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3 years ago

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A pharmaceutical company has developed a test for a scarce disease that is present in 0.5% of the population. The test is 98% ac

curate in determining a positive result, and the chance of a false positive is 4%. What is the probability that someone who tests positive actually has the disease?

2 answers:

marysya [2.9K]3 years ago

7 0

I'm pretty sure it is 3 percent. I may be wrong.

ioda3 years ago

5 0

The probability is 3%

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Janet has a bag of marbles. The bag contains 10 blue, 5 green, and 1 orange marble. She will randomly select 1 marble from the b

Musya8 [376]

Answer:

1/16

Step-by-step explanation:

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3 years ago

Forty percent of the 30 members of the drama club were members of the senior class

andriy [413]

10%of30=3

3x4=12

A. 40%of 30=12 senior class members.

B. 60% of the class were not senior class members

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2 years ago

Read 2 more answers

How do i solve that question?

yawa3891 [41]

a) The solution of this <em>ordinary</em> differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the <em>ordinary</em> differential equation is $-\frac{1}{x}$ .

The <em>particular</em> solution of the <em>ordinary</em> differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ <em>, </em>then the integration constant is $C = -2$ .

The solution of this <em>ordinary</em> differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the <em>ordinary</em> differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The <em>particular</em> solution of the <em>ordinary</em> differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

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3 years ago

Identify the mean of each distribution. Distribution A Mean = Distribution B Mean = Distribution C Mean =.

mariarad [96]

Answer:

50

100

200

B,C,A

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Write -33/12 as a mixed number in reduced form. -2 -1 -2 -2

AysviL [449]

-33/12. Lets forget the negative symbol.

12 goes into 33, 2 times with 9 remaining.

So now get the negative sign back.

It will be -2 9/12 but you can reduce that to -2 3/4.

Hope this helps :)

7 0

3 years ago