Answer:
45% probability that a randomly selected customer saw the advertisement on the internet or on television
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a customer saw the advertisement on the internet.
B is the probability that a customer saw the advertisement on television.
We have that:
In which a is the probability that a customer saw the advertisement on the internet but not on television and 
is the probability that the customers saw the advertisement in both the internet and on television.
By the same logic, we have that:
12% saw it on both the internet and on television.
This means that 
20% saw it on television
This means that 
37% of customers saw the advertisement on the internet
This means that 
What is the probability that a randomly selected customer saw the advertisement on the internet or on television
45% probability that a randomly selected customer saw the advertisement on the internet or on television
Answer:
32
Step-by-step explanation:
Short answer: 2 * sqrt(13)
Remark
There are a number of ways to look at this. I'll pick the easiest.
Step one
Factor 52 until there are no more prime factors to be used.
52 = 2 * 26
52 = 2 * 2 * 13. That's as far as you can go.
Rule
For every 2 equal prime factors, 1 of them can be taken out side of the root sign. The other one disappears.
sqrt(52) = sqrt(2*2* 13) = 2*sqrt(13)
Answer 2sqrt(13) <<<< answer
Answer:
The probability of both points falling in the same row or column is 7/19, or approximately 37%
Step-by-step explanation:
The easiest way to solve this is to think of it rephrased as "what is the probability that your second point will be in the same row or column as your first point". With that frame of reference, you can simply consider how many other points are left that do or do not fall in line with the selected one.
After selecting one, there are 19 points left.
The row that the first one falls in will have 3 remaining empty points.
The column will have 4 remaining empty points.
Add those up and you have 7 possible points that meet the conditions being checked.
So the probability of both points falling in the same row or column is 7/19, or approximately 37%
The answer is 53 to the power of 6
