Answer :
In general, the exponential growth function is given by the formula below
[tex]f(x)=a(1+r)^x[/tex]Where a and r are constants, and x is the number of time intervals.
In our case, n=0 for 1960; therefore, 1968 is n=8,
[tex]\begin{gathered} f(8)=a(1+r)^8 \\ \text{and} \\ f(8)=1.6 \\ \Rightarrow1.6=a(1+r)^8 \end{gathered}[/tex]And 1976 is n=16
[tex]\begin{gathered} f(16)=a(1+r)^{16} \\ \text{and} \\ f(16)=2.3 \\ \Rightarrow2.3=a(1+r)^{16} \end{gathered}[/tex]Solve the two equations simultaneously, as shown below
[tex]\begin{gathered} \frac{1.6}{(1+r)^8}=a \\ \Rightarrow2.3=\frac{1.6}{(1+r)^8}(1+r)^{16} \\ \Rightarrow2.3=1.6(1+r)^8 \\ \Rightarrow\frac{2.3}{1.6}=(1+r)^8 \\ \Rightarrow(\frac{2.3}{1.6})^{\frac{1}{8}}=(1+r)^{}^{} \\ \Rightarrow r=(\frac{2.3}{1.6})^{\frac{1}{8}}-1 \\ \Rightarrow r=0.0464078 \end{gathered}[/tex]Solving for a,
[tex]\begin{gathered} r=0.0464078 \\ \Rightarrow a=\frac{1.6}{(1+0.0464078)^8}=1.113043\ldots \end{gathered}[/tex]a) Thus, the equation is
[tex]\Rightarrow f(n)=1.113043\ldots(1+0.0464078\ldots)^n[/tex]b) 1960 is n=0; thus,
[tex]f(0)=1.113043\ldots(1+0.0464078\ldots)^0=1.113043\ldots[/tex]The answer to part b) is $1.113043... per hour
c)1996 is n=36
[tex]\begin{gathered} f(36)=1.113043\ldots(1+0.0464078\ldots)^{36} \\ \Rightarrow f(36)=5.6983\ldots \end{gathered}[/tex]The model prediction is above $5.15 by $0.55 approximately. The answer is 'below'
