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kakasveta [241]
3 years ago
10

1. Jason travelled from Florida to Mississauga He travelled 3/10 of the journey by lorry, 2/5 of the journey by taxi and the res

t by train. What fraction of the journey did he travel by train 2. Which of the fractions 13/20, 3/5, 3/4 and 7/10 is the greatest?
Mathematics
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

1. He traveled 3/ 10 by  train

2. 3/4 is the greatest

Step-by-step explanation:

3/ 10 = Iorry

2/5 = taxi

? = train

2=4

5=10

4+ 3 = 7

10 + 10=  10

13/20

12/20

15/20

17/20

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Any expression is formed of terms separated by addition or subtraction.
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BartSMP [9]

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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and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

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

The probability that the bulb fails the first year is

\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

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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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