A large shipment of batteries (several million) contains 4% that are defecttive.It 60 batteries are selected randomly,What is the probability that at most two are defective?

0.5676

Step-by-step explanation:

We have been given these informations from this question ,

probability of 4% = 0.04

q = 1 - p = 1 - 0.04 = 0.96

Sample size n = 60

We solve further by Using binomial probability formula,

P(X = x) = (nCx)*px*(1 - p)^n - x

We have:

(60C0)x 0.04⁰x(0.96⁶⁰+(60C1)x0.04¹x(0.96)⁵⁹+(60C2)x0.04²x (0.96)⁵⁸

From this we would get the probability to be = 0.5676

A) 20 in

B) 26 in and it got bigger by 1.3 times

C) 40 in and it got bigger by 2 times

Step-by-step explanation:

A) so the formula for finding perimeter is P= 2(L+W) so its (3+7)*2 = 20

B) if you double 3, 3*2=6 then it is 2*(6+7) = 26. 26/20= 13/20= 1.3 times more

C) if you double 3, 3*2=6 and 7, 7*2= 14 then it is (14+6)*2 = 40. 40/20=2/1 = 2 times more.

Also i couldnt solve D for you but try to solve it using what I wrote below.

Hope it helps :)