An engineer made a development in the production process which will save his company 10,000$ per year. The following equation is used to calculate the savings in 8 years at i-10% per year. Select one: O a P-10,000(P/A,8,10%) Ob F-10,000(F/A,10%,8) Oc A-10,000(A/F,10%,8) O d. F-10,000(F/P10%,8)

ENGR.ECONOMIC ANALYSIS
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Chapter1: Making Economics Decisions
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An engineer made a development in the production process which will save his company 10,000$ per year. The following equation is
used to calculate the savings in 8 years at i-10% per year.
Select one:
O a. P-10,000(P/A,8,10%)
O b. F-10,000(F/A,10%,8)
Oc. A 10,000(A/F,10%,8)
O d. F-10,000(F/P,10%,8)
Alternative x has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y has a
life of 3 years. Comparing the alternatives using present worth analysis at a MARR of 12%, the equation of PWX is: Select one:
Select one:
O a. PWX=-20,000-9000(P/A,12 %, 15) + 5000(P/F,12 %, 15)
O b. PWX=-20,000-9000(P/A,12 %,3)+ 5000(P/F,12%,3)
OC PWX=-20,000-9000(P/A,12%, 5) + 5000(P/F,12%,5)
O d. none of these answers.
Transcribed Image Text:An engineer made a development in the production process which will save his company 10,000$ per year. The following equation is used to calculate the savings in 8 years at i-10% per year. Select one: O a. P-10,000(P/A,8,10%) O b. F-10,000(F/A,10%,8) Oc. A 10,000(A/F,10%,8) O d. F-10,000(F/P,10%,8) Alternative x has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y has a life of 3 years. Comparing the alternatives using present worth analysis at a MARR of 12%, the equation of PWX is: Select one: Select one: O a. PWX=-20,000-9000(P/A,12 %, 15) + 5000(P/F,12 %, 15) O b. PWX=-20,000-9000(P/A,12 %,3)+ 5000(P/F,12%,3) OC PWX=-20,000-9000(P/A,12%, 5) + 5000(P/F,12%,5) O d. none of these answers.
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