a) The initial number of infections refers to the time t = 0, not 1. At the start, there are
[tex]P(0) = \dfrac{700}{1 + 49e^{-0}} = \dfrac{700}2 = \boxed{350}[/tex]
infected students.
b) Here you look for the time t such that P(t) = 600. We have
[tex]\dfrac{700}{1+49e^{-0.3t}} = 600[/tex]
[tex]\dfrac{700}{600} = 1+49e^{-0.3t}[/tex]
[tex]\dfrac76 = 1+49e^{-0.3t}[/tex]
[tex]\dfrac16 = 49e^{-0.3t}[/tex]
[tex]\dfrac1{294} = e^{-0.3t}[/tex]
[tex]\ln\left(\dfrac1{294}\right) = \ln\left(e^{-0.3t}\right)[/tex]
[tex]-\ln(294) = -0.3t \ln(e)[/tex]
[tex]\ln(294) = \dfrac3{10}t[/tex]
[tex]t = \dfrac{10}3 \ln(294) \approx 18.945[/tex]
Rounding up, the school will close after around 19 days.
