- #1
nokia8650
- 219
- 0
Question:
The number of bacteria present in a culture at time t hours is modeled by the
continuous variable N and the relationship
N = 2000e^(kt)
where k is a constant.
Given that when t = 3, N = 18 000, find
(a) the value of k to 3 significant figures
(b) how long it takes for the number of bacteria present to double, giving your
answer to the nearest minute
k is found to be 0.73241 to 5s.f
For the next part, i simply let N=18000*2=36000, and solved to find t=3.94 therefore it takes 3.94-3 = 0.94 hours for the number to double. This obviously yields the correct answer, however the markscheme let N=4000 subsequently solved to give 0.94 immediately (without having to subtract 3). Can someone please explain the logic behind the markschemes method?
Thanks
The number of bacteria present in a culture at time t hours is modeled by the
continuous variable N and the relationship
N = 2000e^(kt)
where k is a constant.
Given that when t = 3, N = 18 000, find
(a) the value of k to 3 significant figures
(b) how long it takes for the number of bacteria present to double, giving your
answer to the nearest minute
k is found to be 0.73241 to 5s.f
For the next part, i simply let N=18000*2=36000, and solved to find t=3.94 therefore it takes 3.94-3 = 0.94 hours for the number to double. This obviously yields the correct answer, however the markscheme let N=4000 subsequently solved to give 0.94 immediately (without having to subtract 3). Can someone please explain the logic behind the markschemes method?
Thanks