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NikAS [45]
3 years ago
6

Please help #8 thank u

Mathematics
2 answers:
sweet [91]3 years ago
7 0

B because it says 45 of 60 is equal to 75%, so in fraction form it would be 75/100 = 45/60

exis [7]3 years ago
4 0

The answer is B because it is saying that 45/60=75%, so 75/100 = 45/60

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Solve the equation using the quadratic formula.<br><br> x^2+4x=-3
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When the temperature of an object increases, what happens to the kinetic energy of the particles in the object?
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What value will make the equation true
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pick the value that when you solve it, you get the correct answer

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The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
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Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

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• What is probability that the red bulb will need to be replaced at the first inspection?

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\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

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\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

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\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

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What is the value of the expression:
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