Math Problem Statement
P1 = 225417 GR = 2.3 growth_factor = 1 + (GR / 100)
years = range(2006, 2035) population_projection = []
for year in years: P1 = P1 * growth_factor population_projection.append((year, round(P1)))
population_projection
Solution
It looks like you've provided the Python code that calculates the population projection from 2006 to 2034 using the given growth rate and initial population. The code works as expected to generate the projected population for each year.
Here's the output from running that code:
- 2006: 230,602
- 2007: 235,905
- 2008: 241,331
- 2009: 246,882
- 2010: 252,560
- 2011: 258,369
- 2012: 264,312
- 2013: 270,391
- 2014: 276,610
- 2015: 282,972
- 2016: 289,480
- 2017: 296,138
- 2018: 302,949
- 2019: 309,917
- 2020: 317,045
- 2021: 324,337
- 2022: 331,797
- 2023: 339,428
- 2024: 347,235
- 2025: 355,222
- 2026: 363,392
- 2027: 371,750
- 2028: 380,300
- 2029: 389,047
- 2030: 397,995
- 2031: 407,149
- 2032: 416,513
- 2033: 426,093
- 2034: 435,893
This projection shows how the population is expected to grow each year under the given conditions.
Would you like to explore this further, or do you have other questions?
Here are 5 related questions you might consider:
- How does the population change over shorter periods, like 5 or 10 years?
- What is the expected population in a specific year, such as 2022 or 2030?
- How sensitive is the population projection to changes in the growth rate?
- What would be the population in 2034 if the growth rate was different, say 3% instead of 2.3%?
- How does this method compare to other population projection methods, like exponential growth?
Tip: When making long-term projections, it’s helpful to review and adjust the growth rate periodically to account for changing circumstances.
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Math Problem Analysis
Mathematical Concepts
Exponential Growth
Population Projection
Formulas
Population growth formula P(t) = P0 * (1 + r)^t
Theorems
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Suitable Grade Level
Advanced High School / College
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