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Marina86 [1]
2 years ago
13

the number of subscribers f(t) to a newspaper after t years is shown by the equation below f(t) = 75(.95)^t

Mathematics
1 answer:
Julli [10]2 years ago
4 0
The number of subscribers y to a newspaper after t years is shown by the equation , where 75 is the initial number of subscribers to a newspaper and t is the time period is years.
The above equation can be rewrite as
which is equivalent to the exponential decay function with rate r=0.05=5%
⇒ The number of subscribers to a newspaper decreased by 5% every year.
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One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P
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Step-by-step explanation:

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Solving the diferential equation

\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}

P(0)=  \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)=  \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447

Then the model for the population infected at time t is:

P(t)=\frac{1}{1-0.99e^{t/447}}

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P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533

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