Respuesta :
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
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
For more information, refer to the link given below:
https://brainly.com/question/25277954
