Answer:
[tex]P(t) = 104.267\ millions[/tex]
Step-by-step explanation:
The exponential growth formula is
[tex]P(t) = Ae^{kt}[/tex]
Where
A is the main coefficient and represents the initial population.
e is the base
k is the growth rate
t is time in years.
Let's call t = 0 to the initial year 1993.
At t = 0 the population was 94 millions
Therefore we know that:
[tex]P(0) = 94[/tex] millions
After t = 6 years, in 1996, the population was 99 millions.
Then [tex]P(6) = 99[/tex] million.
Now we use this data to find the variables a, and k.
[tex]P(0) = 94 =Ae ^{k(0)}\\\\94 = A(e ^ 0)\\\\A = 94[/tex].
Then:
[tex]P(6) = 94e^{k(6)}\\\\99 = 94e ^{6k}\\\\\frac{99}{94} = e^{6k}\\\\ln(\frac{99}{94}) = 6k\\\\k = \frac{ln(\frac{99}{94})}{6}\\\\k =0.008638[/tex]
Finally the function is:
[tex]P(t) = 94e^{0.008638t}[/tex]
In the year 2005:
[tex]t = 2005-1993\\\\t= 12\ years[/tex]
So the population after t=12 years is:
[tex]P(t) = 94e^{0.008638(12)}[/tex]
[tex]P(t) = 104.267\ millions[/tex]
Answer:
104 million
Step-by-step explanation:
We are given that a country's population in 1993 was 94 million and in 1999, it was 99 million.
We are to estimate its population in 2005.
Let P(t) denote the population at t years after 1993, then:
P(0) = 94 million
P(6) = 99 million
[tex]P(t) = P(0) e^{(k t) }[/tex]
So [tex]P(6) = P(0) e^{(6 k) }[/tex]
[tex]99 = 94 e^{(6 k) }[/tex]
[tex]e^{(6 k)} = \frac{99}{94}[/tex]
[tex]6k=ln(\frac{99}{94} )[/tex]
[tex]k=ln(\frac{\frac{99}{94}}{6} )[/tex]
[tex]k=0.0086375[/tex]
Now since we have found the value of [tex]k[/tex], we can estimate the population in 2005:
t = 2005 - 1993 = 12
P (12) = [tex]P(0) e^{( 12 k)} = 94 e^{( 12 (0.0086375) ) }[/tex] = 104.266 million
