Respuesta :
[tex]L(x)=\frac{325000000}{1+3249999e^{-0.2x}}=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}}[/tex]
Where x is the time, as you know
As for step 2:
First, we determine half of the total population that can get infected, this is simply:
[tex]\frac{325000000}{2}=1.625\times10^8[/tex]I'll use scientific notation since it is much easier to work in this way
[tex]\Rightarrow L(half_{population})=1.625\cdot10^8=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}}[/tex][tex]\begin{gathered} \Rightarrow1+3.249999\cdot10^6e^{-0.2x}=\frac{3.25\cdot10^8}{1.625\cdot10^8}=2 \\ \Rightarrow3.249999\cdot10^6e^{-0.2x}=1 \\ \Rightarrow e^{-0.2x}=3.249999\cdot10^{-6} \end{gathered}[/tex]Remember the next property of the exponential function:
[tex]\ln (e^x)=x;x\in\mathfrak{\Re }[/tex][tex]\begin{gathered} \Rightarrow-0.2x=\ln (3.249999\cdot10^{-6}) \\ \Rightarrow x=-\ln (3.249999\cdot10^{-6})\frac{1}{0.2}\approx6.32 \end{gathered}[/tex]So, in 6.32 days it is expected that half the population will be infected.
Regarding Part C, which is quite similar to Part B, we only need to exchange 50% by 80%
80% of the population corresponds to:
[tex]\text{Total}_{\text{population}}\cdot0.8=2.6\cdot10^8[/tex][tex]\Rightarrow L(80percent)=2.6\cdot10^8=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}}[/tex][tex]\begin{gathered} \Rightarrow1+3.249999\cdot10^6e^{-0.2x}=\frac{3.25\cdot10^8}{2.6\cdot10^8}=1.25 \\ \Rightarrow e^{-0.2x}=1.25\cdot3.249999\cdot10^{-6} \\ \Rightarrow x=-\frac{1}{0.2}\ln (1.25\cdot3.249999\cdot10^{-6}) \end{gathered}[/tex]Finally,
[tex]x\approx62.07[/tex]The point of no return will be reached in 62 days, approximately
Now, regarding part 4
So, 100 people left
If there are only 100 people, then there are 3.25x10^8-100 zombies.
[tex]\Rightarrow L(100people)=3.25\cdot10^8-100=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}}[/tex][tex]\begin{gathered} \Rightarrow3.25\cdot10^8-100=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}} \\ \Rightarrow1+3.249999\cdot10^6e^{-0.2x}=\frac{3.25\cdot10^8}{3.25\cdot10^8-100}\approx1.00000031 \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow3.249999\cdot10^6e^{-0.2x}=0.00000031=3.1\cdot10^{-7} \\ \Rightarrow e^{-0.2x}\approx9.54\cdot10^{-14} \end{gathered}[/tex][tex]\Rightarrow x=-\frac{1}{0.2}\cdot\ln (9.54\cdot10^{-14})\approx149.904[/tex]Therefore, 100 humans will be left after approximately 150 days
About the process, the only thing we did was to substitute the value of L(x) for the corresponding percentage (50%,80%,100 humans left) and solve for x.
You can interpret the function in this way:
[tex]L(x)=\frac{3.25\cdot10^8}{1+3.249999\cdot10^6e^{-0.2x}}=\frac{a}{1+be^{-0.2x}^{}}=a(1+be^{-0.2x})^{-1}[/tex]The only notable factor is the exponential function, as x increases, we obtain:
[tex]\begin{gathered} x\rightarrow\infty \\ \Rightarrow-0.2x\rightarrow-\infty \\ \Rightarrow e^{-0.2x}\rightarrow e^{-\infty}\rightarrow0 \end{gathered}[/tex]However, that function is never zero. It's only a limit
So, there will always remain at least a fraction of a human left.
That from a mathematical point of view, the physical reality would be different.
Otras preguntas
