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Butoxors [25]
2 years ago
7

plssssssssssssssssssssss help Six friends, four boys and two girls, went to a movie theater. They wanted to sit in a way so no g

irl sits on either first or last chair. How many such arrangements are possible?
Mathematics
2 answers:
Airida [17]2 years ago
7 0

Answer:

672 ways

Step-by-step explanation:

The total number of ways the friends were supposed to sit without restrictions would have been = 6!

= 720 possible ways.

If the girs should sit on the first or last sit

The possible arrangement= 2!4!

= 2*24

= 48

But now the two girls should be arranged in such a way that the don't sit on either the first or the last chair.

So the possible way now is= way without restriction - way if the girls are to sit at either first or last sit

= 720-48

= 672

tatyana61 [14]2 years ago
5 0

Answer:

672.

Step-by-step explanation:

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Ivahew [28]

Answer:

B, 14.6 units.

Step-by-step explanation:

The answer is be because:

the diameter = radius times 2

radius= 7.3

diameter=7.3*2

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Therefore the answer is, B

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The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv
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Answer:

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

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So on this case the 95% confidence interval would be given by (12.150;16.690)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=14.42 represent the sample mean

\mu population mean (variable of interest)

\sigma=6.95 represent the population standard deviation

n=36 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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we know that 1 hour = 60mins.

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