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maria [59]
3 years ago
13

A construction worker earned $64 in a week. If he worked eight hours in one week, how much money did he earn per hour?

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0

Answer:

X=8

Step-by-step explanation:

64 Divided by Eight equals Eight.

Because he worked eight hours that week and earned $64 so divide that up by eight then you got your answer.

Klio2033 [76]3 years ago
6 0
9.14 you divide 64 by 7 since that’s a week plus you multiply 9.14 by 7 it give you 63.98
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Step-by-step explanation:

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3 years ago
How many solutions exist for the given equation?
zmey [24]
<h3> Answer </h3>

x = 2

<h3>Step by step</h3>

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1/2x + 12/2 = 4x - 1

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1/2x - 8/2x = -7

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8 0
3 years ago
The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 res
Rina8888 [55]

Answer:

z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179    

p_v =2*P(z>1.179)=2*[1-P(Z    

Step-by-step explanation:

Data given and notation    

\bar X=4000 represent the average score for the sample    

s=12000 represent the sample standard deviation    

n=200 sample size    

\mu_o =39000 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

z 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 apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  \mu = 39000    

Alternative hypothesis :\mu \neq 39000    

Compute the test statistic  

The sample size is large enough to assume the distribution for the statisitc normal. The statistic for this case is given by:    

z=\frac{\bar X-\mu_o}{\frac{s}{\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".    

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

z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179    

Give the appropriate conclusion for the test  

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

p_v =2*P(z>1.179)=2*[1-P(Z    

Conclusion    

If we compare the p value and a significance level given for example \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, then the true population mean for the salary not differs significantly from the value of 39000.    

6 0
3 years ago
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GenaCL600 [577]
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Answer:

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Step-by-step explanation:

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