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anonymous
 5 years ago
A car was valued at $31,000 in the year 1990. The value depreciated to $15,000 by the year 2003.
A) What was the annual rate of change between 1990 and 2003?
= Round the rate of decrease to 4 decimal places.
B) What is the correct answer to part A written in percentage form?
= %.
C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2007 ?
value = $ Round to the nearest 50 dollars.
anonymous
 5 years ago
A car was valued at $31,000 in the year 1990. The value depreciated to $15,000 by the year 2003. A) What was the annual rate of change between 1990 and 2003? = Round the rate of decrease to 4 decimal places. B) What is the correct answer to part A written in percentage form? = %. C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2007 ? value = $ Round to the nearest 50 dollars.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you using exponentials?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if so \[ A=31e^{rt}\] is your model, in thousands. you know \[15=31e^{r\times 13}\] \[\frac{15}{31}=e^{r\times 13}\] \[ln(\frac{15}{31})=13r\] \[\frac{ln(\frac{15}{31})}{13}=r\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0someone else tried helping me out with something similar to this but i didnt know how he got the %

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got r = .0558 which is 5.58% by moving the decimal over twice.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks, and to get the last answer i count from 1990? or start from 2003?
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