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3 years ago

15

A line segment has endpoints P (3, 6) and Q (12, 18) and is dilated so that its new endpoints are P’ (2, 4) and Q’ (8, 12). What

is the scale factor? If the length of PQ is 15, what is the length of P’Q’?
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2 answers:

just olya [345]3 years ago

7 0

Answer:

Scale Factor ≈ 0.67

P'Q' Length = 10

Step-by-step explanation:

To find the scale factor, a relationship must be found between the original points, and the dilated points. To do this, simply divide the new points by the old points.

P to P' : $\frac{2}{3}=0.6667$ AND $\frac{4}{6}=0.6667$

Q to Q' : $\frac{8}{12}=0.6667$ AND $\frac{12}{18}=0.6667$

Therefore, the scale factor is 0.6667 ≈ 0.67

The length of P'Q' can be calculated with the distance formula:

$d^{2}=(y_{1}-y_{2} )^{2} +(x_{1}-x_{2} )^{2} \\d^{2} =(4-12)^{2} + (2-8)^{2} \\d^{2} =(-8)^{2} +(-6)^{2} \\d^{2} =64+36\\d=\sqrt{100} =10$

Alternatively, you could multiply the length of PQ with the scale factor to find P'Q':

15*0.6667=10

Fed [463]3 years ago

4 0

It should be 14, because 8 - 2 = 6 and 12 - 4 = 8

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a) We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The Group 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The Group 2, of size n2=11 has a mean of -0.273 and a standard deviation of 5.901.

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$M_d=M_1-M_2=5-(-0.273)=5.273$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288$

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The critical t-value for a 95% confidence interval and 19 degrees of freedom is t=1.96.

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$MOE=t\cdot s_{M_d}=1.96 \cdot 3.288=6.444$

Then, the lower and upper bounds of the confidence interval are:

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The sample 1, of size n1=11 has a mean of -0.273 and a standard deviation of 5.901.

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The degrees of freedom for this test are:

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$P-value=P(t>1.604)=0.063$

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The null hypothesis failed to be rejected.

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