Answer:
A - ![f(t)=80(1.5)^t](https://tex.z-dn.net/?f=f%28t%29%3D80%281.5%29%5Et)
B - Graph given below
C - Number of trouts in 5th week are 607.2
D - Population of trouts will exceed 500 on the 5th week.
Step-by-step explanation:
We are given that,
The number of trout increases by a factor of 1.5 each week and the initial population of the trout is observed to be 80.
Part A: So, the explicit formula representing the situation is,
, where f(t) represents the population of trouts after 't' weeks.
Part B: The graph of the function can be seen below.
It can be seen that the function is an exponential function.
Part C: It is required to find the number of trouts in the 5th week.
So, we have,
![f(5)=80(1.5)^5](https://tex.z-dn.net/?f=f%285%29%3D80%281.5%29%5E5)
i.e. ![f(5)=80\times 7.59](https://tex.z-dn.net/?f=f%285%29%3D80%5Ctimes%207.59)
i.e. f(5) = 607.2
Thus, the number of trouts in 5th week are 607.2
Part D: We are given that the trout population exceeds 500.
It is required to find the week in which this happens.
So, we have,
![50](https://tex.z-dn.net/?f=50%3C80%281.5%29%5Et)
i.e. ![(1.5)^t>\frac{500}{80}](https://tex.z-dn.net/?f=%281.5%29%5Et%3E%5Cfrac%7B500%7D%7B80%7D)
i.e. ![(1.5)^t>6.25](https://tex.z-dn.net/?f=%281.5%29%5Et%3E6.25)
i.e. ![t\log 1.5>\log 6.25](https://tex.z-dn.net/?f=t%5Clog%201.5%3E%5Clog%206.25)
i.e. ![t\times 0.1761>0.7959](https://tex.z-dn.net/?f=t%5Ctimes%200.1761%3E0.7959)
i.e. ![t>\frac{0.7959}{0.1761}](https://tex.z-dn.net/?f=t%3E%5Cfrac%7B0.7959%7D%7B0.1761%7D)
i.e. t > 4.5
As, t represents the number of weeks. So, to nearest whole, t = 5.
Thus, the population of trouts will exceed 500 on the 5th week.