Let's go step by step with this linear regression problem.

Problem Overview:

• The truck was bought in 2020.
• In 2023, its value is $30,500.
• The company predicts the truck's value will depreciate linearly until 2033, at which point the salvage value will be $15,500.

The goal is to:

1. Complete the table.
2. Compute the slope of the linear function.
3. Write the equation of the line that models the truck's value over time.

Step a: Completing the Table

The table needs to show:

• $t$ (the number of years since 2020)
• $V$ (the value of the truck)

We know:

• In 2020 ($t = 0$), the truck's value was $30,500.
• In 2023 ($t = 3$), the value is still $30,500.
• In 2033 ($t = 13$), the value will be $15,500.

So, the table would look like this:

Step b: Compute the Slope of the Linear Function

The slope $m$ of the line is calculated as the change in value $\Delta V$ divided by the change in time $\Delta t$:

$m = \frac{\Delta V}{\Delta t} = \frac{15,500 - 30,500}{13 - 0} = \frac{-15,000}{13} = -1153.85 \, \text{dollars per year}$

So, the slope is $-1153.85$.

Step c: Write the Equation of the Linear Function

The linear function follows the form:

$V(t) = mt + b$

Where:

• $m = -1153.85$ (the slope)
• $b = 30,500$ (the initial value when $t = 0$)

Thus, the equation of the line is:

$V(t) = -1153.85t + 30,500$

Would you like more details on any step, or do you have additional questions?

Here are 5 related questions for practice:

1. What would be the value of the truck in 2025 based on this linear model?
2. How long will it take for the truck's value to drop below $10,000?
3. If the truck was purchased for $35,000 instead of $30,500, how would the slope and equation change?
4. How would a different depreciation model, such as exponential, affect the equation for $V(t)$?
5. Can this linear model accurately predict the value of a truck beyond 2033?

Tip: When interpreting the slope, remember that a negative slope indicates depreciation, meaning the value is decreasing over time.