1 1/2 times a number equals 6

Mathematics

2 answers:

olga nikolaevna [1]3 years ago

8 0

$1\frac{1}{2}x=6\\x=4$

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babunello [35]3 years ago

3 0

Answer:

The number would be 4.

Step-by-step explanation:

There is an unknown number in the equation. Let that unknown number be x.

$1 \frac{1}{2} * x = 6$

1. Convert the fraction into a mixed fraction or a decimal.

$\frac{3}{2} = 6$

2. Mutiply both sides by 2

$2*\frac{2}{3} = 6 * 2$

$3x=12$

3. Divide by 3 to get x by itself.

$\frac{3x}{3} = \frac{12}{3}$

$x=4$

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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

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