Thirty-six percent of U. S. adults have postponed medical checkups or procedures to save money. You randomly select nine U. S. adult. Find the probability that the number of U. S. adults out of the nine who have postponed medical checkups or procedures to save money is exactly 3.

0.2693

Step-by-step explanation:

Thus is a binomial probability problem :

p = 0.36

1 - p = 0.64

n = 9

x = 3

Using the binomial probability relation :

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

P(x = 3) = 9C3 * 0.36^3 * 0.64^6

P(x = 3) = 84 * 0.36^3 * 0.64^6

P(x = 3) = 0.2693

c. 6tan80

step-by-step explanation: