Answer:
The description again for the problem is listed throughout the section below on explanations.
Step-by-step explanation:
A 2012 survey conducted a week since Voting day because the local paper in Columbus asked voters whatever individual they might vote for the state attorney. 37% of respondents said that they'd vote for both the dem candidate. In reality, 41 percent voted for both the Democratic nominee on Elections Day.
The 37% is supported by a survey as well as being a factual estimate. The sample proportion is denoted by "P". Therefore,
⇒ P = 0.37
The specific proportion becomes supplied as a factor of 41% = 0.41. Since the importance of proportion is real. The proportion of community is represented as p or π
Hence p = 0.41.
Consider such events:
A - slip with number 3 is chosen;
B - the sum of numbers is 4.
You have to count 
Use formula for conditional probability:
1. The event 
consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event is
2. The event 
consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event is
3. Then
Answer: 
2508(1+0.04)^24 = 6,428. trust me, found it to 6,019.20.
b.
Answer:
Step-by-step explanation:
Answer:
We have to prove,
(A \ B) ∪ ( B \ A ) = (A U B) \ (B ∩ A).
Suppose,
x ∈ (A \ B) ∪ ( B \ A ), where x is an arbitrary,
⇒ x ∈ A \ B or x ∈ B \ A
⇒ x ∈ A and x ∉ B or x ∈ B and x ∉ A
⇒ x ∈ A or x ∈ B and x ∉ B and x ∉ A
⇒ x ∈ A ∪ B and x ∉ B ∩ A
⇒ x ∈ ( A ∪ B ) \ ( B ∩ A )
Conversely,
Suppose,
y ∈ ( A ∪ B ) \ ( B ∩ A ), where, y is an arbitrary.
⇒ y ∈ A ∪ B and x ∉ B ∩ A
⇒ y ∈ A or y ∈ B and y ∉ B or y ∉ A
⇒ y ∈ A and y ∉ B or y ∈ B and y ∉ A
⇒ y ∈ A \ B or y ∈ B \ A
⇒ y ∈ ( A \ B ) ∪ ( B \ A )
Hence, proved......
