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STALIN [3.7K]
3 years ago
7

In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of

age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that an adult over 40 years of age is diagnosed as having cancer?
Mathematics
1 answer:
makkiz [27]3 years ago
3 0

Answer: Our required probability is 0.406.

Step-by-step explanation:

Since we have given that

Probability of selecting an adult over 40 years of age with cancer = 0.05

Probability of a doctor correctly diagnosing a person with cancer as having the disease = 0.78

Probability of incorrectly diagnosing a person without cancer as having the disease = 0.06

Let A be the given event i.e. adult over 40 years of age with cancer. P(A) = 0.05.

So, P(A')=1-0.05 = 0.95

Let C be the event that having cancer.

P(C|A)=0.78

P(C|A')=0.06

So, using the Bayes theorem, we get that

P(A|C)=\dfrac{P(A).P(C|A)}{P(A).P(C|A)+P(A')P(C|A')}\\\\P(A|C)=\dfrac{0.78\times 0.05}{0.78\times 0.05+0.06\times 0.95}\\\\P(A|C)=0.406

Hence, our required probability is 0.406.

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Kryger [21]

Answer:

Infinite amount of solutions.

General Formulas and Concepts:

  • Order of Operations: BPEMDAS
  • Regular + Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

6y + 4 - 3y - 7 = 3(y - 1)

<u>Step 2: Solve for </u><em><u>y</u></em>

  1. Combine like terms:                         3y - 3 = 3(y - 1)
  2. Distribute 3:                                       3y - 3 = 3y - 3
  3. Subtract 3y on both sides:               -3 = -3

Here we see that there will be infinite amount of solutions. We can plug in any number <em>y</em> and it will render the equation true.

8 0
2 years ago
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HEY CAN ANYONE HELP ME!
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Answer:

Given: In triangle ABC and triangle DBE where DE is parallel to AC.

In ΔABC and ΔDBE

DE || AC   [Given]

As we know, a line that cuts across two or more parallel lines.  In the given figure, the line AB is a transversal.

Line segment  AB is transversal that intersects two parallel lines.  [Conclusion from statement 1.]

Corresponding angles theorem: two parallel lines are cut by a transversal, then the  corresponding angles are congruent.

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\angle BDE \cong \angle BAC  and

\angle BEC \cong \angle BCA

Reflexive property of equality states that if angles in geometric figures can be congruent to themselves.

by Reflexive property of equality:

\angle B \cong \angle B  

By AAA (Angle Angle Angle) similarity postulates states that all three pairs of corresponding angles are the same then, the triangles are similar

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\triangle ABC \sim \triangle DBE

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Statement                                                                   Reason

3.\angle BDE \cong \angle BAC      Corresponding angles theorem

and  \angle BEC \cong \angle BCA        

5. \triangle ABC \sim \triangle DBE   AAA similarity postulates    

6. BD over BA                                                    Definition of similar triangle



7 0
2 years ago
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