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3 years ago

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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

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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Step-by-step explanation:

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

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

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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,

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

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

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