Exercise 4.51. Suppose customer arrivals at a post office are modeled by a Poisson process N with intensity λ > 0. Let T1 be the time of the first arrival. Let t > 0. Suppose we learn that by time t there has been precisely one arrival, in other words, that Nt = 1. What is the distribution of T1 under this new information? In other words, find the conditional probability P(T1 ≤ s|Nt = 1) for all s ≥ 0. Hint. You should see a familiar distribution.

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

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the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer

d i think

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d

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i used the pitagoras theoreme

something over 10 so ? /10

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