A chocolate bar measures 50 mm wide, 90 mm long, and 3 and 1 over 2 mm high. Will's Wrapping Company is making a wrapper to cover the chocolate bar. Use the correct net and determine how much paper will be needed to make the wrapper.

The net of a square prism with a height of 50 millimeters, a length of 90 millimeters, and a width of 3 and 1 over 2 millimeters.
(4 points)

Will's Wrapping Company needs 4,640 mm2 of paper to wrap the chocolate bar.

Will's Wrapping Company needs 4,990 mm2 of paper to wrap the chocolate bar.

Will's Wrapping Company needs 9,980 mm2 of paper to wrap the chocolate bar.

Will's Wrapping Company needs 19,960 mm2 of paper to wrap the chocolate bar.

im pretty sure its

(8,15)

step-by-step explanation:

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.

y - 25 = 2 ( x - 10 ) should be the answer