Answer:
1) Part A
The function that models the situation is P = 8,300·(1 - 0.12)ⁿ
Part B
2626 people
Part C
1575 people
2) Part A
The function that models the situation is A = 342.000·(1 - 0.06)ⁿ
Part B
The value of the hose in 2025 is approximately $208,472.6
Part C
The value of the hose in 2032 is approximately $135,189.8
Explanation:
1) Part A
The given parameters are;
Martianville population in 2012, P₀ = 8,300 people
The percentage annual decrease in the population, r = -12% = -0.12
The formula for the growth of a population is presented as follows;
P = P₀·(1 + r)ⁿ
Where;
P = The population amount at a later time
P₀ = The initial amount of the population
r = The rate of the population growth
n = The number of years
Therefore;
The function that models the situation is P = 8,300·(1 - 0.12)ⁿ
Part B
The population that will attend Martianville in 9 years is given as follows;
P₉ = 8,300 × (1 + (-0.12))⁹ = 8,300 × (1 - 0.12)⁹ ≈ 2626.77
Therefore, by rounding down to the next whole person, the population that will attend Martianville in 9 years, P₉ = 2626 people
Part C
The population that will attend Martianville in 13 years is given as follows;
P₁₃ = 8,300 × (1 - 0.12)¹³ ≈ 1575 people
2) Part A
The price at which the house was bought, A₀ = $342,000
The rate of decrease of the value of the house = 6% therefore, r = -0.06
The function that models the situation is presented as follows;
A = A₀·(1 + r)ⁿ
Therefore;
The function that models the situation is A = 342.000·(1 - 0.06)ⁿ
Part B
The number of years between 2025 and 2017 = 2025 - 2017 = 8 years
The value of the hose in 2025, A₂₀₂₅ = 342.000·(1 - 0.06)⁸ = 208,472.576981
The value of the hose in 2025, A₂₀₂₅ ≈ $208,472.6
Part C
The number of years between 2032 and 2017 = 2032 - 2017 = 15 years
The value of the hose in 2032, A₂₀₃₂ = 342.000·(1 - 0.06)¹⁵ = 135,189.795176
The value of the hose in 2025, A₂₀₃₂ ≈ $135,189.8