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Ainat [17]
3 years ago
11

Nina ran 12 4/5 laps in 1/4 of an hour . At this rate , how many laps could she run in a hour ?

Mathematics
1 answer:
slega [8]3 years ago
8 0

\text{ Answer: There are }51 \frac{1}{5}\ laps.

Explanation:

Since we have given that

Number of laps Nina ran is given by

12\frac{4}{5}

Time taken by Nina to ran the lap is given by

\frac{1}{4}

Now, we use the unitary method ,

In\ \frac{1}{4}\ hour,\text{ Nina ran }12\frac{4}{5}=\frac{64}{5}\ laps\\\\In\ 1\ hour,\text{ Nina ran }\frac{\frac{64}{5}}{\frac{1}{4}}=\frac{256}{5}\ laps

Now, we change it into mixed fraction,

\frac{256}{5}=51\frac{1}{5}\ laps

Hence,

\text{ there are }51 \frac{1}{5}\ laps.

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D. -14.88 meters

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4.96(3)=14.88
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3 years ago
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kozerog [31]

Answer:

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3 0
3 years ago
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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.

8 0
3 years ago
The price of a pair of jeans was $60 after a 50% markup. What was the price of the jeans before the markup?
zhenek [66]

Answer:

40

Step-by-step explanation:

We have an original price p

We mark them up by 50%

p + p*50%

Changing to decimal form

p+.50p = 1.5p

The new price is 60 dollars

1.5p = 60

Divide each side by 1.5

1.5p/1.5 = 60/1.5

p =40

The original price is 40

5 0
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yarga [219]

Answer:

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Step-by-step explanation:

3 0
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