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NeX [460]
3 years ago
5

Allen needs to prepare a compound for the patient to pick up tomorrow. He needs to mix of drug A, with of drug B and of drug C.

What is the total amount of the drugs mixed together?
Determine the lowest common denominator
Show work
Mathematics
1 answer:
Phoenix [80]3 years ago
5 0

Answer:

76/15

Step-by-step explanation:

Add the drug combinations together

Drug A + B + C=2/5 + 5/3 + 9/3

The L.C.M of the denominators is 15

=6+25+45/15

=76/15

Or

=5 1/15

Allen needs to prepare 76/15 compound for the patient

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561.985429 rounded to the hundred-thousandths is
kirill115 [55]

Answer:

Step-by-step explanation:

561.98543

The 9 is greater than 5, so add one to the number before it

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2. A printing machine makes 240 books
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Answer: 8 books/ hour

Explanation:

240 / 30 = 8 books/ hour

I hope this helped!

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Aleksandr-060686 [28]

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6.5in

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find the area of the shaded regions

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A = 2×2

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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.

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3 years ago
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