9514 1404 393
Answer:
r = 1/9
Step-by-step explanation:
First of all, solve the equation for r:
y = rx
y/x = r . . . . . . . divide by x
__
Since r is a constant, it will be the same for any corresponding pairs of x and y. It is convenient to choose both x and y as integers, as in the third table entry.
r = y/x = 5/45
r = 1/9 . . . . . . . . . reduced fraction
_____
<em>Additional comment</em>
It is not a bad idea to check to see that this works with other values of x and y. For the first line of the table, we have x = 11:
y = rx = (1/9)(11) = 11/9 = 1 2/9 . . . . matches the table value
Answer:
Step-by-step explanation:
We can obtain that the dimensions of the rectangle are 7 by 5.
Since A = lw, the total area is 35.
The purpose of the tensor-on-tensor regression, which we examine, is to relate tensor responses to tensor covariates with a low Tucker rank parameter tensor/matrix without being aware of its intrinsic rank beforehand.
By examining the impact of rank over-parameterization, we suggest the Riemannian Gradient Descent (RGD) and Riemannian Gauss-Newton (RGN) methods to address the problem of unknown rank. By demonstrating that RGD and RGN, respectively, converge linearly and quadratically to a statistically optimal estimate in both rank correctly-parameterized and over-parameterized scenarios, we offer the first convergence guarantee for the generic tensor-on-tensor regression. According to our theory, Riemannian optimization techniques automatically adjust to over-parameterization without requiring implementation changes.
Learn more about tensor-on-tensor here
brainly.com/question/16382372
#SPJ4
Answer:
6.025 as a mixed number is 6 1/40
By definition of covariance,
![\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5B%28X-%5Cmathbb%20E%5BX%5D%29%28Y-%5Cmathbb%20E%5BY%5D%29%5D)
![\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5BXY-%5Cmathbb%20E%5BX%5DY-X%5Cmathbb%20E%5BY%5D%2B%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%5D%3D%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D)
We have
![\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5B%28aX-b%29%28cY-d%29%5D%3D%5Cmathbb%20E%5BacXY-adX-bcY%2Bbd%5D)
![=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%3Dac%5Cmathbb%20E%5BXY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
![\mathbb E[aX-b]=a\mathbb E[X]-b](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%3Da%5Cmathbb%20E%5BX%5D-b)
![\mathbb E[cY-d]=c\mathbb E[Y]-d](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BcY-d%5D%3Dc%5Cmathbb%20E%5BY%5D-d)
![\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%5Cmathbb%20E%5BcY-d%5D%3Dac%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
Putting everything together, we find the covariance reduces to
![\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28aX-b%2CcY-d%29%3Dac%28%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%29%3Dac%5Cmathrm%7BCov%7D%28X%2CY%29)
as desired.
