The population of an Asian country is growing at the rate of 0.7% annually. If there were 3,942.000 residents in the city in 1995. Find how many to the nearest ten thousand) are living in that city in 2000. Use y = 3,942,000(2.7)0.0074 a) 370,000 b) 4,000,000 c) 4.160,000 d) 4.320,000

Answer:b) 4,000,000Step-by-step explanation:Let the population is measured since 1995,Given,The initial population, P = 3,942,000,Annual rate of growing, r = 0.7% = 0.007,If y represents the population after t yearsSo, the population after t years would be,[tex]y=Pe^{rt}[/tex][tex]y=3942000(2.7)^{0.007x}[/tex]Therefore, the population after 5 years,[tex]y=3942000(2.7)^{0.007\times 5}=3942000(2.7)^{0.035}=4081448.78924\approx 4000000[/tex]Hence, the population in 2000 would be approximately 40,00,000.Option 'b' is correct.