coordinate plane with triangles lmn and pqr with l at negative 3 comma 2, m at 1 comma 6, n at 1 comma 2, p at negative 2 comma 3, q at 0 comma 5, and r at 0 comma 3
which set of transformations would prove δlmn ~ δpqr?
dilate δpqr by the scale factor of 2 from point r, and translate δp′q′r′ by the rule (x + 0, y − 1).
dilate δpqr by the scale factor of 2 from point q, and translate δp′q′r′ by the rule (x + 1, y + 0).
translate δpqr by the rule (x + 1, y + 1), and dilate δp′q′r′ by a scale factor of 2 from point p.
translate δpqr by the rule (x + 1, y − 1), and dilate δp′q′r′ by a scale factor of 2 from point r.

it would be:

c.parallelogram.

lo98765432wqasdfghjkloi87654567890

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.

the given 6 = w/p is equivalent to 6p = w, where w is the total area of the walls and p is the area covered by a single can of paint.

knowing that 6 cans of paint will be needed, 6p = w becomes 6(6) = w = 36.   this tells us that the total area to be painted is 36 square units.