Answer:
The addition property of equality
Step-by-step explanation:
According to the addition property of equality, if we add the same number to each side of an equation, the equality is still true.
We are given the equation
x – 3 = 7, and we must solve for x.
We want to get x by itself, so we <em>add 3 to each side</em>.
x – 3 + 3 = 7 + 3
x = 10
We have just used the addition property of equality.
We have the following given
p1 - probability for outcome 1
p2 - probability for outcome 2
p3 - probability for outcome 3
v1 - amount of money that you will win or lose for outcome 1
v2 - amount of money that you will win or lose for outcome 2
v3 - amount of money that you will win or lose for outcome 3
Therefore,
p1v1 + p2v2 + p3v3 is the average money you win or lose in playing the game.
Answer:
216 cookies
Step-by-step explanation:
From the question,
Let the total cookies in the first place be x cookies.
Their ratio = 5:12:9
Total = 5+12+9 = 26.
Weiming share = (5/26)x = 5x/26
If weiming sold 28 cookies
⇒ (5x/26)-28
⇒ (5x-728)/28
Hence Weiming new share = (5x-728)/28 cookies
The new ratio is 1:8:6
The total cookies becomes = x-28
Therefore,
(1/15)(x-28) = (5x-728)/28
(x-28)/15 = (5x-728)/28
Crossmultiply
28(x-28) = 15(5x-728)
28x-784 = 75x-10920
Collect like terms
28x-75x = -10920+784
-47x = -10136
x = -10136/-47
x = 215.7
x ≈ 216 cookies
Given:
Either has a school certificate or diploma or even both = 20 people
Having school certificates = 14
Having diplomas = 11
To find:
The number of people who have a school certificate only.
Solution:
Let A be the set of people who have school certificates and B be the set of people who have diplomas.
According to the given information, we have
![n(A)=14](https://tex.z-dn.net/?f=n%28A%29%3D14)
![n(B)=11](https://tex.z-dn.net/?f=n%28B%29%3D11)
![n(A\cup B)=20](https://tex.z-dn.net/?f=n%28A%5Ccup%20B%29%3D20)
We know that,
![n(A\cup B)=n(A)+n(B)-n(A\cap B)](https://tex.z-dn.net/?f=n%28A%5Ccup%20B%29%3Dn%28A%29%2Bn%28B%29-n%28A%5Ccap%20B%29)
![20=14+11-n(A\cap B)](https://tex.z-dn.net/?f=20%3D14%2B11-n%28A%5Ccap%20B%29)
![20=25-n(A\cap B)](https://tex.z-dn.net/?f=20%3D25-n%28A%5Ccap%20B%29)
Subtract both sides by 25.
![20-25=-n(A\cap B)](https://tex.z-dn.net/?f=20-25%3D-n%28A%5Ccap%20B%29)
![-5=-n(A\cap B)](https://tex.z-dn.net/?f=-5%3D-n%28A%5Ccap%20B%29)
![5=n(A\cap B)](https://tex.z-dn.net/?f=5%3Dn%28A%5Ccap%20B%29)
We need to find the number of people who have a school certificate only, i.e.
.
![n(A\cap B')=n(A)-n(A\cap B)](https://tex.z-dn.net/?f=n%28A%5Ccap%20B%27%29%3Dn%28A%29-n%28A%5Ccap%20B%29)
![n(A\cap B')=14-5](https://tex.z-dn.net/?f=n%28A%5Ccap%20B%27%29%3D14-5)
![n(A\cap B')=9](https://tex.z-dn.net/?f=n%28A%5Ccap%20B%27%29%3D9)
Therefore, 9 people have a school certificate only.