The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this might be an overstatement of the breaking weight. So we decide to take a random sample of 36 of these rubber bands and record the weight required to break each of them. The mean breaking weight of our sample of 36 rubber bands is 6.6 lbs. Assume that the standard deviation of the breaking weight for the entire population of these rubber bands is 2 lbs. True or false

The statement is True.

Step-by-step explanation:

In this case we need to determine whether the rubber bands in a package of a particular brand of rubber band can hold a weight of 7 lbs or less.

A one-sample test can be used to perform the analysis.

The hypothesis can be defined as follows:

H₀: The mean weight the rubber bands can hold is 7 lbs, i.e. μ = 7.

Hₐ: The mean weight the rubber bands can hold is less than 7 lbs, i.e. μ < 7.

The information provided is:

As the population standard deviation is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

The test statistic value is -1.20.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

*Use a z-table for the probability.

The p-value of the test is 0.1151.

The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.

Thus, it can be concluded that the mean weight the rubber bands can hold is 7 lbs.

Hence, the statement is True.

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)