Refer to unions and intersections of relations. Since relations are subsets of Cartesian products, their unions and intersections can be calculated as for any subsets. Given two relations R and S from A to B, R ∪ S = {(x, y) ∈ A × B | (x, y) ∈ R or (x, y) ∈ S}
R ∩ S = {(x, y) ∈ A × B | (x, y) ∈ R and (x, y) ∈ S}.

Exercise

Let A = { –1, 1, 2, 4} and B = {1, 2} and define relations R and S from A to B as follows: For all (x, y) ∈ A × B,
x R y ⇔ |x | = |y| and
x S y ⇔ x – y is even.

Required:
State explicitly which ordered pairs are in A × B, R, S, R ∪ S, and R ∩ S.

4

step-by-step explanation:

8

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.