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Conservationists tagged 120 black-nosed rabbits in a national forest in 1990. In 1991, they tagged 240 black-nosed rabbits in the same range. If the rabbit population follows the exponential law, how many rabbits will be in the range 9 years from 1990?

Answer :

The total number of rabbits in the range 9 years from​ 1990 is 61414 and this can be determined by using the given data.

Given :

  • Conservationists tagged 120 ​black-nosed rabbits in a national forest in 1990.
  • In 1991​, they tagged 240 ​black-nosed rabbits in the same range.
  • The rabbit population follows the exponential​ law.

The below formula can be used in order to determine the total number of rabbits that will be in the range of 9 years from​ 1990.

[tex]\rm A=A_0 \times e^{kt}[/tex] --- (1)

where A is the initial value, [tex]\rm A_0[/tex] is the previous value, k is the exponential constant, and t is the time.

Now, substitute the values of the known terms in the above formula in order to determine the value of 'k'.

[tex]\rm 240 = 120e^{k\times 1}[/tex]

[tex]2=e^{k}[/tex]

Take log on both sides in the above expression.

ln2 = k

k = 0.6931

Now, at t = 9 the expression (1) becomes:

[tex]\rm A = 120e^{0.6931\times 9}[/tex]

[tex]\rm A=120e^{6.2379}[/tex]

A = 61414 rabbits

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https://brainly.com/question/25277954

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Rewritten by : Barada

Answer:

Hence after period of 9 years from 1990 t0 1999 will be 61438 rabbits.

Step-by-step explanation:

Given:

Population for rabbit obeys exponential law.

120 at 1990 and 240 1991 ...after 1 year time period

To Find:

After 9 year time period how many rabbits will be there.

Solution:

Exponential law goes on present value and various value and time period and defined as ,

let Y be present value Y0 previous year value and k exponential constant and t be time period.

So

Y=Y0e^(kt)

Here Y=240 ,Y0=120 t=1 year time period

So

240=120e^(k*1)

240/120=e^k

2=e^k

Now taking log on both side, [natural log]

ln(2)=ln(e^k)

ln(2)=kln(e)

k=ln(2)

k=0.6931

For t=9 year of time period

Y0=120, t=9 ,k=0.6931

Y=Y0e^(k*t)

Y=120*e^(0.6931*9)

=120e^6.2383

=61438.48

=61438 rabbits