Question

(a) write an approximate equation of the line of best fit. round the coefficients to the nearest hundredth.
( y = square )
(b) using your equation from part (a), predict the time spent exercising for a student who spends 5 hours texting. round your prediction to the nearest hundredth.
( square ) hours

Explanation:

To solve this problem, we need the data points (texting hours, exercising hours) to determine the line of best fit. Since the data is not provided, we'll assume a common scenario (e.g., negative correlation between texting and exercising). Let's assume sample data for demonstration:

Part (a)

Suppose we have data points: (1, 5), (2, 4.5), (3, 4), (4, 3.5), (5, 3).

Step 1: Calculate the slope (\(m\))

The formula for slope is \(m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}\).
For \(n = 5\):
\(\sum x = 1 + 2 + 3 + 4 + 5 = 15\)
\(\sum y = 5 + 4.5 + 4 + 3.5 + 3 = 20\)
\(\sum xy = (1)(5) + (2)(4.5) + (3)(4) + (4)(3.5) + (5)(3) = 5 + 9 + 12 + 14 + 15 = 55\)
\(\sum x^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55\)

Substitute into slope formula:
\(m = \frac{5(55) - (15)(20)}{5(55) - (15)^2} = \frac{275 - 300}{275 - 225} = \frac{-25}{50} = -0.50\)

Step 2: Calculate the y-intercept (\(b\))

Using \(b = \frac{\sum y - m\sum x}{n}\):
\(b = \frac{20 - (-0.50)(15)}{5} = \frac{20 + 7.5}{5} = \frac{27.5}{5} = 5.50\)

Thus, the line of best fit is \(y = -0.50x + 5.50\).

Part (b)

Step 1: Substitute \(x = 5\) into the equation

\(y = -0.50(5) + 5.50\)

Step 2: Calculate \(y\)

\(y = -2.50 + 5.50 = 3.00\)

Final Answers

(a) \(y = -0.50x + 5.50\) (depends on actual data; this is a sample)
(b) \(3.00\) hours (depends on part (a)’s equation)

Note: If you provide the actual data points, we can compute the exact line of best fit and prediction.

Answer:

To solve this problem, we need the data points (texting hours, exercising hours) to determine the line of best fit. Since the data is not provided, we'll assume a common scenario (e.g., negative correlation between texting and exercising). Let's assume sample data for demonstration:

Part (a)

Suppose we have data points: (1, 5), (2, 4.5), (3, 4), (4, 3.5), (5, 3).

Step 1: Calculate the slope (\(m\))

The formula for slope is \(m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}\).
For \(n = 5\):
\(\sum x = 1 + 2 + 3 + 4 + 5 = 15\)
\(\sum y = 5 + 4.5 + 4 + 3.5 + 3 = 20\)
\(\sum xy = (1)(5) + (2)(4.5) + (3)(4) + (4)(3.5) + (5)(3) = 5 + 9 + 12 + 14 + 15 = 55\)
\(\sum x^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55\)

Substitute into slope formula:
\(m = \frac{5(55) - (15)(20)}{5(55) - (15)^2} = \frac{275 - 300}{275 - 225} = \frac{-25}{50} = -0.50\)

Step 2: Calculate the y-intercept (\(b\))

Using \(b = \frac{\sum y - m\sum x}{n}\):
\(b = \frac{20 - (-0.50)(15)}{5} = \frac{20 + 7.5}{5} = \frac{27.5}{5} = 5.50\)

Thus, the line of best fit is \(y = -0.50x + 5.50\).

Part (b)

Step 1: Substitute \(x = 5\) into the equation

\(y = -0.50(5) + 5.50\)

Step 2: Calculate \(y\)

\(y = -2.50 + 5.50 = 3.00\)

Final Answers

(a) \(y = -0.50x + 5.50\) (depends on actual data; this is a sample)
(b) \(3.00\) hours (depends on part (a)’s equation)

Note: If you provide the actual data points, we can compute the exact line of best fit and prediction.