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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

john is knitting a blanket. he records the length he knits each day. how many more inches does john need to knit so the blanket

sammy [17]

Answer:

60 - 16 = 44

He has to knit 44 more inches so the blanket is 6o inches long.

Hope this helps <3

6 0

3 years ago

Trimble Graphic Design receives $2,250 from a client billed in a previous month for services provided. Which of the following ge

krok68 [10]

Answer:

Dr Cash/Bank $2,250

Cr Account Receivable $2,250

Step-by-step explanation:

At the time the transaction was carried out, that is the previous month, the accounting entries was a debit to account receivable (client account) and a credit to sales account. which is as shown below:

Dr Account Receivable $2,250

Cr Sales $2,250

Being sales on credit to a client.

The client account will have an outstanding debit balance of $2,250 until payment is made. once payment is made, the following entries will be made to ensure no outstanding balance in the client account.

Dr Cash/Bank $2,250

Cr Account Receivable $2,250.

7 0

3 years ago

The sum of an infinite geometric series is 450, while the common ratio of the series is 4/ 5 . What is the first term of the ser

pogonyaev

Answer:

answer is 90 for first term

Step-by-step explanation:

Let the terms be

First term x

We will use the formula s∞=x/1−r to find the sum of an infinite geometric series, where −1<r<1.

We know the sum and the common ratio, so we'll be solving for x where r =4/5

s∞=x/1−r

450=x/1−4/5

450=x/1/5

450=5x

x=90

this is the first term x1 = 90

we know that common ratio is 4/5, so multiplying the first term by factor 4/5 to get the second term

90 x 4/5= 72 second term

7 0

3 years ago

Read 2 more answers

Given the regular polygon, what is the measure of each numbered angle?

ANEK [815]

Answer:

Answer A

Step-by-step explanation:

Approach 1

B, C, and D are wrong because angles 1 and 2 are congruent.

Therefore your answer of both being 60°

Approach 2

Let x = angle 1.

360/x = 6

x = 60°.

Therefore Angle 1 is 60°

An interior angle of a hexagon is 120°

Let y = angle 2.

y is half of 120°

So 120/y = 2

y = 60°

4 0

2 years ago