lexber
  • lexber
how do i make a line of best fit?
Algebra
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You can use the method of least squares to find a regression line: https://en.wikipedia.org/wiki/Least_squares
anonymous
  • anonymous
Of course it depends on what kind of "best fit" you're looking for. Do you have a data set you're trying to model with linear function?
lexber
  • lexber
i need to make a line of best fit out of this and write the approximate slope but im not understanding how to do it.

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anonymous
  • anonymous
You're looking for a line of the form \(\hat{y}=\beta_0+\beta_1x\), and \(\beta_1\) is the slope you want in the end. I don't know where you're at in terms of the theory behind regression, so for the sake brevity, you can determine the slope of the regression line using this formula: \[\beta_1=\frac{\displaystyle\sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}\]where \(\bar{x}=\displaystyle\frac{1}{n}\sum_{i=1}^nx_i\) (the average of all the x-coordinates of the given points). The same goes for \(\bar{y}\), just replace \(x_i\) with \(y_i\). The plot gives you the points \((x_i,y_i)\) you need to compute the averages. So for example, the three bottom-leftmost points are \((2.5,1)\), \((5,2)\), and \((7.5,2)\). Your averages would be \[\bar{x}=\frac{2.5+5+7.5}{3}=5\quad\quad\quad\bar{y}=\frac{1+2+2}{3}\approx1.67\] Then the best fit line FOR THESE THREE POINTS ONLY would be \[\begin{align*}\beta_1&=\frac{(2.5-5)(1-1.67)+(5-5)(2-1.67)+(7.5-5)(2-1.67)}{(2.5-5)^2+(5-5)^2+(7.5-5)^2}\\[1ex] &=0.2\end{align*}\] You have 17 points to account for, but this is basically a template for what you have to do.

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