Question

the rates of change in population for two cities are ( p(t)=45 ) for alphaville and ( q(t)=102e^{0.05t} ) for betaburgh, where ( t ) is the number of years since 1990, and ( p ) and ( q ) are measured in people per year. in 1990, alphaville had a population of 5000, and betaburgh had a population of 3500. answer parts a) through c). a) determine the population models for both cities. the population model for alphaville is ( p(t)=45t + 5000 ). the population model for betaburgh is ( q(t)=2040e^{0.05t}+1460 ). b) what were the populations of alphaville and betaburgh, to the nearest hundred, in 2000? the population of alphaville in 2000 was 5500 people. (round to the nearest hundred as needed.) the population of betaburgh in 2000 was 4800 people. (round to the nearest hundred as needed.) c) sketch the graph of each citys population model and estimate the year in which the two cities have the same population. choose the correct graph below.

Explanation:

Step1: Determine the value of \(t\) for 2000

Since \(t\) is the number of years since 1990, for the year 2000, \(t = 2000 - 1990=10\).

Step2: Calculate the population of Alphaville in 2000

We have the population model \(P(t)=45t + 5000\). Substitute \(t = 10\) into the formula:
\(P(10)=45\times10+5000=450 + 5000=5450\approx5500\) (rounded to the nearest hundred).

Step3: Calculate the population of Betaburgh in 2000

We have the population model \(Q(t)=2040e^{0.05t}+1460\). Substitute \(t = 10\) into the formula:
First, calculate \(e^{0.05\times10}=e^{0.5}\approx1.6487\).
Then \(Q(10)=2040\times1.6487+1460\).
\(2040\times1.6487 = 2040\times(1 + 0.6487)=2040+2040\times0.6487=2040 + 1323.348=3363.348\).
\(Q(10)=3363.348+1460=4823.348\approx4800\) (rounded to the nearest hundred).

Answer:

The population of Alphaville in 2000 was \(5500\) people. The population of Betaburgh in 2000 was \(4800\) people.