+0

​(xi – x)2 where xi = 1, 2, 3, 4, 5

0
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$$\epsilon$$(xi – x)2 where xi = 1, 2, 3, 4, 5

Guest Sep 24, 2017
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+91517
+1

(xi – x)2 where xi = 1, 2, 3, 4, 5

Mmm

Do you mean

$$\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\quad?$$

$$\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\\ =(1-5)^2+(2-5)^2+(3-5)^2+(4-5)^2+(5-5)^2\\ =16+9+4+1+0\\ =30$$

Melody  Sep 24, 2017

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