Question: The estimates for two alternatives are to be compared on the basis of their perpetual equivalent annual worth. At an interest rate of 10% per

(a) AWY1 = 90,000(0.10) 4000 + 15,000(0.10)
(b) AWY1 = 90,000(0.10) 4000 + 15,000(AF,10%,6)
(c) AWY1 = 90,000(0.10) 4000 15,000(PF,10%,3)(0.10) + 15,000(0.10)
(d) AWY1 = 90,000(AP,10%,6) 4000 + 15,000(AF,10%,6)
Alternative Y1 X1 First cost, $ Annual cost, $/year Salvage value, $ - 50,000 -90,000 - 10,000 -4000 13,000 3 15,000 6. Life, years
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