Answer:
heyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy
Step-by-step explanation:
Answer:
i) Probability that both candidates employed are women = 5/14
ii) Probability that the second candidate is a woman = 5/8
iii) Probability that the first candidate is a woman given that second one is a woman = 4/5
Step-by-step explanation:
Let the probability that a man is employed be P(M) = 3/8
Probability that a woman is employed P(W) = 5/8
a) Probability that both candidates employed are women = (5/8) × (4/7) = 5/14
b) Probability that the second candidate is a woman = (probability that first candidate is a man and second candidate is a woman) + (probability that first candidate is a woman & second candidate is a woman)
= (3/8)(5/7) + (5/8)(4/7) = (15/56) + (20/56) = 35/56 = 5/8
c) Probability that the first candidate is a woman given that second one is a woman
Given that the second candidate was a women, means that the first candidate-women was selected among other four women.
Probability = (4/8)/(5/8) = 4/5
This is a way to work it out
Answer:
![-3y^{2} -y+11](https://tex.z-dn.net/?f=-3y%5E%7B2%7D%20-y%2B11)
Step-by-step explanation:
Collect like terms
![-3y^{2} +7y-8y+11](https://tex.z-dn.net/?f=-3y%5E%7B2%7D%20%2B7y-8y%2B11)
Collect like terms again and you get ![-3y^{2} -y+11](https://tex.z-dn.net/?f=-3y%5E%7B2%7D%20-y%2B11)
Hope this helps ʕ•ᴥ•ʔ
Answer:
a) The half life of the substance is 22.76 years.
b) 5.34 years for the sample to decay to 85% of its original amount
Step-by-step explanation:
The amount of the radioactive substance after t years is modeled by the following equation:
![P(t) = P(0)(1-r)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%281-r%29%5E%7Bt%7D)
In which P(0) is the initial amount and r is the decay rate.
A sample of a radioactive substance decayed to 97% of its original amount after a year.
This means that:
![P(1) = 0.97P(0)](https://tex.z-dn.net/?f=P%281%29%20%3D%200.97P%280%29)
Then
![P(t) = P(0)(1-r)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%281-r%29%5E%7Bt%7D)
![0.97P(0) = P(0)(1-r)^{0}](https://tex.z-dn.net/?f=0.97P%280%29%20%3D%20P%280%29%281-r%29%5E%7B0%7D)
![1 - r = 0.97](https://tex.z-dn.net/?f=1%20-%20r%20%3D%200.97)
So
![P(t) = P(0)(0.97t)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%280.97t%29%5E%7Bt%7D)
(a) What is the half-life of the substance?
This is t for which P(t) = 0.5P(0). So
![P(t) = P(0)(0.97t)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%280.97t%29%5E%7Bt%7D)
![0.5P(0) = P(0)(0.97t)^{t}](https://tex.z-dn.net/?f=0.5P%280%29%20%3D%20P%280%29%280.97t%29%5E%7Bt%7D)
![(0.97)^{t} = 0.5](https://tex.z-dn.net/?f=%280.97%29%5E%7Bt%7D%20%3D%200.5)
![\log{(0.97)^{t}} = \log{0.5}](https://tex.z-dn.net/?f=%5Clog%7B%280.97%29%5E%7Bt%7D%7D%20%3D%20%5Clog%7B0.5%7D)
![t\log{0.97} = \log{0.5}](https://tex.z-dn.net/?f=t%5Clog%7B0.97%7D%20%3D%20%5Clog%7B0.5%7D)
![t = \frac{\log{0.5}}{\log{0.97}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B0.5%7D%7D%7B%5Clog%7B0.97%7D%7D)
![t = 22.76](https://tex.z-dn.net/?f=t%20%3D%2022.76)
The half life of the substance is 22.76 years.
(b) How long would it take the sample to decay to 85% of its original amount?
This is t for which P(t) = 0.85P(0). So
![P(t) = P(0)(0.97t)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%280.97t%29%5E%7Bt%7D)
![0.85P(0) = P(0)(0.97t)^{t}](https://tex.z-dn.net/?f=0.85P%280%29%20%3D%20P%280%29%280.97t%29%5E%7Bt%7D)
![(0.97)^{t} = 0.85](https://tex.z-dn.net/?f=%280.97%29%5E%7Bt%7D%20%3D%200.85)
![\log{(0.97)^{t}} = \log{0.85}](https://tex.z-dn.net/?f=%5Clog%7B%280.97%29%5E%7Bt%7D%7D%20%3D%20%5Clog%7B0.85%7D)
![t\log{0.97} = \log{0.85}](https://tex.z-dn.net/?f=t%5Clog%7B0.97%7D%20%3D%20%5Clog%7B0.85%7D)
![t = \frac{\log{0.85}}{\log{0.97}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B0.85%7D%7D%7B%5Clog%7B0.97%7D%7D)
![t = 5.34](https://tex.z-dn.net/?f=t%20%3D%205.34)
5.34 years for the sample to decay to 85% of its original amount