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Biologists notice an alarming trend in the walleye population in Lake Erie. Despite efforts, the population is decreasing by 15% each year. They estimate the walleye population was 26 million at the beginning of the current year. Write a sequence that lists the population for the first 4 years using 26 million as the first term.



Walleye populations are considered to be in a rehabilitation state when they are between 15 and 20 million. In what year will the population reach this level? In addition to a harvest restriction, what intervention could be used?


Using the formula [tex]s_n = \frac{a^1 - a^1 (1 + r)^n}{1-(1+r)}[/tex] the sum of the population for the first 10 years? Remember it is decreasing by 15% each year!!

Respuesta :

Answer:

Part A

The population for the first four years are;

Year 1; 26,000,000,

Year 2; 22,100,000

Year 3;  18,785,000

Year 4; 15,967,250

Part B

i)The third year

ii) Provision of spawning environment and the reduction of predators

Part C

The population after 10 years is approximately  139,208,436

Step-by-step explanation:

Part A

The given parameters are;

The rate at which the population is decreasing = 15%

The population of the walleye at the beginning of the current year, P = 26 million

Therefore, we have;

The population of subsequent year = 0.85 × The population of the current year

Therefore, we have a geometric series, with a first term, a = 26 million, and a common ratio, r = 0.85, given as follows;

aₙ = a + ar + ar² + ...

Therefore the first 4 years using 26 million as the first term are;

a₁ = 26 × 10⁶  

a₂ = 26 × 10⁶ × 0.85 = 22,100,000

a₃ = 26 × 10⁶ × 0.85² = 18,785,000

a₄ = 26 × 10⁶ × 0.85³ = 15,967,250

The population for the first four years are;

26,000,000, 22,100,000, 18,785,000, 15,967,250

Part B

i) Given that the walleye population are considered to be in rehabilitation state when they are between 15 and 20 million, we have;

The population in the third year = 18,750,000 < 20,000,000

Therefore, the population enters rehabilitation state in the 3rd year

ii) Other intervention method includes provision of a adequate environment for spawning and the removal of predators

Part C

The formula for the sum of the geometric progression is presented as follows;

[tex]S_n = \dfrac{a^1 - a^1 \cdot (1 + r)^n}{1 - (1 + r)}[/tex]

Therefore, we have;

[tex]S_n = \dfrac{26 - 26 \cdot (1 - 0.15)^{10}}{1 - (1 - 0.15)} = 139.208436581[/tex]

The population after 10 years, Sₙ = 139.208436581 × 10⁶ ≈ 139,208,436

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