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siniylev [52]
3 years ago
12

Circle A has center of (4, 5), and a radius of 3 and circle B has a center of (1, 7), and a radius of 9. What steps will help sh

ow that circle A is similar to circle B? (6 points) Translate circle A using the rule (x + 3, y - 2). Сь Dilate circle A by a scale factor of 3. Oc Rotate/circle A 90° about the center, D Reflect circle A over the x-axis,​
Mathematics
2 answers:
Katarina [22]3 years ago
8 0

Answer:

Dilate Circle A by a scale factor of 3

Step-by-step explanation:

Circle A has a radius of 3, Circle B has a radius of 9. By dilating Circle a by a scale factor of 3, it will make it the same size as Circle B, showing that the two circles are similar.

Rotating and reflecting does nothing because both circles are in the same quadrant and rotating a circle does nothing to the placement.

Translating the circle (x+3, y-2) does not put it in the same place as circle B.

Tanzania [10]3 years ago
7 0

Answer:

The other person is correct. B: Dilate circle A by a scale factor of 3.

Step-by-step explanation:

i took the test

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erma4kov [3.2K]

Answer:

<em>m<ZXY - m<WXZ   = m<WXY   </em>

Step-by-step explanation:

given:

point Z is in the interior of <WXY creating XZ.

<em>if m<WXZ = 20 </em>

<em>  m<WXY = 100 </em>

find:

what would equation would you set up to find the missing angle measure?

solution:

<em>Since Z is interior of <WXY </em>

<em> m<ZXY - m<WXZ   = m<WXY </em>

<em><WXY =  100  - 20  = 80° </em>

therefore, the equation would be  <em>m<ZXY - m<WXZ   = m<WXY  </em>

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2 years ago
According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women
blagie [28]

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : P(X=r) =^nC_r p^r q^{n-r}

At x = 2

P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}

P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}

P(X=2) =8.8009 \times 10^{-7}

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So,  P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}

 P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})

P(X=r) =9.9439 \times 10^{-6}

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=0.2360

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Answer:

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NeX [460]
Move all terms that don't contain y to the right side and solve.
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3 years ago
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