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: 34,381
Step-by-step explanation:
You have to subtract 52,835 by the sum 87,216 which will get you the answer of 34,381
Answer:Pois(ln(200))
Step-byy-step explanation:
Let N be the number of received calls in a day
That is
N∼Pois(λ).
0.5% = 0.5/100 = 1/200 of no calls
P(N=0)=e^−λ=1/200
-λ=e^(1/200)
λ=in(200)
Our number of calls in a day is distributed Pois(ln(200)).








On simplifying, we get

<h2>Therefore, correct option is 1st option.</h2>
Convert all into a fraction and you will get 115/100+1/50+47/40
So your answer is 469/200 or 2.345 as a decimal