Jason estimates that his car loses 12% of its value every year. The initial value is $12,000. Which best describes the
The Correct Answer and Explanation is:
Correct Answer:
The car’s value after ttt years can be modeled by the equation:V(t)=12000×(0.88)tV(t) = 12000 \times (0.88)^tV(t)=12000×(0.88)t
Explanation
When an object loses a fixed percentage of its value each year, this is called exponential decay. The general formula for exponential decay is:V(t)=V0×(1−r)tV(t) = V_0 \times (1 – r)^tV(t)=V0×(1−r)t
Where:
- V(t)V(t)V(t) is the value after ttt years,
- V0V_0V0 is the initial value,
- rrr is the decay rate (as a decimal),
- ttt is the number of years.
In Jason’s case:
- The initial value of the car V0=12,000V_0 = 12,000V0=12,000 dollars,
- The annual rate of depreciation r=12%=0.12r = 12\% = 0.12r=12%=0.12.
So the formula becomes:V(t)=12000×(1−0.12)t=12000×(0.88)tV(t) = 12000 \times (1 – 0.12)^t = 12000 \times (0.88)^tV(t)=12000×(1−0.12)t=12000×(0.88)t
This means that each year, the car retains 88% of its value from the previous year. For example:
- After 1 year:
V(1)=12000×0.88=10560V(1) = 12000 \times 0.88 = 10560V(1)=12000×0.88=10560 - After 2 years:
V(2)=12000×(0.88)2=9292.8V(2) = 12000 \times (0.88)^2 = 9292.8V(2)=12000×(0.88)2=9292.8 - After 3 years:
V(3)=12000×(0.88)3≈8177.66V(3) = 12000 \times (0.88)^3 \approx 8177.66V(3)=12000×(0.88)3≈8177.66
This model accurately shows that the car’s value declines more slowly over time — a hallmark of exponential decay. The biggest drop happens early, and the decline gets smaller each year.
In summary, the best way to describe the car’s depreciating value is with the equation:V(t)=12000×(0.88)tV(t) = 12000 \times (0.88)^tV(t)=12000×(0.88)t
This exponential equation provides a reliable estimate of the car’s value for any year ttt, accounting for a consistent 12% annual decrea
