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3 years ago

The answer is 5/3, but I don’t know the solution.

Mathematics

1 answer:

VLD [36.1K]3 years ago

8 0

Check the picture below.

let's bear in mind that the segment MN is simply the sum of MQ + QN, and since M and N are midpoints, they cut that respective section into two equal halves.

$\bf \begin{cases} PQ=5QR\\ MN = \frac{PQ}{2}+\frac{QR}{2} \end{cases}\qquad \qquad \cfrac{PQ}{MN}\implies \cfrac{5QR}{\frac{PQ}{2}+\frac{QR}{2}}\implies \cfrac{5QR}{\frac{PQ+QR}{2}} \\\\\\ \cfrac{\frac{5QR}{1}}{\frac{PQ+QR}{2}}\implies \cfrac{5QR}{1}\cdot \cfrac{2}{PQ+QR}\implies \cfrac{10QR}{\underline{PQ}+QR}\implies \cfrac{10QR}{\underline{5QR}+QR}$

$\bf \cfrac{~~\begin{matrix} 10QR \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 6QR \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \cfrac{5}{3}$

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Step-by-step explanation:

1) 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 for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

We have the following distribution for the random variable:

$X \sim N(\mu , \sigma=0.45)$

And by the central theorem we know that the distribution for the sample mean is given by:

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

2) Confidence interval

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12$

The sample deviation calculated $s=2.268$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=8-1=7$

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that $t_{\alpha/2}=2.36$

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

$12-2.36\frac{2.268}{\sqrt{8}}=10.108$

$12+2.36\frac{2.268}{\sqrt{8}}=13.892$

So on this case the 95% confidence interval would be given by (10.108;13.892)

$10.108 < \mu < 13.892$

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