Suppose you roll a fair die. Let X be the value of the roll. What is the Moment Generating Function of X?

μ₁`= 1/6

μ₂=  5/36

Step-by-step explanation:

The rolling of a fair die is described by the binomial distribution, as  the

The moment generating function of the binomial distribution is derived as below

M₀(t) = E (e^tx)

        = ∑ (e^tx) (nCx)pˣ (q^n-x)

        = ∑ (e^tx) (nCx)(pe^t)ˣ (q^n-x)

        = (q+pe^t)^n

the expansion of the binomial is purely algebraic and needs not to be interpreted in terms of probabilities.

We get the moments by differentiating the M₀(t) once, twice with respect to t and putting t= 0

μ₁`= E (x) = [ d/dt (q+pe^t)^n]  t= 0

            = np

μ₂`=  E (x)² =[ d²/dt² (q+pe^t)^n]  t= 0

              = np +n(n-1)p²

μ₂=μ₂`-μ₁` =npq

in similar way the higher moments are obtained.

μ₁`=1(1/6)= 1/6

μ₂= 1(1/6)5/6

   = 5/36

not solvable

step-by-step explanation:

we need more information, or at least an

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.