I need help i need the statement and reason

Mathematics

1 answer:

Marina86 [1]2 years ago

6 0

For number one
S: supplement.
R: angle1+angle8 =18
#2
S: supplement
R: angle 3+ angle 6=180

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The temperature was 19 degrees Fahrenheit on Monday and 31 degrees Fahrenheit on Tuesday. How much did the temperature rise?

Lera25 [3.4K]

Answer:

31-19=12°F

12°F

Step-by-step explanation

If it was 19 degrees on Monday and 31 degrees on Tuesday you have to see how much did it go up in temp. So..

the difference between 19 and 31 is 12, so that means it went up by 12°F.

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3 years ago

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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

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