:)pleaseandthankyou!A used car was purchased in July 2000 for $11,900. If the car depreciates 13% of its value each year, what is?
If it depreciates 13%, then it will have 87% of its value a year later.
July 2000 = $11,900
July 2001 = $11,900 x 0.87 = $10,353
July 2002 = $10,353 x 0.87 = $9,007.11
July 2003 = $9.007.11 x 0.87 = $7,836.19
Rounded to the nearest dollar:
$7,836
Another way to write this is:
P(t) = Po * (1-r)^t
Here Po is the original value, r is the depreciation rate (0.13) and t are the number of years.
P(3) = 11900 * (0.87)^3
P(3) = 7836.1857
Rounding...
P(3) = $7,836
Same answer either way!A used car was purchased in July 2000 for $11,900. If the car depreciates 13% of its value each year, what is?
Doh! Nearest hundred! Oh well, what's $36 here or there? I'll just take it as my commission and you can keep the $7800. :-)
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A used car was purchased in July 2000 for $11,900. If the car depreciates 13% of its value each year, what is?11900*0.87^3
=7836.19
[13]
The depreciated value in July 2003
=$11,900*(87/100)^3
=$ 11,900*658,503/1000000
=$7,836 (approax)
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