Answer:
8% probability that he or she actually has the disease
Step-by-step explanation:
We use the Bayes Theorem to solve this question.
Bayes Theorem:
Two events, A and B.

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
If a randomly chosen person is given the test and the test comes back positive for conditionitis, what is the probability that he or she actually has the disease?
This means that:
Event A: Test comes back positive.
Event B: Having the disease.
Test coming back positive:
2% have the disease(meaning that P(B) = 0.02), and for those, the test comes positive 98% of the time. This means that 
For the 100-2 = 98% who do not have the disease, the test comes back positive 100-77 = 23% of the time.
Then

Finally:

8% probability that he or she actually has the disease
Answer: whats the question here?
this is false
(6+8)^2= 196
6^2+8^2=100
They are not equal so this is a false statement
Note that multip. 10 by 2 results in 20; mult. 20 by 2 results in 40, and so on. So the common ratio is 2.
24:36
9:12
12:16
Because yes