Ken and Paul have been mowing lawns to earn extra money. Their jobs earn $65 per week. Ken worked for x weeks and earned an extra $18. Paul worked for y weeks. From the expressions below, select one that determines the total amount that Ken and Paul earned mowing lawns. ? (5 points) a 83x + 65y b 65x − 18 + 65y c 60x + 83y d 65x + 18 + 65y

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.

the five is the numerator

answer: i dont get it ur bad

step-by-step explanation: boi

okay, what do you want us to do?

step-by-step explanation: