Side AC corresponds to side<br> Angle BAC corresponds to angle

Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is a

Svetradugi [14.3K]

Answer:

0.9009

Step-by-step explanation:

Let X be the event that Current will flow

Let Y be the event that the first relay which is 1 is closed

Thus, we can say that every element of Y is in X, but X possesses more elements. Thus, Y ⊂ X.

Thus, we can say that;

P(X ∩ Y) = P(Y)

Thus, given that current flowed, the probability that relay 1 functioned will be expressed as;

P(Y | X) = (P(Y ∩ X))/P(X)

From earlier, we saw that P(X ∩ Y) = P(Y). Thus;

P(Y | X) = P(Y)/P(X)

From the question, P(Y) = 0.9

Since there are 3 relays, then we have;

P(X) = 1 - 0.1³ = 0.999

Thus;

P(Y | X) = 0.9/0.999

P(Y | X) = 0.9009

4 0

3 years ago

Assume that power surges occur as a Poisson process with rate 3 per hour. These events cause damage to a computer, thus a specia

lidiya [134]

Answer:

a) 0.30327

b) 0.07582

Step-by-step explanation:

This is a Poisson distribution problem

Poisson distribution probability function is given as

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 3 power surges per hour

x = variable whose probability is required

a) If a single disturbance in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of a single disturbance (power surge) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 1 disturbance per 10 minutes

P(X = 1) = (e^-0.5)(0.5¹)/1! = (e⁻⁰•⁵)(0.5)/1!

= 0.30327

b) If two disturbances in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of two disturbances (power surges) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 2 disturbance per 10 minutes

P(X = 2) = (e^-0.5)(0.5²)/2! = (e⁻⁰•⁵)(0.25)/2!

= 0.07582

Hope this Helps!!!

3 0

3 years ago

Order from greatest to least:show your work❗️<br><br> 12,-4,8,-3,0

Aneli [31]

The answer would be 12, 8, 0, -3, -4.

The answer is that because 12 is a positive number so that's higher than negative, 8 goes second because its positive and higher than the others, 0 is third because negatives numbers are LOWER than 0, not higher, -3 is next because higher negative numbers are always lower, not greater than smaller numbers because -8 is LOWER than -4 since its on the left of a number line.

6 0

3 years ago

Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94. The average of her last t

Bingel [31]

Answer: The average of the first two test scores is 97.

Step-by-step explanation: Given that Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94 and the average of her last three test scores is 92.

We are to find the average score of her first two tests.

Let a1, a2, a3, a4 and a5 be teh scores of Kristle in first , second, third, fourth and fifth tests respectively.

Then, according to the given information, we have

$\dfrac{a1+a2+a3+a4+a5}{5}=94\\\\\Rightarrow a1+a2+a3+a4+a5=94\times5\\\\\Rightarrow a1+a2+a3+a4+a5=470~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$\dfrac{a3+a4+a5}{3}=92\\\\\Rightarrow a3+a4+a5=92\times3\\\\\Rightarrow a3+a4+a5=276~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Subtracting equation (ii) from equation (i), we get

$(a1+a2+a3+a4+a5)-(a3+a4+a5)=470-276\\\\\Rightarrow a1+a2=194\\\\\Rightarrow \dfrac{a1+a2}{2}=\dfrac{194}{2}\\\\\Rightarrow \dfrac{a1+a2}{2}=97.$

Thus, the average of the first two test scores is 97.

6 0

3 years ago

Matt has 3 dark t-shirts and 6 white t-shirts. First, he will close his eyes to select a

valina [46]

Answer:

around 73-89%

Step-by-step explanation:

Hope this helps!

8 0

3 years ago

Side AC corresponds to side Angle BAC corresponds to angle