Solution:

This is an exponential decay function. Start by making a table with the values we start with when [tex]$t=0$[/tex]. Then, calculate the value of the car by reducing it by [tex]$85\%$[/tex] for each passing year (since the car loses value, it retains [tex]$85\%$[/tex] of its value: [tex]$100\% - 15\% = 85\%$[/tex]).

\begin{tabular}{|l|l|}
\hline
Time (Years) & Value (Thousands) \\
\hline
0 & 32 \\
\hline
1 & 27.2 \\
\hline
2 & 23.12 \\
\hline
\end{tabular}



Answer :

Sure, let's break down the solution step-by-step and create the complete table as needed.

### Step-by-Step Solution:

1. Initial Value of the Car (Time = 0 years):
The car's initial value is \[tex]$32,000 or 32 (thousands). 2. Value After 1 Year: Each year, the car retains \( 85\% \) of its previous year's value. To find the value after 1 year, multiply the initial value by 0.85. \[ 32 \times 0.85 = 27.2 \] 3. Value After 2 Years: To find the car's value after 2 years, multiply the value after 1 year by 0.85. \[ 27.2 \times 0.85 = 23.12 \] 4. Value After 3 Years: To find the car's value after 3 years, multiply the value after 2 years by 0.85. \[ 23.12 \times 0.85 = 19.652 \] ### Summary Table: Here is the completed table showing the car's value at the start and for each year up to 3 years. \begin{tabular}{|c|c|} \hline \textbf{Time (Years)} & \textbf{Value (Thousands)} \\ \hline 0 & 32.0 \\ \hline 1 & 27.2 \\ \hline 2 & 23.12 \\ \hline 3 & 19.652 \\ \hline \end{tabular} So, the car's value declines over the years as follows: \$[/tex]32,000 at the start, \[tex]$27,200 after 1 year, \$[/tex]23,120 after 2 years, and \$19,652 after 3 years.

Note: In practice, values in financial calculations are often rounded to two decimal places. The slight variations you see (such as 23.119999999999997 instead of 23.12) are results of floating-point arithmetic precision limits in computing.