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Big Ideas Math Integrated I, 2016 View details

5. Analyzing Lines of Fit

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Exercise 5 Page 196

Make a table of the residual values and then graph them on a scatter plot.

No, see solution.

Practice makes perfect

Let's begin by making a table of the residual values.

$x$	$y$	$y = 4 x - 5$	$y -$ value From Model	Residual
$- 4$	$- 18$	$4 (- 4) - 5$	$- 21$	$- 18 - (- 21) = 3$
$- 3$	$- 13$	$4 (- 3) - 5$	$- 17$	$- 13 - (- 17) = 4$
$- 2$	$- 10$	$4 (- 2) - 5$	$- 13$	$- 10 - (- 13) = 3$
$- 1$	$- 7$	$4 (- 1) - 5$	$- 9$	$- 7 - (- 9) = 2$
$0$	$- 2$	$4 (0) - 5$	$- 5$	$- 2 - (- 5) = 3$
$1$	$0$	$4 (1) - 5$	$- 1$	$0 - (- 1) = 1$
$2$	$6$	$4 (2) - 5$	$3$	$6 - 3 = 3$
$3$	$10$	$4 (3) - 5$	$7$	$10 - 7 = 3$
$4$	$15$	$4 (4) - 5$	$11$	$15 - 11 = 4$

Now we can create a scatter plot using the given $x -$ values and our residuals. Remember, if the model is a good fit for the data, the scatter plot will be evenly distributed above and below the $x -$ axis. Also, there will be no apparent patterns.

This line of fit does not model the data well. It is not evenly distributed above and below the $x -$ axis. The residual scatter plot shows that every residual is positive.

Subchapter links

Communicate Your Answer p.191

Monitoring Progress p.193-195

Exercises p.196-198

Exercise 5 Page 196

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