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Fiesta28 [93]
3 years ago
15

Popcorn was donated for the school fair by 3 different popcorn vendors. They donated a total of 996 bags of popcorn. Each vendor

donated the same number of bags. How many bags of popcorn did each vendor donate? Each vendor donated bags of popcorn
Mathematics
2 answers:
Marysya12 [62]3 years ago
5 0

Answer:

let the of bag of vendors be X

3donate X bag to give 996

3×x=996

3x=996

divide both side by 3

3x\3=996\3

x=332

each vendor donated 332bags of popcorn

grin007 [14]3 years ago
3 0

Answer:

Step-by-step explanation:

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juin [17]

Answer:

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the commercial improved the mean purchase potential rating.

Step-by-step explanation:

If we calculate the difference in ratings we have:

After Before Difference

1 6 -5

5 2 3

6 4 2

3 7 -4

7 4 3

4 3 1

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7 7 0

5 8 -3

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We will calculate the mean and standard deviation of the difference to test them later.

The sample size is n=12.

The mean is:

M=\dfrac{1}{12}\sum_{i=1}^{12}((-5)+3+2+(-4)+3+1+2+(-1)+1+0+(-3)+0)\\\\\\ M=\dfrac{-1}{12}=-0.083

The standard deviation is:

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{12}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}\cdot [(-5-(-0.083))^2+(3-(-0.083))^2+...+(0-(-0.083))^2]}\\\\\\

s=\sqrt{\dfrac{1}{11}\cdot [(24.17)+(9.51)+...+(1.17)+(0.01)+(8.51)+(0.01)]}\\\\\\s=\sqrt{\dfrac{78.92}{11}}=\sqrt{7.174}\\\\\\s=2.678

The null hypothesis states that the mean rating "after" would be less than or equal to the mean rating "before." The alternative hypothesis states that the mean rating difference is greater than 0.

Then, the null and alternative hypothesis:

H_0: \mu\leq0\\\\H_a:\mu> 0

The significance level is 0.05.

The sample has a size n=12.

The sample difference is d=-0.083.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.678.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2.678}{\sqrt{12}}=0.773

Then, we can calculate the t-statistic as:

t=\dfrac{\bar d}{s/\sqrt{n}}=\dfrac{-0.083}{0.773}=-0.108

The degrees of freedom for this sample size are:

df=n-1=12-1=11

This test is a right-tailed test, with 11 degrees of freedom and t=-0.108, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>-0.108)=0.542

As the P-value is greater than the significance level, the effect is not significant. The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the commercial improved the mean purchase potential rating.

5 0
3 years ago
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Answer: 149°F

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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