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nata0808 [166]
4 years ago
11

Cory just started his first job and does not have much money saved. Each month he needs to pay his rent and make payments on his

car. What kind of bank account would be most appropriate for him at this point? Select the best answer from the choices provided. a.checking account b.savings account c.money market account d.certificate of deposit
Mathematics
1 answer:
LenKa [72]4 years ago
8 0
Given that Cory does not have much money saved he will be using her regular incomes to pay his rent and car, so the most appropiate bank account is a checking account.

That will let him to deposit his incomes and make the payments with checks.

It does not make sense opening an account that pays interests because the money will be spent promptly.

Answer: option a. checking account.
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] It is claimed that 42% of US college graduates had a mentor in college. For a sample of college graduates in Colorado, it was
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Answer:

z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930  

The p value for this case would be given by:

p_v =P(z>3.930)=0.0000443  

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

Step-by-step explanation:

Information given

n=1045 represent the random sample selected

X=502 represent the college graduates with a mentor

\hat p=\frac{502}{1045}=0.480 estimated proportion of college graduates with a mentor

p_o=0.42 is the value that we want to test

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:  

Null hypothesis:p \leq 0.42  

Alternative hypothesis:p > 0.42  

The statistic is given by:

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

Replacing the info we got:

z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930  

The p value for this case would be given by:

p_v =P(z>3.930)=0.0000443  

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

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3 years ago
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