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Harlamova29_29 [7]
3 years ago
6

Two pumps connected in parallel fail independently of one another on any given day. The probability that only the older pump wil

l fail is .10, and the probability that only the newer pump will fail is .05. What is the probability that the pump- ing system will fail on any given day (which happens if both pumps fail)?
Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

Answer : Probability that the pumping system will fail on any given day = 0.005

Explanation:

Probability that only older pump fail = P(A) = 0.10

Probability that only newer pump fail = P(B) = 0.05

Since they are independent events ,

then the conditions will be

P(A∩B) = P(A). P(B)

             = 0.10 ×0.05

             =  0.005

Hence, the probability will fail on any given day = 0.005

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Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
Can someone help me with these 3 questions please
Alex_Xolod [135]
#1 is 2 #2 is 6 i dont know 3
3 0
3 years ago
Pls help ( if you can’t read it it says, a toddler weighs 30.0 pounds. If 1.00 Newton equals the weight of 0.228 pounds, what is
gizmo_the_mogwai [7]

Answer:

131.6 Newtons

Step-by-step explanation:

30 ÷ 0.228 =131.6 Newtons

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3 years ago
Read 2 more answers
8x+16=14x<br><br><br>checking my awnser
Firdavs [7]

8x + 16 = 14x

To solve this equation, move everything containing the variable you want to solve for on one side of the equation and everything else to the other side of the equation.

Let's start by subtracting 8x from both sides to have the variable x all on the right side of the equation. Once you subtract 8x from both sides, your equation will now look like this:

16 = 6x

To solve for the variable x, you want to divide both sides of the equation by 6 to isolate x and therefore find your answer. Divide both sides by 6 and your equation should look like this:

16/6 = x

Simplifying 16/6 into a mixed number ⇒ 2 2/3

Your answer is x = 16/6 or 2 2/3.

8 0
3 years ago
You are planning a survey of starting salaries for recent computer science majors. In a recent survey by the National Associatio
nydimaria [60]

Answer:

A sample size of 554 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

Standard deviation is known to be $12,000

This means that \sigma = 12000

What sample size do you need to have a margin of error equal to $1000, with 95% confidence?

This is n for which M = 1000. So

M = z\frac{\sigma}{\sqrt{n}}

1000 = 1.96\frac{12000}{\sqrt{n}}

1000\sqrt{n} = 1.96*12

Dividing both sides by 1000:

\sqrt{n} = 1.96*12

(\sqrt{n})^2 = (1.96*12)^2

n = 553.2

Rounding up:

A sample size of 554 is needed.

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