A fast-food company is interested in knowing the probability of whether a customer viewed an advertisement for their new special on the internet or on television. They found that 37% of customers saw the advertisement on the internet, 20% saw it on television, and 12% saw it on both the internet and on television. 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

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

his first public project he is asked to create a small scale drawing. the garden is a right triangle with base 25m and height 30m. are you trying to solve for the last side/hypotenuse? the irrational form of the answer would be 5\sqrt{61}

answer: a) 56% women

b)44% men

step-by-step explanation: