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3 years ago

4. Two families went camping. The Becker family was 4 miles west of the campsite lake. The Dunn family

Mathematics

1 answer:

tresset_1 [31]3 years ago

8 0

They were 1 mile away from each other.

4-3=1 mile

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A rumor spreads through a small town. Let y ( t ) be the fraction of the population that has heard the rumor at time t and assum

Ivan

Answer:

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solution

$y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

The rumor reaches 80% at 8.48 days.

Step-by-step explanation:

We know

y(t): proportion of people that heard the rumor

y'(t)=ky(1-y), rate of spread of the rumor

Differential equation

$\frac{dy}{dt} =ky(1-y)$

Solving the differential equation

$\frac{dy}{y(1-y)}=k\cdot dt \\\\\int \frac{dx}{y(1-y)} =k \int dt \\\\-ln(1-\frac{1}{y} )+C_0=kt\\\\1-\frac{1}{y} =Ce^{-kt}\\\\\frac{1}{y} =1-Ce^{-kt}\\\\y=\frac{1}{1-Ce^{-kt}}$

Initial conditions:

$y(0)=0.2\\y(3)=0.4\\\\y(0)=0.2=\frac{1}{1-Ce^0}\\\\1-C=1/0.2\\\\C=1-1/0.2= -4\\\\\\y(3)=0.4=\frac{1}{1+4e^{-3k}} \\\\1+4e^{-3k}=1/0.4\\\\e^{-3k}=(2.5-1)/4=0.375\\\\k=ln(0.375)/(-3)=0.327\\\\\\y=\frac{1}{1+4e^{-0.327t}}$

Value of constant k=0.327 days^(-1)

At what time the rumor reaches 80%?

$y(t)=0.8=\frac{1}{1+4e^{-0.327t}} \\\\1+4e^{-0.327t}=1/0.8=1.25\\\\e^{-0.327t}=(1.25-1)/4=0.0625\\\\t=ln(0.0625)/(-0.327)=8.48$

The rumor reaches 80% at 8.48 days.

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3 years ago