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raden_zaikaria
 one year ago
Best ResponseYou've already chosen the best response.0yes... insyaAllah

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

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

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

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

hba
 one year 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.

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0is this statistics?

chihiroasleaf
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0If i new it i wouldn't have been asking :/

chihiroasleaf
 one year 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 \(b= \bar X\)

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0what do you get by expanding it?

phi
 one year 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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