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amid [387]
3 years ago
15

Lee worked a total of 30 hours at two jobs last week her combined pay from the jobs was $240 she earned $9 per hour working at t

he movie theater and $6 per hour working in her aunt's store. How many hours did she spend working in her aunt's store last week?
Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Answer:she spent 10 hours working in her aunt's store last week.

Step-by-step explanation:

Let x represent the number of hours that Lee spent working at the movie theater.

Let y represent the number of hours that Lee spent working at her aunt's store.

Lee worked a total of 30 hours at the movie theater and her aunt's store. This means that

x + y = 30

Her combined pay from the jobs was $240. She earned $9 per hour working at the movie theater and $6 per hour working in her aunt's store. This means that

9x + 6y = 240 - - - - - - - - - -1

Substituting x = 30 - y into equation 1, it becomes

9(30 - y) + 6y = 240

270 - 9y + 6y = 240

- 9y + 6y = 240 - 270

- 3y = - 30

y = - 30/ - 3 = 10 hours

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A soft-drink machine at a steakhouse is regulated so that the amount of drink dispensed is normally distributed with a mean of 2
AnnZ [28]

Answer:

z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6    

p_v =P(z>0.6)=0.274  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=203 represent the mean height for the sample  

\sigma=15 represent the population standard deviation

n=9 sample size  

\mu_o =200 represent the value that we want to test

\alpha=0.05 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 higher than 200, the system of hypothesis would be:  

Null hypothesis:\mu \leq 200  

Alternative hypothesis:\mu > 20  

If we analyze the size for the sample is < 30 but we  know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-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:  

z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6    

P-value

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

p_v =P(z>0.6)=0.274  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

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Lerok [7]

Given:

The value of the solid's surface area is equal to the value of the solid's volume.

Length(l) = 9 units

width(b) = 4 units

Height(h) = x units

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The value of x.

Solution:

Solid's surface area is

Area=2(lb+bh+hl)

Area=2(9\times 4+4\times x+x\times 9)

Area=2(36+4x+9x)

Area=2(36+13x)

Area=72+26x

Volume of solid is

Volume=l\times b\times h

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Solid's surface area = Volume of solid

72+26x=36x

72=36x-26x

72=10x

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7.2=x

Therefore, the value of x is 7.2.

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