I believe suspected outliers are points that lie between a lower/upper, inner and outer fence line. 

lower inner fence line = lower quartile - 1.5*(inner quartile range)
lower outer fence line = lower quartile - 3*(inner quartile range)

And outside the lower and upper outer fence lines are outliers (or as I just found out - highly suspect outliers).

No offense intended, I appologize if it sounded that way.  However using Maple on this problem ...

We have twelve 360 degree rotations of the minute hand equaling one 360 degree rotation of the hour hand

eq:=(12*360)/(1*360)=m/h
                
solve(eq,h); #number of degree's the hour hand moves
               
solve(subs(m=90,eq)); #the minute hand movement through 90 degrees will cause the hour hand to move
                

evalf(%,2);
                   7.5

Thus the hour hand has moved 7.5 degrees from it's 90 degree position and is 7.5 degrees away from the 15 minute position of the minute hand.

The harder part of the question is making it into a great maple application.

I wonder if some of these 7 city learning questions are interesting enough for Maple?

I thought they were the same, thanks for clearing that up.  Makes more sense now.

For Maple, it appears quick updates to some bugs are possible without re-distributing the core version. 

I thought they were the same, thanks for clearing that up.  Makes more sense now.

For Maple, it appears quick updates to some bugs are possible without re-distributing the core version. 

Seems to be a wording trick.  One release per version as wikipedia shows except version 8 which shows 8.0.4

**edit add** my misunderstanding that wolframalpha was mathematica

Thank you!  I am sure this is appreciated by all.

Thank you!  I am sure this is appreciated by all.

I believe he means to separate the + and - terms that don't reside inside brackets.  So the extracted piece you show would look like this

-3/32*e*cosfJ^2*sinfK^2*sindincl^2*cos(2*ele-2*g-anom)
-3/32*e*cosfJ^2*sinfK^2*sindincl^2*cos(2*ele-2*g+anom)
+3/16*e*cosfJ^2*sinfK^2*sindincl^2*cos(-anom+

It's all relative.  Ironic how relativity crept in.  Lanczos' Ph.D. thesis  (1921) was on relativity theory.  He even served as an assistant to Albert Einstein. 

In fact it might interest you that Cornelius Lanczos pioneered work on the Fast Fourier Transform which is an important tool used in Financial Economics :)

They work fine in M12. 

I can only conclude that either the equations weren't exactly as shown or there were some prior interfering calculations. 

@dot4xdash - if you use restart; and then try it again do you get the same results?

Ah, okay fair enough on the comments. 

I do like seeing all the bugs and their workarounds though.

Of course the regression is easily corrected, no mountain.  May I ask what is wrong with my contributions?

**edit deleted irrelavent content

Actually there is no need to change the variable it does work with x on 16.02.  You just need to adjust the x range to -4..2

So it appears you can't just arbitrarily pick a range (which should work) but you must have a range of x that is suitable.  The x range is dependent on the y range chosen and vice versa.

DEplot(diff(y(x), x) = x+y(x), y(x), x = -4 .. 2, y(x) = -5 .. 5, [[0, 0]], linecolor = black)

However using the x range -5..5 will not produce the solution curve unless you increase your y range to something like y=-5..130

Changing the the function y to y(t) instead works

Also note if your y range is not large enough for the solution it will not display so either remove the y range or choose a suitable larger range that will allow the solution to be visible.  If you want the smaller y range the x sorry t range will have to be smaller. 

So changing to this will help

A:=DEplot(y'(t)=t+y(t),y(t),t=-5..5,[[0,0]],linecolor=black);

display(A)