Answer:
[tex]N(t)=84(1+0.0405)^t[/tex].
Step-by-step explanation:
We have been given that [tex]N(11)=130[/tex] and [tex]N(0)=84[/tex]. We are asked to write N as a discrete exponential function of t of the form [tex]N(t)=A(1+r)^t[/tex].
Using our given information, we can set an equation as:
[tex]N(0)=A(1+r)^0[/tex]
[tex]84=A*1[/tex]
[tex]84=A[/tex]
Using our given information, we will get:
[tex]N(11)=A(1+r)^{11}[/tex]
Substitute the given values:
[tex]130=84(1+r)^{11}[/tex]
[tex]\frac{130}{84}=\frac{84(1+r)^{11}}{84}[/tex]
[tex]1.5476190476190476=(1+r)^{11}[/tex]
Switch sides:
[tex](1+r)^{11}=1.5476190476190476[/tex]
Take 11th root of both sides:
[tex]\sqrt[11]{(1+r)^{11}}=\sqrt[11]{1.5476190476190476}[/tex]
[tex]1+r=1.04050024747304[/tex]
[tex]r=1.04050024747304-1[/tex]
[tex]r=0.04050024747304[/tex]
[tex]r\approx 0.0405[/tex]
Therefore, our required function would be [tex]N(t)=84(1+0.0405)^t[/tex].