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Learn to recognize relationships in data and find the equation of a line of best fit. Course 3 12-7 Lines of Best Fit.

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Presentation on theme: "Learn to recognize relationships in data and find the equation of a line of best fit. Course 3 12-7 Lines of Best Fit."— Presentation transcript:

1 Learn to recognize relationships in data and find the equation of a line of best fit. Course Lines of Best Fit

2 When data show a correlation, you can estimate and draw a line of best fit that approximates a trend for a set of data and use it to make predictions. To estimate the equation of a line of best fit: calculate the means of the x-coordinates and y-coordinates: (x m, y m ) draw the line through (x m, y m ) that appears to best fit the data. estimate the coordinates of another point on the line. find the equation of the line. Course Lines of Best Fit

3 Plot the data and find a line of best fit. Additional Example 1: Finding a Line of Best Fit Plot the data points and find the mean of the x- and y-coordinates. x m = = y m = = x y (x m, y m )= 6, 4 Course Lines of Best Fit

4 A line of best fit is a line that comes close to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line. Remember! Course Lines of Best Fit

5 Additional Example 1 Continued Draw a line through 6, 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line. 2 3 Course Lines of Best Fit

6 Find the slope. y – y 1 = m(x – x 1 )Use point-slope form. y – 4 = (x – 6) Substitute. y – 4 = x – y = x The equation of a line of best fit is. 2 3 y = x Additional Example 1 Continued m = = = 6 – 4 8 – Course Lines of Best Fit

7 Plot the data and find a line of best fit. Check It Out: Example 1 Plot the data points and find the mean of the x- and y-coordinates. x m = = 2 – – y m = = 1 – – x–1026–38 y–1037–74 (x m, y m ) = (2, 1) Course Lines of Best Fit

8 Check It Out: Example 1 Continued Draw a line through (2, 1) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line. Course Lines of Best Fit

9 Find the slope. y – y 1 = m(x – x 1 )Use point-slope form. y – 1 = (x – 2) 9 8 Substitute. y – 1 = x – The equation of a line of best fit is. y = x – Check It Out: Example 1 Continued m = = 10 – 1 10 – y = x – Course Lines of Best Fit

10 Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in Is it reasonable to make this prediction? Explain. Additional Example 2: Sports Application Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107). x m = = Year Distance (ft) y m = = (x m, y m ) = (5, 103) Course Lines of Best Fit

11 Additional Example 2 Continued Draw a line through (5, 103) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line. Course Lines of Best Fit

12 m = = Find the slope. y – y 1 = m(x – x 1 )Use point-slope form. y – 103 = 0.8(x – 5)Substitute. y – 103 = 0.8x – 4 y = 0.8x + 99 The equation of a line of best fit is y = 0.8x Additional Example 2 Continued Course Lines of Best Fit Since 1990 represents year 0, 2006 represents year 16.

13 Substitute. y = y = 0.8(16) + 99 The equation predicts a winning distance of about 112 feet for the year A toss of about 112 feet is a reasonable prediction. y = Additional Example 2 Continued Course Lines of Best Fit Add to find the distance.

14 Predict the winning weight lift in Check It Out: Example 2 Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170). x m = = y m = = Year Lift (lb) (x m, y m ) = (6, 132) Course Lines of Best Fit

15 Check It Out: Example 2 Continued Draw a line through (5, 132) the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line. Years since 1990 weight (lb) Course Lines of Best Fit

16 m = = – – 5 Find the slope. y – y 1 = m(x – x 1 )Use point-slope form. y – 132 = 4(x – 5)Substitute. y – 132 = 4x – 20 y = 4x The equation of a line of best fit is y = 4x Since 1990 represents year 0, 2010 represents year 20. Check It Out: Example 2 Continued Course Lines of Best Fit

17 Substitute and add to find the winning weight lift. y = 192 y = 4(20) The equation predicts a winning weight lift of about 192 lb for the year A weight lift of 193 lbs is a reasonable prediction. Check It Out: Example 2 Continued Course Lines of Best Fit


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