Answer:
the answer is E, right?
Step-by-step explanation:
Answer:
The change to the face 3 affects the value of P(Odd Number)
Step-by-step explanation:
Analysing the question one statement at a time.
Before the face with 3 is loaded to be twice likely to come up.
The sample space is:
![S = \{1,1,2,2,2,3\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7B1%2C1%2C2%2C2%2C2%2C3%5C%7D)
And the probability of each is:
![P(1) = \frac{n(1)}{n(s)}](https://tex.z-dn.net/?f=P%281%29%20%3D%20%5Cfrac%7Bn%281%29%7D%7Bn%28s%29%7D)
![P(1) = \frac{2}{6}](https://tex.z-dn.net/?f=P%281%29%20%3D%20%5Cfrac%7B2%7D%7B6%7D)
![P(1) = \frac{1}{3}](https://tex.z-dn.net/?f=P%281%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
![P(2) = \frac{n(2)}{n(s)}](https://tex.z-dn.net/?f=P%282%29%20%3D%20%5Cfrac%7Bn%282%29%7D%7Bn%28s%29%7D)
![P(2) = \frac{3}{6}](https://tex.z-dn.net/?f=P%282%29%20%3D%20%5Cfrac%7B3%7D%7B6%7D)
![P(2) = \frac{1}{2}](https://tex.z-dn.net/?f=P%282%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
![P(3) = \frac{n(3)}{n(s)}](https://tex.z-dn.net/?f=P%283%29%20%3D%20%5Cfrac%7Bn%283%29%7D%7Bn%28s%29%7D)
![P(3) = \frac{1}{6}](https://tex.z-dn.net/?f=P%283%29%20%3D%20%5Cfrac%7B1%7D%7B6%7D)
P(Odd Number) is then calculated as:
![P(Odd\ Number) = P(1) + P(3)](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%20P%281%29%20%2B%20P%283%29)
![P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7D)
Take LCM
![P(Odd\ Number) = \frac{2+1}{6}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B2%2B1%7D%7B6%7D)
![P(Odd\ Number) = \frac{3}{6}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B3%7D%7B6%7D)
![P(Odd\ Number) = \frac{1}{2}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D)
After the face with 3 is loaded to be twice likely to come up.
The sample space becomes:
![S = \{1,1,2,2,2,3,3\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7B1%2C1%2C2%2C2%2C2%2C3%2C3%5C%7D)
The probability of each is:
![P(1) = \frac{n(1)}{n(s)}](https://tex.z-dn.net/?f=P%281%29%20%3D%20%5Cfrac%7Bn%281%29%7D%7Bn%28s%29%7D)
![P(1) = \frac{2}{7}](https://tex.z-dn.net/?f=P%281%29%20%3D%20%5Cfrac%7B2%7D%7B7%7D)
![P(2) = \frac{n(2)}{n(s)}](https://tex.z-dn.net/?f=P%282%29%20%3D%20%5Cfrac%7Bn%282%29%7D%7Bn%28s%29%7D)
![P(2) = \frac{3}{7}](https://tex.z-dn.net/?f=P%282%29%20%3D%20%5Cfrac%7B3%7D%7B7%7D)
![P(3) = \frac{n(3)}{n(s)}](https://tex.z-dn.net/?f=P%283%29%20%3D%20%5Cfrac%7Bn%283%29%7D%7Bn%28s%29%7D)
![P(3) = \frac{1}{7}](https://tex.z-dn.net/?f=P%283%29%20%3D%20%5Cfrac%7B1%7D%7B7%7D)
![P(Odd\ Number) = P(1) + P(3)](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20P%281%29%20%2B%20P%283%29)
![P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B2%7D%7B7%7D%20%2B%20%5Cfrac%7B1%7D%7B7%7D)
Take LCM
![P(Odd\ Number) = \frac{2+1}{7}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B2%2B1%7D%7B7%7D)
![P(Odd\ Number) = \frac{3}{7}](https://tex.z-dn.net/?f=P%28Odd%5C%20Number%29%20%3D%20%5Cfrac%7B3%7D%7B7%7D)
Comparing P(Odd Number) before and after
--- Before
--- After
<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>
Answer: 5
Step-by-step explanation:
First we should do the first step: 7+2=9, The they lose 4 which then 9-4
which would equal 5
(9m - 6)7
Distribute 7 to both sides
63m - 42
Add 42 to get 63m by itself
63m = 42
Divide both sides by 63
m = 42/63 or simplified, 2/3