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Tems11 [23]
4 years ago
13

PLEASE HELP ITS AN EMERGENCY!!

Mathematics
2 answers:
Veseljchak [2.6K]4 years ago
8 0

Answer:

m∠B=m∠R

Step-by-step explanation:

If you make the triangles congruent, you'll see many things like:

∠A=∠P

∠B=∠R

∠C=∠S

Line AB=Line RP

Line BC=Line RS

Line CA=Line SP

Out of the options, only m∠B=m∠R is true.

nignag [31]4 years ago
4 0

Answer:

m∠b = m∠r

Step-by-step explanation:

The triangle was reflected across line m.

A reflection is a rigid transformation. A rigid transformation does not change the measurement or size of a shape, so the pre-image and the image are <em>congruent</em>.

The angles on the pre-image corresponds to the angles on the image. This also is for line segments.

This means (for angle measurements):

  • Angle A is congruent to Angle P
  • Angle C is congruent to Angle S
  • Angle B is congruent to Angle R

Option D should be the correct answer.

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The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a
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Answer and explanation:

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The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

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P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)

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P(C|E)=\frac{P(E|C)P(C)}{P(E)}

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P(C|E)=\frac{0.032}{0.098}

P(C|E)=0.3265

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Probability of no printer problem given poor quality is

P(A|E)=\frac{P(E|A)P(A)}{P(E)}

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P(A|E)=\frac{0}{0.098}

P(A|E)=0

Probability of misaligned paper given poor quality is

P(B|E)=\frac{P(E|B)P(B)}{P(E)}

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P(B|E)=\frac{0.006}{0.098}

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P(D|E)=\frac{P(E|D)P(D)}{P(E)}

P(D|E)=\frac{0.6\times 0.1}{0.098}

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P(D|E)=0.6122

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