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2 years ago

Select the most precise and correct statement about perpendicular lines from among those that are given.

Mathematics

1 answer:

san4es73 [151]2 years ago

8 0

The most precise and correct statement about perpendicular lines from among those that are given is; A: AB is perpendicular to CD if the two line intersect such that ∠AOD is a right angle triangle.

<h3>How to Interpret Perpendicular Lines?</h3>

In geometry, perpendicular lines are defined as two lines that meet or intersect each other at right angles (90°).

Now, from the given image, we are told that lines AB and CD Intersect at point O. Looking at the options;

A) This option is true because if the intersect at right angles, then according to the definition above, they are perpendicular.

B) This option is incorrect because if they never meet, then they can never intersect.

C) This option is incorrect because the equality has to be at 90 degrees.

Read more about Perpendicular Lines at; brainly.com/question/1202004

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a triangle has side lengths of 13, 15, and 17. determine whether the triangle is right, acute, or obtuse.

Len [333]

Square of longest side = 17^2 = 289

15^2 = 225
13^2 = 169

total 394 289 is less than this so its acute

3 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

$P(X \geq 2) = 0.93803$

5 0

3 years ago

8 thank you so much!!!

777dan777 [17]

Answer:

C) (-4, 2)

Step-by-step explanation:

It is easier to begin by putting the y equation back into the x equation so it would be rewritten as:

x= 2 (-2x - 6) - 8

Since you replace the y for the given y value

Distribute and you get

x = -4x -12 -8

Add 4x to both sides to get rid of it on the right and balance the equation

5x = -12-8 which is -20

5x = -20

divide by 5 gives you the value -4 for the x

Now simply put that value in for the y equation

y= (-2 (-4)) -6

y = 8 - 6

y = 2

8 0

3 years ago

Read 2 more answers

You invested $2,300 in a stock. Your account now has a value of $2,643. Your percentage gain on the investment (rounded to the n

solong [7]

I hope this helps you

7 0

3 years ago

Bouncing Baby Company sells their cribs for the following amounts to stores:

goblinko [34]

Answer:

C and E

Step-by-step explanation:

Ik u prolly dont care about the explanation but ill just say it

the ratio is 1:132 or 1 crib is $132. So 40 x 132 = 5280 and 80 x 132 = 10560

For the test im doing the answer is C and D yw ;D

4 0

3 years ago