When commercial aircraft are inspected, wing cracks are reported as nonexistent, detectable, or critical. The history of a particular fleet indicates that 70% of the planes inspected have no wing cracks, 25% have detectable wing cracks, and 5% have critical wing cracks. Five planes are randomly selected. Find the probability that at least one plane has critical cracks.

0.226

Step-by-step explanation:

Given that:

No wing crack (x) = 70%

Detectable wing crack (y) = 25%

Critical wing crack (z) = 5%

Number of samples (n) = 5

Probability that atleast a plane has critical crack:

1 - p(no critical crack)

Using the multinomial distribution calculator :

P(5,0,0) + p(4,1,0) + p(3,2, 0) + p(2,3,0) + p(1,4,0) + p(0,5,0)

From the calculator :

P(5,0,0) + p(4,1,0) + p(3,2, 0) + p(2,3,0) + p(1,4,0) + p(0,5,0) = 0.7741

1 - 0.7741 = 0.2259

= 0.226

hows it hangin lol

step-by-step explanation:

to find the unit price, we must figure out the cost of one item (in this case one dish towel).   to do this, we divide the total price ($7.92) by the number of items bought (for this problem it is 8, because there are eight dish towels in one pack).  

$7.92/8 = $0.99

therefore, the unit price, or the price of one dish towel, is $0.99.

hope this !