A young doctor is working at night in an emergency room. Emergencies come in at times of a Poisson process with rate 0.5 per hour. The doctor can only get to sleep when it has been 36 minutes (.6 hours) since the last emergency. For example, if there is an emergency at 1:00 and a second one at 1:17 then she will not be able to get to sleep until at least 1:53, and it will be even later if there is another emergency before that time Required:
Compute the long-run fraction of time she spends sleeping, by formulating a renewal reward process in which the reward in the ith interval is the amount of time she gets to sleep in that interval.

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

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step-by-step explanation:

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