A food manufacturer uses an extruder (a machine that produces bite-size cookies and snack food) that yields revenue for the firm at a rate of $200 per hour when in operation. However, the extruder breaks down an average of two times every day it operates. If Y denotes the number of breakdowns per day, and suppose Y follows a Poisson distribution. The daily revenue generated by the machine is R=1600−50Y^2. Find the expected daily revenue for the extruder.

$1300

Step-by-step explanation:

λ = average = 2 per day, Y denotes the number of breakdowns per day,

But E(Y) = λ, E[Y(Y - 1)] = λ²

E(Y²) = E[Y(Y - 1)] + E(Y), hence:

E(Y²) = λ² + λ

The daily revenue is R=1600−50Y². The expected daily revenue for the extruder E(R) is:

E(R) = E(1600−50Y²) = 1600 - 50E(Y²)

E(R) = 1600 - 50(λ² + λ)

Substituting:

E(R) = 1600 - 50(2² + 2)

E(R) = 1600 - 300 = 1300

E(R) = $1300

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.