Answer :
To solve this you need to find the unknown constants B and k.
The information given:
R(0) = 5
R(1) = 149
will allow us to do this.
First plug in initial point R(0) = 5 and solve for B (e^0 = 1)
[tex] \frac{3000}{1+ B} = 5 [/tex]
[tex]1+B = \frac{3000}{5} = 600 [/tex]
[tex] B = 599 [/tex]
Next plug in R(1) = 149 and solve for k (actually only solve for e^k)
[tex] \frac{3000}{1+599e^k} = 149 [/tex]
[tex]e^k = \frac{3000-149}{(599)(149)} [/tex]
Finally we can solve for t when R(t) = 2700.
Plug in the fraction for e^k since e^(kt) = (e^k)^t
[tex]\frac{3000}{1 +599( \frac{3000-149}{(599)(149)})^t } = 2700[/tex]
[tex]599 (\frac{3000-149}{(599)(149)})^t = \frac{3000}{2700} - 1 = \frac{1}{9}[/tex]
[tex]t = \frac{ln ( \frac{1}{(599)(9)}) }{ln(\frac{3000-149}{(599)(149)})} = 2.495[/tex]
The information given:
R(0) = 5
R(1) = 149
will allow us to do this.
First plug in initial point R(0) = 5 and solve for B (e^0 = 1)
[tex] \frac{3000}{1+ B} = 5 [/tex]
[tex]1+B = \frac{3000}{5} = 600 [/tex]
[tex] B = 599 [/tex]
Next plug in R(1) = 149 and solve for k (actually only solve for e^k)
[tex] \frac{3000}{1+599e^k} = 149 [/tex]
[tex]e^k = \frac{3000-149}{(599)(149)} [/tex]
Finally we can solve for t when R(t) = 2700.
Plug in the fraction for e^k since e^(kt) = (e^k)^t
[tex]\frac{3000}{1 +599( \frac{3000-149}{(599)(149)})^t } = 2700[/tex]
[tex]599 (\frac{3000-149}{(599)(149)})^t = \frac{3000}{2700} - 1 = \frac{1}{9}[/tex]
[tex]t = \frac{ln ( \frac{1}{(599)(9)}) }{ln(\frac{3000-149}{(599)(149)})} = 2.495[/tex]
The logistic function is [tex]S[/tex] shaped curve which use to determine the statistical distribution over the energy states of a system.
The time will take for 2,700 student to hear the rumor is 2.495 days.
Given:
The given logistic function is,
[tex]R(t)=\frac{3000}{1+Be^{kt}}[/tex]
As per the question, at initial stage 5 students heard the rumor. At initial stage [tex]t=0[/tex].
Calculate the constant [tex]B[/tex] at initial stage.
[tex]R(t)=\frac{3000}{1+Be^{kt}}\\R(0)=\frac{3000}{1+Be^{k\times 0}}\\R(0)=\frac{3000}{1+B}[/tex]
Substitute [tex]R(0)=5[/tex].
[tex]\begin{aligned}5=\frac{3000}{1+B}\\1+B= \frac{3000}{5}\\B=599\\\end[/tex]
As per the question, after 1 day 149 students heard the rumor. After 1 day stage [tex]t=1[/tex].
[tex]R(t)=\frac{3000}{1+Be^{kt}}\\R(1)=\frac{3000}{1+Be^{k\times 1}}\\R(1)=\frac{3000}{1+B^k}[/tex]
Substitute [tex]R(1)=149[/tex] and [tex]B=599[/tex].
[tex]\begin{aligned}149=\frac{3000}{1+599e^k}\\e^k=\frac{(3000-1149)}{(599)(149)} \\\end{aligned}[/tex]
Calculate the time take to hear rumor by 2,7000 students.
[tex]\begin {aligned}2700=\frac{3000}{1+599(\frac{3000-149}{599\times 149})^t }\\\frac{3000}{2700}-1 =599(\frac{3000-149}{599\times 149})^t }\\t=\frac{ln(\frac{1}{(599)(9)})}{ln(\frac{3000-149}{599\times 149}) }\\t=2.495\:\rm days\\\end{aligned}[/tex]
Thus, the time will take for 2,700 student to hear the rumor is 2.495 days.
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