A diet plan claims that women have a mean weight of 145 lbs. Assume that the weights of women have a standard deviation of 30.86 lb. A sample of 40 weights of women has a mean weight of 153.2 lb. Find the P-value and, using a 0.05 significance level, state the conclusion about the null hypothesis. 0.0465; reject the null hypothesis 0.0930; fail to reject the null hypothesis

0.0930; fail to reject the null hypothesis

Step-by-step explanation:

Sample Statistics

Population Statistics

Claim

The mean weight of women is 145lbs

Counterclaim:

The mean weight of women is not equal to 145lbs

Null hypothesis

Alternative hypothesis

Significance Level

for a 2-tailed test at a 95% confidence level

Test Statistic

Corresponding p-value

The test is 2-tailed because our alternate hypothesis, , already implies that the population mean, , could be either less than or greater than 145lb. Since 0.09286>0.05, we are within the 95% confidence level, so there's insufficient evidence in our sample data to suggest that we reject the null hypothesis. This means that it's 9.3% more likely that the null hypothesis is true than the alternate hypothesis is. Thus, the sample data do not differ significantly from the expected mean of 145lb.

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation: