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Consumer products are required by law to contain at least as much as the amount printed on the package. For example, a bag of po

Troyanec [42]

Answer:

The average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802

Step-by-step explanation:

We are given that

Standard deviation, $\sigma=0.2$ ounces

We have to find the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.

$P(x\geq 10)=0.99$

Assume the bag weight distribution is bell-shaped

Therefore,

$P(\frac{x-\mu}{\sigma}\geq 10)=0.99$

We know that

$z=\frac{x-\mu}{\sigma}$

Using the value of z

Now,

$\frac{10-\mu}{0.2}=0.99$

$10-\mu=0.99\times 0.2$

$\mu=10-0.99\times 0.2$

$\mu=9.802$

Hence, the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802

5 0

3 years ago

PLEASE HELP ASAP!!!!!!!!!!!

matrenka [14]

C. 1/3 of about hour or 20 Min

7 0

3 years ago

A psychological experiment was conducted to investigate the length of time (time delay) between the administration of a stimulus

aleksklad [387]

Answer:

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

The significance level for this test is 0.05.

Step-by-step explanation:

In this case, we want to prove if there is significant evidence that the mean differs from 1.6 seconds. That is the same as having evidence to reject the the hypothesis $H_0:\mu= 1.6$ (the null hypothesis always have the equal sign).

We have to test the null hypothesis

$H_0:\mu= 1.6$

The significance level for this test is 0.05

Calculation of the t-statistic:

$t=\frac{M-\mu}{s/\sqrt{n}} =\frac{2.2-1.6}{0.57/\sqrt{36}} =\frac{0.6}{0.063}= 9.47$

If we look up in a t-table for t=9.47 and df=(36-1)=35, we get this value appears with a probability of zero. A large number like that is very unlikely to happen.

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

6 0

3 years ago

Use a net to find the surface area of the right triangular prism shown below:

AnnZ [28]

I don't exactly know what a net is, but I can figure out the surface area (maybe). When you put this pattern together, you have to make a couple of assumptions.
First of all the 2 triangles are connected to a base that's 8 by 13.
Base
The area of the base = 8 * 13 = 104 square feet.

Two Triangles.
The area of any triangle is
Area = 1/2 b * h
Area = 1/2 *12 * 5
Area = 30 square feet. There are 2 triangles so ...
The total area of both is 60 square feet.

Overlap
Now we come to the tough part. how big is it from the first dotted line to the second one?
The answer is 5. But why? Well the 5 comes from the the length of the side that has to be covered by this pattern. The 5 is part of the triangle.

How far across is the dotted line? The answer is 8. Why again? That's the width of the whole prism or pattern.
So the area between the two dotted lines is 5 * 8 = 40

Finally the area over the 12.
The question is how far is it from the second dotted line to the last boundary? The answer is 12. And it's width is 8
So the area = 12 * 8 = 96

Total area
96 + 60 + 40 + 104 = 300 square feet.

4 0

3 years ago

If x = 4 and y = 7 , evaluate the following expression: 100 − 3 ( 3 y − 4 x

ivolga24 [154]

Answer:

100-3*3*7-4*4

97

97*21-16

97*5

=455

3 0

3 years ago

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