The answer is letter C.) 603.40
The minimum required score can solved using the equation
<span>MRS = mean + (zscore)(SD)
where z score of 0.03 is 1.88
</span>MRS <span>= </span>500 + (1.88)(55)
MRS = <span>603.40
</span>
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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
Answer:
f(2)=3×2+5
=11
Step-by-step explanation:
plz make me brainliest
Answer: I believe the answer is B