Answer
a) $y-51.7=1.09\left( x-20 \right)$ or $y-62.6=1.09\left( x-30 \right)$.
b) $f\left( x \right)=1.09x+29.9$.
c) The percentage of never married American males in $2015$ is $68.05\%$.
Work Step by Step
(a)
Then,
$\begin{align}
& m=\frac{62.6-51.7}{30-20} \\
& =1.09
\end{align}$
It is given that the line passes through the points:
$\begin{align}
& y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
& y-{{y}_{2}}=m\left( x-{{x}_{2}} \right)
\end{align}$
Then,
$\begin{align}
& y-51.7=1.09\left( x-20 \right) \\
& y-62.6=1.09\left( x-30 \right)
\end{align}$
Therefore, the equation of the line is $y-51.7=1.09\left( x-20 \right)$ or $y-62.6=1.09\left( x-30 \right)$.
(b)
Consider any one of the given equations,
$\begin{align}
& y-51.7=1.09\left( x-20 \right) \\
& y=1.09x-21.8+51.7 \\
& y=1.09x+29.9
\end{align}$
Where, $y$ is also a function of $x$. We can write $y=f\left( x \right)$.
Therefore, the equation of the line by use of function notation is $f\left( x \right)=1.09x+29.9$.
(c)
Consider the linear function,
$f\left( x \right)=1.09x+29.9$
Where, $x$ is the number of years after $1980$ and $f\left( x \right)$ is the percentage of never married American males.
In $2015$ ,
$\begin{align}
& x=2015-1980 \\
& =35
\end{align}$
Substitute the value of $x$ in the given equation,
$\begin{align}
& f\left( x \right)=1.09\cdot 35+29.9 \\
& =68.05
\end{align}$
Therefore, the percentage of never married American males in $2015$ is $68.05\%$.
