Answer:
Part 1) [tex]k=\frac{1}{16}\ \frac{trains}{min}[/tex]
Part 2) [tex]160\ min[/tex]
Part 3) [tex]30\ trains[/tex]
Step-by-step explanation:
Part 1) Write a constant of proportionality equation for this relationship
Let
y ----> the number of trains
x ----> the time minutes
we know that
A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form [tex]y/x=k[/tex] or [tex]y=kx[/tex]
In this problem we have
For x=48 min, y=3 trains
[tex]k=y/x[/tex] ----> [tex]k=\frac{3}{48}[/tex]
Simplify
[tex]k=\frac{1}{16}[/tex]
The units of the constant of proportionality are [tex]\frac{trains}{min}[/tex]
so
[tex]k=\frac{1}{16}\ \frac{trains}{min}[/tex]
The linear equation is
[tex]y=\frac{1}{16}x[/tex]
Part 2) Given the relationship is the same, how many minute have passed after 10 trains have left the station?
For y=10 trains
substitute the value of y in the equation and solve for x
[tex]10=\frac{1}{16}x[/tex]
[tex]x=10(16)=160\ min[/tex]
Part 3) Given the relationship is the same, how many trains have left the station after 8 hours?
For x=8 hours
substitute the value of x in the equation and solve for y
But first convert hours to minutes
remember that
[tex]1\ h=60\ min[/tex]
[tex]8\ h=8(60)=480\ min[/tex]
substitute
[tex]y=\frac{1}{16}(480)=30\ trains[/tex]
