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

hbaBest 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
 one year ago

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

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

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

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

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

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

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

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

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

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

phiBest ResponseYou've already chosen the best response.1
show \[ \sum_{i=1}^{n}({X}_{i}\bar{X} )^2=\sum_{i=1}^{n}{X_i}^2n\bar{X}^2\] 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 \[ = \sum_{i=1}^{n} {X}_{i}^2  n \bar{X}^2 \]
 one year ago
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