Answer:
here i finished!
hope it helps yw!
Step-by-step explanation:
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs
Answer:
The sample size to obtain the desired margin of error is 160.
Step-by-step explanation:
The Margin of Error is given as

Rearranging this equation in terms of n gives
![n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2](https://tex.z-dn.net/?f=n%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2)
Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as
![n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n](https://tex.z-dn.net/?f=n_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM_2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%2F2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B2%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D2%5E2%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4n)
As n is given as 40 so the new sample size is given as

So the sample size to obtain the desired margin of error is 160.
About 500 times since there is a 50/50 chance
i think that is right