Respuesta :
The number of fish decreases by x% each year. x% can also be written as 0.01x.
If the total number of fish in a lake is A, after one year the number of fish will be:
[tex] A_{1}=A(1-0.01x) [/tex]
After two years the number of fish will be:
[tex] A_{2}=A(1-0.01x)^{2} [/tex]
So, the general formula for the number of fish after n years can be written as:
[tex] A_{n}=A(1-0.01x)^{n} [/tex]
It is given that after 8 years, the number of fish is reduced to half. So we can write:
[tex]0.5A=A(1-0.01x)^{8} \\ \\ 0.5=(1-0.01x)^{8} \\ \\ 0.917=1-0.01x \\ \\ 0.01x=0.083 \\ \\ x=8.3[/tex]
This means the value of x is 8.3%. So, the number of fish in a lake decreases by 8.3% each year
If the total number of fish in a lake is A, after one year the number of fish will be:
[tex] A_{1}=A(1-0.01x) [/tex]
After two years the number of fish will be:
[tex] A_{2}=A(1-0.01x)^{2} [/tex]
So, the general formula for the number of fish after n years can be written as:
[tex] A_{n}=A(1-0.01x)^{n} [/tex]
It is given that after 8 years, the number of fish is reduced to half. So we can write:
[tex]0.5A=A(1-0.01x)^{8} \\ \\ 0.5=(1-0.01x)^{8} \\ \\ 0.917=1-0.01x \\ \\ 0.01x=0.083 \\ \\ x=8.3[/tex]
This means the value of x is 8.3%. So, the number of fish in a lake decreases by 8.3% each year
Answer: Hence, the value of x is 8.3%.
Step-by-step explanation:
Since we have given that
The number of fish in a lake decreases by x% each year.
Given : The number of fish halves in 8 years.
Let N be the number after 8 years.
So, our equation becomes:
[tex]\dfrac{N}{2}=N(1-\dfrac{x}{100})^8\\\\\dfrac{1}{2}=(1-0.01x)^8\\\\0.5=(1-0.01x)^8\\\\\sqrt[8]{0.5}=1-0.01x\\\\0.917=1-0.01x\\\\0.917-1=-0.01x\\\\-0.0829=-0.01x\\\\\dfrac{0.0829}{0.01}=x\\\\8.3\%=x[/tex]
Hence, the value of x is 8.3%.
