The number of hours between successive train arrivals at the station is uniformly distributed on (0, 1). passengers arrive according to a poisson process with rate 7 per hour. suppose a train has just left the station. let x denote the number of people who get on the next train. what are the e[x] and var[x]?

The value of .

The value of .

Step-by-step explanation:

Consider the provided information.

The number of hours between successive train arrivals at the station is uniformly distributed on (0, 1). Passengers arrive according to a Poisson process with rate 7 per hour.

Let X denote the number of people who get on the next train.

Part (A)

Here, X = N(T), N(t) ∼ Poisson(λt),  and λ = 7

Therefore, E(N(T)|T) = λT

Now find E(X) as shown below

Hence, the value of .

Part (B)

Now we need to find VAR[X]

Hence, the value of

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

what is the your question?