question 7 1 pts a towns population grows according to ( p ( t ) = 4000 e ^ { 0.03 t } ). what will the…

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?
Answer
Question 7
Explanation:
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).
Answer:
(5400)
Question 8
Explanation:
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:
(78242)