Answer:
D. 2 3 • 3 2 • 5
Step-by-step explanation:
ig I need at least 20 characters to submit my answer. :)
Answer:
e. 2x - 7y = 31.
Step-by-step explanation:
If you want a line parallel to the equation, the line must have the same slope.
2x - 7y = 9
-7y = -2x + 9
y = 2/7x - 9/7
You are looking for a line where the slope is 2/7.
a. Slope is 7/2.
b. Slope is -7/2.
c. Slope is 7/2.
d. Slope is...
y + 2 = 3/4x + 3
y = 3/4x + 1
Slope is 3/4.
e. Slope is -2/-7, which is 2/7. Just make sure to check whether it passes through (5, -3) by substituting those points into the equation.
2 * 5 - 7 * -3 = 31
10 --21 = 31
10 + 21 = 31
And there's your answer: e.
Hope this helps!
Step-by-step explanation:
the leading coefficient means the coefficient (factor) of the term with the highest exponent of the variable (typically x).
with sufficiently large values of this variable (x - going far enough to the right) this term will "win" in value against any other term of the polynomial expression.
and therefore the sign of its factor (coefficient) will determine, if the curve will go up or down.
a positive factor (coefficient) will make the value of this term and therefore of the whole polynomial larger and larger, making the curve going up to +infinity.
a negative factor (coefficient) will make the value of this term and therefore of the whole polynomial smaller and smaller (more negative and more negative), making the curve going down to -infinity.
Answer:
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 probability of success remains constant for all trials, p.
- the successive trials are all independent
- the experiment is repeated a fixed number of times
- there are two outcomes success, p, and failure ,q.
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
