Assume that a simple random sample has been selected and test the given claim. use the p-value method for testing hypotheses. identify the null and alternative hypotheses, test statistic, p-value, and state the final conclusion that addresses the original claim. the ages of actresses when they won an acting award is summarized by the statistics nequals79, xequals35.9 years, and sequals11.1 years. use a 0.01 significance level to test the claim that the mean age of actresses when they win an acting award is 32 years.

alpha = 0.05

xbar = 42.5

s = 5.9

n = 15

df = n-1 = 15 - 1 = 14

we will use the t distribution because n > 30 is not true and we don't know sigma.

hypothesis:

h0: mu = 32.6

h1: mu =/= 32.6

claim is in the null hypothesis h0. this is a two tailed test. the null will be rejected if the test statistic is not between the critical values.

t = (42.5 - 32.6)/(5.9/sqrt(15))

t = 6.49873476736499

t = 6.4987

critical values (alpha = 0.05, df = 14)

the test statistic 6.4987 is not between the critical values -2.145 and 2.145. so we reject the null hypothesis. there is enough statistically significant evidence to reject the null.

this means that the population mean mu is not 32.6. the initial claim is false.

d

step-by-step explanation:

d

(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