Question

Jaime bought a car in 2005 for$\mathrm{\backslash (}28500$. By 2008, the car was worth$\mathrm{\backslash )}23700$it. Based on a linear model, what will the value of the car be in 2012?

A.$\mathrm{\backslash (}17300$

B.$\mathrm{\backslash )}18100$

C.$\mathrm{\backslash (}17550$

D.$\mathrm{\backslash )}18475$

Short answer

The correct option is represented by A.

Step-by-step solution

Step 1. Determine the formula that can be used.

The standard form associated with the slope and the intercept form related to the line that is straight$y=mx+c$ in which the slope is represented by $m$ and the $y$-intercept is represented by $c$ can be used to write the required equation.

Step 2. Take the help of the obtained method. 

Determine the depreciation associated with the car for each year.

$$\begin{gathered} \frac{28500-23700}{3}=\frac{4800}{3} \\ =1600 \end{gathered}$$

Take the help of obtained standard form represented by $y=mx+c$in which the required values can be filled by using the obtained depreciation, the initial price of the car, $t$which represents the time in years and $C$which will represent the cost.

$C=2850+1600t$

Step 3. Write the conclusion.

The obtained equation is $C=2850+1600t$so the required value can be determined by using the gap of years between 2005 and 2012 in place of time into the obtained equation so,

$$\begin{gathered} C=28500-1600\times 7 \\ =17300 \end{gathered}$$

The obtained value is the same as in option A.