(a) `3/10`+`2/10`+`7/10`

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 is

5 1/7-{3 3/10-:[2 4/5-7/10]}

lim_(x->oo){7/10+29/10^2+133/10^3+.......+(5^n+2^n)/10^n} equals

The sum of 1/10+23/10^3+23/10^5+23/10^7+.... as a recurring decimal fraction is

(7) / (5) (1+ (1) / (10 ^ (2)) + (1.3) / (1.2) (. 1) / (10 ^ (4)) + .... oo) =

If (10)^(9)+2(11)^(1)(10)^(8)+3(11)^(2)(10)^(7)+....+10(11)^(9)=k(10)^(9), then k is equal to (1)(121)/(10)(2)(441)/(100) (3) 100(4)110

3xx10^(5) + 4xx10^(3) + 7xx10^(2) + 5 is equal to

The mean of 3,10,10,4,7,10,5 is

Simplify : ((1)/(12) + ((-3)/( 4)) + (7)/(8)) xx (3 (2)/(5) - (7)/(10) + ((-2)/( 15)) - 10 (1)/(30))