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professor190 [17]
3 years ago
10

Which of the following facts, if true, would allow you to prove that lines l and m are parallel ?

Mathematics
2 answers:
Ymorist [56]3 years ago
6 0

We are given lines l and m and a transversal line to those two given lines.

We need to explain which of the following facts, if true, would allow you to prove that lines l and m are parallel.

Let us describe each given options.

A)m<2+m<4=180 degrees.

m<2 and m<4 makes a line pair. So, it doesn't prove if lines l and m are parallel.

<h3>B)m<1=m<8</h3><h3>m<1 and m<8 are Alternate Exterior Angles and when Alternate Exterior Angles equal lines are parallel.</h3>

C)m<6+m<3=180 degrees

This statement can't be true. Alternate interior angles can't be sum upto 180 degrees.

D)m<7=m<6

Those are vertical angles, that doesn't make the l and m parallel.

<h3>Therefore, only B option is correct option.</h3><h3>B)m<1=m<8.</h3>
ale4655 [162]3 years ago
3 0
B becuase they are alternate exterior angles
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Hello!

To solve this, we must solve the equation..

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It can't be D because 5^2/3 = 2.92...

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A system consists of two components C1 and C2, each of which must be operative in order for the overall system to function. Let
hram777 [196]

Answer:

The reliability of the first system to work is 0.72 whereas the reliability of the second system to work is 0.98.As the reliability of the second system is more than the first one so the second system is more reliable.

Step-by-step explanation:

For first system as given in the attached diagram gives,

P(W_1W_2)=P(W_1) \times P(W_2)

As the systems are independent.

The given data indicates that

  • P(W_1) is given as 0.9
  • P(W_2) is given as 0.8

Now the probability of the system is given as

P(W_1W_2)=P(W_1) \times P(W_2)\\P(W_1W_2)=0.9 \times 0.8\\P(W_1W_2)=0.72

So the reliability of the first system to work is 0.72.

For the second system is given as

P(W_1W_2)=1-P(W_1'W_2')

Where

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P(W_1'W_2')=P(W_1') \times P(W_2')\\P(W_1'W_2')=(1-P(W_1)) \times (1-P(W_2))\\P(W_1'W_2')=(1-0.9) \times (1-0.8)\\P(W_1'W_2')=(0.1) \times (0.2)\\P(W_1'W_2')=0.02

So now the probability of the second system is given as

P(W_1W_2)=1-P(W_1'W_2')\\P(W_1W_2)=1-0.02\\P(W_1W_2)=0.98

So the reliability of the second system to work is 0.98.

As the reliability of the second system is more than the first one so the second system is more reliable.

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2 years ago
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