Mathematics: Choose all options that apply. Which of the following are considered whole numbers? a) 0 b) 219 Oc) 6.75 d) 1/2 Oe) 2

(a) (b) (c) 80% of the population will have heard the rumor in about 7.4 hrs.Step-by-step explanation:We know that a rumor spreads according to the equationwhere is the proportion of the population that knows the rumor at time , and and are positive constants.(a) To find you must:Use this fact,Apply this identity, to find (b) To find the rate of speed of the rumor you must find the derivative (c) To find the time that will take for 80% of the population to hear the rumor, you must substitute a =...

Choose all options that apply. Which of the following

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31.07.2021

Mathematics

(a) $\lim_{t \to \infty} \frac{1}{1+ae^{-kt}}=1$

(b) $\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)=\frac{ake^{-kt}}{\left(1+ae^{-kt}\right)^2}$

Step-by-step explanation:

We know that a rumor spreads according to the equation

$p(t)=\frac{1}{1+ae^{-kt}}$

where p(t) is the proportion of the population that knows the rumor at time , and and are positive constants.

(a) To find $\lim_{t \to \infty} p(t)$ you must:

Use this fact,

$\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0$

$\lim_{t \to \infty} \frac{1}{1+ae^{-kt}}\\\\\frac{\lim _{t\to \infty \:}\left(1\right)}{\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)}$

Apply this identity, to find $\lim _{t\to \infty \:}\left(1)$

$\lim _{x\to a}c=c$

$\lim _{t\to \infty \:}\left(1\right)=1$

$\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=\lim _{t\to \infty \:}\left\left(1\right)+\lim _{t\to \infty \:}\left(ae^{-kt}\right)\\\\\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=1+a\cdot \lim _{t\to \infty \:}\left(e^{-kt}\right)\\\\\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=1+0$

$\frac{\lim _{t\to \infty \:}\left(1\right)}{\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)}=\frac{1}{1+0} =1$

(b) To find the rate of speed of the rumor you must find the derivative $\frac{dp}{dt}$

$\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)\\\\\frac{d}{dt}\left(\left(1+ae^{-kt}\right)^{-1}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\\frac{d}{du}\left(u^{-1}\right)\frac{d}{dt}\left(1+ae^{-kt}\right)\\\\\left(-\frac{1}{u^2}\right)\left(-ake^{-kt}\right)\\\\\mathrm{Substitute\:back}\:u=\left(1+ae^{-kt}\right)\\\\\left(-\frac{1}{\left(1+ae^{-kt}\right)^2}\right)\left(-ake^{-kt}\right)$

$\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)=\frac{ake^{-kt}}{\left(1+ae^{-kt}\right)^2}$

(c) To find the time that will take for 80% of the population to hear the rumor, you must substitute a = 10, k = 0.5, and p(t) = 0.8 into $p(t)=\frac{1}{1+ae^{-kt}}$ and solve for t

$\frac{1}{1+10e^{-0.5t}}=0.8\\\\\frac{1}{1+10e^{-0.5t}}\left(1+10e^{-0.5t}\right)=0.8\left(1+10e^{-0.5t}\right)\\\\1=0.8\left(1+10e^{-0.5t}\right)\\\\0.8\left(1+10e^{-0.5t}\right)=1\\\\0.8\left(1+10e^{-0.5t}\right)\cdot \:10=1\cdot \:10\\\\8\left(1+10e^{-0.5t}\right)=10\\\\\frac{8\left(1+10e^{-0.5t}\right)}{8}=\frac{10}{8}\\\\1+10e^{-0.5t}=\frac{5}{4}\\\\1+10e^{-0.5t}-1=\frac{5}{4}-1\\\\10e^{-0.5t}=\frac{1}{4}\\\\\frac{10e^{-0.5t}}{10}=\frac{\frac{1}{4}}{10}$

$e^{-0.5t}=\frac{1}{40}\\\\\ln \left(e^{-0.5t}\right)=\ln \left(\frac{1}{40}\right)\\\\-0.5t\ln \left(e\right)=\ln \left(\frac{1}{40}\right)\\\\-0.5t=\ln \left(\frac{1}{40}\right)\\\\t=2\ln \left(40\right) \approx 7.377 \:hrs$

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Choose all options that apply. Which of the following are considered whole numbers? a) 0 b) 219 Oc) 6.75 d) 1/2 Oe) 2

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