Let Pr (X=2) = 1. Define µ2n = E(X-µ)^2n, µ = E(X). Then:
Because Pr (X=2) = 1 is a sure event. E(X) = 2. Further, X - µ = 0 always. So, µ2n = 0 irrespective of what n is.
13/3 – (23/6) = ? – (22/9)
[(82× 162)/12] - 28 = ?
(630 ÷ 35) × 2 + 144 = ? × 2
23% of 8040+ 42% of 545 = ?%of 3000
1120 / √x = 80 Then x = ?
5555 ÷ 11 ÷ 5 = 100 + ?
{640 of 8 - (12)2 of 6} - 900 = ?
(60/15) × 25 + 15 2 – 18% of 200 = ? 2
540240 ÷ 24 ÷ 25 =?
7.4 × 8.2 + 3.5 × 4.5 = ? + 11.5 × 2.5