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3 years ago

Find the circumference and area r = 5 m

Mathematics

1 answer:

stira [4]3 years ago

4 0

Step-by-step explanation:

Let's get our relevant base equations listed out:

d = 2r

area = pi*r²

c = πd

Thus, d=2*5m = 10m, c=π*10m=10π m, and area = 25m²*π

Feel free to ask further questions!

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0.75 is the square root of what value?

lina2011 [118]

0.75^2 = 0.5625.....sq rt 0.5625 = 0.75

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2 years ago

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The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of si

OverLord2011 [107]

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

$P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!$

a) The probability that exactly 3 arrivals occur during a particular hour is:

$P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\$

b) The probability that <em>at least</em> 3 people arrive during a particular hour is:

$P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938$

c) In this case, t=0.25, so we recalculate the parameter as:

$\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5$

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

$E(x)=\lambda=1.5$

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3 years ago

Math home work I have two questions

andreyandreev [35.5K]

Answer:

2d = 58 (If you are just finding d then d = 29)

Step-by-step explanation:

69 + 53 = 122

180 - 122 = 58

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3 years ago

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Question 13 letter A and B

dybincka [34]

A) 2.50/5= 0.5,
B) 0.5*25=$12.50

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2 years ago

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A local country club has a membership of 600 and operates facilities that include an 18-hole championship golf course and 12 ten

shusha [124]

Answer:

0.45 = 45% probability that the member uses the golf course but not the tennis courts

Step-by-step explanation:

I am going to solve this question using the events as Venn sets.

I am going to say that:

Event A: Uses the golf courses.

Event B: Uses the tennis courts.

5% use neither of these facilities.

This means that $P(A \cup B) = 1 - 0.05 = 0.95$

75% use the golf course, 50% use the tennis courts

This means, respectively, by:

$P(A) = 0.75, P(B) = 0.5$

Probability that a member uses both:

This is $P(A \cap B)$ . We have that:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

$P(A \cap B) = 0.75 + 0.5 - 0.95 = 0.3$

What is the probability that the member uses the golf course but not the tennis courts?

This is $P(A - B)$ , which is given by:

$P(A - B) = P(A) - P(A \cap B)$

$P(A - B) = 0.75 - 0.3 = 0.45$

0.45 = 45% probability that the member uses the golf course but not the tennis courts

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3 years ago