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podryga [215]
3 years ago
14

Please help i suck at these

Mathematics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

1st Option;

j = 4.5

k = 2

Step-by-step explanation:

Let's solve for "j" first:

=> We know that by the definition of midpoint segment theorem we can say;

3j = 5j - 9

0 = 5j - 3j - 9

0 = 2j - 9

0 + 9 = 2j

9 = 2j

9/2 = j

4.5 = j

=> Now that we have j-value we use the same method to solve for k-value;

6k = k + 10

6k - k = 10

5k = 10

k = 10/5

k = 2

Therefore;

j = 4.5

k = 2

<u>So the first option would be correct!</u>

Hope this helps!

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Find the y-intercept of the following equation. Simplify your answer.<br> 10x + y = 8
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Answer: 8

Explanation:

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Thus, the y-intercept is 8
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The probability of an event happening is 23%. What is the complement of the event?
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In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given
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Answer:

a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

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.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so \mu = 1.3

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is P(X > 5)

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

So

P(X \leq 5) + P(X > 5) = 1

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980

P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324

P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949

Finally

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051

There is a 0.51% probability of more than five quakes.

5 0
3 years ago
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