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3 years ago

Values for 24/25 ÷ 4/5.

Mathematics

2 answers:

lions [1.4K]3 years ago

5 0

The answer is 1.2 hope this helped ;)

astraxan [27]3 years ago

5 0

Answer:

1.2

Step-by-step explanation:

24/25 ÷ 4/5

Change the numbers from division to decimals. PS: this is the same thing just written much easier.

0.96/0.8

1.2

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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

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 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 X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

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

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
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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4 years ago

You mix the letters A, C, Q, U, A, I, N, T, A, N, C, E throughly. Without looking you select one letter. Find the probability of

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3 years ago

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Answer:

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Step-by-step explanation:

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Isolate x by dividing both sides by 111

x=55.5

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Answer:

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Step-by-step explanation:

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1 year ago

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Answer:

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Step-by-step explanation:

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