R=S*0.5^(t/8)
<span>R is the remaining amount </span>
<span>S is the starting amount (500) </span>
<span>0.5^ is for the HALF in half-life </span>
<span>t/8 show that every 8 ts (every 8 hours), it will be halved once </span>
<span>...so plug in 500mg for the general solution... </span>
<span>R=(500)*(0.5)^(t/8) </span>
<span>... plug in 24h to solve for after 24h </span>
<span>R=(500)*(0.5)^(24/8) </span>
<span>R=(500)*(0.5)^(3) </span>
<span>R=(500)*(0.125) </span>
<span>R=(0.0625) </span>
<span>...therefore there with be 0.0625 mg of the dose remaining</span>
The probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328
<h3>How to determine the probability?</h3>
The given parameters are:
- Sample size, n = 100
- Population proportion, p = 82%
Start by calculating the mean:
![\mu = np](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np)
![\mu = 100 * 82\%](https://tex.z-dn.net/?f=%5Cmu%20%3D%20100%20%2A%2082%5C%25)
![\mu = 82](https://tex.z-dn.net/?f=%5Cmu%20%3D%2082)
Calculate the standard deviation:
![\sigma = \sqrt{\mu(1 - p)](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cmu%281%20-%20p%29)
![\sigma = \sqrt{82 * (1 - 82\%)](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B82%20%2A%20%281%20-%2082%5C%25%29)
![\sigma = 3.84](https://tex.z-dn.net/?f=%5Csigma%20%3D%203.84)
Within ± 0.02 of the population proportion are:
![x_{min} = 82 * (1 - 0.02) = 80.38](https://tex.z-dn.net/?f=x_%7Bmin%7D%20%3D%2082%20%2A%20%281%20-%200.02%29%20%3D%2080.38)
![x_{max} = 82 * (1 + 0.02) = 83.64](https://tex.z-dn.net/?f=x_%7Bmax%7D%20%3D%2082%20%2A%20%281%20%2B%200.02%29%20%3D%2083.64)
Calculate the z-scores at these points using:
![z = \frac{x - \mu}{\sigma}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D)
So, we have:
![z_1 = \frac{80.36 - 82}{3.84} = -0.43](https://tex.z-dn.net/?f=z_1%20%3D%20%5Cfrac%7B80.36%20-%2082%7D%7B3.84%7D%20%3D%20-0.43)
![z_2 = \frac{83.64 - 82}{3.84} = 0.43](https://tex.z-dn.net/?f=z_2%20%3D%20%5Cfrac%7B83.64%20-%2082%7D%7B3.84%7D%20%3D%200.43)
The probability is then represented as:
P(x ± 0.02) = P(-0.43 < z < 0.43)
Using the z table of probabilities, we have:
P(x ± 0.02) = 0.3328
Hence, the probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328
Read more about probability at:
brainly.com/question/25870256
#SPJ1
Answer: C, 9 units
Step-by-step explanation:
(-4, -3) is new point
The answer is C just trust me