Chapter 7, Problem 20CT

a.

To determine

To calculate: The decay factor and explain its meaning

Answer to Problem 20CT

The decay factor is 0.15 means the car’s value decrease 15% per year

Explanation of Solution

Given:

  $y=30,000{\left(0.85\right)}^x$ is the function showing car’s value after x years

Concept Used:

The depreciation function for the value of any purchase after x units of time is:

  $\begin{array}{l}y=P{\left(1\pm r\right)}^x\\ Where,\\ P\to Initialvalueofthepurchase\\ r\to decayfactor/growthfactor\end{array}$

Negative sign for the decrease in product value or positive sign for the increase in product value

Calculation:

  $\begin{array}{l}y=30,000{\left(0.85\right)}^x\\ y=30,000{\left(1-0.15\right)}^x\end{array}$

Here, the decay factor is 0.15 means the car’s value decrease 15% per year

b.

To determine

To calculate: Car’s value after 1 year

Answer to Problem 20CT

Car’s value after 1 year is 25,500 dollars

Explanation of Solution

Given:

  $y=30,000{\left(0.85\right)}^x$

Calculation:

  $\begin{array}{l}y=30,000{\left(0.85\right)}^x\\ y=30,000{\left(0.85\right)}^1\\ y=30,000\left(0.85\right)\\ y=30,000\left(0.85\right)\\ \boxed{y=25,500dollars}\end{array}$

c.

To determine

To calculate: Car’s value after 4 years

Answer to Problem 20CT

Car’s value after 4 years is 15,660 dollars approximately

Explanation of Solution

Given:

  $y=30,000{\left(0.85\right)}^x$

Calculation:

  $\begin{array}{l}y=30,000{\left(0.85\right)}^x\\ y=30,000{\left(0.85\right)}^4\\ y=30,000\left(0.522\right)\\ \boxed{y=15,660.18dollars}\end{array}$