Suppose an actual census showed that 18.4% of the households in California have incomes in excess of $50,000.
The probability that the sample proportion will be 0.22 or greater for a random sample of 750 households is:

a) 0.0055. b) 0.9945. c) 0.9949. d) 0.0041. e) None

This is a Normal Distribution  problem.

P = 0.184  (18.4% as a decimal)

n = 750

Standard Deviation = Sqrt (P*(1-p) /n) = Sqrt(0.184*0.816/750)  = 0.014149

Normal Distribution  = (0.22-0.184)/0.0141 49 = 0.036/0.0141 = 2.544

2.544 on the Standard Normal Table  = 0.9945

Probability sample is greater: 1 - 0.9945 = 0.0055

Answer is a) 0.0055                

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.