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3 years ago

Can someone help me please?

Mathematics

1 answer:

Finger [1]3 years ago

5 0

Answer: x = 4

y = - 3

Step-by-step explanation:

The given system of simultaneous equations is expressed as

x-3y = 13 ------------------------1

2x+4y=-4--------------------------2

We would apply the method of substitution in solving the equations. From equation 1, we would make x to stand alone by adding 3y to the left hand side and the right hand side of the equation. It becomes

x - 3y + 3y = 13 + 3y

x = 13 + 3y

Substituting x = 13 + 3y into equation 2, it becomes

2(13 + 3y) + 4y = - 4

26 + 6y + 4y = - 4

26 + 10y = - 4

Subtracting 26 from the left hand side and the right hand side of the equation, it becomes

26 - 26 + 10y = - 4 - 26

10y = - 30

Dividing the left hand side and the right hand side of the equation by 10, it becomes

10y/10 = - 30/10

y = - 3

Substituting y = - 3 into x = 13 + 3y, it becomes

x = 13 + 3 × - 3

x = 13 - 9

x = 4

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3 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%.

7 0

3 years ago

What is the distance between point A and point B?what is the ordered pair for each of the points plotted above?

Alekssandra [29.7K]

Answer:

A: (0,0), B: (3, -4), 5 units

Step-by-step explanation:

A: (0,0), B: (3, -4)

To find the distance, use the distance formula. Square-root((0-3)^2+(0-(-4))^2) ---> Square-root((-3)^2 + 4^2) ---> Square-root(9+16) ---> Square-root(25) ---> 5units.

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3 years ago

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