Question 978607: A car was valued at $31,000 in the year 1991. The value depreciated to $14,000 by the year 2007.
Use the compound interest formula S=P(1+r)^t to answer the following questions.
A) What was the annual rate of change between 1991 and 2007?
r = Round the rate of decrease to 4 decimal places.
B) What is the correct answer to part A written in percentage form?
r = %.
C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2010 ?
value = $ Round to the nearest 50 dollars.
I have tried this following direct steps and i am not getting the right answer, i dont know if im completely doing it wrong or just rounding incorrectly. please help!
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 14000=31000(1+r)^16
dividing 14,000 by 31000=0.45161, and that equals (1+r)^16
ln of both sides
-0.79493=16 ln(1+r)
dividing by 16
-0.04968=ln (1+r)
raise both to e power
0.95153=(1+r)
r=-0.04847= -0.0485
minus 4.85%
14000=31000(0.95153)^19 ;;now the 19th power
$12,061.15 or $12,100.
I didn't round at the end, but if you are close to the answer, it is likely a rounding issue.
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