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A new car costs $18000 and loses 11% of its value each year.

(a) Find the depreciation (loss of value) in dollars in the third year. First year depreciation: $ 1980.00 Second year depreciation: $ 1762.20 Third year depreciation: $
(b) Give a formula for dn, the depreciation in the nth year. Hint: It might be helpful to note that if 11% of the value is lost, then 89% of the value remains. dn =

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Answer:

Third year depreciation: $ 12689,44

Formula for dn, dn= Ci x0,89^n

Ci is the initial cost

Explanation:

Spreadsheet is attached with the calculus and the probation formula.  

In the traditional method each year we reduce the value by 11%

So, is  dn=C(n-1)*0,11

C(n-1) is the carrying value

Also,  we obtain the same result with the following formula

dn= Ci x0,89^n

Ci is the initial car cost

As the depreciation is 11%, 89% is the value that remains.

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