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Crank
3 years ago
12

PLEASE HELP ME WITH THIS ASAP Melissa put her cat, Herman, on a diet and kept track of his weight. At the end of 4 weeks, she re

corded Herman's weight change as −412 ounces. She noticed that he had lost the same amount of weight each week.
What is Herman's weight change each week of the diet?

Enter your answer as a simplified mixed number, in the box.
I am giving the last of my points for this and will mark brainliest
Mathematics
2 answers:
castortr0y [4]3 years ago
4 0

Answer:

The answer should be 103.

Step-by-step explanation:

If Herman has been put on the diet for 4 weeks and has lost the same amount of weight each week you could divide 412 by 4 and get 103, not sure about the simplified mixed number though. Sorry.

forsale [732]3 years ago
3 0
103 ounces. It’s just 412 divided by the four weeks that passed.
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A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r
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Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

p_v =P(z  

If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

Step-by-step explanation:

1) Data given and notation  

n=180 represent the random sample taken  

X=125 represent the number of americans between 17 to 24 that not qualify for the military

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 is the value that we want to test  

\alpha=0.05 represent the significance level  

Confidence=95% or 0.95  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

Null hypothesis: p\geq 0.75  

Alternative hypothesis:p < 0.75  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.  

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

4) Statistical decision  

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The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

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Answer:

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