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Scrat [10]
3 years ago
12

Cindy worked 23 hours, she earned $126.5 whats her rate of pay​

Mathematics
2 answers:
Annette [7]3 years ago
7 0

Answer:

5.5

Step-by-step explanation:

you do 126.5/23. this is because rate of pay x hours worked=earnings, so you flip that to get rate of pay =earnings/hours, so you just plug it in to get the first equation, 126.5/23.

balu736 [363]3 years ago
6 0

Answer:

Step-by-step explanation:

Divide $126.5/23 hours to get the rate of pay, which is $5.5 per hour.

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Y=x+2<br> 3x+3y=6<br> solving systems by substitution
motikmotik

Answer:6

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6

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6

6

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3 years ago
Is my answer choice wrong or right?
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Gr8 job

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3 years ago
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In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt
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Answer:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

df=n-1=44-1=43  

p_v =P(t_{(43)}>1.436)=0.079  

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

\bar X=13.88 represent the sample mean

s=3.14 represent the sample standard deviation

n=44 sample size  

\mu_o =13.2 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:  

Null hypothesis:\mu \leq 13.2  

Alternative hypothesis:\mu > 13.2  

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

P-value

The degrees of freedom are:

df=n-1=44-1=43  

Since is a right tailed test the p value is:  

p_v =P(t_{(43)}>1.436)=0.079  

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

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