In comparing alternatives I and J by the present worth method, the equation that yields the present worth of alternative J is: Alternative IAlternative J Initial cost, $ -150,000 -250,000 Annual income, $ per year 20,000 40,000 Annual expenses, $ per year -9,000 -14,000 Salvage value, $ 25,000 35,000 Life, years 3 6. The interest rate is 15% per year. a) PW J=-250,000+40,000(P/A,15%,6)+35,000*(P/F,15%,6) b) PW J=-250,000+26,000(P/A,15%,6)+35,000*(P/F,15%,6) c) PW J=-250,000-26,000(P/A,15%,6)+35,000*(P/F,15%,6) d) PW J=-250,000-26,000(P/A,15%,6)-35,000*(P/F,15%,6)

ENGR.ECONOMIC ANALYSIS
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Author:NEWNAN
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Chapter1: Making Economics Decisions
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Can some one please help me to answer each question correctly? please and thank you.
In comparing alternatives I and J by the present worth method, the equation that
yields the present worth of alternative J is:
Alternative I Alternative J
Initial cost, $
|- 150,000
-250,000
Annual income, $ per year
20,000
40,000
Annual expenses, $ per year
-9,000
-14,000
Salvage value, $
Life, years
25,000
35,000
3
6.
The interest rate is 15% per year.
a) PW J=-250,000+40,000(P/A,15%,6)+35,000*(P/F,15%,6)
b) PW J=-250,000+26,000(P/A, 15%,6)+35,000*(P/F,15%,6)
c) PW J=-250,000-26,000(P/A,15%,6)+35,000*(P/F,15%,6)
d) PW J=-250,000-26,000(P/A,15%,6)-35,000*(P/F,15%,6)
Transcribed Image Text:In comparing alternatives I and J by the present worth method, the equation that yields the present worth of alternative J is: Alternative I Alternative J Initial cost, $ |- 150,000 -250,000 Annual income, $ per year 20,000 40,000 Annual expenses, $ per year -9,000 -14,000 Salvage value, $ Life, years 25,000 35,000 3 6. The interest rate is 15% per year. a) PW J=-250,000+40,000(P/A,15%,6)+35,000*(P/F,15%,6) b) PW J=-250,000+26,000(P/A, 15%,6)+35,000*(P/F,15%,6) c) PW J=-250,000-26,000(P/A,15%,6)+35,000*(P/F,15%,6) d) PW J=-250,000-26,000(P/A,15%,6)-35,000*(P/F,15%,6)
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