The test to detect the presence of a liver disorder is 98% accurate for a person who has the disease and 97% accurate for a person who does not have the disease. if 3.5% of the people in a given population actually have the disorder, what is the probability that a randomly chosen person tests positive?

The probability that a randomly chosen person tests positive is 0.06325

Step-by-step explanation:

if we chose a person randomly and the test is positive, there are two possibilities:

1. The person actually have a disease: the probability of this case is given by the multiplication of the probability of having the disease and the probability that the test detect the disorder. That can be written as:

0.035*0.98=0.0343

where 0.98 is the probability that the test detect the disease given that a person have a disease.

2. The person doesn't have a disease: the probability of this case is given by the multiplication of the probability of doesn't have the disease and the probability that the test said that the person have the disorder. That can be written as:

0.965*(1-0.97)=0.02895

Where (1-0.97) is the probability that the test detect the disease given that a person don't have a disease.

Finally the probability that a randomly chosen person tests positive is the sum of both cases and it calculated as:

P=0.0343+0.02895=0.06325

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.