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3 years ago

1.3/4 - 5 3/7 2.-9 3/10 - 2/11

1 answer:

3 0

$1)\\\\\frac{3}{4}-5\frac{3}{7}\\\\\frac{3}{4}-\frac{38}{7}\\\\\frac{3}{4}*\frac{7}{7}-\frac{38}{7}*\frac{4}{4}\\\\\frac{21}{28}-\frac{152}{28}\\\\\frac{-131}{28}\\\\-4\frac{19}{28}\\\\\\2)\\\\-9\frac{3}{10}-\frac{2}{11}\\\\\frac{-93}{10}-\frac{2}{11}\\\\\frac{-93}{10}*\frac{11}{11}-\frac{2}{11}*\frac{10}{10}\\\\\frac{-1023}{110}-\frac{20}{110}\\\\\frac{-1043}{110}\\\\-9\frac{53}{110}$

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Step-by-step explanation:

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4 years ago

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2% of the population suffers from the terrible disease conditionitis. The test for conditionitis comes back positive 98% of the

vlabodo [156]

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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.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

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 $P(A|B) = 0.98$

For the 100-2 = 98% who do not have the disease, the test comes back positive 100-77 = 23% of the time.

Then

$P(A) = 0.02*0.98 + 0.98*0.23 = 0.245$

Finally:

$P(B|A) = \frac{0.02*0.98}{0.245} = 0.08$

8% probability that he or she actually has the disease

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4 years ago

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