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2 years ago

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

Mathematics

2 answers:

Ymorist [56]2 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.

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]2 years ago

3 0

B becuase they are alternate exterior angles

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Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]

the area of each trapezoid is (v(t1)+v(t2))/2 times width

for the first interval
the average between 0 and 0.4 is 0.2
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average between 0.4 and 1 is 0.7
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average betwen 1.0 and 1.5 is 1.25
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2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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2 years ago

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Answer:A

Step-by-step explanation:

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2 years ago

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$v(r)=4P\eta\ell(R^2-r^2)$

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The average velocity of the blood is given by

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2 years ago

Polygon MNOPQ is dilated by a scale factor of 0.8 with the origin as the center of dilation, resulting in the image M’N’O’P’Q’.

N76 [4]

Answer:
slope of M'N' = 1

Explanation:
First, we will need to get the coordinates of points M' and N':
We are given that the dilation factor (k) is 0.8
Therefore:
For point M':
x coordinate of M' = k * x coordinate of M
x coordinate of M' = 0.8 * 2 = 1.6
y coordinate of M' = k * y coordinate of M
y coordinate of M' = 0.8 * 4 = 3.2
Therefore, coordinates of M' are (1.6 , 3.2)

For point N':
x coordinate of N' = k * x coordinate of N
x coordinate of N' = 0.8 * 3 = 2.4
y coordinate of N' = k * y coordinate of N
y coordinate of N' = 0.8 * 5 = 4
Therefore, coordinates of M' are (2.4 , 4)

Then, we can get the slope of M'N':
slope = (y2-y1) / (x2-x1)
For M'N':
slope = (3.2-4) / (1.6-2.4)
slope = 1

Hope this helps :)

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2 years ago