The maintenance expense on a piece of machinery is estimated to be as follows: Year Maintenance Cost 1 2 3 $150 $300 $450 If the interest rate is 8%, the equivalent annual maintenance cost is? 4 $600

Essentials Of Investments
11th Edition
ISBN:9781260013924
Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Chapter1: Investments: Background And Issues
Section: Chapter Questions
Problem 1PS
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1. Use the provided chart to solve this problem. Show your work.

1. The maintenance expense on a piece of machinery is estimated to be as follows:
Year
Maintenance Cost
1
2
3
$150 $300 $450
4
$600
If the interest rate is 8%, the equivalent annual maintenance cost is?
Transcribed Image Text:1. The maintenance expense on a piece of machinery is estimated to be as follows: Year Maintenance Cost 1 2 3 $150 $300 $450 4 $600 If the interest rate is 8%, the equivalent annual maintenance cost is?
factor name
single payment
compound amount
single payment
present worth
uniform series
sinking fund
capital recovery
uniform series
compound amount
uniform series
present worth
uniform gradient
present worth
uniform gradient
future worth
uniform gradient
uniform series
converts
P to F
F to P
F to A
P to A
A to F
A to P
G to P
G to F
G to A
symbol
(F/P, i%, n)
(P/F, 1%, n)
(A/F, i%, n)
(A/P, i%, n)
(F/A, 1%, n)
(P/A, i%, n)
(P/G,i%, n)
(F/G,1%, n)
(A/G, i%, n)
formula
(1 + i)"
(1 + i)-¹
i
(1 + i)" - 1
i(1+i)"
(1+i)n-1
(1+i)n-1
i
(1+i)n-1
i(1+i)n
1-
i
(1+i)n-1
1² (1+i)n
n
i(1+i)n
(1+i)n-1
n
i
n
(1+i)n-1
Transcribed Image Text:factor name single payment compound amount single payment present worth uniform series sinking fund capital recovery uniform series compound amount uniform series present worth uniform gradient present worth uniform gradient future worth uniform gradient uniform series converts P to F F to P F to A P to A A to F A to P G to P G to F G to A symbol (F/P, i%, n) (P/F, 1%, n) (A/F, i%, n) (A/P, i%, n) (F/A, 1%, n) (P/A, i%, n) (P/G,i%, n) (F/G,1%, n) (A/G, i%, n) formula (1 + i)" (1 + i)-¹ i (1 + i)" - 1 i(1+i)" (1+i)n-1 (1+i)n-1 i (1+i)n-1 i(1+i)n 1- i (1+i)n-1 1² (1+i)n n i(1+i)n (1+i)n-1 n i n (1+i)n-1
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