QUESTION IMAGE
Question
- this scatter plot shows the relationship between the number of hours a pool has been draining and the water level remaining in the pool.
pool water level vs. time draining
the y-intercept of the estimated line of best fit is at (0, b). enter the approximate value of the b in the first response box:
enter the approximate slope of the estimated line of best fit in the second response box:
- this scatter plot shows the relationship between the number of sweatshirts sold and the temperature outside.
sweatshirt sales vs. temperature
the y-intercept of the estimated line of best fit is at (0, b). enter the approximate value of the b in the first response box:
enter the approximate slope of the estimated line of best fit in the second response box:
Problem 8: Pool Water Level vs. Time Draining
Step 1: Find the y - intercept (b)
The y - intercept is the value of y when x = 0. From the scatter plot, when the number of hours draining (x) is 0, the water level (y) is approximately 60. So, \( b\approx60 \).
Step 2: Find the slope of the line of best fit
We can use two points on the line of best fit. Let's take the points \((0, 60)\) and \((10, 25)\) (approximate points from the line). The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Substituting the values, we get \( m=\frac{25 - 60}{10 - 0}=\frac{- 35}{10}=- 3.5\) (approximate).
Step 1: Find the y - intercept (b)
The y - intercept is the value of y when x = 0. From the scatter plot, when the temperature (x) is 0, the number of sweatshirts sold (y) is approximately 250. So, \( b\approx250 \).
Step 2: Find the slope of the line of best fit
We can use two points on the line of best fit. Let's take the points \((0, 250)\) and \((50, 150)\) (approximate points from the line). The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Substituting the values, we get \( m=\frac{150 - 250}{50 - 0}=\frac{- 100}{50}=- 2\) (approximate).
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The approximate value of \( b \) is 60. The approximate slope is - 3.5.