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2 years ago

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There are three workstations available having steady-state probabilities of 0.99, 0.95, 0.85 of being available on demand. What

is the probability that at least two of the three will be available at any given time?

1 answer:

faust18 [17]2 years ago

3 0

Answer:

99.065% probability that at least two of the three will be available at any given time.

Step-by-step explanation:

We have these following probabilities:

99% probability of the first workstation being available

95% probability of the second workstation being available

85% probability of the third workstation being avaiable

Two being available:

We can have three outcomes

First and second available, third not. So

0.99*0.95*0.15 = 0.141075

First and third available, second not. So

0.99*0.05*0.85 = 0.042075

Second and third available, first not. So

0.01*0.95*0.85 = 0.008075

Adding them all

P(2) = 0.141075 + 0.042075 + 0.008075 = 0.191225

Three being available:

P(3) = 0.99*0.95*0.85 = 0.799425

What is the probability that at least two of the three will be available at any given time?

P = P(2) + P(3) = 0.191225 + 0.799425 = 0.99065

99.065% probability that at least two of the three will be available at any given time.

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.

7 0

2 years ago

Please help thank you and show all work.

Daniel [21]

Question 1:

Look at the first attached image.

We want to solve for the height of the smaller building. Let's start by making a right triangle with angle of 15 degrees and adjacent side of 51 m.

Let's solve for the other leg. Multiply the tangent with 51 m to get the length of the other leg.

$51*tan(15)$
$\approx 14$

Now, subtract that from 207 to get the height of the smaller building.

$207-14=193$

That's your answer.

Question 2:

The angle of elevation is congruent to the angle of depression because they are alternate interior angles.

Thus, you can choose the 1st and 4th choices as your answer.

Question 3:

Look at the second attached image.

Angle E is congruent to angle D for the same reasons mentioned in the last question.
We can create an equation.

$3x+1=2(x+8)$

Distributive property

$3x+1=2x+16$

Subtract both sides by 1 and 2x

$x=15$

Since E and D are congruent, just plug this value into one of the equations for E or D. You get 46.

The third choice is your answer. Hope this helps! :)

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Step-by-step explanation:

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Explanation:

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2 years ago

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