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Archy [21]
2 years ago
5

Ruben earns $4 an hour as a cashier and $5.50 and hour as a waiter. If he works 5 days per week, how much does he earn if he wor

ks 4 hours a days as a cashier and 3 hours a day as a waiter?
Mathematics
1 answer:
34kurt2 years ago
7 0
So to find how much Ruben will make if he works 4 hours a day as a cashier and 3 hours a day as a waiter for 5 days per week first we'll work on his cashier job. So he makes 4$ an hour as a cashier and her works 4 hours every day for 5 days, multiply 4 by 4 and you'll get 16, so he makes 16 dollars per day as a cashier. now that we know how much he makes a day we simply have to multiply 16 by 5 since he works for 5 days every week. 16*5 is 80, so Ruben makes 80 dollars working 4 hours a day for 5 days. Now we move onto his waiter job. he makes 5.50 an hour as a waiter and her works 3 hours a day. So we multiply 3 by 5.50 to get 16.5. Next, we multiply 16.5 by 5 since he also works 5 days a week as a waiter. 16.5*5 is equal to 82.5. now we add 82.5 and 80 to get 162.5. SO Ruben makes 162.5$ working both jobs 5 days a week 
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Find an equation of the line parallel to y= 1/3x +1 and that passes through the point (3,5)
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2 years ago
Yˆ=2.391x+57.420 which number is the y intercept?
BaLLatris [955]

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3 years ago
Read 2 more answers
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
Rasek [7]

Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}

\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

t_{11,0.025} = 2.201

The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0
3 years ago
George plans to cover his circular pool for the upcoming winter season. The pool has a diameter of 20 feet and the cover extends
Verizon [17]

Answer:

Area of Pool cover: 110.25π

Length of rope: 21π

Step-by-step explanation:

To find the area of a circle the formula is πr^2.

1. take 10.5, which is half of the diameter and multiply it by itself then multiply by pi(π). Answer is 110.25π

To find the length of the rope(circumference) the formula is 2πr.

1. Take r, which we established as 10.5 then multiply it by 2, which is 21. Then multiply by π. Answer is 21π

____________________________________________________________

If ur teacher wants you to use 3.14 as pi then these are your answers:

Area of Pool cover: 346.185, round up to 346.19 if needed

Length of rope: 65.94

8 0
3 years ago
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