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BabaBlast [244]
2 years ago
5

Show all work below to answer the following for a certain city that had a population of 150,000 in the year 2010. In 2020, the p

opulation was 185,000.
a. Find the constant rate of continuous growth, k, for this population. (Please round to 5 decimal places)

b. When will the population reach 250,000 people?
Mathematics
1 answer:
Gelneren [198K]2 years ago
7 0

The constant rate of continuous growth, k, for this population is equal to 2.11935%. And the population  will reach 250,000 people in 24.36 years.

For solving this question, you should apply the Population Growth Equation.

<h3>Population Growth Equation</h3>

The formula for the Population Growth Equation is:

      P_f=P_o*(1+\frac{R}{100} )^t        

Pf= future population

Po=initial population

r=growth rate

t= time (years)

STEP 1 - Find the constant rate of continuous growth, k, for this population.

For this exercise, you have:

Pf= future population= 185,000 in 2020.

Po=initial population =150,000 in 2010.

r=growth rate= ?

t= time (years)=2020-2010=10

Then,

P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 185000=150000\cdot \left(1+\frac{R}{100}\:\right)^{10}\\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{185000}{150000} \\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{37}{30}\\ \\ R=100\sqrt[10]{\frac{37}{30}}-100=2.11935\%

STEP 2 - Find the <em>t</em>  for population 250,000 people.

P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 250000=150000\cdot \left(1+\frac{2.11935}{100}\:\right)^{10}\\ \\ \left(1+\frac{2.11935}{100}\right)^{10}=\frac{250000}{150000} \\ \\ \left(1+\frac{2.11935}{100}\right)^t=\frac{5}{3}\\ \\ t\ln \left(1+\frac{2.11935}{100}\right)=\ln \left(\frac{5}{3}\right)\\ \\ t=\frac{\ln \left(\frac{5}{3}\right)}{\ln \left(\frac{102.11935}{100}\right)}\\ \\ t=24.36

Read more about the population growth equation here:

brainly.com/question/25630111

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