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mel-nik [20]
1 year ago
15

On a recent road trip, Mr. Yost drove 210 miles in 3 1/2 hours. Find both the miles driven per hour and the hours driven per mil

e.
Mathematics
1 answer:
amid [387]1 year ago
7 0

Miles driven per hour is 60 miles per hour

Hours driven per mile is 0.01667 hours per mile

<em><u>Solution:</u></em>

Given that,

On a recent road trip, Mr. Yost drove 210 miles in 3 1/2 hours

Therefore,

Miles driven = 210 miles

Time\ taken = 3\frac{1}{2}\ hour = \frac{7}{2} = 3.5\ hour

To find: miles driven per hour and the hours driven per mile

<h3><u>Miles driven per hour</u></h3>

\frac{miles}{hour} = \frac{210}{3.5}\\\\\frac{miles}{hour} = 60\ miles\ per\ hour

<h3><u>Hours driven per mile</u></h3>

\frac{hours}{miles} = \frac{3.5}{210}\\\\\frac{hours}{miles} = 0.01667\ hours\ per\ miles

Thus both the miles driven per hour and the hours driven per mile are found

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Answer:

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Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

\hat p=\frac{67}{150}=0.447 estimated proportion of applicants who request financial aid

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\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

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.

Calculate the statistic  

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

z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

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

p_v =P(z>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \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 applicants who request financial aid is significantly higher than 0.35

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Answer:

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