I’m assuming there is a picture you are missing.
The answer would be choice a they are both functions
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:
y - 3 = (5/3)(x - 6)
Step-by-step explanation:
The usual first step is to determine the slope of the given line.
In this case the slope is -3/5.
A line perpendicular to this given line would have a slope of 5/3, which is the negative reciprocal of -3/5.
This new line goes thru (6, 3) and has a slope of 5/3. Thus, the equation of the line in point-slope form is
y - 3 = (5/3)(x - 6)
Double check that you have copied the problem down correctly. The slope of the line represented by -3x - 5y = 17 is neither 2 nor -2; it is -3/5, and so the slope of a line perpendicular to -3x - 5y = 17 is +5/3.
