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MA_775_DIABLO [31]
2 years ago
5

In a large corporate computer network, user log-ons to the system can be modeled as a Poisson RV with a mean of 25 log-ons per h

our. (20pts) (a) What is the probability that there are no logons in an interval of 6 minutes? (b) What is the probability that the distance between two log-ons be more than one hour?
Mathematics
1 answer:
Eva8 [605]2 years ago
5 0

Answer:

F(t<0.1 ) = 0.91791

Step-by-step explanation:

Solution:

- Let X be an exponential RV denoting time t in hours from start of interval to until first log-on that arises from Poisson process with the rate λ = 25 log-ons/hr. Its cumulative density function is given by:

                               F(t) = 1 - e ^ ( - 25*t )                    t > 0

A) In this case we are interested in the probability that it takes t = 6/60 = 0.1 hrs until the first log-on. F ( t < 0.1 hr ), we have:

                            F(t<0.1 ) = 1 - e ^ ( - 25*0.1 )

                            F(t<0.1 ) = 0.91791

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central angle of a circle whose radius is 8 cm and intercepted arc length is 7.2 cm is 0.9 radians

Step-by-step explanation:

We need to find the central angle of a circle whose radius is 8 cm and intercepted arc length is 7.2 cm.

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The formula used is:

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Putting values:

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So, central angle of a circle whose radius is 8 cm and intercepted arc length is 7.2 cm is 0.9 radians

Keywords: central angle of circle

Learn more about central angle of circle at

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PLEASE HELP!
melamori03 [73]

Answer:

The answer is C.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Therefore the value of the secant of this angle would be the reciprocal of the cosine of the angle, that is: sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}

6 0
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