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Mathematics, 26.03.2020 22:28 matthewlucas8613

An initial population of fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are fish in the lake after years. (a)Let be the time (in years) since the initial population is introduced, and let be the number of fish at time . Write a formula relating to . Use exact expressions to fill in the missing parts of the formula. Do not use approximations. (b)How many fish are there years after the initial population is introduced

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Answer from: musicaljay8320

Given question is incomplete; find the complete question,

An initial population of 80 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 216 the lake after 12 years.

(a) Let t be the time (in years) since the initial population is introduced and let y be the number of fish at time t. Write a formula relating y to t. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.

(b) How many fish are there 3 years after the initial population is......? Do not round any intermediate computations, and round your answer to the nearest whole number.

Expression for the population growth will be, $y=80e^{0.08277t}$ and there will be 103 fishes in the pond after 3 years.

(a). Given in the question,

Initial population of fish = 80Final population of fish = 216Time taken to reach the final population = 12 years

Expression for the population after time 't' is given by,

$y=P_0e^{rt}$

Here, P_0= Initial population

Duration

Expression for the population,

$y=P_0e^{rt}$

$216=80.e^{12r}$

$e^{12r}=\frac{216}{80}$

$12r[\text{ln(e)}]=\text{ln}(2.7)$

$r=\frac{0.99325}{12}$

r=0.08277

Expression for the population growth will be,

$y=80e^{0.08277t}$

(b). For t = 3 years,

$y=80.e^{0.08277\times 3}$

$=80\times 1.28186$

=102.55

$\approx103$

Therefore, expression for the population growth will be $y=80.e^{0.08277t}$ and population of fish after 3 years will be nearly 103.

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Answer from: miyiab7666

(a) $y=80e^{\dfrac{ln(2.7)t}{12}$

(b)301

Step-by-step explanation:

By the exponential population model,the population P(t) at any time t is given as: $P(t)=P_{0}e^{rt}$ where $P_{0}$ is the initial population.

P(12)=y=216 at t=12 years

$P_{0}=80$

$P(t)=P_{0}e^{rt}$

$216=80e^{12r}\\\frac{216}{80} =e^{12r}$

Taking the natural logarithm of both sides

12r=ln(2.7)

$r=\frac{ln (2.7)}{12}$

$y=80e^{\dfrac{ln(2.7)t}{12}$

(b) After 16 years

At t=16

$y=80e^{\dfrac{ln(2.7)t}{12}}\\y=80e^{\dfrac{ln(2.7)X16}{12}}\\y=300.77$

After 16 years, there are approximately 301 fishes in the lake.

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Answer from: Quest

It creates dogx^2 and then flip your answer and multiply by 2

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Answer from: Quest

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step-by-step explanation:

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