 = $500 $500
=
19.
(1 + 0.05) 1.05 1.1025 = $453.51
2 2
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( )
20. $1000 (1+ 0.02) = $1000 1.023 = $1000 (1.061208 ) = $1061.21
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 (1+ 0.04)2 − 1  1.042 − 1  0.0816 
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21. $100 0.04  = $100 0.04  = $100 0.04  = $204.00
     
 1   1 − 1 
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1 − (1 + 0.03)2  1.0609  1 − 0.942596 
22. $300  = $300  = $300  = $574.04
 0.03   0.03   0.03 
   
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   

Concept Questions (Section 1.2)
1. You must retain at least one more figure than you require in the answer. To achieve four-
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figure accuracy in the answer, you must retain a minimum of five figures in the values
used in the calculations. B)
2. We want six-figure accuracy in the answer. Therefore, values used in the calculations
must be accurate to at least seven figures. B)
3. We want seven-figure accuracy in the answer. Therefore, values used in the calculations
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must retain at least eight figures. C)
4. To be accurate to the nearest 0.01%, an interest rate greater than 10% must have four-
figure accuracy. Therefore, five figures must be retained in numbers used in the
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calculations. C)




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