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3 years ago

Explain how to compare 0.7 and 5/8

1 answer:

6 0

In order to compare you have to find which one is greater. In this case 0.7 is greater than 5/8 because 5/8 as a decimal is 0.62. 0.7 is greater than 0.62 by 0.08

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Find the 92nd term of the arithmetic sequence 11, 19, 27,

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Answer:

739

Step-by-step explanation:

a sub 92 = 11 + (92 - 1) * 8

a sub 92 = 11 + 728

a sub 92 = 739

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3 years ago

Write the word sentence as an equation then solve the difference between a number c and 17 is 3

Digiron [165]

Answer:

c-17=3

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Step-by-step explanation:

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3 years ago

How many sides do 1980 have

mel-nik [20]

Answer:

13 sides

Step-by-step explanation:

1980 = (n − 2)×180 Formula for number of sides

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3 years ago

3(j+7)? a n s w e r please

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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

hichkok12 [17]

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.

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

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

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