The probability according to the given scenario is:
(a) <em>0.0395</em>
(b) <em>0.95</em>
(c) <em>0.076</em>
The given values are:
(a)
The probability that the b/w has tuberculosis as well as having (+) test will be:
→ ![P(A \cap B) = P(B/A).P(A)](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%20P%28B%2FA%29.P%28A%29)
By substituting the values, we get
→ ![= 0.79\times 0.05](https://tex.z-dn.net/?f=%3D%200.79%5Ctimes%200.05)
→ ![= 0.0395](https://tex.z-dn.net/?f=%3D%200.0395)
(b)
The probability that a person does not have tuberculosis will be:
→ ![P(A') = 1-P(A)](https://tex.z-dn.net/?f=P%28A%27%29%20%3D%201-P%28A%29)
→ ![=1-0.05](https://tex.z-dn.net/?f=%3D1-0.05)
→ ![= 0.95](https://tex.z-dn.net/?f=%3D%200.95)
(c)
The person does not have tuberculosis as well as have a (+) test, the probability will be:
→ ![P(A' \cap B) = P(A').P(B/A')](https://tex.z-dn.net/?f=P%28A%27%20%5Ccap%20B%29%20%3D%20P%28A%27%29.P%28B%2FA%27%29)
→ ![= 0.95\times 0.08](https://tex.z-dn.net/?f=%3D%200.95%5Ctimes%200.08)
→ ![= 0.076](https://tex.z-dn.net/?f=%3D%200.076)
Learn more:
brainly.com/question/14816227
Answer:
![y\leq 4](https://tex.z-dn.net/?f=y%5Cleq%204)
Step-by-step explanation:
Start with the given
![2y\leq 8\\](https://tex.z-dn.net/?f=2y%5Cleq%208%5C%5C)
Divide the two on both sides to get
![y\leq 4](https://tex.z-dn.net/?f=y%5Cleq%204)
If we begin with the first term, 2, mult. it by 4 and then subtract 3, we get 5 (not 4, as shown).
If we begin with 4, mult. it by 4 and then subtract 3, we get 13. This agrees with the terms of the given sequence.
If we begin with 13, mult. it by 4 and then subtract 3, we get 49. This agrees with the terms of the given sequence.
Remove the first term, 2, and then the remaining terms follow the given procedure for finding terms.