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attashe74 [19]
2 years ago
11

Bentley is working two summer jobs, making $18 per hour lifeguarding and making $8 per hour clearing tables. In a given week, he

can work no more than 15 total hours and must earn at least $180. If Bentley worked 5 hours clearing tables, determine the minimum number of whole hours lifeguarding that he must work to meet his requirements. If there are no possible solutions, submit an empty answer.
Mathematics
1 answer:
Mariulka [41]2 years ago
8 0

Answer:

8 hours

Step-by-step explanation:

First, let's find how much Bentley gets from clearing tables.

5 hours • $8 per hour = $40 total

We can take $40 off his $180 total, leaving the rest for when we figure out how much he needs to work lifeguarding.

$180 needed - $40 earned = $140 left to earn

We know that Bentley makes $18 an hour lifeguarding, so in order to get the number of hours he would need to work in order to make $140, we can divide it by $18.

$140 needed ÷ $18 per hour = 7.77...

We need to convert this to a whole number because we're talking about hours worked, so Bentley would need to work 8 hours lifeguarding to make $140 more. He can only work 15 hours total, so let's check if this exceeds his limit.

5 hours clearing tables + 8 hours lifeguarding = 13 hours < 15 hours max

It is possible to make $180 total from this work schedule, so your answer would be 8 hours.

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Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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