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3 years ago

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A rectangular yard has a total perimeter of 220 feet. The widths of the yard are each 30 feet shorter than the lengths of the ya

rd. What are the dimensions of the yard?

1 answer:

ziro4ka [17]3 years ago

5 0

Answer:

L = 70 W = 40

Step-by-step explanation:

P = 220 ft

W = L - 30

I started guessing numbers for L that would work. When I did this, I made sure that all of the sides added up to 220 ft.

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What is the quotient of 13divide 8

Ahat [919]

Answer:

1

Step-by-step explanation:

mr clean

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2 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

The last time Reid checked, his dog weighed 30 kilograms. Reid just measured his dog again and found out that it weighs 10% less

Vedmedyk [2.9K]

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3 years ago

Find the value of a answers: a:15 b:14 c:19 d:16

Paul [167]

Answer:

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A grocer wants to make a 10-pound mixture of cash tans peanuts that he can sell for $3.29 per pound. If cashews cost $5.60 per p

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Label each nut with a variable, c = cashews, p = peanuts.....
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Now you 2 have a system of 2 equations and 2 unknowns
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3 years ago