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#### A tractor has a first cost of \$40,000, a monthly op- erating cost of \$1500, and a salvage value of \$12,000 in 10 years

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A tractor has a first cost of \$40,000, a monthly op- erating cost of \$1500, and a salvage value of \$12,000 in 10 years. The MARR is 12% per year. An identical tractor can be rented for \$3200 per month (operating cost not included). If n is the minimum number of months per year the tractor must be used in order to justify its purchase, the relation to find n is represented by: (a) -40,000(A/P,1%,10) - 1500n + 12,000(A/F,1% 10) = 3200n (b) -40,000(A/P,12%,10) - 1500n + 12,000(A/F,12%,10) = 3200n (c) -40,000(A/P,1%,120) - 1500n + 12,000(A/F,1%,120) = 3200n (d) -40,000(A/P,11.4%,10) - 1500n + 12,000(A/F,11.4%,10) = 3200n
3) (3pts) 13.58: No change (Determine the number of month for breakeven) *Note: Any answer choice shown in the textbook (multiple choices of 13.58) is NOT correct.

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