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drek231 [11]
3 years ago
15

Robert needs to cut four shelves from a board that is 2.5 meters long. The second shelf is 18 centimeters longer than twice the

length of the first shelf. The third shelf is 12 centimeters shorter than the first shelf. The remaining shelf is 4 centimeters longer than the first shelf. If Robert must use the entire 2.5 meter board for the shelves, what is the length of the second shelf, in centimeters?
Mathematics
1 answer:
USPshnik [31]3 years ago
7 0
   1st shelf: 48cm
 2nd shelf: 114cm
   3rd shelf: 36cm
   4th shelf: 52cm
Total length: 250cm or 2.5m 
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Answer:

a) \bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

b) ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n=80 represent the sample size  

Solution to the problem

Part a

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

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

For this case we can calculate the mean like this:

\bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

Part b

For this case is the sample size is doubled the margin of error would be:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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Answer:

Step-by-step explanation:

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According to the given statement

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Then for this purpose, we all know that the

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According to the this law

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Disclaimer: This question was incomplete. Please find the full content below.

Question:

Law of Sines and the Ambiguous Case.

In ∆ ABC, a =35, c = 25, and m < A = 60*

How many distinct triangles can be drawn given these measurements?

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