https://file.aiquickdraw.com/mathbot/file/1726787354883cpueqhae.jpg

Let's break down the problem in the image:

### Question 6:
It asks for the equation of the trend line for the data set. The given choices are:
- A: \( y = \frac{4}{5}x + 94 \)
- B: \( x = -\frac{4}{5}y + 94 \)
- C: \( x = \frac{4}{5}y + 94 \)
- D: \( y = -\frac{4}{5}x + 94 \)

The correct answer (checked in the image) is **A: \( y = \frac{4}{5}x + 94 \)**, where \( y \) represents the distance driven (in miles) and \( x \) is the time in minutes.

### Question 7:
It asks to predict Maria's distance from her house after driving for 45 minutes. Using the equation from question 6 (\( y = \frac{4}{5}x + 94 \)):

- Substitute \( x = 45 \) into the equation:

\[
y = \frac{4}{5}(45) + 94
\]

Let me calculate this for you.After driving for 45 minutes, Maria will be approximately **130 miles** from her house.

Would you like further details or have any additional questions?

Here are 5 follow-up questions to consider:
1. How is the slope of the trend line \( \frac{4}{5} \) related to the rate of change in distance?
2. How can you interpret the y-intercept (94) in this context?
3. What would happen to the equation if Maria's initial distance from the house was different?
4. How could you use this equation to predict Maria's distance after any given amount of time?
5. What would the graph of this equation look like, and how can you interpret its features?

**Tip:** The slope in a linear equation represents the rate of change—in this case, how much distance is covered per minute.

Predict the distance Maria is from her house after driving for 45 minutes based on the linear trend line equation provided.

Learn how to predict Maria's distance from her house after driving for 45 minutes by solving the linear equation y = 4/5x + 94. This problem covers key algebraic concepts such as slope-intercept form and linear equations, ideal for Grade 9-10 students.

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