Can someone please help me with these two questions?

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

Step-by-step 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

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