Baseball ticket price trend from 1997 to 2006 with linear regression line.

Since I can't view or access uploaded images directly, I can’t see the specific data or scatter plots you're referring to. However, based on your description, I can help you solve each of the three problems (5, 6, and 7) if you provide the necessary data — such as:

- The coordinates from the scatter plot for problem 5 (Baseball ticket prices),
- The table of foodborne illness cases for problem 6,
- The table of animal longevity for problem 7.

But since you've mentioned that the image contains these details, I’ll walk through how to solve each problem step-by-step, assuming typical data patterns seen in such tasks. If you can supply the actual numbers, I can give exact answers.

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🔹 Problem 5: BASEBALL – Ticket Prices (1997 to 2005)

Given:
Use points (1997, 17.60) and (2005, 38.75) to write a linear equation in slope-intercept form.

#### a) Write the slope-intercept form of the equation

The slope-intercept form is:
> $ y = mx + b $

Where:
- $ m $ = slope
- $ b $ = y-intercept
- $ x $ = year
- $ y $ = price of ticket

##### Step 1: Find the slope ($ m $)
$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{38.75 - 17.60}{2005 - 1997} = \frac{21.15}{8} = 2.64375
$$

So, $ m \approx 2.64 $

##### Step 2: Use point-slope form to find $ b $
Using point (1997, 17.60):
$$
y = mx + b \\
17.60 = 2.64375(1997) + b \\
b = 17.60 - 2.64375 \times 1997
$$

Let’s compute:
$$
2.64375 \times 1997 = 5279.2125 \\
b = 17.60 - 5279.2125 = -5261.6125
$$

So, the equation is:
> $ y = 2.64375x - 5261.6125 $

Or rounded:
> $ y = 2.64x - 5261.61 $

✔ Answer (a): $ y = 2.64x - 5261.61 $

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#### b) Predict the price in 2009

Plug $ x = 2009 $ into the equation:
$$
y = 2.64(2009) - 5261.61 \\
= 5298.76 - 5261.61 = 37.15
$$

So, predicted price in 2009 is $37.15

Now, is this extrapolation or interpolation?

- The original data goes from 1997 to 2005.
- 2009 is after 2005 → outside the range
→ This is extrapolation

✔ Answer (b): $37.15; Extrapolation

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🔹 Problem 6: DISEASE – Foodborne Illness Cases

Assume the table gives years and number of cases.

Let’s suppose the data looks like this (typical example):

| Year | Cases |
|------|-------|
| 2000 | 100 |
| 2002 | 120 |
| 2004 | 140 |
| 2006 | 160 |
| 2008 | 180 |

(Note: You should replace with actual values from your table.)

#### a) Draw scatter plot and determine relationship

Plotting these would show a positive linear trend — as years increase, cases increase.

✔ Relationship: Positive correlation; likely linear.

#### b) Line of best fit and equation

Use two points, say (2000, 100) and (2008, 180)

Slope:
$$
m = \frac{180 - 100}{2008 - 2000} = \frac{80}{8} = 10
$$

Now use point (2000, 100):
$$
y = 10x + b \\
100 = 10(2000) + b \Rightarrow b = 100 - 20000 = -19900
$$

Equation: $ y = 10x - 19900 $

✔ Answer (b): $ y = 10x - 19900 $

> Note: This assumes linear trend. In real data, it might not be perfectly linear, but for school-level problems, we assume linearity.

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🔹 Problem 7: ANIMAL LONGEVITY

Suppose the table shows:

| Average Lifespan (years) | Max Lifespan (years) |
|--------------------------|------------------------|
| 10 | 20 |
| 15 | 30 |
| 20 | 40 |
| 25 | 50 |
| 30 | 60 |

Again, adjust based on your actual data.

#### a) Scatter plot and relationship

Plot average vs max lifespan.

You'll likely see a strong positive linear relationship.

Draw a line of best fit.

Use two points, e.g., (10, 20) and (30, 60)

Slope:
$$
m = \frac{60 - 20}{30 - 10} = \frac{40}{20} = 2
$$

Use point (10, 20):
$$
y = 2x + b \\
20 = 2(10) + b \Rightarrow b = 0
$$

So, equation: $ y = 2x $

✔ Answer (a): $ y = 2x $, where $ y $ = max lifespan, $ x $ = average lifespan

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#### b) Predict max longevity for an animal with average lifespan of 33 years

Use $ y = 2x $:
$$
y = 2(33) = 66 \text{ years}
$$

Is this extrapolation or interpolation?

- If your data ranges from 10–30 years, then 33 is outside the range → extrapolation

✔ Answer (b): 66 years; Extrapolation

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✔ Summary Answers

#### Problem 5: Baseball Tickets
a) Equation: $ y = 2.64x - 5261.61 $
b) Price in 2009: $37.15; Extrapolation

#### Problem 6: Disease Cases
a) Positive linear relationship
b) Equation: $ y = 10x - 19900 $ (example — depends on actual data)

#### Problem 7: Animal Longevity
a) Equation: $ y = 2x $ (example — depends on data)
b) Max longevity: 66 years; Extrapolation

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📌 Important Note: These solutions are based on assumed data. Please provide the actual numbers from your tables or scatter plot so I can give precise answers.

Would you like to type out the data from the image? I’ll calculate exactly!