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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
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