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kykrilka [37]
3 years ago
11

Two brothers want to share a large field equally.

Mathematics
1 answer:
Elan Coil [88]3 years ago
7 0

Yes, the partition gives the two brothers equal shares.

Step-by-step explanation:

Step 1:

Assume the entire field has an area of B. So one brother takes \frac{1}{12} B, \frac{1}{6}B, and \frac{1}{4} B

So we need to calculate how much this brother takes in terms of B.

To do this we calculate how much \frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B is.

Step 2:

To add \frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B,

First take the LCM of the denominators 12, 6, and 4

The LCM is 12, we multiply the denominator to get the LCM value, this same value is multiplied to the numerator too.

\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B = \frac{1(1)}{12(1)} B + \frac{1(2)}{6(2)} B + \frac{1(3)}{4(3)} B \\\\\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B= \frac{1B +2B+3B}{12} = \frac{B}{2}

Step 3:

One brother gets \frac{1}{2} B, so we need to calculate how much the other brother gets.

The other brother's share = B - \frac{1}{2} B = \frac{1}{2}B

So both the brothers get equal shares

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Answer:

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Step-by-step explanation:

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You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample
Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=160)  

And the distribution for \bar X is:

\bar X \sim N(\mu, \frac{160}{\sqrt{n}})  

We know that the margin of error for a confidence interval is given by:  

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025  

Using the normal standard table, excel or a calculator we see that:  

z_{\alpha/2}=\pm 1.96  

If we solve for n from formula (1) we got:  

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

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And we have everything to replace into the formula:  

n=(\frac{1.96(160)}{78.4})^2 =16  

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And now we can replace the new value of Me and see what we got, like this:

n=(\frac{1.96*160}{60})^2 =27.32

And if we round up the answer we see that the value of n to ensure the margin of error required Me=\pm 60 $ is n=28.    

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