P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)
The brobability of a binomial distribution is given by

where: n = 4

Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439
Answer:

Step-by-step explanation:
![\frac{1}{\sqrt[5]{7}}=\frac{1}{7^{1/5}}=7^{-1/5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%5B5%5D%7B7%7D%7D%3D%5Cfrac%7B1%7D%7B7%5E%7B1%2F5%7D%7D%3D7%5E%7B-1%2F5%7D)
Answer:
(A) 9
Step-by-step explanation:
Substitute 2 for X

The answer is a it’s pretty easy