Answer:
itself was just<em><u> </u></em><em><u>an</u></em><em><u> </u></em><em><u>awesome</u></em><em><u> </u></em><em><u>game</u></em><em><u> </u></em><em><u>that</u></em><em><u> </u></em><em><u>had</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>game</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>game</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>my</u></em><em><u> </u></em><em><u>friends</u></em><em><u>.</u></em>
Answer:
117
Step-by-step explanation:
216/3=72
216-72=144
144-26=117
Answer:
0.00002 = 0.002% probability of actually having the disease
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is

In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Positive test
Event B: Having the disease
Probability of having a positive test:
0.05 of 1 - 0.000001(false positive)
0.99 of 0.000001 positive. So

Probability of a positive test and having the disease:
0.99 of 0.000001. So

What is the probability of actually having the disease

0.00002 = 0.002% probability of actually having the disease
Answer is B.
hope it helps