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ikadub [295]
3 years ago
6

When Cheryl goes Mountain climbing she rests five minutes for every 15 minutes that she climbs. If Cheryl combined time is 2 hou

rs how many minutes does she rest
Mathematics
2 answers:
castortr0y [4]3 years ago
7 0
Cheryl would rest for 30 mins in the 2 hour period
laila [671]3 years ago
5 0
40 minutes would be the answer.

120 minutes = 2 hours.

If she rests every 15 minutes, divide 120 by 15 and get 8.
She takes 5 minute rests, so times 5 by the 8, and you get 40.

Hope this helped.
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.67×.3=.201
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Three point one four times eight squared
Margaret [11]

Answer:

200.96

Step-by-step explanation:

Hope it helps! Good luck with math.

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Svetradugi [14.3K]
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3 years ago
g A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estim
kipiarov [429]

Answer:

a) n= 1045 computers

b) n= 442 computers

c) A. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

Step-by-step explanation:

Hello!

The variable of interest is

X: Number of computers that use the new operating system.

You need to find the best sample size to take so that the proportion of computers that use the new operating system can be estimated with a 99% CI and a margin of error no greater than 4%.

The confidence interval for the population proportion is:

p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

Z_{1-\alpha /2}= Z_{0.995}= 2.586

a) In this item there is no known value for the sample proportion (p') when something like this happens, you have to assume the "worst-case scenario" that is, that the proportion of success and failure of the trial are the same, i.e. p'=q'=0.5

The margin of error of the interval is:

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p)}{n} }

(\frac{d}{Z_{1-\alpha /2}})^2 = \frac{p'(1-p')}{n}

n * (\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')

n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2

n=[0.5(1-0.5)]*(\frac{2.586}{0.04} )^2= 1044.9056

n= 1045 computers

b) This time there is a known value for the sample proportion: p'= 0.88, using the same confidence level and required margin of error:

n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2

n= [0.88*0.12]*(\frac{{2.586}}{0.04})^2= 441.3681

n= 442 computers

c) The additional information in part b affected the required sample size, it was drastically decreased in comparison with the sample size calculated in a).

I hope it helps!

4 0
4 years ago
The quotient of 37.2/6 is?
Dimas [21]

Answer:

6.2

Just normal division

4 0
3 years ago
Read 2 more answers
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