Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
The average squared distance between the points is their variance
The average squared distance is 8/3
How to determine the average squared distance?
The given parameters are:
R={(x,y): 0<=x<=2, 0<=y<=2}
The point = (2,2).
f(x,y) = (x-2)² + (y-2)²
The squared distance is calculated as:
D² = f(x,y) = (x-2)² + (y-2)²
Where the area (A) is:
A = xy
Substitute the maximum values of x and y in A = xy
A = 2 * 2
A = 4
The minimum values of x and y are 0.
So, the limits for the integrals are 0 to 2
The integral becomes
[tex]D\² = \int \int (x-2)\² + (y-2)\² dx dy[/tex]
Expand
[tex]D\² = \int \int x\² -4x + 4 + (y-2)\² dx dy[/tex]
Evaluate the inner integral with respect to x from 0 to 2
[tex]D\² = \int \frac 13x^3 -2x^2 + 4x + x(y-2)\² |\limits^2_0 dy[/tex]
Expand
[tex]D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy[/tex]
Simplify
[tex]D\² = \int \frac 83 + (2)(y-2)\² dy[/tex]
Expand
[tex]D\² = \int \frac 83 + 2y^2-8y + 8 \ dy[/tex]
Evaluate the integral with respect to y from 0 to 2
[tex]D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0[/tex]
Expand
[tex]D\² = \frac 83*2 + \frac 23*2^3- 4*2^2 + 8*2[/tex]
D² = 32/3
The average squared distance (AD²) is calculated as:
AD² = D²/A
So, we have:
AD² = 32/3 [tex]\div[/tex] 4
Evaluate the quotient
AD² = 32/12
Simplify
AD² = 8/3
Hence, the average squared distance is 8/3
Read more about average distance or variance at:
https://brainly.com/question/15858152
The average squared distance between the points of R = {(x,y): 0<=x<= h, 0<=y<=k } is 8/3.
How do determine the average squared distance?
From the given parameters R = {(x,y): 0<=x<= h, 0<=y<=k }
The squared distance from point (h, k) is calculated as
D² = f(x,y) = (x-2)² + (y-2)²
Where the area (A) will be
A = xy
Substitute the maximum values of x and y
A = 2 × 2 = 4
The minimum values of x and y are 0.
So, The integral becomes
[tex]D^{2} = \int \int(x-2)^2 +(y-2)^2 dx dy\\\\D^{2} = \int \int\limits^2_0 (x)^2 -4x + 4+(y-2)^2 dx dy\\\\D^{2} = \int1/3(x)^3-2x^{2} +4x + 4+x(y-2)^2 dy\\\\D^{2} = \int8/3+2y^{2} -8y +8 dy[/tex]
[tex]D^2 = {8/3y +2/3y^2 -4y^2+8y[/tex]
By expanding we get,
D^2 = 32/3
The average squared distance (AD²) is calculated as:
AD² = D²/A
AD² = 32/3 4
AD² = 32/12
AD² = 8/3
Hence, the average squared distance is 8/3
Read more about average distance;
brainly.com/question/15858152
#SPJ4
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
