Suppose that 10% of people are left handed. If 6 people are selected at random, what is the probability that exactly 2 of them are left handed?

0.0984

Step-by-step explanation:

Given that :

Percentage that are left handed = 10%

P(left handed) = probability of success = 0.1

Number of selection (n) = 6

Number of left hands in selection (x)= exactly 2

P(6, 2) = 6C2 * 0.1^2 * (1 - 0.1)^(6-2)

P(6, 2) = 15 * 0.01 * 0.6561

P(6, 2) = 0.098415

answer: idk

step-by-step explanation:

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!