Need to know if I got the correct answer for #29. If it’s wrong can you help me solve it please and thank you.

Mathematics

1 answer:

Pie3 years ago

6 0

Answer:

-x² -2x -5 -12/(x-2)

Step-by-step explanation:

The attachment shows the synthetic division. You can find the remainder by evaluating the dividend for x=2:

-2³ -2 -2 = -12 . . . . matches the above answer, not [A] or [C] shown

_____

You can always multiply out your answer to see if it matches the problem.

(-x^2 -2x +3)(x -2) +3 = -x^3 +7x -3 . . . . not the dividend you started with

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Solve the following equation 5x-15=50

denis-greek [22]

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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

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