12. An engineer must choose between two technically equivalentalternatives, A and B. Based upon the data in the table below and aninterest rate of 9%, which alternative should be selected?A B$3,000 $6,000$500First CostAnnual Maintenance$300End of Useful Life SalvageValue$700$1,00015Useful Lifeyears yearsa. Choose A; we want to maximize the costsO b. Choose B; we want to minimize the costsO c. Choose A; we want to minimize the costsO d. Choose B; we want to maximize the costsO e. I don't wanna chooseO other:

Alternative A:EUAC = 3,000(A/P, 9%, 5) + 500 − 700(A/F, 9%, 5)= 3,000(0.2571) + 500 " 700(0.1671) =…

12. An engineer must choose between two technically equivalent alternatives, A and B. Based upon the data in the table below and an interest rate of 9%, which alternative should be selected? A B $3,000 $6,000 $500 $300 First Cost Annual Maintenance End of Useful Life Salvage Value $700 $1,000 15 Useful Life years years O a. Choose A; we want to maximize the costs O b. Choose B; we want to minimize the costs Oc. Choose A; we want to minimize the costs O d. Choose B; we want to maximize the costs O e.I don't wanna choose

12. An engineer must choose between two technically equivalent alternatives, A and B. Based upon the data in the table below and an interest rate of 9%, which alternative should be selected? A B $3,000 $6,000 $500 $300 First Cost Annual Maintenance End of Useful Life Salvage Value $700 $1,000 15 Useful Life years years O a. Choose A; we want to maximize the costs O b. Choose B; we want to minimize the costs Oc. Choose A; we want to minimize the costs O d. Choose B; we want to maximize the costs O e.I don't wanna choose

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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