Respuesta :
Answer:
[tex] 11000= 18000 (1-0.08)^t[/tex]
And solving for t we got:
[tex] \frac{11}{18} = 0.92^t[/tex]
And if we apply natural logs we got:
[tex] ln(\frac{11}{18})= t ln (0.92)[/tex]
And then the value of t would be:
[tex] t = 5.906[/tex]
So then after 5.906 or 6 years we will have approximately 11000 or less for the population
Step-by-step explanation:
For this case we can use the following model:
[tex] P(t) = A (1+r)^t [/tex]
Where A = 18000 the initial value, r = -0.08 since is a decreasing rate and t the number of years. And we want to find the value of t until we have 11000 or lower and we can set up the following equation:
[tex] 11000= 18000 (1-0.08)^t[/tex]
And solving for t we got:
[tex] \frac{11}{18} = 0.92^t[/tex]
And if we apply natural logs we got:
[tex] ln(\frac{11}{18})= t ln (0.92)[/tex]
And then the value of t would be:
[tex] t = 5.906[/tex]
So then after 5.906 or 6 years we will have approximately 11000 or less for the population
