Suppose that the probability of having a particular disease is 0.08. Suppose that the probability of testing positive for the disease is 0.94 given that a person has the disease and 0.08 given that the person does not have the disease. Given that a person tests positive for the disease, what is the probability that they actually have the disease? (Answer to four decimal places).

0.5054

Step-by-step explanation:

This is a question on conditional probability.

We solve using Baye's Theorem of conditional probability

From the question

The probability of having a particular disease = 0.08.

The probability of not having a particular disease = 1 - 0.08 = 0.92

The probability of testing positive for the disease is given that a person has the disease = 0.94

The probability of testing positive given that the person does not have the disease = 0.08

Given that a person tests positive for the disease, the probability that they actually have the disease is

= (0.08 × 0.94)/(0.08 × 0.94) + (0.08 × 0.92)

= 0.0752/0.0752 + 0.0736

=0.0752/ 0.1488

= 0.5053763441

≈ Approximately to 4 decimal places = 0.5054

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15