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aleksandrvk [35]
3 years ago
9

Please answer missing boxes in order please ASAP will give 10 points

Mathematics
1 answer:
Naily [24]3 years ago
7 0

Answer:

326 x 52 is the factors of the problem.

Step-by-step explanation:

To solve this problem, we need to figure out what number multiplied by what numbers is 652 and 16300. The first product (652) is made by 2 and a number. We can divide 652 by 2 to figure out this number. 652 / 2 = 326. The missing digit of the first factor is 2 making the factor 326. However, the second factor has a missing digit, too. 326 multiplied by a number has to equal 16,300. We can also solve this with division. 16,300 / 326 = 50. The missing digit has a value of 50 making the factor 52. 326 x 52 is the factors of the problem.

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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
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Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

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What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

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We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

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