Solve and check please help ty

The probability that a person with certain symptoms has hepatitis is 0.75. The blood test used to confirm this diagnosis gives p

emmasim [6.3K]

Answer:

0.981

Step-by-step explanation:

We have to find the probability that an individual who has the symptoms and who reacts positively to the test has hepatitis

Let $E_1$ be the event that denotes the symptoms has hepatitis and $E_2$ denotes the symptoms have not hepatitis

Let A be the event the denotes the blood test result positive

We have to find the value of $P(E_1/A)$

We have $P(E_1)=0.75$

$P(E_2)=1-0.75=0.25$

$P(A/E_1)=86%=0.86$

$P(A/E_2)=5%=0.05$

Using formula

$P(E_1/A)=\frac{P(E_1)\times P(A/E_1)}{P(E_1)\cdotP(A/E_1)+P(E_2)\cdotP(A/E_2}$

Substitute the values in the given formula

Then,we get

$P(E_1/A)=\frac{0.75\times .86}{0.75\times0.86+0.25\times 0.05}$

$P(E_1/A)=\frac{0.645}{0.6575}=0.9809$

$P(E_1/A)=0.981$

Hence, the probability that an individual who has the symptoms and who reacts positively to the test actually has hepatitis =0.981

5 0

2 years ago

Factor the Higher degree polynomial<br><br> 5y^4 + 11y^2 + 2

kotykmax [81]

$\bf 5y^4+11y^2+2\implies 5(y^2)^2+11y^2+2\implies (5y^2+1)(y^2+2)$

8 0

3 years ago

Read 2 more answers

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

Ostrovityanka [42]

Answer:

2.25

Step-by-step explanation:

The computation of the number c that satisfied is shown below:

Given that

$f(x) = \sqrt{x}$

Interval = (0,9)

According to the Rolle's mean value theorem,

If f(x) is continuous in {a,b) and it is distinct also

And, f(a) ≠ f(b) so its existance should be at least one value

i.e

$f^i(c) = \frac{f(b) - f(a)}{b -a }$

After this,

$f(x) = \sqrt{x} \\\\ f^i(x) = \frac{1}{2}x ^{\frac{1}{2} - 1} \\\\ = \frac{1}{2}x ^{\frac{-1}{2}$

$f^i(x) = \frac{1}{{2}\sqrt{x} } = f^i(c) = \frac{1}{{2}\sqrt{c} } \\\\\a = 0, f (a) = f(o) = \sqrt{0} = 0 \\\\\ b = 9 , f (b) = f(a) = \sqrt{9} = 3\\$

After this,

Put the values of a and b to the above equation

$f^i(c) = \frac{f(b) - f(a)}{b - a} \\\\ \frac{1}{{2}\sqrt{c} } = \frac{3 -0}{9-0} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{3}{9} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{1}{3} \\\\ \sqrt[2]{c} = 3\\\\\sqrt{c} = \frac{3}{2} \\\\ c = \frac{9}{4}$