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Suppose there was a cancer diagnostic test was 95% accurate both on those that do and those do not have the disease. If 0.4% of

Musya8 [376]

Complete Question

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.

Answer:

The probability is $P(C | A) = 0.0042$

Step-by-step explanation:

From the question we are told that

The probability that the test was accurate given that the person has cancer is

$P(A | C) = 0.95$

The probability that the test was accurate given that the person do not have cancer is

$P(A | C') = 0.90$

The probability that a person has cancer is

$P(C) = 0.004$

Generally the probability that a person do not have cancer is

$P(C') = 1- P(C)$

=> $P(C') = 1- 0.004$

=> $P(C') = 0.996$

Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as

$P(C | A) = \frac{P(A | C) * P(C)}{P(A|C) * P(C) + P(A| C') * P(C')}$

=> $P(C | A) = \frac{ 0.95 * 0.004 }{ 0.95 * 0.004 + 0.90 * 0.996}$

=> $P(C | A) = 0.0042$

7 0

2 years ago

a windscreen wiper of a vehicle of length 30 cm clears out an angle of 180 degrees what is the area of the screen cleared take p

zavuch27 [327]

Answer:

Therefore the area of the screen cleared is 1414.28 cm².

Step-by-step explanation:

Given:

a windscreen wiper of a vehicle of length 30 cm

Consider this length as Radius of Semicircle, r = 30 cm

clears out an angle of 180 degrees implies Semicircle Shape

pie=22/7

To Find:

Area of the screen cleared

= ?

Solution:

We have

$\textrm{Area of Semicircle}=\frac{1}{2}\pi(radius)^{2}$

Substituting r and pie we get

$\textrm{Area of Semicircle}=\frac{1}{2}\times \frac{22}{7}(30)^{2}=1414.28\ cm^{2}$

Therefore the area of the screen cleared is 1414.28 cm².

7 0

3 years ago

Determine the maximum or minimum value of the following quadratic function. LaTeX: f\left(x\right)=-3\left(x+5\right)^2-1f ( x )

Genrish500 [490]

Answer:

Step-by-step explanation:

$y=-3(x^2+10x+25)-1\\ \\ y=-3x^2-30x-76\\ \\ dy=-6x-30\\ \\ d^2y=-6\\ \\ \text{Since the second derivative is always negative, when the first derivative is equal to zero the function will be at a global maximum.}\\ \\ dy=0=-6x-30\\ \\ x=-5\\ \\ f(-5)=-1\\ \\ \text{So the global maximum occurs at the point (-5,-1)$