F(x)=(x+1) ^2 . Determine for each x-value whether it is in the domain of f or not.

what is the your question?

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.

1st integer x

2nd integer x+1

3rd integer x+2

4th integer x+3

x+x+1 +x+2+x+3 =74

combine like terms

4x+6 = 74

subtract 6 from each side

4x = 68

x = 17

the integers are 17,18,19,20