There is a difference because in the spring, there may be more days of rain, therefore that would increase the mean for the spring compared to the other days.

5 0

3 years ago

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The differential equation in Example 3 of Section 2.1 is a well-known population model. Suppose the DE is changed to dP dt = P(a

LuckyWell [14K]

Answer:

Decreases

Step-by-step explanation:

We need to determine the integral of the DE;

$dP/dt=P(aP-b)$

$dP=P(aP-b)dt$

$1/(dP^2-bP)dP=dt$

We can solve this by integration by parts on the left side. We expand the fraction 1/P²:

$1/(d-b/P)\cdot{P^2} dP$

let

$u=d-b/P$

$du/dP=b/P^2$

$dP=$ $\int\limits {P^2/b} \, du$

$P=lnu/b$

Substitute u in:

$P=ln(d-b/P)/b$

Therefore the equation is:

$ln(d-b/P)/b=t$

We simplify:

$d-b/P=e^b^t$

$P=b/(d-e^b^t)$

As t increases to infinity P will decrease

6 0

3 years ago

For items 5 - 7 consider the score distribution of 15

Vika [28.1K]

The median is 79. Hope this helps!

7 0

3 years ago

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