If `barx=1/n sum_(i=1)^n x_i` then prove that `sum_(i=1)^n (x_i-barx)=0`

If sum_(i=1)^(n)sin x_(i)=n then sum_(i=1)^(n)cot x_(i)=

If sum_(i=1)^(n)sin x_(i)=n then sum_(i=1)^(n)cot x_(i)=

sum_(i=1)^n costheta_i=n then sum_(i=1)^n sintheta_i

Prove that identity : sum_(i=1)^(n) (x_i-bar x)^2 = sum_(i=1)^(n) x_i^2-n bar x^2= sum_(i=1)^(n) x_i^2 -(sum_(i=1)^(n) x_i)^2/n .

If sum_(i=1)^n (x_i -a) =n and sum_(i=1)^n (x_i - a)^2 =na then the standard deviation of variate x_i

Let x_1, x_2, ... ,x_n be n observations and barX be their arithmetic mean. The standard deviation is given by (a) sum_(i=1)^n(x_i- barX )^2 (b) 1/nsum_(i=1)^n(x_i- barX )^2 (c) sqrt(1/nsum_(i=1)^n(x_i- X )^2) (c) 1/nsum_(i=1)^n x_i^2- barX ^2

If sum_(i=1)^(2n)cos^(-1)x_i=0 then find the value of sum_(i=1)^(2n)x_i

If sum_(i=1)^(2 n) cos ^(-1) x_(i)=0, then sum_(i=1)^(2 pi) x_(i)=

sum_(i=1)^(n)cos theta_(i)=n then sum_(i=1)^(n)sin theta_(i)