Certainly! Let's go through the steps to solve this problem:
### Step 1: Identify the initial amount [tex]\( a \)[/tex] and the depreciation rate [tex]\( r \)[/tex].
Given:
- Initial value of the boat, [tex]\( a = \$45,000 \)[/tex]
- Depreciation rate, [tex]\( r = 20\% \)[/tex] per year (expressed as a decimal, [tex]\( r = 0.20 \)[/tex])
### Step 2: Write the exponential decay model using [tex]\( y = a(1 - r)^t \)[/tex].
We substitute the given values for [tex]\( a \)[/tex] and [tex]\( r \)[/tex] into the formula.
[tex]\[ y = 45000(1 - 0.20)^t \][/tex]
This simplifies to:
[tex]\[ y = 45000(0.80)^t \][/tex]
### Step 3: Find the value of the boat after 4 years.
Substitute [tex]\( t = 4 \)[/tex] into the model:
[tex]\[ y = 45000(0.80)^4 \][/tex]
### Step 4: Calculate the result.
The value of the boat after 4 years is [tex]\( y = 45000(0.80)^4 \)[/tex].
Using the known result:
[tex]\[ y = 18432.000000000004 \][/tex]
So, the value of the boat after 4 years is approximately \[tex]$18,432.
### Graphing the Function
To verify the solution, we can graph the function \( y = 45000(0.80)^t \).
1. On the horizontal axis (\( t \)), we plot the number of years.
2. On the vertical axis (\( y \)), we plot the value of the boat.
\[
\begin{array}{|c|c|}
\hline
t (\text{years}) & y (\text{value of the boat}) \\
\hline
0 & 45000 \\
1 & 45000 \cdot (0.80)^1 = 36000 \\
2 & 45000 \cdot (0.80)^2 = 28800 \\
3 & 45000 \cdot (0.80)^3 = 23040 \\
4 & 45000 \cdot (0.80)^4 = 18432 \\
\hline
\end{array}
\]
Plot these points and draw the curve to show the exponential decay model.
### Summary
The value of the speedboat after 4 years is approximately \(\$[/tex]18,432\) when it depreciates at a rate of 20% per year.
