The population of a city is growing at an average rate of 3% per year. In 1990 , the population was 45000

a. N(y) = 45,000*1.03^(y-1990)
N(y) = population in year y,
45,000 = poppulation in 1990,
1.03 = yearly growth factor,
y = year y

b. 2,007−1,990=17 years
N(17)=45,000×1.03^(17)=74,378.14

c. 2N=N*1.03^ t
2=1.03^t
t=log(2)÷log(1.03)=23.4
23.4+1,990=2,013.4

d. 2=(1+r)^10
r=2^(1/10)-1)*100=7.18%

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pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-04-15T14:47:58+02:00

Population growth is an example of an exponential growth problem and can be solved using P(t) = Poe^rt.

What is an exponential function?

An exponential function is one in which can be represented as in this case using; P(t) = Poe^rt

P(t) = population at time t

Po = initial population

r = population growth rate

t = time taken

So we can write;

P(t) = 45000e^0.03t

In 2007 which is 17 years after;

P(t) = 45000e^0.03(17)

P(t) = 74938

The year the population will double is obtained from;

90000= 45000e^0.03t

90000/ 45000 = e^0.03t

2 = e^0.03t

ln 2 = 0.03t

t = ln 2/0.03

t = 23 years

If the population took 10 years to double, the growth rate is obtained from;

90000= 45000e^10r

90000/ 45000 = e^10r

ln 2 = 10r

r = ln 2/10

r = 0.069 or 6.9%

Missing parts:

The population of a city is growing at an average rate of 3% per year. In 1990, the population was 45 000.

a) Write an equation that models the growth of the city. Explain what each part of the equation represents.

b) Use your equation to determine the population of the city in 2007.

c) Determine the year during which the population will have doubled.

d) Suppose the population took only 10 years to double. What growth rate would be required for this to have happened?

Learn more about population growth: https://brainly.com/question/13805601