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Tanzania [10]
1 year ago
8

The point halfway between two endpoints of a line segment

Mathematics
1 answer:
r-ruslan [8.4K]1 year ago
8 0

The midpoint of a line segment is the point that is halfway between the two ends of the line segment. A midway separates a line segment into two equal parts.

This is further explained below.

<h3>What is a line segment?</h3>

Generally, In the field of geometry, a line segment is defined as the portion of a line that is enclosed by two unique points on the line.

Alternatively, we may say that a line segment is the portion of a line that is between two points. A line does not have any endpoints and may stretch endlessly in any direction, however, a line segment has two endpoints that are fixed or definite in some way.

In conclusion, The term "midpoint" refers to the point along a line segment that is situated exactly midway between the beginning and ending points. A midway separates a line segment into two equal parts.

Read more about the line segment

brainly.com/question/25727583

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THE STATEMENT IS TRUE.


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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Step-by-step explanation:

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Answer:

a) 9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b) 4.284 years are expected to have 4 ​hurricanes.

c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

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 sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

6.9 per year.

This means that \mu = 6.9

a. Find the probability​ that, in a​ year, there will be 4 hurricanes.

This is P(X = 4).

So

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

P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}

P(X = 4) = 0.0952

9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b. In a 45​-year ​period, how many years are expected to have 4 ​hurricanes?

For each year, the probability is 0.0952.

Multiplying by 45

45*0.0952 = 4.284.

4.284 years are expected to have 4 ​hurricanes.

c. How does the result from part​ (b) compare to a recent period of 45 years in which 4 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?

The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

5 0
3 years ago
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