Question
A car was valued at $38,000 in the year 2003. The value depreciated to $11,000 by the year 2009. Assume that the car value continues to drop by the same percentage.
-What will the value be in the year 2013?
-What will the value be in the year 2013?
Answers
let the rate of depreciation be r
38000(1-r)^6 = 11000
(1-r)^6 = 11/38
1-r = (11/38)^(1/6)
1-r = .8133326
r = .186667
rate of depreciation is 18.67 %
so in 2012 it will be
38000(.8133326)^10 = $4813.55
38000(1-r)^6 = 11000
(1-r)^6 = 11/38
1-r = (11/38)^(1/6)
1-r = .8133326
r = .186667
rate of depreciation is 18.67 %
so in 2012 it will be
38000(.8133326)^10 = $4813.55
A new car was valued at $43,000, and it's value depreciated to $15,000 over the next 6 years. Use the exponential equation for depreciation to answer the following questions.
) What was the annual percent rate of depreciation, to 2 decimal places? Your answer will be a positive number.
) What was the annual percent rate of depreciation, to 2 decimal places? Your answer will be a positive number.
ctm
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