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3 years ago

Using similar triangles if KLM~ PQR with a scale factor of 3:5, find the perimeter of PQR (With work please)

Mathematics

2 answers:

Blizzard [7]3 years ago

7 0

Hi,

KLM = 12, 15, 6

PQR = (12 + 4 + 4), (15 + 5 + 5), (6 + 2 + 2)
PQR = 20, 25, 10

PQR Perimeter = 55

Hope this helps.
r3t40

damaskus [11]3 years ago

6 0

Answer: 55 units

Step-by-step explanation:

Given: ΔKLM~ ΔPQR with a scale factor of 3:5

⇒ $\frac{sides\ of\ \triangle{KLM}}{sides\ of\ \triangle{PQR}}=\frac{3}{5}$

Also ΔKLM~ ΔPQR and the corresponding sides of similar triangles are proportional therefore,

$\\\Rightarrow\frac{KL}{PQ}=\frac{LM}{QR}=\frac{KM}{PR}=\frac{3}{5}$

$\\\Rightarrow PQ=\frac{KL\times5}{3}=\frac{6\times5}{3}=10$

Similarly,

$QR=\frac{LM\times5}{3}=\frac{15\times5}{3}=25\\PR=\frac{KM\times5}{3}=\frac{12\times5}{3}=20$

Now, Perimeter of ΔPQR= $PQ+QR+PR=25+20+10=55\ units$

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Consider testing H^0: u=20 against H^a: u<20 where u is the mean number of latex gloves used per week by all hospital employe

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Part a: P-value

The first step is calculate the degrees of freedom, on this case:

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Since is a one sided test the p value would be:

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Part b: Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can conclude that the true mean is not lower than 20 at 1% of signficance.

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Data given and notation

$\bar X=19.1$ represent the sample mean

$s=11.8$ represent the sample standard deviation

$n=46$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

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We need to conduct a hypothesis in order to check if the mean is lower than 20, the system of hypothesis would be:

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Alternative hypothesis: $\mu < 20$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

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We can replace in formula (1) the info given like this:

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