Suppose that ​thank-you notes are written and envelopes are addressed.​ Accidentally, the notes are randomly inserted into the envelopes and mailed without checking the addresses. What is the probability that all the notes will be inserted into the correct​ envelopes?

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

yx = 36

step-by-step explanation:

cos f = opposite side/ hypotenuse

cos f = yw / yx

we know cos f = 4/9 and yw =16

4/9 = 16/yx

multiply each side by yx

4/9 yx = 16/ yx * yx

4/9 yx = 16

multiply each side by 9/4

9/4 * 4/9 yx = 16 *9/4

yx = 36