Identify the common external tangent. line r line s segment AB

Mathematics

2 answers:

Oxana [17]3 years ago

8 0

The external tangent is line s

spin [16.1K]3 years ago

8 0

Answer:

line s.

Step-by-step explanation:

First let's see what a tangent line is.

A line that touches a circle at only one point is called the tangent line of the circle.

Here we have to find the common external tangent.

Which means the line has to touch the two circle at only point on each circle.

If you look at the given figure. The line "s" touches the big circle at the point J and the smaller circle at the point K.

Therefore, the external tangent line is "line "s"

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Step-by-step explanation:

Since we have given that

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

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

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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

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