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natulia [17]
3 years ago
9

Help i need step by step 15 points

Mathematics
2 answers:
soldi70 [24.7K]3 years ago
4 0

Answer:

With what exactly can i help u with??

ad-work [718]3 years ago
4 0

Answer:?

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
an actor gains 20pounds for a part and then losses 15 pounds during the filming of the movie to go along with the story.who do y
guajiro [1.7K]
Let the actor's initial weight be 'x'

He is then gaining 20 pounds. The clue is in the word 'gaining' as it means 'increasing' or in Mathematics can be related as 'plus'

The expression is: x+20

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Last value is x + 20
The weight now is x + 20 - 15 = x + 5
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Simplify the expression:<br> 7r2+5r2+9r
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Step-by-step explanation:

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Mrrafil [7]

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3 years ago
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A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

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

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0
3 years ago
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