Question

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.

Answer 1

are you using exponentials?

Answer 2

believe so :)

Answer 3

if so \[ A=31e^{-rt}\] is your model, in thousands. you know
\[15=31e^{-r\times 13}\]
\[\frac{15}{31}=e^{-r\times 13}\]
\[ln(\frac{15}{31})=-13r\]
\[\frac{ln(\frac{15}{31})}{-13}=r\]

Answer 4

someone else tried helping me out with something similar to this but i didnt know how he got the %

Answer 5

i got r = .0558 which is 5.58% by moving the decimal over twice.

Answer 6

ok thanks, and to get the last answer i count from 1990? or start from 2003?

Answer 7

start from 1990