Answer:
$1863
Step-by-step explanation:
The model for exponential decay (depreciation) is [tex]y=y_0e^{-rt}[/tex].
r is the depreciation rate (which can be found first)
t is the time since the initial purchase
y is the value of the car at time t
[tex]y_0[/tex] is the initial purchase price (t = 0)
Year 0 is 1995.
2000 is year 5.
So, [tex]6800=20000e^{-r\cdot5}\\\\\frac{6800}{20000}=e^{-5r}\\[/tex]
To solve for r take the logarithm of both sides.
[tex]-5r=\ln\left({\frac{6800}{20000}\right)\\\\r=-\frac{1}{5}\ln\left({\frac{6800}{20000}\right) \approx 0.215762[/tex]
Now find the value of the car in 2006, which is year t = 11.
[tex]y=20000e^{-0.215762\cdot11} \approx 1863[/tex]
