## 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.

1. anonymous

are you using exponentials?

2. anonymous

believe so :)

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

start from 1990