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Fynjy0 [20]
3 years ago
14

Tess ran 4 3 3 4 ​ start fraction, 4, divided by, 3, end fraction kilometers each day for 2 22 days. Cam ran 2 3 3 2 ​ start fra

ction, 2, divided by, 3, end fraction of a kilometer each day for 5 55 days. Who ran more kilometers in total? Choose 1 answer: Choose 1 answer: (Choice A) A Tess (Choice B) B Cam (Choice C) C They ran the same number of total kilometers. Stuck
Mathematics
1 answer:
mestny [16]3 years ago
7 0

We have been given that Tess ran \frac{4}{3} kilo-meters each day for 2 days.

Let us find total distance traveled by Tess in 2 days by multiplying \frac{4}{3} by 2 as:

\text{Distance traveled by Tess in 2 days }=\frac{4}{3}\times 2

\text{Distance traveled by Tess in 2 days }=\frac{4\times 2}{3}

\text{Distance traveled by Tess in 2 days }=\frac{8}{3}

We are also told that Cam ran  \frac{2}{3} kilo-meters each day for 5 days.

Let us find total distance traveled by Cam in 5 days by multiplying \frac{2}{3} by 5 as:

\text{Distance traveled by Cam in 5 days }=\frac{2}{3}\times 5

\text{Distance traveled by Cam in 5 days }=\frac{2\times 5}{3}

\text{Distance traveled by Cam in 5 days }=\frac{10}{3}

Since \frac{10}{3} km is greater than \frac{8}{3} km, therefore, Cam ran more kilo-meters in total and option B is the correct choice.

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{The number, "3", occurs FOUR (4) times.  The number, "4", occurs ONE (1) time.  The number, "5", occurs TWO (2) times.  The number, "6", occurs ONE (1) time.  The number, "9", occurs ONE (1) time.}.
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Answer:

Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

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The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

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Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

\frac{500-200}{200} = 4e^{-0.04t}.

This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

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