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3 years ago

Show work also Plss help smsjsjs

Mathematics

2 answers:

choli [55]3 years ago

8 0

Answer:

Step-by-step explanation:

A simple fraction (also known as a common fraction or vulgar fraction, where vulgar is Latin for "common") is a rational number written as a/b or {\displaystyle {\tfrac {a}{b}}}

, where a and b are both integers. As with other fractions, the denominator (b) cannot be zero.

Andrew [12]3 years ago

8 0

Answer:

13 5/6 miles

Step-by-step explanation:

Monday --> 5 1/2 miles

Wednesday --> 8 1/3 miles

5 1/2 = 11/2

8 1/3 = 25/3

11/2 + 25/3 = 33/6 + 50/6 = 83/6 = 13 5/6 miles

Hope this helps!

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 99.9% of people

aleksklad [387]

Answer:

Part a: The probability that someone tests positive and does not have the disease is given as P(F'|E) is 0.6667

Part b: The probability that someone tests negative and have the disease is given as P(F|E') is 0.000000103

Step-by-step explanation:

The event that the test is positive is E and the patient has disease is F

The probability of a person having disease is P(F)=1/10000=0.0001

The probability of a person having disease and test positive is P(E|F)=0.999

The probability of a person not having disease and testing positive is P(E|F')=0.0002

Now by using complement rule, the value of people not having the disease is P(F')=1-P(F)=1-0.0001=0.9999

Part a:

The probability that someone tests positive and does not have the disease is given as P(F'|E) by Bayes theorem is as

$P(F'|E)=\dfrac{P(E|F')P(F')}{P(E|F)P(F)+P(E|F')P(F')}$

By substitution of the values

$P(F'|E)=\dfrac{P(E|F')P(F')}{P(E|F)P(F)+P(E|F')P(F')}\\P(F'|E)=\dfrac{(0.0002)(0.9999)}{(0.9999)(0.0001)+(0.0002)(0.9999)}\\P(F'|E)=\dfrac{0.00019998}{0.00029997}\\P(F'|E)=0.6667$

The probability that someone tests positive and does not have the disease is given as P(F'|E) is 0.6667

Part b

The probability that someone tests positive and does not have the disease is given as P(F|E') by Bayes theorem is as

$P(F|E')=\dfrac{P(E'|F)P(F)}{P(E'|F)P(F)+P(E'|F')P(F')}$

The probabilities are calculated as

P(E'|F)=1-P(E|F)=1-0.999=0.001

P(E'|F')=1-P(E|F')=1-0.0002=0.9998

By substitution of the values

$P(F|E')=\dfrac{P(E'|F)P(F)}{P(E'|F)P(F)+P(E'|F')P(F')}\\P(F|E')=\dfrac{(0.001)(0.0001)}{(0.001)(0.0001)+(0.9998)(0.9999)}\\P(F|E')=\dfrac{0.0000001}{0.99970012}\\P(F|E')=0.000000103$

The probability that someone tests negative and have the disease is given as P(F|E') is 0.000000103

6 0

4 years ago