Answer:
(40 x 11) +1 .5(11 x 10) = 165
Step-by-step explanation:
Answer:
V = 38659.68 cm^3
Step-by-step explanation:
If we pretend A barrel is a cylinder we use V=πr^2h for the volume. We know h, but we need to find r
The question tells us the circumference, well how do we find circumference? C = 2πr, so we can use this to find r.
C = 2πr
56.52 = 2 * 3.14 * r Here divide both sides by 2*3.14
56.52/(2*3.14) = r
9 cm = r
Now we can just plug into the volume.
V=πr^2h
V = 3.14 * 9^2 * 152
V = 38659.68 cm^3
I feel I should mention this is all true if the circumference is in centimeters, since the question didn't specify.
The best interpretation of 2 in the given context is that; 2 represents the total length of fabric in yards purchased when no costumes were made.
<h3>Slope Intercept Form</h3>
We are given the relationship between the number of costumes that Kai made, x, and the total length of fabric that he purchased y, in yards as; y - 5x = 2
Where;
- x is number of costumes made
- y is total length of fabrics purchased.
Now, the general slope intercept form of an equation is given by; y = mx + c
Where c is the y-intercept which is the point where x is zero.
Thus, our equation can be rewritten as;
y = 5x + 2
Comparing with y = mx + c, we can say that c = 2
This means that at x = 0, y = 2.
In conclusion, the value 2 represents the length of yards purchased when no costumes were made.
Read more about slope-intercept form at;brainly.com/question/1884491
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
