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romanna [79]
3 years ago
5

Given f(x)=x^3 and g(x)= 1-5x^2

Mathematics
1 answer:
Allisa [31]3 years ago
3 0
Im doing this for points lol
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Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p
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Answer:

  • <em>m</em> = \frac{1}{20}
  • <em>μ</em> = 20
  • <em>σ </em>= 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable <em>X</em> follows an exponential distribution with parameter <em>m</em>.

The decay parameter is, <em>m</em>.

The probability distribution function of an Exponential distribution is:

f(x)=me^{-mx}\ ;\ m>0, x>0

<u>Given</u>: The decay parameter is, \frac{1}{20}

<em>X</em> is defined as the distance people are willing to commute in miles.

  • The decay parameter is <em>m</em> = \frac{1}{20}.
  • The mean of the distribution is: \mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20.
  • The standard deviation is: \sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20

Compute the probability that a person is willing to commute more than 25 miles as follows:

P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865

Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.

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