Parallel lines have the same slope. Since our reference line has a slope of -1, so will a line parallel to it. Thus far we have:
y=-x+b, using point (7,1) we can solve for b
1=-7+b
8=b so the parallel line passing through (7,1) is:
y=-x+8
The correct given equation is:
v = [3 (h + 1)^2.5 + 580 h – 3] / 10 h
So to solve for the instantaneous velocity at t = 1, we
must set h = 0. However we cannot do that since h is in denominator and a
number divided by a denominator is infinite. Therefore we must set h to
something almost zero. In this case, h = 0.0000001, so that:
at t = 1
v = [3 (0.0000001 + 1)^2.5 + 580 (0.0000001) – 3] / 10 *
0.0000001
<span>v = 58.75 ft/s</span>
Basically degrees of freedom are related to sample size (n-1). If the df increases, it also stands that the sample size is increasing; the graph of the t-distribution will have skinnier tails, pushing the critical value towards the mean.
