CNN/USA today poll in April 2005 reported that 75% of adult Americans were satisfied with the job the nations major airlines were doing. 10 adult Americans were selected at random and the number who are satisfied with the airlines is recorded what is the probability that at least one of the American was satisfied with the job the nations airlines were

0.99999

Step-by-step explanation:

Percentage that were satisfied, Probability of success, p = 0.75

Number of trials, n = 10

1-p = 0.25

Probability that atleast one is satisfied ;

P(x ≥ 1) = p(x=1)+p(x=2)+p(x=3)+..p(x=10)

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

P(x ≥ 1) = 0.99999

Using the binomial distribution calculator to save time :

P(x ≥ 1) = 0.99999

it should be a and d. lmk if i'm wrong xd

answer: h

step-by-step explanation: