Answer
Yes, $\qquad m_{1}m_{2}$
Work Step by Step
$f\circ g(x)=f[g(x)]=m_{1}g(x)+b_{1}$
$=m_{1}(m_{2}x+b_{2})+b_{1}$
$=(m_{1}m_{2})x+(m_{1}b_{2}+b_{1})$
which has the form $F(x)=Mx+B$,
so its graph is a line.
The slope is $M=m_{1}m_{2}$, the y-intercept is $B=(m_{1}b_{2}+b_{1})$
