Hi there what you need is lagrange multipliers for constrained minimisation. It works like this,
V(X)=α2σ2X¯1+β2\sigma2X¯2
Now we want to minimise this subject to α+β=1 or α−β−1=0.
We proceed by writing a function of alpha and beta (the paramters you want to change to minimse the variance of X, but we also introduce another parameter that multiplies the sum to zero constraint. Thus we want to minimise
f(α,β,λ)=α2σ2X¯1+β2σ2X¯2+λ(\alpha−β−1).
We partially differentiate this function w.r.t each parameter and set each partial derivative equal to zero. This gives;
∂f∂α=2ασ2X¯1+λ=0
∂f∂β=2βσ2X¯2+λ=0
∂f∂λ=α+β−1=0
Setting the first two partial derivatives equal we get
α=βσ2X¯2σ2X¯1
Substituting 1−α into this expression for beta and re-arranging for alpha gives the result for alpha. Repeating the same steps but isolating beta gives the beta result.
Lagrange multipliers and constrained minimisation crop up often in stats problems. I hope this helps!And gosh that was a lot to type!xd
Answer:X/3X=3
We move all terms to the left:
X/3X-(3)=0
Domain of the equation: 3X!=0
X!=0/3
X!=0
X∈R
We multiply all the terms by the denominator
X-3*3X=0
Wy multiply elements
X-9X=0
We add all the numbers together, and all the variables
-8X=0
X=0/-8
X=0
Step-by-step explanation:
The perpendicular equation is y = -3/2x - 4.
You can find this by first realizing that perpendicular lines have opposite and reciprocal slopes. So since it starts at 2/3 we flip it and make it a negative and the new slope is -3/2. Now we can use that and the point to get the y intercept using slope intercept form.
y = mx + b
5 = (-3/2)(-6) + b
5 = 9 + b
-4 = b
And now we can use our new slope and new intercept to model the equation.
y = -3/2x - 4
Ok ok im gonna with C but if it’s wrong i am sooo sorry
