Answer:
The solution of the equations are -6 and 1
Step-by-step explanation:
* <em>Lets explain how to solve the problem</em>
- We want to find the solution of the equation (x + 2) (x + 3) = 12
- <em>At first lets use the Foil method to multiply the two brackets</em>
(x + 2) (x + 3) = (x)(x) + (x)(3) + (2)(x) + (2)(3)
(x + 2) (x + 3) = x² + 3x + 2x + 6 ⇒ add the like term
(x + 2) (x + 3) = x² + 5x + 6
∵ (x + 2) (x + 3) = 12
∴ x² + 5x + 6 = 12
- Subtract 12 from both sides
∴ x² + 5x - 6 = 0
- <em>Factorize the left hand side</em>
∵ x² = (x)(x)
∵ -6 = 6 × -1
∵ 6x + -1x = 5x
∴ (x + 6)(x - 1) = 0
- <em>Lets use the zero product property </em>
∵ (x + 6)(x - 1) = 0
∴ x + 6 = 0 ⇒ <em>OR</em> ⇒ x - 1 = 0
∵ x + 6 = 0
- Subtract 6 from both sides
∴ x = -6
∵ x - 1 = 0
- Add 1 to both sides
∴ x = 1
∴ The solution of the equations are -6 and 1
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.
Answer:
n-315=1245
Step-by-step explanation:
n stands for the original amount of the loan. Margo has paid $315, so the original number minus what Margo has paid, and the equality is the amount that Margo still owes.
The answer is 52%. 65/125=0.52 * 100 = 52%
Answer:
Seestep by step explanation
Step-by-step explanation:
First Selection:
First branch P₁ = 12/18 P₁ = 4/6 P₁ = 2/3 (To pick a red one)
Second Branch P₂ = 6/18 P₂ = 1/3 ( To pick a black one)
Second Selection
We already have a red checker the we have 11 red checkers and 6 black checkers, then
First branch P₃ = 11/17
Second Branch P₄ = 6/17
But in case that our first checker was black then we have
First branch P₅ = 12/17
Second Branch P₆ = 5/17