Answer:

And we can round this to the nearest up integer and we got 110194.
Explanation:
The natural growth and decay model is given by:
(1)
Where P represent the population and t the time in years since 1970.
If we integrate both sides from equation (1) we got:


And if we apply exponentials on both sides we got:

And we can assume

And we have this model:

And for this case we want to find

By 1990 we have t=20 years since 1970 and we have this equation:

And we can solve for
like this:
(1)
By 2019 we have 49 years since 1970 the equation is given by:
(2)
And replacing
from equation (1) we got:
We can divide both sides by 143000 we got:

And if we apply ln on both sides we got:

And then k =-0.01303024661[/tex]
And replacing into equation (1) we got:

And we can round this to the nearest up integer and we got 110194.