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Bogdan [553]
3 years ago
15

Given v^2 = u^2 − 2as. Solve for u, when v = 10, a = 5 and s = 2.

Mathematics
1 answer:
iogann1982 [59]3 years ago
3 0

Answer:

u ≈ 11

Step-by-step explanation:

10² = u² - 2(5)(2)

100 = u² - 20

120 = u²

u = \sqrt{120}

u = \sqrt{4}\sqrt{30} = 2\sqrt{30} which is 10.95

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(4.2.24) One design for a system requires the installation of two identical components. The system will work if at least one of
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Answer:

<em>The second design has a greater probability of functioning than the first one</em>

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<u>Probabilities</u>

The probability that a component works is 0.9. We need to evaluate two options to find the arrangement that provides the greatest probability for the system to function.

The first option is to install two identical components and consider if at least one of them works, the system will work. Let's call W to the event in which one component works, and N if it doesn't.

Choosing two components, we can have the following outcomes

\Omega=\{WW,WN,NW,NN\}

The first 3 of the outcomes will make the system work. Only the last one is not acceptable. Knowing the probability of the component to work is p=0.9, the negated probability is 1-0.9=0.1. The probability of NN is

P(NN)=0.1^2=0.01

Thus, the probability that the system works is 1-0.01=0.99 or 99%

Now for the second option, we have 4 components. There are 24 possible outcomes from which we can accept only those who have at least two W's, that is {WWNN,WWWN,WWWW} or any equivalent combination regardless of the order of appearance.

To calculate the probability of such arrangements, we'll use the binomial distribution formula

\displaystyle P(n,k,p)=\binom{n}{k}p^kq^{n-k}

Where q=1-p, n=4 and k is the number of W in the combination. We have

\displaystyle P(4,2,0.9)=\binom{4}{2}0.9^2q^{2}=0.0486

\displaystyle P(4,3,0.9)=\binom{4}{3}0.9^3q^{1}=0.2916

\displaystyle P(4,4,0.9)=\binom{4}{4}0.9^4q^{0}=0.6561

The total probability is

P(WWNN,WWWN,WWWW)=0.9963=99.63%

The second design has a greater probability of functioning than the first one

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3 years ago
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