Question

Compound Interest. Pre calc. Ignore what I have written so far.

Answer 1

a) The initial number of infections refers to the time t = 0, not 1. At the start, there are

$P(0) = \dfrac{700}{1 + 49e^{-0}} = \dfrac{700}2 = \boxed{350}$

infected students.

b) Here you look for the time t such that P(t) = 600. We have

$\dfrac{700}{1+49e^{-0.3t}} = 600$

$\dfrac{700}{600} = 1+49e^{-0.3t}$

$\dfrac76 = 1+49e^{-0.3t}$

$\dfrac16 = 49e^{-0.3t}$

$\dfrac1{294} = e^{-0.3t}$

$\ln\left(\dfrac1{294}\right) = \ln\left(e^{-0.3t}\right)$

$-\ln(294) = -0.3t \ln(e)$

$\ln(294) = \dfrac3{10}t$

$t = \dfrac{10}3 \ln(294) \approx 18.945$

Rounding up, the school will close after around 19 days.