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3 years ago

Six oranges cost 5.34 how much does it cost for one orange

Mathematics

1 answer:

Mila [183]3 years ago

6 0

6/5.4=1.1111

hop this helps

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218,457 to the nearest hundred thousands

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-3/2

Step-by-step explanation:

Slope is defined as $\frac{rise}{run}$ ,

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The line goes over 4, making the run: 4

So -6/4 or -3/2

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The simple interest on $600.00 at 5% per year for two years is?

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2 years ago

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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

Darya [45]

Answer:

The differential equation for the model is

$\frac{dP}{dt}=kP(1-P)$

The model for P is

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

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

$\frac{dP}{dt}=kP(1-P)$

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

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:

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

Now, we can calculate t for P(t)=90,000

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

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

8 0

3 years ago