Let x represent time in hours; and y represent Fee in $.
Using the data points given, I am able to construct a line of best fit with the equation of y = 100x^-0.998 with a r^2 value of 1.
In that case would the coefficient "100" have a unit of $/hr or $/(hr^-0.998)?
3 answers
the slope has units of y/x, so $/hr
But the equation is not linear; it is a power series (see x^-0.998)
In that case, your "line" of best fit is not correct. It needs to be s straight line. Maybe you should use a semi-log graph, so you are plotting y against logx
then your function is log100 - 0.998 logx
Label the axes y and logx to plot the straight line with slope -0.998
and the units will still be $/hr. The log paper is just used to transform an exponential function into a linear one. Read up on it if you want more info.
then your function is log100 - 0.998 logx
Label the axes y and logx to plot the straight line with slope -0.998
and the units will still be $/hr. The log paper is just used to transform an exponential function into a linear one. Read up on it if you want more info.
1 answer