Sub in the values and solve for W
50=3W+14
50-14=3W
36=3W
12=W
The team won 12 games.
12+14=26 (games drawn and won)
35-26=9
Therefore they lost 9 games
Answer:
x+((4 + 1/3))=(-(2 + 5/6))
6x = -43
x≈ -7.166667
First, let's cancel out the x by multiplying 2x + 18y = -9 by -2.
-2 ( 2x + 18y = -9) = -4x -36y = 18
Then, we combine the two equations.
-4x + 4x = 0
18y - 36y = -18y
-27 + 18 = -9
Our new equation is -18y = -9.
Now, divide both sides by -18.
-18y / -18 = y
-9/ -18 = 1/2
y = 1/2
We can plug in a value for y since y = 1/2 now.
Let's use 2x + 18y = -9
Plug in y.
2x + 18(1/2) = -9
2x + 9 = -9
Then, subtract 9 from both sides.
2x = -18
Divide by 2.
2x/2 = x
-18/2 = -9
x = -9
Lastly, we can plug in both x and y values to see it works.
2(-9) + 18(1/2) = -9
-18 + 9 = -9
Therefore, the values of x and y does work.
x = -9
y = 1/2
To find c, you must isolate it.
To do this, you must divide both sides by 5/7, since that is being multiplied by c and you must do the inverse to it to cancel it out in order to leave c by itself.
5/7c ÷ 5/7 = c
13/14 ÷ 5/7
To divide fractions, follow these steps:
Step 1- Turn the second fraction, 5/7 in this case, into its reciprocal. This means swapping the places of the numerator and denominator.
5/7 reciprocal = 7/5
Step 2- multiply the original first fraction and reciprocal second fraction.
13/14 • 7/5
13 • 7 = 91
14 • 5 = 70
13/14 ÷ 5/7 = 91/70
Step 3- Simplify if possible.
91/70
Since 70 can go into 90, you can turn this into a mixed number.
1 and 21/70
Now simplify 21/70.
Both can be divided by 7.
21 ÷ 7 = 3
70 ÷ 7 = 10
So simplified, 91/70 equals 1 and 3/10.
As a decimal, this is 1.3.
So the answer is c = 1.3, or 1 and 3/10.
Hope this helps :)
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.
