Please help me now I hace 5 mins

2. Mr Alison fill ups his car at the gas station. He also gets a car wash at the station and visits with the manager, Then he dr

e-lub [12.9K]

Answer:

Hmm graph 3, i think.

Step-by-step explanation:

7 0

3 years ago

Husam works at a music store.Last week,he sold 6 more than 3 times the number of CDs that she sold this week.Husam told a total

lana [24]

Answer:

84 CDs last week and 26 CDs this week

Step-by-step explanation:

6 0

3 years ago

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The question is in the picture

SSSSS [86.1K]

The 100th term would be 505.

7 0

3 years ago

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To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

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

$p_v =P(z>2.491)=0.0064$

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

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

$p_o=0.35$ is the value that we want to test

$\alpha$ 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

5 0

3 years ago

A trip takes 7 3/4 hours and 2/3 of the trip is done how much longer to the end of the trip

mylen [45]

Total trip time: 7 3/4
time spent: 2/3
time left: total time - time spent
= 7 3/4 - 2/3
=31/4 - 2/3
by taking lcm:
= (63-8)/12
= 55/12
= if in decimals: 4.583
= if in mixed numbers: 4 7/12
so the time left is 4 7/12 hours

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3 years ago

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