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san4es73 [151]
3 years ago
8

A population of 2000 starfish has a yearly per capita population growth rate of 0.012. By next year, how do you expect populatio

n size to have changed?
Assume that the population grows exponentially according to the equation d N d t = r N

It will increase by 50 starfish
It will increase by 24 starfish
It will increase by 100 starfish
It will increase by 200 starfish
Mathematics
1 answer:
Alona [7]3 years ago
4 0

Answer:

It will increase by 24 starfish

Step-by-step explanation:

The population is modeled by the following differential equation:

\frac{dN}{dt} = rN

Which has the following solution:

N(t) = N(0)e^{rt}

In which N(0) is the initial population and r is the growth rate.

A population of 2000 starfish has a yearly per capita population growth rate of 0.012.

This means that N(0) = 2000, r = 0.012

By next year, how do you expect population size to have changed?

This is N(1).

N(t) = N(0)e^{rt}

N(1) = 2000*e^{0.012}

N(1) = 2024

2024 - 2000 = 24

So the correct answer is:

It will increase by 24 starfish

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In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given
hichkok12 [17]

Answer:

a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so \mu = 1.3

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is P(X > 5)

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

So

P(X \leq 5) + P(X > 5) = 1

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725

P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543

P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303

P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980

P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324

P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949

Finally

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051

There is a 0.51% probability of more than five quakes.

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3 years ago
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