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Scrat [10]
3 years ago
6

The football team has 50 jerseys. There are 15 medium-sized jerseys. What percent of the jerseys are medium sized jerseys?

Mathematics
1 answer:
icang [17]3 years ago
8 0
15/50 = 3/30 = 1/10 1/10 = 10% That is about as simple as I could put it.
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Hopefully this is what you mean. Have a nice day!

Step-by-step explanation:

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A cardboard box is open at one end and it's shaped like a square prism missing one of its Square bases the volume of the prism i
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What 3x4x5x6+7= <br> single numbers
Lana71 [14]

Answer:

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Step-by-step explanation:

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2 years ago
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In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt
ankoles [38]

Answer:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

df=n-1=44-1=43  

p_v =P(t_{(43)}>1.436)=0.079  

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

\bar X=13.88 represent the sample mean

s=3.14 represent the sample standard deviation

n=44 sample size  

\mu_o =13.2 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:  

Null hypothesis:\mu \leq 13.2  

Alternative hypothesis:\mu > 13.2  

The statistic for this case is:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Calculating the statistic

Replacing the info given we got:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

P-value

The degrees of freedom are:

df=n-1=44-1=43  

Since is a right tailed test the p value is:  

p_v =P(t_{(43)}>1.436)=0.079  

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

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