Asked by mwk please help
On a college campus of 10,000 students, a single student relurned to campus infected by a discase. The spread of the disease through the student body is given by y = 10000/1+9999e^0.99t where y is the total number infected at time r (in days). (a) How many are infected after 4 days? (b) The school will shut down if 50% of the students are ill. During what day will it close?
Answers
Answered by
oobleck
well, you have the formula, so plug in your numbers.
But I dislike the formula, since it starts at 1 and then decreases. I suspect you meant
10000/(1+9999e^(-0.99*4))
That starts at 1 and then increases.
(a) 10000/(1+9999e^(-0.99*4)) = 0.019065 = 52
(b) 10000/(1+9999e^(-0.99t)) = 5000
1+9999e^(-0.99t) = 2
e^(-.99t) = 1/9999
-.99t = -ln9999
t = 9.3
But I dislike the formula, since it starts at 1 and then decreases. I suspect you meant
10000/(1+9999e^(-0.99*4))
That starts at 1 and then increases.
(a) 10000/(1+9999e^(-0.99*4)) = 0.019065 = 52
(b) 10000/(1+9999e^(-0.99t)) = 5000
1+9999e^(-0.99t) = 2
e^(-.99t) = 1/9999
-.99t = -ln9999
t = 9.3
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