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)?
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 $/hr. The log paper is just used to transform an exponential function into a linear one. Read up on it if you want more info.
To determine the unit of the coefficient "100" in the given equation y = 100x^(-0.998), we need to consider the units of both variables.
In this case, x represents time in hours, so its unit is simply hours (hr).
Now, let's examine the exponent in the equation, which is -0.998. The exponent represents the power to which x is raised. In this case, x is raised to the power of -0.998.
To simplify the units, we can rewrite the equation as follows:
y = 100/hr^0.998 * x^-0.998
Now, if we simplify the expression 100/hr^0.998, we can write it as:
100/(hr^0.998)
This expression does not have any negative exponent, so we don't need to worry about that. Based on the expression, we can conclude that the coefficient "100" has a unit of dollars per hour ($/hr).
Therefore, the correct interpretation of the coefficient "100" in this equation is dollars per hour ($/hr).