Asked by need help asap!! Thanks!!!
(need help answering #3 only!!! for the first two problems I included the answers)
The population of bacteria in one cubic centimeter of the blood of a sick person has been modeled by the function P(t)=115t(0.75t) where t is the time, in days, since the person became ill.
Use your calculator to graph the function.
Then use the graph to answer the following questions:
1. To the nearest day, when is the bacteria population at a maximum? Day: 3
2. What is the maximum population? Round your answer to one decimal place. Population: 147.059
3. Estimate how fast the population is changing 14 days after the onset of the illness. Round your answer to two decimal places. Rate of Change:
The population of bacteria in one cubic centimeter of the blood of a sick person has been modeled by the function P(t)=115t(0.75t) where t is the time, in days, since the person became ill.
Use your calculator to graph the function.
Then use the graph to answer the following questions:
1. To the nearest day, when is the bacteria population at a maximum? Day: 3
2. What is the maximum population? Round your answer to one decimal place. Population: 147.059
3. Estimate how fast the population is changing 14 days after the onset of the illness. Round your answer to two decimal places. Rate of Change:
Answers
Answered by
mathhelper
I have a valid suspicion that P(t)=115t(0.75t)
is supposed to be P(t)=115t(0.75^t), your calculations reflected that.
P'(t) = (115t)(.75^t)(ln.75) + 115(.75^t)
= 0 for a max
(115t)(.75^t)(ln.75) = -115(.75^t)
tln.75 = -1
t = -1/ln.75 = 3.48 , or 3 to the nearest day, you had that, Great!
2) you are correct, I had the same correct to 3 decimals
3) Don't know what your course understands by "estimate".
To me it means: do your work in your head with as little "paperwork" as
possible. Definitely no calculator. I don't know how that would be possible here.
rough work:
(100t)(3/4)^14 (ln.75) + 100(3/4)^t
actual answer: put t = 14 into the derivative
is supposed to be P(t)=115t(0.75^t), your calculations reflected that.
P'(t) = (115t)(.75^t)(ln.75) + 115(.75^t)
= 0 for a max
(115t)(.75^t)(ln.75) = -115(.75^t)
tln.75 = -1
t = -1/ln.75 = 3.48 , or 3 to the nearest day, you had that, Great!
2) you are correct, I had the same correct to 3 decimals
3) Don't know what your course understands by "estimate".
To me it means: do your work in your head with as little "paperwork" as
possible. Definitely no calculator. I don't know how that would be possible here.
rough work:
(100t)(3/4)^14 (ln.75) + 100(3/4)^t
actual answer: put t = 14 into the derivative
Answered by
need help asap!! Thanks!!!
Thank you so much for your help!!!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.