Answer:
that last answer is right
nothing to worry about it
Domain and range:
The domain of a function is all the values the the input assumes.
The range of a function is given by all the values that the output assumes.
In a graphic:
The domain is given by the x-values, the horizontal axis.
The range is given by the y-values, the vertical axis.
In this question:
x, in the horizontal axis, assumes values between -4 and 2.
The closed circle at x = -4 and x = 2 indicates that these values are part of the interval.
Thus, the domain of the graphed function is: , and the correct answer is given by option D.
For more about domain and range of functions, you can check: brainly.com/question/21853810?referrer=searchResults
The uniqueness of LIMITS of a function theorem. Recall from the LIMIT of a function page that for a function where is a cluster point of , then if such that if and then...... then if are both LIMITS of at , that is and , then . PROOF: Let be a function and let be a cluster point of. I hope this helps.
Answer:
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
Step-by-step explanation:
Let the random variable Y have a geometric distribution g (y;p) = pq y-¹
The m.g.f of the geometric distribution is derived as below
By definition , M₀ (t) = E (e^ ty) = ∑ (e^ ty )( q ^ y-1)p ( for ∑ , y varies 1 to infinity)
= pe^t ∑(e^tq)^y-1
= pe^t/1- qe^t, where qe^t <1
In order to differentiate the m.g.f we write it as
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
M₀` (t) = pe^-t (e^-t -q) ^ -2 and
M₀^n(t) = 2pe^-2t (e^-t -q) ^ -3 - pe^-t (e^-t -q) ^ -2
Hence
E (y) = p (1-q)-² = 1/p
E (y²) =2 p (1-q)-³ - p (1-q)-²
= 2/p² - 1/p and
σ² = [E (y²) -E (y)]²
= 2/p² - 1/p - (1/p)²
= q/p²
4x(3+2)-6x1= 14 is the answer