Answer:
$27,436
Step-by-step explanation:
Recall that an exponential function or equation take the form of [tex]y=a(b)^x[/tex], where a is the initial value, b is the growth/decay rate
(growth: b > 1; decay: 0 < b < 1), and x is the time that passes.
The problem mentions "depreciating", which means decreasing. If it depreciates at a rate of 5%, then it decreases the car's value at 5% each year that passes.
The value of b, however, will be 0.95 rather than 0.05.The car's overall value decreases by 5% yearly, meaning the car's value is 95% of its value the previous year.
Think of sales such as "50% off" and "25% off" seen in shops; the larger the percentage, the cheaper the product. A customer wouldn't think too much of a "5% off" sale compared to a "75% off" one, as the price would decrease slightly.
So, b is now identified!
The problem mentions that the car was worth $32,000 in 2010 and is asking for its value in 2013. This must mean that the car's value at 2010 is the initial value or a!
Repeating what's been said in "Finding a" the problem gives us two measures of time: 2010 and 2013. 2010 is the initial time, and 2013 is the final time.
Since we're finding the value of the car at 2013, where 3 years elapsed between the car's initial value and the value we're trying to find, x must equal 3!
Using the exponential equation and plugging in the known values, the car's value in 2013 is,
[tex]32,000(0.95)^3=27,436[/tex].