Acity's bus's delayed time (x) is uniformly distributed between [0,y] where 0 < = y < = 1 and is uniformly disturbed. suppose you observe a delay time x=x.

a) compute the posterior distribution of y given the data observed and find an approximate value of y by calculating e(y | x=x)

b) suppose the first delay you observe is x(1) = 0.5. the next delay is independently drawn by the same distribution (x(2) ~ uniform [0,y]. given what you have observed from the first delay, what is the probability that the second delay will be more than a 0.5 hour delay?

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

- "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."

3

step-by-step explanation: