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3 years ago

What equation best represents the circumference of a circle with a radius of 6 feet? A y = 2π B y = 6π C y = 12π D y = 16π

Mathematics

1 answer:

sleet_krkn [62]3 years ago

6 0

Answer:

12π

Step-by-step explanation:

The circumference is the distance you need to walk to do 1 full rotation of a shape. The formula to calculate the circumference of a circle will be

circumference =2πr

where π is a constant that equal to 3.14 or 22/7 and r is the radius of the circle (radius = 6 feet).

The circumference of this circle will be:

circumference =2πr

circumference =2π(6)

circumference =12π

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Big Reward! (Answer the right answer not just comment).

Zina [86]

A: One solution- a=11 b=5
B: Has no solution- a=10 b=5
C: Infinite solution- a=10 b=-8

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3 years ago

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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

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

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

<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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3 years ago

Someone please help me I only have until January 18th to turn this in. Question: Mr Lopez picks up 2 pallets of dog food to deli

miv72 [106K]

Answer:

Mr. Lopez delivers 1.5 tons of dog food to the animal shelter.

Step-by-step explanation:

Mr Lopez picks up 2 pallets of dog food to deliver to the animal shelter. Each pallet contains 40 bags of dog food, and the dog food weighs 37 1/2 pounds pet bag.

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Hence, Mr. Lopez delivers 1.5 tons of dog food to the animal shelter.

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3 years ago

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Brut [27]

Answer:

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Step-by-step explanation:

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