# R-squared equal to 0 [closed]

R2=

Solution says it is equal to 0 only if b1 is equal to 0. But if the sum of deviations of a variable from the mean is always equal to 0, wouldn't this formula somehow suggest R2 is always 0?

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Which "variable" are you referring to? And how is the sum of deviations from the mean relevant?--It does not appear anywhere in the formula. – whuber Oct 21 '12 at 17:03
It seems to me that sigma(x-xbar) in the numerator is the sum of the deviations of the independent variable from its mean – user14386 Oct 21 '12 at 17:07
You are overlooking the squares. This distinction was made in an answer to an earlier question of yours, which you accepted, so I presume you read and understood it. – whuber Oct 21 '12 at 17:12
Thank you, I realized the difference now. – user14386 Oct 21 '12 at 18:21

## closed as not a real question by whuber♦Oct 21 '12 at 21:31

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