Question

question 7 1 pts a towns population grows according to ( p ( t ) = 4000 e ^ { 0.03 t } ). what will the population be after 10 years? question 8 1 pts a citys population decreases at a continuous rate of 2.8% per year. if the current population is 90,000, what will it be in 5 years?

Explanation:

Question 7

Step1: Substitute \(t = 10\) into the formula

Given \(P(t)=4000e^{0.03t}\), when \(t = 10\), we have \(P(10)=4000e^{0.03\times10}\).

Step2: Simplify the exponent

\(0.03\times10 = 0.3\), so \(P(10)=4000e^{0.3}\).

Step3: Calculate the value of \(e^{0.3}\)

Using a calculator, \(e^{0.3}\approx1.34986\).

Step4: Multiply to find \(P(10)\)

\(P(10)=4000\times1.34986 = 5399.44\approx5400\).

Step1: Use the continuous - decay formula

The formula for continuous decay is \(P(t)=P_0e^{-rt}\), where \(P_0 = 90000\), \(r=0.028\), and \(t = 5\).

Step2: Substitute the values into the formula

\(P(5)=90000e^{-0.028\times5}\).

Step3: Simplify the exponent

\(-0.028\times5=-0.14\), so \(P(5)=90000e^{- 0.14}\).

Step4: Calculate the value of \(e^{-0.14}\)

Using a calculator, \(e^{-0.14}\approx0.86936\).

Step5: Multiply to find \(P(5)\)

\(P(5)=90000\times0.86936 = 78242.4\approx78242\).

Answer:

\(5400\)

Question 8