Explanation Given: The principal of loan P is $ 37 , 948 , the rate of interest is 6.5% compounded annually and there are 10 annual payments. Explanation: Use the formula for the Amortization payments: R = P i 1 − ( 1 + i ) − n Here, P is the principal of loan, R is the periodic payment, i is the interest rate per period, and n is the number of periods. As 10 annual payments are made, so n = 10 , and the interest rate per period will be as follows: i = 6.5 % = 0.065 Substitute P = $ 37 , 948 , i = 0.065 , and n = 10 to obtainthe value of each annual payment as follows: R = P i 1 − ( 1 + i ) − n = 37 , 948 × 0.065 1 − ( 1 + 0.065 ) − 10 = 37 , 948 × 0.065 1 − 1

Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

11th Edition

ISBN: 9780133886818

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 5.3, Problem 54E

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An amortization schedule for a loan of $37,948 at an interest of 6.5% compounded annually to be paid in 10 years.

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Chapter 5.3, Problem 53E

Chapter 5.3, Problem 55E

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Chapter 5 Solutions

Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

Ch. 5.1 - 1. What factors determine the amount of interest...Ch. 5.1 - In your own words, describe the maturity value of...Ch. 5.1 - 3. What is meant by the present value of...Ch. 5.1 - We calculated the loan in Example 2(b) assuming...Ch. 5.1 - Find the simple interest. $25,000 at 3% for 9...Ch. 5.1 - Find the simple interest. 6. $4289 at 4.5% for...Ch. 5.1 - Find the simple interest. 7. $1974 at 6.3% for...Ch. 5.1 - Find the simple interest. 8. $6125 at 1.25% for...Ch. 5.1 - Find the simple interest. Assume a 360-day year....Ch. 5.1 - Find the simple interest. Assume a 360-day...

Ch. 5.1 - Find the maturity value and the amount of simple...Ch. 5.1 - Find the maturity value and the amount of simple...Ch. 5.1 - Find the maturity value and the amount of simple...Ch. 5.1 - Find the maturity value and the amount of...Ch. 5.1 - 15. Explain the difference between simple...Ch. 5.1 - Prob. 16E Ch. 5.1 - 17. What is the difference between t and n? 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Delinquent Taxes An accountant for a...Ch. 5.1 - Savings A $1500 certificate of deposit held for 75...Ch. 5.1 - Prob. 52E Ch. 5.1 - Cash Advance Fees According to an advertisement,...Ch. 5.1 - Prob. 54E Ch. 5.1 - Investments Suppose $10,000 is invested at an...Ch. 5.1 - 56. Investments In Exercise 55, notice that as...Ch. 5.1 - Prob. 57E Ch. 5.1 - Prob. 58E Ch. 5.1 - Student Loan Upon graduation from college, Warren...Ch. 5.1 - Comparing Investments Two partners agree to invest...Ch. 5.1 - Prob. 61E Ch. 5.1 - Prob. 62E Ch. 5.1 - Negative Interest Under certain conditions, Swiss...Ch. 5.1 - Negative Interest Under certain conditions,...Ch. 5.1 - Savings On January 1, 2010, Jack deposited $1000...Ch. 5.1 - Prob. 66E Ch. 5.1 - Prob. 67E Ch. 5.1 - Prob. 68E Ch. 5.1 - Prob. 69E Ch. 5.1 - 70. 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An amortization schedule for a loan of $37,948 at an interest of 6.5% compounded annually to be paid in 10 years.

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An amortization schedule for a loan of $37,948 at an interest of 6.5% compounded annually to be paid in 10 years.

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Chapter 5 Solutions