A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 14% of the employees needed corrective shoes, 21% needed major dental work, and 3% needed both corrective shoes and major dental work. What is the probability that an employee selected at random will need either corrective shoes or major dental work

0.32 = 32% probability that an employee selected at random will need either corrective shoes or major dental work.

Step-by-step explanation:

We solve this question treating these events as Venn probabilities.

I am going to say that:

Event A: Needing corrective shoes.

Event B: Needing major dental work.

14% of the employees needed corrective shoes

This means that

21% needed major dental work

This means that

3% needed both corrective shoes and major dental work.

This means that

What is the probability that an employee selected at random will need either corrective shoes or major dental work?

This is:

So, from the values given by the exercise:

0.32 = 32% probability that an employee selected at random will need either corrective shoes or major dental work.

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

i believe the answer is d.

hope it you! : d