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4 years ago

How to do question 2

Mathematics

1 answer:

jek_recluse [69]4 years ago

5 0

Answer:

i wish i could help you but i haven't learned that.Sorry.hope u find whatchu looking for.

Step-by-step explanation:

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PLS HELP LOOK AT PIC

zepelin [54]

31 trust In me I learned this last week

7 0

3 years ago

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

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

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

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

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

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

We have

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

hence,

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

• 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

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

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago

Need help quick please

Travka [436]

Answer:

line 2

Step-by-step explanation:

on line 2 the two minuses at the end turn into a + (where the -2 is there should be a +2) because - x - = +

4 0

3 years ago

The sum of 2 numbers is 90 and one of them is 2 times the other what are the 2 numbers?

bekas [8.4K]

……………………………………………………

8 0

3 years ago

Read 2 more answers

(5 points)

Schach [20]

Answer:

The original Slope is y=-1/3x-2

The perpendicular slope would be y=3x+b

Step-by-step explanation:

Original Slope:

You already have the y-interecpet, (0,-2), and you know two points (0,-2) and (-6,0), so you can find the slope with (y1-y2)/(x1-x2) so in this case, it would be (0-(-2))/(-6-0)= -1/3 and you can find y=1/3x-2.

Perpendicular Slope:

The slope of any perpendicular line would be the reciprocal of the original. So the reciprocal of -1/3 is 3. The b can be any real number because no matter the y-intercept the slope would still be perpendicular

3 0

3 years ago

How to do question 2 ​

How to do question 2