(a) A = P[i(1 + i)n / (1 + i)n - 1]
= 10000[0.05(1 + 0.05)5 / (1 + 0.05)5 - 1]
= 10000 [0.0638 / 0.2763]
= $2,309.08
(b) A = P[i(1 + i)n / (1 + i)n - 1]
= 5500[0.097(1 + 0.097)4 / (1 + 0.097)4 - 1]
= 5500 [0.1405 / 0.4482]
= $1,724.12
(c) A = P[i(1 + i)n / (1 + i)n - 1]
= 8500[0.025(1 + 0.025)3 / (1 + 0.025)3 - 1]
= 8500 [0.0269 / 0.0769]
= $2,973.34
(d) A = P[i(1 + i)n / (1 + i)n - 1]
= 30000[0.085(1 + 0.085)20 / (1 + 0.085)20 - 1]
= 30000 [0.4345 / 4.1120]
= $3,170
3.22 What equal annual payment series is required to repay the following present amounts? (a) $10,000...
only answer b,c,d thanks . What equal-payment series is required to repay the following present amounts? a. $10,000 in 4 years at 10% interest compounded annually with 4 annual payments. b. $5,000 in 3 years at 12% interest compounded semiannually with 6 semiannual payments. c. $6,000 in 5 years at 8% interest compounded quarterly with 20 quarterly payments. d. $80,000 in 30 years at 9% interest compounded monthly with 360 monthly payments.
Problem 2-39 (book/static) What equal annual payment series is required to repay the following present amounts? (a) $15,000 in six years at 3.5% interest compounded annually (b) $7,500 in seven years at 7.5% interest compounded annually (c) $2,500 in five years at 5.25% interest compounded annually (d) $12,000 in 15 years at 6.25% interest compounded annually (a) The annual payment is S. (Round to the nearest dollar.) Enter your answer in the answer box and then click Check Answer equipment...
Determine the equal, annual, end-of-year payment required over the life of the following loans to repay them fully during the stated term. Loan Principal ($) Interest Rate (%) Term of Loan (Yrs) Annual Payment ($) A 110,000 8 17 B 20,000 14 10 Loan Principal ($) Interest Rate (%) Term of Loan (Yrs) Annual Payment ($) A 10,000 8 17 (Round to the nearest cent.)
determine the annual payment of 5% compounded annually to repay a loan of $ 100,000,000 in 20 years please show every single step!!
b) What is the equal payment series for 10 years that is equivalent to a payment series of $18,000 at the end of the first year, decreasing by $1,500 each year over 10 years? Interest is 9% compounded annually.
.29 What is the equal payment series for 12 years that is equivalent to a payment series of $15,000 at the end of the first year, decreasing by $1000 each year over 12 years? Interest is 8% compounded annually.
What is an annuity? Select one: a. present worth of a series of equal payments. b. a single payment. c. a series of payments that changes by a constant amount from one period to the next. d. a series of equal payments over a sequence of equal periods. e. a series of payments that changes by the same proportion from one period to the next. Question 2 The present worth factor Select one: a. gives the future value equivalent to...
You borrow $100,000 today. You will repay the loan with 5 equal annual payments starting next year. Each payment is equal to $20,000 In addition to these payments, you will make a "balloon payment" in year 5 . If the interest rate on the loan is 2% APR, compounded annually, how big is the balloon payment? Group of answer choices $6,304 $6,960 $5,731 $6,327
Answer the following questions: a. What equal annual series of payments must be paid to accumulate $12,000 in 13 years at 5%, compounded annually? b. Part of the income a machine generates is put into a fund to finance the purchase of a new machine. If $1,500 is invested annually at 7% interest (compound) how many years before the fund becomes $25,000? c. What is the future value of the following series of payments: $1,000 for 5 years at 8.25%?
Kelsey owes $10,000 today but cannot make the payment. Instead she is to repay this amount in two equal payments, one in 3 months from today and one in 9 months from today. If the focal date is 9 months from today and money is worth 4.5% compounded quarterly. The amount of the equal payments is (6 marks) a. $5,170.66 b. $5,112.81 C. $5,042.13 d. $5,141.73 Julia decides to purchase a new home and takes out a mortgage for $250,000....