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Strike441 [17]
3 years ago
8

4. In football, the team that has the ball has fourchances to gain at least ten yards. If they don't gain at least ten yards, th

e other team gets the ball. Positive numbers represent a gain and negative numbers represent a loss. Select allof the sequences offour plays that result in the team getting to keep the ball.
A. 8, -3, 4, 21
B. 30, -7, -8, -12
C. 2, 16, -5, -3
D. 5, -2, 20, -1
E. 20, -3, -13, 2
Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0

Answer:

A , C, D

Step-by-step explanation:

Check A:

8 - 3 + 4 + 21 = 30         30 > 10 , so they keep the ball

Check B:

30 - 7 - 8 - 12 = 3          3 < 10 , so they don't keep the ball

Check C:

2 + 16 - 5 - 3 = 10         10 = 10 , so they keep the ball

Check D:

5 - 2 + 20 - 1 = 22       22 > 10 , so they keep the ball

Check E:

20 - 3 - 13 + 2 = 6         6 < 10 , so they don't keep the ball

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

castortr0y [4]3 years ago
4 0
A, C, and D

Reason:
A = 30 > 10
B = 3 < 10
C = 10 = 10
D = 22 > 10
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V=254.47in^3

EXPLANATION

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V=\pi \: {r}^{2} h

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A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in
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Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

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Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

n=10 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =0.0217 represent the sample variance

\sigma^2_0 =0.12^2= 0.0144 represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 0.0144

Alternative hypothesis: \sigma^2 \neq 0.0144

The statistic can be calculated like this;

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54

The degrees of freedom are:

df=n-1=10-1=9

Now we can calculate the p value using the alternative hypothesis:

p_v =2*P(\chi^2 >13.54)=0.279

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

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