Baseball ticket prices over time, illustrating a trend from 1997 to 2006.

Let's solve each problem step by step.

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Problem 5: BASEBALL

We are given a scatter plot of average baseball ticket prices from 1997 to 2006 and asked to:

#### a. Use the points (2001, 17.60) and (2002, 18.75) to write the slope-intercept form of the equation for the line of fit.

Step 1: Find the slope (m)

The slope formula is:

$$
m = \frac{y_2 - y_1}{x_2 - x_1}
$$

Using the points:
- $ (x_1, y_1) = (2001, 17.60) $
- $ (x_2, y_2) = (2002, 18.75) $

$$
m = \frac{18.75 - 17.60}{2002 - 2001} = \frac{1.15}{1} = 1.15
$$

So, the slope is 1.15 dollars per year.

Step 2: Use point-slope form to find the equation

Point-slope form:
$$
y - y_1 = m(x - x_1)
$$

Using point (2001, 17.60):

$$
y - 17.60 = 1.15(x - 2001)
$$

Now convert to slope-intercept form: $ y = mx + b $

$$
y = 1.15(x - 2001) + 17.60
$$
$$
y = 1.15x - 1.15 \times 2001 + 17.60
$$
$$
1.15 \times 2001 = 2301.15
$$
$$
y = 1.15x - 2301.15 + 17.60
$$
$$
y = 1.15x - 2283.55
$$

✔ Equation of the line of fit:
$$
\boxed{y = 1.15x - 2283.55}
$$

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#### b. Use your equation to tell the price of a ticket in 2009. Is this extrapolation or interpolation?

Plug $ x = 2009 $ into the equation:

$$
y = 1.15(2009) - 2283.55
$$
$$
= 2309.35 - 2283.55 = 25.80
$$

So, the predicted price in 2009 is $25.80

Now, determine if this is interpolation or extrapolation:

- The original data goes from 1997 to 2006
- We are predicting for 2009, which is after the last data point

➡️ This is extrapolation.

✔ Answer:
Predicted price in 2009: $25.80
This is extrapolation.

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Problem 6: DISEASE – U.S. Foodborne Botulism Cases

Given data:

| Year | 2001 | 2002 | 2003 | 2004 | 2005 |
|------|------|------|------|------|------|
| Cases | 39 | 28 | 20 | 16 | 18 |

#### a. Draw a scatter plot and determine what relationship, if any, exists.

We would plot:
- X-axis: Year (2001 to 2005)
- Y-axis: Number of cases (39, 28, 20, 16, 18)

Plotting these points:
- (2001, 39)
- (2002, 28)
- (2003, 20)
- (2004, 16)
- (2005, 18)

Observations:
- From 2001 to 2004: cases decrease
- In 2005, cases increase slightly from 16 to 18

Overall trend: Decreasing, then slight rise at end → general downward trend with a small rebound.

➡️ Relationship: There appears to be a negative correlation between year and number of cases (but not perfectly linear).

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#### b. Draw a line of fit and write the slope-intercept form of the equation.

We’ll use two points that seem to represent the overall trend.

Let’s pick:
- (2001, 39)
- (2004, 16)

Why? These are on the decreasing part.

Slope:
$$
m = \frac{16 - 39}{2004 - 2001} = \frac{-23}{3} \approx -7.67
$$

Now use point-slope form with (2001, 39):

$$
y - 39 = -7.67(x - 2001)
$$

Convert to slope-intercept:

$$
y = -7.67x + 7.67 \times 2001 + 39
$$
$$
7.67 \times 2001 = 15347.67
$$
$$
y = -7.67x + 15347.67 + 39 = -7.67x + 15386.67
$$

✔ Approximate equation:
$$
\boxed{y = -7.67x + 15386.67}
$$

> Note: This is an estimate. A better fit might use least squares, but since we're doing it manually, this is reasonable.

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Problem 7: ZOOS – Animal Longevity

Given table:

| Avg. | 12 | 25 | 15 | 8 | 35 | 40 | 41 | 20 |
|------|----|----|----|----|----|----|----|----|
| Max. | 47 | 50 | 40 | 20 | 70 | 77 | 61 | 54 |

#### a. Draw a scatter plot and determine the relationship.

Plot:
- X-axis: Average longevity (years)
- Y-axis: Maximum longevity (years)

Points:
- (12, 47), (25, 50), (15, 40), (8, 20), (35, 70), (40, 77), (41, 61), (20, 54)

Visual inspection:
- As average increases, maximum tends to increase.
- Some variation (e.g., (41, 61) is below trend)
- Overall, positive correlation — animals with higher average lifespans tend to have higher maximum lifespans.

✔ Relationship: Positive correlation

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#### b. Draw a line of fit and write the slope-intercept form.

Choose two representative points:
- (8, 20)
- (40, 77)

These cover the range.

Slope:
$$
m = \frac{77 - 20}{40 - 8} = \frac{57}{32} \approx 1.78125
$$

Use point (8, 20) to find intercept:

$$
y - 20 = 1.78125(x - 8)
$$
$$
y = 1.78125x - 1.78125 \times 8 + 20
$$
$$
1.78125 \times 8 = 14.25
$$
$$
y = 1.78125x - 14.25 + 20 = 1.78125x + 5.75
$$

✔ Equation:
$$
\boxed{y = 1.78x + 5.75} \quad \text{(rounded)}
$$

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#### c. Predict the maximum longevity for an animal with average longevity of 33 years. Is this interpolation or extrapolation?

Use the equation:
$$
y = 1.78(33) + 5.75 = 58.74 + 5.75 = 64.49
$$

So, predicted maximum longevity ≈ 64.5 years

Now check if this is interpolation or extrapolation:

- The average values in the data range from 8 to 41
- 33 is within this range

➡️ So, this is interpolation

✔ Answer:
- Predicted maximum longevity: ~64.5 years
- This is interpolation

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✔ Final Answers Summary:

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Problem 5: Baseball

a. Equation: $ \boxed{y = 1.15x - 2283.55} $

b. Price in 2009: $ \boxed{25.80} $; This is extrapolation

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Problem 6: Disease

a. Scatter plot shows negative trend (cases decrease over time)

b. Approximate line of fit: $ \boxed{y = -7.67x + 15386.67} $

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Problem 7: Zoos

a. Scatter plot shows positive correlation between average and maximum longevity

b. Line of fit: $ \boxed{y = 1.78x + 5.75} $

c. Predicted max longevity for avg = 33: $ \boxed{64.5} $ years; This is interpolation

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Let me know if you'd like the scatter plots drawn or more precise regression lines!