The Food Marketing Institute shows that 17\%17% of households spend more than $$100$ per week on groceries. Assume the population proportion is p = .17p=.17 and a simple random sample of 800800 households will be selected from the population. a. Show the sampling distribution of \bar{p} pˉ , the sample proportion of households spending more than $$100$ per week on groceries.
b. What is the probability that the sample proportion will be within \pm .02±.02 of the population proportion?
c. Answer part (b) for a sample of 16001600 households.

step-by-step explanation:

the answer is d. 75%

468

step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0