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Yes Max, Mr Wong is YOU.  The future you!  The prophesy foretells us all, you are to have 10 grandchildren !!

Arnolde1234 has said - "Max will multiply" 

You always were good with numbers   

I do not want to doubt your word Mistress Ginger but when did Sir CPhill find the Roman Zero??

He had Rosala, Heather, LL and the great general Titanium Rome all searching on his behalf but I do not recall the capure of the non-directional integer  :///

Still perhaps it is best he divert his attention into writing the digits of pi in roman numerals ....  wait, this does sound somewhat impossible !!  It be an challenge of circular direction!!       

THANKS for the trip down memory lane Ginger.  That WAS a GREAT thread :)

This forum had a GREAT childhood.  It is a teenager now and it is still fun but we had a WONDERFUL infancy!

LancelotLink and all his namesakes were SO much fun to have about I wonder where they are all hiding now?

If they didn't get washed away in the great floods then they must be in an inebriated stupor under the toll bridge. They were all far too keen on banana daiquiris !!

Maybe they found Morgan Tud's rum supply!!   I wonder if they are also sharing it with Nauseated?

Top of the morning to you, Sir CPhill:

LancelotLink and Naus were both persuaded by Morgan Tud’s prophecy that you would indeed, one day, intercept Sisyphus’s bolder and extract the Roman Zero and display the precious Roman knowledge of nothing.

3(12-2x)=11x+2?

36-6x=11x+2

34=17x

x=34/17

x=2

3(12-2x)=11(x-2)+24?

36-6x=11x-22+24

36-2=11x+6x

34=17x

x=2

5(x+2)-(3x+8)=32

5x+10-3x-8=32

2x+2=32

2x=30

x=15

Yes well, his spit was ALWAYS his secret ingredient.

Nothing can imitate the spit of a genetically enhanced chimp !! 

\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\

\(\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{\left(2*1\right)^4\cdot 2m^{2+-1}n^{4+3}}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{2^4\cdot 2^1m^{1}n^{7}}{m^{-4}n^4}}\\ =\sqrt[3]{\dfrac{2^5m^{1--4}n^{7-4}}{1}}\\ =\sqrt[3]{2^5m^{5}n^{3}}\\ =\sqrt[3]{2^3*2^2m^{3}m^2n^{3}}\\ =\sqrt[3]{2^3m^{3}n^{3}*2^2m^2}\\ =2mn\sqrt[3]{4m^2}\\\)

This is the same as Max's answer  :)

Ah yes . . .  the ever changing baseline of moral codes.  I wonder what the formula is.  Whatever it is, it probably follows an infinite descent model. Mine does. A few years ago, I would never have been a consort to a troll.  Now, it’s hard to imagine not being one. 

One thing about Lancelot, his liquor is always high quality. You can’t get a better banana daiquiri than one made by a genetically enhanced chimp. 

   360 x 12 ÷ 2

= 360 ÷ 2

= 180

180RM (30rm*6months)

I don’t think Lancelot was too disappointed by his loss.  On Election Day, he was distracted by a malfunctioning PORV (whatever that is) on a used, portable, nuclear reactor that he bought to power his banana and peanut processing plant.  His board of directors made wagers on whether Lancelot or the reactor would melt down first. 

To pay for the replacement parts, Lancelot sold “glow-in-the-dark bananas” and retooled the production line to make Manhattans. He called it “The Manhattan Project.” He also smuggled mangos across state boarders. The authorities caught him during his final run. Five weeks later, LL appended his mango-smuggling fiasco to a post about random probabilities of typing-monkeys. 

It depends on what the question is :)

Just add 0.9 to both sides.   Easy peazy :)))

Here is a you tube clip on it :)

1+1/87*85,9851

1+1/87*85,9851 = 1.9883344827586207

cos((3π/2) - ϴ)

\(cos((3π/2) - \theta)\\ =\text{change to degrees}\\ =cos(270-\theta)\\ =cos(180+(90-\theta))\\ =-cos(90-\theta)\\ =-sin(\theta)\)

I think Max is right in that not all this working is really needed.  

But it is easier for me to think through this way :)

Mr. Wong has 10 grandchildren. Assuming that the gender of each child is determined independently and with equal likelihood of male and female, what is the probability that Mr. Wong has more grandsons than granddaughters or more granddaughters than grandsons?

1 - (probability of 5 girls and 5 boys)

10C5(0.5)^5*(0.5)^5  =  10C5*(0.5)^10 = 0.24609375

1-0.24609375 = 0.75390625 = 75%

0.52173913043 how would you round this number to a whole number

\(0.52173913043=52173913043\times 10^{-11}\)   !

If you want to round to the nearest whole integer, that will be 1, since 0.52 is closer to 1 than 0.

However depending on the amount of significant figures needed this may be rounded to 0.52 for 2 significant figures or 0.5218 for four significant figures.

Example:

\(2\log_2(x+1) - \log_2x = 2\\ \log_2(x+1)^2 -\log_2x=2\\ \log_2\left(\dfrac{(x+1)^2}{x}\right) = 2\\ \dfrac{(x+1)^2}{x} = 4\\ x^2 + 2x + 1 = 4x\\ x^2 - 3x + 1 = 0\\ x = \dfrac{-(-3)\pm \sqrt{(-3)^2-(4)(1)(1)}}{2(1)}=\dfrac{3\pm \sqrt5}{2}\\ x = \dfrac{3+\sqrt5}{2} \text{ or }\dfrac{3-\sqrt5}{2}\text{(rejected)}\\ x = \dfrac{3+\sqrt5}{2}\)

Try to do your own question now!!

Ok.......

I shouldn't have answered this question.......

No Max: Probability can get very complicated very quickly:

The probability of having 5 boys and 5 girls is computed as follows:

10! /[ 5! x 5! x 2^10] =0.24609375 x 100 =~24.61%

\(\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\ =\left(\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}\right)^{1/3}\\ =\left(2^5\cdot m^5n^3\right)^{1/3}\\ =\sqrt[3]{32}\cdot m^{5/3}n\\ =2\sqrt[3]{4}\cdot m^{5/3}n\)

Ok....... I am not very familiar with probability...... Calculus is my thing!!

No. 1 above should read " equal probability of 5 boys and 5 girls"