A tractor has a first cost of $40,000 and a salvage value of $15,000 in year 10. The MARR is 12% per year. An identical tractor can be rented for $32,000 per year. Annual :operating cost is same for both options. If x is the minimum number of years the tractor must be used in order to justify its purchase, the relation to find x is represented by 15,000x + 12,000(A/F, 12%, 10) = -32,000x - (A/P, 12%, 10)40,000- O 15,000x + 12,000(A/F, 12%, 10) = 32,000x - (A/P, 12%, 10)40,000- O 12,000x + 15,000(A/F, 10%, 12) = 40,000x – (A/P, 10%, 12)32,000- O 12,000x + 15,000(A/F, 10%, 12) = -40,000x – (A/P, 10%, 12)32,000- O 15,000(A/F, 12%, 10) = -32,000x + (A/P, 12%, 10)40,000- O 15,000(A/F, 12%, 10) = 32,000x + (A/P, 12%, 10)40,000- O None of them O
A tractor has a first cost of $40,000 and a salvage value of $15,000 in year 10. The MARR is 12% per year. An identical tractor can be rented for $32,000 per year. Annual :operating cost is same for both options. If x is the minimum number of years the tractor must be used in order to justify its purchase, the relation to find x is represented by 15,000x + 12,000(A/F, 12%, 10) = -32,000x - (A/P, 12%, 10)40,000- O 15,000x + 12,000(A/F, 12%, 10) = 32,000x - (A/P, 12%, 10)40,000- O 12,000x + 15,000(A/F, 10%, 12) = 40,000x – (A/P, 10%, 12)32,000- O 12,000x + 15,000(A/F, 10%, 12) = -40,000x – (A/P, 10%, 12)32,000- O 15,000(A/F, 12%, 10) = -32,000x + (A/P, 12%, 10)40,000- O 15,000(A/F, 12%, 10) = 32,000x + (A/P, 12%, 10)40,000- O None of them O
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Transcribed Image Text:السؤال 1
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10 درجات
A tractor has a first cost of $40,000 and a salvage value of $15,000 in year 10. The MARR is 12% per year. An identical tractor can be rented for $32,000 per year. Annual
:operating cost is same for both options. If x is the minimum number of years the tractor must be used in order to justify its purchase, the relation to find x is represented by
15,000x + 12,000(A/F, 12%, 10) = -32,000x – (A/P, 12%, 10)40,000- O
15,000x + 12,000(A/F, 12%, 10) = 32,000x – (A/P, 12%, 10)40,000- O
12,000x + 15,000(A/F, 10%, 12) = 40,000x – (A/P, 10%, 12)32,000- O
12,000x + 15,000(A/F, 10%, 12) = -40,000x – (A/P, 10%, 12)32,000- O
15,000(A/F, 12%, 10) = -32,000x + (A/P, 12%, 10)40,000- O
15,000(A/F, 12%, 10) = 32,000x + (A/P, 12%, 10)40,000- O
None of them O
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