Its a very similar problem to the word problem you posted here http://www.mapleprimes.com/questions/151971-Number-Theory

using Kitonum's code , I get 2 answers

[HUNSEN = 126356, STRONG = 379068], [HUNSEN = 318928, STRONG = 956784]

ps, are you a paid up member of the CPP?

 

tot := 0;
AD := 30;
SP := 7;
for i from SP by AD to 1007 do
if isprime(i) = true then tot := tot+1; print([tot, i])
else break end if end do;

where AD is arithmetic difference (guess), SP is start prime (guess).

there must be a better way

 

 @Carl Love made the point about the searches needing to be case sensitive....

@Joe Riel thx for worksheet. Sorry to bang on about this. maybe i didn't fully explain the nuances of the problem.

Given the convention should be

1) one element per word (identified randomly or otherwise- if it exists)

2) keep the original case of the word, except for the 1 or 2 letter element (bold),

example:

i/p: "anna aaron dean Mitte zt michael sherry au  Lynne"

o/p: "anNa aAron deaN MitTe zt micHael sherrY Au  LynNe"

do you mind modifying your workbook to reflect the above?

Thanks Carl appreciate that

The original question is this:

Q116. A player sits down at a roulette table with $20. He bets $1 at a time on either red or black. Either bet pays even money and has a probability of 9/19 of winning. What is the probability that the player wins $10 before losing all of his $20?

http://mathproblems.info/working.php

That solves it!

The enclosed worksheet is my interpretation of the physical problem. DROPvLEAP.mw

I'm not sure how you got k=Pi*27/1625/2. I stuffed up one of my parameters in the original dt, but i have corrected it so now your dt is closer to my dt.

absolutely Kitonum, i need to blow the dust off my calculus 101 notes. better than my 'sledgehammer" approach

 

subs.mw

my expression doesn't have differential operators so why would invoking DEools work? It doesn't seem to...

anyway i have another one. I want to subs x^2=z into the expression (x^n)^4/x^2-(x^n)^4*R/x^4+1/(4*n^2) to yield z^(2*n-1)-R*z^(2*n-2)+(1/4)*n^2. 

> algsubs(1/x^2 = 1/z, algsubs(x^2 = z, (x^n)^4/x^2-(x^n)^4*R/x^4+1/(4*n^2))) doesn't do it, the x^n in the expression is the sticking point.

 

Cost as function of time t (months)

Utilizing the ceil command (rounds up to nearest integer):

> Feq := 3.67*ceil((1/3)*t)+0.151e-3*(13*W*(4*hr/day))*(365*t*(1/12))*day/(W*hr);

                                    
                       3.67ceil(  t/3) + 0.2388316667 t

check:                       

> eval(Feq, t = 12);
     
                           17.54598000

> Ieq := (3.29*(1/2))*ceil((1/12)*t)+0.151e-3*(40*W*(4*hr/day))*(365*t*(1/12))*day/(W*hr);

                                                
                   1.645000000 ceil(  t / 12) + 0.7348666667 t
                                                 

check:
> eval(Ieq, t = 12);
                                
                      10.46340000

> plot([Feq, Ieq], t = 0 .. 24, title = Cost v Time(M), legend = ["Fluorescent", "Incandescent"]);

 

so you're telling me you have to anticipate beforehand the simplification involves tan.

in mathematica when you execute Sin[x]/Cos[x] you get Tan[x]. simple.

....but not to maple apparently.

 

cool bananas

ok, how would i modify the above  recursion formula to calculate the number of partitions into triples from 

 permute([1, 1, 1, 1, 0, 0, 0, 0 ,0 ,0 ,0 ,0], 12);

Thanks Robert and Joe.

I don't really undersand the methodology fully Robert, but it works. Brilliant!

thanks Doug,pagan,acer,Joe 

Given a normal distribution with mu=2500 and s=583, the Z value =(x-2500)/583.The area under the curve (probability) is given by Int(1/sqrt(2*Pi)*exp(-t^2/2),t=-Z..Z); Given p = 0.95, what is x? solve(int(1/sqrt(2*Pi)*exp(-t^2/2),t=-(x-mu)/s..(x-mu)/s)=.95,x); x= 3642.96 This means Z= 1.96 which agrees with tables. However if one uses the inbuilt Maple function with p=0.95 with(stats): x:=statevalf[icdf,normald[mu,s]](0.95); x= 3458.95 This means Z= 1.64 which implies p=0.90 Why is there a difference?