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3 years ago

Please help me with this question below

Mathematics

1 answer:

BabaBlast [244]3 years ago

4 0

Answer:

381 cubic inches

Step-by-step explanation:

To find the volume of the cube, use the volume formula for a rectangular prism V = l*w*h and for a triangular prism V = 1/2 l*w*h.

Rectangular prism

Substitute l = 16, w = 7, and h = 3.

V = 16*7*3 = 336 cubic inches

Triangular prism

Substitute l = 5, w = 3 and h = 6.

V = 1/2*5*3*6 = 45 cubic inches

Together the volume will be 336 + 45 = 381.

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Rewrite the following equation in log form: 2^2 = 4

Finger [1]

We have to write

$2^2 =4$

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So we will get

$If \ 2^2 =4, \ then \ 2 = log_{2} 4$

$log_{2}4=2$

And that's the required log form .

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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.

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

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 =

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

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3 years ago