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 the sum of the population for the first 10 years? Remember it is decreasing by 15% each year!!

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;

Therefore, we have;

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

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05