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andriy [413]
3 years ago
5

Laraine worked 40 regular hours last week, plus 3 overtime hours at the 1.5 rate. Her pay was $476.15. What was her hourly rate?

Mathematics
1 answer:
Oksanka [162]3 years ago
6 0

Answer:

Her hourly rate is $10.7

Step-by-step explanation:

Here, we are interested in calculating Laraine hourly rate.

In the question, we are told that the overtime rate is 1.5 times the regular hours rate.

Now, let the regular hours rate be $x / hour

For a total of 40 hours, the total money made will be 40 * x = $40x

Now let’s look at the overtime;

Since the overtime is 1.5 times the regular and she had 3 overtime hours;

Money obtained from overtime will be 1.5 * x * 3 = $4.5x

Let’s add this to the regular and equate to the total money made;

4.5x + 40x = 476.15

44.5x = 476.15

x = 476.15/44.5

x = 10.7

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu
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Answer:

a) s^2 =30^2 =900

b) \frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

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c) 23.818 \leq \sigma \leq 41.112

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in  the variability in the number of rooms occupied per day during a particular season of the  year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

s^2 =30^2 =900

Part b

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the significance \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=30.144

\chi^2_{1- \alpha/2}=10.117

And replacing into the formula for the interval we got:

\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

Part c

Now we just take square root on both sides of the interval and we got:

23.818 \leq \sigma \leq 41.112

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Answer:

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c

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