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vlabodo [156]
2 years ago
4

There is considerable evidence to support the theory that for some species there is a minimum population m such that the species

will become extinct if the size of the population falls below m. This condition can be incorporated into the logistic equation by introducing the factor (1−m/P). Thus the modified logistic model is given by the differential equation
dPdt=kP(1−PK)(1−mP)
where k is a constant and K is the carrying capacity.
Suppose that the carrying capacity K=40000, the minimum population m=900, and the constant k=0.1. Answer the following questions.
1. Assuming P≥0 for what values of P is the population increasing. Answer (in interval notation):
2. Assuming P≥0 for what values of P is the population decreasing. Answer (in interval notation):
Mathematics
1 answer:
Simora [160]2 years ago
6 0

Answer:

1. (0, 1/40000) and (1/900, +∞)

2. (1/40000, 1/900).

Step-by-step explanation:

The population is increasing when dP/dt is positive and the population is decreasing when dP/dt is negative.

First, replacing the values of K, k and m, we get that dP/dt are equal to:

dP/dt=0.1P(1-40000P)(1-900P)

To find the intervals when dP/dt is positive, we need to make dP/dt greater than zero as:

dP/dt=0.1P(1-40000P)(1-900P) > 0

So, we have 3 terms in the equation, so:

0.1P > 0 if P > 0

1 - 40000 P > 0   if P < 1/40000

1 - 900 P > 0   if   P < 1/900

We can said that 0.1P is positive if P is positive,  (1 - 40000P) is positive is P is lower than 1/40000 and (1 -900P) is positive is P is lower than 1/900

Therefore, we have the following intervals: (0, 1/40000), (1/40000, 1/900) and (1/900, +∞)

For the interval (0, 1/40000): 0.1P is positive, (1 - 40000P) is positive and (1 -900P) is positive. So we can said that dP/dt is positive.

For the interval (1/40000, 1/900): 0.1P is positive, (1 - 40000P) is negative and (1 -900P) is positive. So we can said that dP/dt is negative.

For the interval (1/900, +∞): 0.1P is positive, (1 - 40000P) is negative and (1 -900P) is negative. So we can said that dP/dt is positive.

Finally, the population increase at values of P between: (0, 1/40000) and (1/900, +∞) and the population decrease at values of P between: (1/40000, 1/900).

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Let

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<u>the answer is</u>

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2 years ago
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antoniya [11.8K]

Answer:

X = 4, 3

Step-by-step explanation:

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hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a
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Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

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B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

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C) The required number of ways is:

60C25 + 60C20 + 60C15

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2 years ago
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Answer:

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Since the printer depreciates at a rate of 5% per year, I believe the stated equation is miss typed. Therefore I'll answer this with the correct equation that would represent that setting:

y(x) = 25,000*0.95^x

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On the second year the value of the printer is:

y(2) = 25,000*0.95^2 = 22,562.5\\

On the third year the value of the printer is:

y(3) = 25,000*0.95^3 = 21,434.38\\

The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.

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