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hba
 3 years ago
Can anyone help me prove this?
hba
 3 years ago
Can anyone help me prove this?

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raden_zaikaria
 3 years ago
Best ResponseYou've already chosen the best response.0yes... insyaAllah

hba
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \sum_{i=1}^{n}({X}_{i}X \prime)^2=\sum_{i=1}^{n}{X_i}^2n X \prime^2 \] X prime is actually x bar.dw:1360770364042:dw

hba
 3 years ago
Best ResponseYou've already chosen the best response.0It's InshaAllah * @raden_zaikaria

hba
 3 years ago
Best ResponseYou've already chosen the best response.0Um,I know we can use this, dw:1360770482862:dw

raden_zaikaria
 3 years ago
Best ResponseYou've already chosen the best response.0so difficult.... is this polynomial system....

hba
 3 years ago
Best ResponseYou've already chosen the best response.0Never studied stats in my life lol. But now i have to :/

hba
 3 years ago
Best ResponseYou've already chosen the best response.0I have a physics quiz tomorrow so i can study for that,Please give me some feedback.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0expand the \[\left( X _{i}\bar{X} \right)^{2}\] then use the properties of sigma

hba
 3 years ago
Best ResponseYou've already chosen the best response.0If i new it i wouldn't have been asking :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use the properties \[ (a+b)^2 = a^2 +2ab + b^2\] let \(a = X_i\) and \(b= \bar X\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you get by expanding it?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1show \[ \sum_{i=1}^{n}({X}_{i}\bar{X} )^2=\sum_{i=1}^{n}{X_i}^2n\bar{X}^2\] where by definition \[ \bar{X}= \frac{1}{n}\sum_{i=1}^{n} X_i \] expand the square to get \[ \sum_{i=1}^{n}({X}_{i}\bar{X} )^2= \sum_{i=1}^{n}({X}_{i}^2  2\bar{X} {X}_{i}+\bar{X}^2)\] or taking the sum term by term \[ = \sum_{i=1}^{n} {X}_{i}^2  2\bar{X}\sum_{i=1}^{n}{X}_{i} + \sum_{i=1}^{n}\bar{X}^2\] the mean \( \bar{X}^2\) is a constant, and summing n copies is \( n\bar{X}^2\) also, by the definition of \( \bar{X} \), \[ n \bar{X}= \sum_{i=1}^{n} X_i \] using these expressions, we get \[ = \sum_{i=1}^{n} {X}_{i}^2  2n \bar{X}^2 + n \bar{X}^2\] or \[ = \sum_{i=1}^{n} {X}_{i}^2  n \bar{X}^2 \]
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