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I think there is no need to stay in 11.2 world, you just need to remember to ask Mathematica to use extended precision for its calculations. Your data coincidentally were given as exact numbers (integers), so it was sufficient to simply use...
For the distribution to be valid, you need to ask for a distribution of a quantity that is a number or vector for a fixed value of r, whereas you are asking for the trajectory. Perhaps you need to sample the function at multiple instances of times,...
The idea is simple. Take a picture on the phone. Write a function `MakePassportPhotoSheet[img]` in Mathematica to export an photo sheet image to print at a local convenience store at size of 4 by 7 inches, so that photos come out of the correct...
Suppose you is able to solve the problem for a fixed size of state-space, i.e. construct the transition matrix and compute the steady state vector. Then you may use...
In the probability speak, the function f is known as a SurvivalFunction. We can define the probability distribution, describing the outage characteristics (either a duration, or an onset, you are not being specific) as [mcode]In[80]:= f =...
The test is supported as one of many in LocationTest: [url=http://wolfram.com/xid/0h2peojsme-lkkzna]http://wolfram.com/xid/0h2peojsme-lkkzna[/url]
Further optimization to `PeakDetect`.  `Take[list, {a,b}]` is always faster that `Part[list,a;;b]`. Instead of using `Piecewise` inside `Compile`, it is better to use `If`. [mcode]PeakDetect3 = Compile[{{data, _Real, 1}, {width, _Integer},...
Using structural reasoning, it easy to build an efficient one: [mcode]algQ[e : Alternatives[_Plus | _Times], z_] := VectorQ[List @@ e, algQ[#, z] &] algQ[HoldPattern[Power][e_, r_?ExactNumberQ], z_] := algQ[e, z] algQ[z_, z_] := True algQ[w_, z_]...
Possibly In[104]:= Cases[list, {__Integer}, Infinity] Out[104]= {{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}}
Dave, RandomChoice[{a, a, a, a, b, b}, 6]will simulate 6 throws of the first die, and RandomChoice[{a, a, a, b, b, c}, 6]will simulate 6 throws of the second die. This will generate 6 pairs of outcomes {die1, die2}...