A producer of 10-ounce bags of pretzels claims that less than 7% of the pretzels in each bag are broken. The 95% confidence interval for the true proportion of broken pretzels in the 10-ounce bags produced by this company is (0.036, 0.067). Is it reasonable to conclude that less than 7% of pretzels in each bag are broken

We are 95% sure that the true proportion of broken pretzels in the 10-ounce bags produced by this company is (0.036, 0.067). Since 0.07 is higher than the upper bound of the interval, it is reasonable to conclude that less than 7% of pretzels in each bag are broken.

Step-by-step explanation:

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

In this question:

We are 95% sure that the true proportion of broken pretzels in the 10-ounce bags produced by this company is (0.036, 0.067). Since 0.07 is higher than the upper bound of the interval, it is reasonable to conclude that less than 7% of pretzels in each bag are broken.

(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

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation: