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timurjin [86]
3 years ago
7

A Given p(x) = X+41, what is (-10)? ​

Mathematics
2 answers:
Maslowich3 years ago
8 0

Answer:

p(-10) = 31

Step-by-step explanation:

Plug in -10 as x into the equation, and solve

p(x) = x + 41

p(-10) = -10 + 41

p(-10) = 31

Murrr4er [49]3 years ago
5 0
<h3><u>Answer</u> :</h3>

Given : p(x) = x + 41

We have to find value of p(-10).

➙ p(x) = x + 41

By substituting x = (-10), we get

➙ p(-10) = (-10) + 41

➙ p(-10) = 41 - 10

➙ <u>p(-10) = 31</u>

<h3>Hope It Helps!</h3>
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b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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\mu = 20.8, \sigma = 4.5

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

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Z = 0.04

Z = 0.04 has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

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Z = 0.71

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Z = \frac{X - \mu}{\sigma}

Z = \frac{18 - 20.8}{4.5}

Z = -0.62

Z = -0.62 has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

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