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2 years ago

2 p

Mathematics

2 answers:

aivan3 [116]2 years ago

5 0

Answer:

5:14

Step-by-step explanation:

The ratio of adults to children using the form 1:n would look like this:

number of adults : number of children

So:

5 adults : 14 children = 5:14

andrey2020 [161]2 years ago

4 0

Answer:

1 : 2.8

Step-by-step explanation:

in such a case what the question is asking you to is to compare the numbers and simplify them with the left hand number.

ratio of adult to children

write it as 5:14

put the 1 :n underneath

5 : 14 what we want to find is the n. among the 2 sides one is complete

1 : n only whole numbers. to get n look at the complete side . if

5:1 14=14 x 1 /5(cross multiplication). or you can

divide both sides by the left hand number(5)

5 and 14 = 1 : 2.8

5 5

the n is what we were trying to find and compare it to 1

this means for every 1 adult there are 2.8 children

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makkiz [27]

Answer:

3+4p+8t

Step-by-step explanation:

We have been given an expression:

3+2(2p+4t)

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2 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

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Answer:

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2 years ago