Question

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Answer 1

A)

`10000 = 43000( {1 + r)}^(13)`

`{(1 + r)}^(13) = (10)/(43)`

`1 + r = \sqrt[13]{ (10)/(43) }`

`r = \sqrt[13]{ (10)/(43) } - 1 = - .1061`

The rate of decrease was .1061.

B) 10.61% decrease

C)

`s = 43000( { \sqrt[13]{ (10)/(43) } )}^(16) = 7141.35`

To the nearest 50 dollars, the value of the car in 2010 will be $7,150.