Acity's buses 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?

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

constant of proportionality = 3 and the equation is c = 3p.

step-by-step explanation:

c = kp   where c = amount of caffeine in mg., p = oz of pepsi and k is the constant of proportionality.

substituting the values given:

24 = 8k

k = 3.

the same value of k will be obtained if we substitute the other given values.

it would be a on edgenuit y

the diagram below is your final answer