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3 years ago

Help please...find x

Mathematics

1 answer:

swat323 years ago

7 0

Because the two labelled line segments are parallel, it follows that the two triangles are similar and corresponding sides occur in the same ratio with one another.

In particular, this means

$\dfrac{2x-5}{(2x-5)+(x+8)}=\dfrac{22.5}{22.5+20}$

Solve for $x$ :

$\dfrac{2x-5}{3x+3}=\dfrac{22.5}{42.5}$

$\implies42.5(2x-5)=22.5(3x+3)$

$\implies85x-212.5=67.5x+67.5$

$\implies17.5x=280$

$\implies\boxed{x=16}$

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$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

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