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2 years ago

13. Check all the true statements.

Mathematics

1 answer:

romanna [79]2 years ago

3 0

Based on the triangle midsegment theorem, the true statements are:

DE║AC

2DE = AC

<h3>What is the Triangle Midsegment Theorem?</h3>

If a line segment joins two sides of triangle and divides them at their midpoint, the divided segments will be congruent, and thus, the line segment joining the two sides is a midsegment of the triangle. According to the triangle midsegment theorem, the midsegment will be parallel to the third side that is not divided and also would be 1/2 its length. That is, the third side is twice the midsegment of that triangle.

In the image given, considering triangle ABC, we can make the following conclusions based on the triangle midsegment theorem:

D and E are midpoints of sides AB and BC respectively because they are divided by segment DE equally.

The midsegment is DE.

DE would be parallel to the third side, line segment AC.

Length of AC would be twice the length of DE

In conclusion, the true statements based on the triangle midsegment theorem are:

DE║AC

2DE = AC

Learn more about the triangle midsegment theorem on:

brainly.com/question/12234706

#SPJ1

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Let X equal the number of typos on a printed page with a mean of 4 typos per page.

timama [110]

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson distribution.

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

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

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

In this problem, we have that $\mu = 4$

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is $P(X \geq 1)$ . Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

$P(X < 1) + P(X \geq 1) = 1$

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

In which

$P(X < 1) = P(X = 0)$ .

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817$

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is $P = P(X = 0) + P(X = 1)$ . So:

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916$

There is a 9.16% probability that a randomly selected page has at most one typo on it.

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