A) 13000 = 26000 (1 - r)^9
B) r * 100%
C) v = 26000 (1 - r)^14
A car was valued at $26,000 in the year 1995. The value depreciated to $13,000 by the year 2004.
A) What was the annual rate of change between 1995 and 2004? Round the rate of decrease to 4 decimals 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 2009 ? Round to the nearest 50 dollars.
5 answers
say value at year x+1 = d * value at year x
then value at year x+ n = d^n * value at year x
or
value at year 2004 = d^(2004 - 1995) * value at year 1995
so
d^9 = 13000/26000 = 0.5
9 log d = log 0.5
log d = log 0.5 / 9 = -0.301 / 9 = -0.03345
d = 10^-0.03345
d = 0.9259 or decreases by .07412 of its value per year
decrease by 100 - 92.59 = 7.412 % per year
value at year 2009 = d^(2009 - 1995) * value at year 1995
= 0.9259^14 * 26000 = 0.3403 * 26,000 = 8,848
then value at year x+ n = d^n * value at year x
or
value at year 2004 = d^(2004 - 1995) * value at year 1995
so
d^9 = 13000/26000 = 0.5
9 log d = log 0.5
log d = log 0.5 / 9 = -0.301 / 9 = -0.03345
d = 10^-0.03345
d = 0.9259 or decreases by .07412 of its value per year
decrease by 100 - 92.59 = 7.412 % per year
value at year 2009 = d^(2009 - 1995) * value at year 1995
= 0.9259^14 * 26000 = 0.3403 * 26,000 = 8,848
A. V = Vo(1-r)^9 = 13,000
26,000(1-r)^9 = 13,000
(1-r)^9 = 0.5
1-r = 0.9259
r = 0.0741.
00
B. .r = 7.41 %.
C. V = 26,000(1-0.0741)^14 =
26,000(1-r)^9 = 13,000
(1-r)^9 = 0.5
1-r = 0.9259
r = 0.0741.
00
B. .r = 7.41 %.
C. V = 26,000(1-0.0741)^14 =
4
5
1 answer