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3 years ago

Express your answer as a polynomial in a standard form.

Mathematics

1 answer:

Andrej [43]3 years ago

7 0

Here you go! Hope this helps!

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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

What are the possible values of x for the following functions?

stiks02 [169]

What can a denominator never be?

0

so, we need to figure what x can't be... the only way to multiply 2 things together to get 0 is if one or both are zero.

so, what x values make our denominator 0?

to figure this out, we need to set (x-1)(x-2)=0

now we split and solve.

$x - 1 = 0 \\ x - 2 = 0$
so when x is 1 or 2 the function doesnt make sense.

but, x can be every other number and it does, so the answer is
ALL REAL NUMBERS not equal to 1 or 2

7 0

3 years ago

Joe earns $425 a week for working 40 hours and an additional $20 an hour for each hour over 40. Last week, Joe made $485. Write

Tasya [4]

Answer:

Joe worked 3 hours overtime.

Step-by-step explanation:

Let t = number of hours overworked

425 + 20 × t = 485

425 + 20t = 485

20t = 485 - 425

= 60

$\frac{20t}{20} = \frac{60}{20}$

t = 3

$\therefore$ Joe worked 3 hours over time.

Hope this helps :)

6 0

3 years ago

If Andrew can paint 1 house in 4 days, which sentence is true? A. Andrew can paint 9 houses in 27 days. B. Andrew can paint 8 ho

mezya [45]

Answer:

The answer is C.

Step-by-step explanation:

I got this by multiplying the houses by the days it would take to paint to paint one house.

9*4=27

8 0

3 years ago

Read 2 more answers

The life, in years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o

Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

$\bf P(X$

So, the probability that a switch fails in the first year is

$\bf P(X$

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and

$\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}$

where

$\bf \binom{100}{k}$ equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

$\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999$

3 0

3 years ago