A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of the employees needed eyeglasses, 15% needed major dental work, and 3% needed both eyeglasses and major dental work. What is the probability that an employee selected at random will need either eyeglasses or major dental work

The Probability = 0.20

Step-by-step explanation:

From the question stated, the first step to take is to find probability that an employee selected at random will need either eyeglasses or major dental work

Solution

Given

Now,

The exams showed tha the number of employees needed eyeglasses = 8%

Employees that needed major dental work = 15%

Employees that needed  both eyeglasses and major dental work =3%

Thus,

The P(needed eyeglasses ) = 8% = 0.08

P(major dental work) = 15% = 0.15

P(eye glasses  and major dental work) = 3% = 0.03

The probability that an employee selected at random will need either eyeglasses or major dental work is  given as

= P(eye glasses ) + P(major dental work) - P(eyeglasses and major dental work)

= 0.08 + 0.15 - 0.03 = 0.20

Therefore the Probability = 0.20

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.