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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.

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  1. anonymous
    • 5 years ago
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    are you using exponentials?

  2. anonymous
    • 5 years ago
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    believe so :)

  3. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    someone else tried helping me out with something similar to this but i didnt know how he got the %

  5. anonymous
    • 5 years ago
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    i got r = .0558 which is 5.58% by moving the decimal over twice.

  6. anonymous
    • 5 years ago
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    ok thanks, and to get the last answer i count from 1990? or start from 2003?

  7. anonymous
    • 5 years ago
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    start from 1990

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