This is Growth and Decay
1. The U.S. populace increases at 0.91% per year. If the population in 2008 was 303,146,000 , what will it be in 2017?
2. A mechanist bought materials for about $125,000. He wants to calculate it's value in ten years. If tye materials depreciate 15% a year, how much will it be worth in ten years
p = 303146000 (1 + .0091)^(2017-2009)
or ... p = 303146000 e^[.0091(2017-2009)]
v = 125000 (1 - .15)^10
To solve these growth and decay problems, you need to apply the appropriate formula and plug in the given values. Here's how you can solve each problem:
1. The population growth problem:
Given:
- Population in 2008 (P₀) = 303,146,000
- Growth rate (r) = 0.91% per year
To find the population in 2017, you need to calculate the final population (P) after 9 years of growth. We can use the formula for exponential growth:
P = P₀ * (1 + r)^n
Where:
- P = Final population
- P₀ = Initial population
- r = Growth rate (in decimal form)
- n = Number of years
Substituting the given values into the formula:
P = 303,146,000 * (1 + 0.0091)^9
Calculating the result will give you the population in 2017.
2. The material depreciation problem:
Given:
- Initial value of materials (V₀) = $125,000
- Depreciation rate (r) = 15% per year
To find the value of the materials in ten years, you need to calculate the final value (V) after 10 years of depreciation. We can use the formula for exponential decay:
V = V₀ * (1 - r)^n
Where:
- V = Final value
- V₀ = Initial value
- r = Depreciation rate (in decimal form)
- n = Number of years
Substituting the given values into the formula:
V = $125,000 * (1 - 0.15)^10
Calculating the result will give you the value of the materials in ten years.
Please note that for both problems, the formulas assume constant growth/decay rates over time.