If `b_i=1-a_i ,n a=sum_(i=1)^n a_i ,n b=sum_(i=1)^n b_i ,t h e nsum_(i=1)^n a_i ,b_i+sum_(i=1)^n(a_i-a)^2=` `a b` b. ` n a b` c. `(n+1)a b` d. `n a b`

sum_(i=1)^(n) sum_(i=1)^(n) i is equal to

Find sum_(i=1)^n sum_(i=1)^n sum_(k=1)^n (ijk)

If sum_(i=1)^(n) cos theta_(i)=n , then the value of sum_(i=1)^(n) sin theta_(i) .

If sum_(r=1)^n I(r)=(3^n -1) , then sum_(r=1)^n 1/(I(r)) is equal to :

The mean deviation for n observations x_1, x_2, ......... , x_n from their mean X is given by (a) sum_(i=1)^n(x_i- X ) (b) 1/nsum_(i=1)^n(x_i- X ) (c) sum_(i=1)^n(x_i- X )^2 (d) 1/nsum_(i=1)^n(x_i- X )^2

The mean deviation for n observations x_1, x_2, ......... , x_n from their mean X is given by (a) sum_(i=1)^n(x_i- X ) (b) 1/nsum_(i=1)^n|x_i- X | (c) sum_(i=1)^n(x_i- X )^2 (d) 1/nsum_(i=1)^n(x_i- X )^2

If n= 10, sum_(i=1)^(10) x_(i) = 60 and sum_(i=1)^(10) x_(i)^(2) = 1000 then find s.d.

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

sum_(i=1)^(2n) sin^(-1)(x_i)=npi then the value of sum_(i=1)^n cos^(-1)x_i+sum_(i=1)^n tan^(-1)x_i= (A) (npi)/4 (B) (2/3)npi (C) (5/4)npi (D) 2npi

If f(x) =Pi_(i=1)^3(x-a_i) + sum_(i=1)_3 a_i - 3x where a_i < a_(i+1) for i = 1,2, then f(x) = 0 have