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erastovalidia [21]
3 years ago
14

Two mechanics worked on a car. The first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. Together they

charged a total of $1525. What was the rate charged per hour by each mechanic if the sum of the two rates was $200 per hour
Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer: the first mechanic charged $105 per hour and the second mechanic charged $95 per hour.

Step-by-step explanation:

Let x represent the the rate charged per hour by the first mechanic.

Let y represent the the rate charged per hour by the second mechanic.

The first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. Together they charged a total of $1525. This means that

10x + 5y = 1525- - - - - - - - -1

if the sum of the two rates was $200 per hour, it means that

x + y = 200

Substituting x = 200 - y into equation 1, it becomes

10(200 - y) + 5y = 1525

2000 - 10y + 5y = 1525

- 10y + 5y = 1525 - 2000

- 5y = - 475

y = - 475/ - 5

y = 95

x = 200 - y = 200 - 95

x = 105

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Answer:

t=\frac{54.1-53}{\frac{1.78}{\sqrt{12}}}=2.14  

df = n-1= 12-1=11

p_v = P(t_{11}>2.14) =0.0278

Step-by-step explanation:

Data provided

\bar X=54.1 represent the sample mean in mg per deciliter of cholesterol level

s=1.78 represent the sample standard deviation

n=12 sample size  

\mu_o =53 represent the value that we want to test  

t would represent the statistic (variable of interest)  

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

System of hypothesis

We need to conduct a hypothesis in order to check if the mean is higher than 53 mg per deciliter, the system of hypothesis would be:  

Null hypothesis:\mu \leq 53  

Alternative hypothesis:\mu > 53  

The statistic is given by:

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

Calculate the statistic  

Replacing into the formula we got:

t=\frac{54.1-53}{\frac{1.78}{\sqrt{12}}}=2.14  

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We need to find first the degrees of freedom:

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The p value for this case since we have a right tailed test is:

p_v = P(t_{11}>2.14) =0.0278

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