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$$S_x{_x} = \sum (x_i -\bar x)^2 = \sum x^2 -n \bar x^2$$

I know $$\bar x = \frac{\sum x }{n}$$

meaning

$$S_x{_x} = \sum (x_i -\frac{\sum x }{n})^2 = \sum x^2 - \sum x\frac{\sum x }{n}$$

but I don't know how to rearrange it.

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Almost there. Just replace the last sum with $\bar{x}$, which you can take it before the other sum. What is left is $\sum x$ which is $n\bar{x}$ Note however that $$S_{xx}=\sum x_i^2-2\sum x_i \bar{x}+\sum\bar{x}^2=\sum x_i^2-2\bar{x}\sum x_i + n\bar{x}^2=\sum x_i^2-2n\bar{x}^2+n\bar{x}^2$$

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