All Answers 358516 Answers

My name is Jeff.......

The numbers would have to end in either 6 and 4  or 4 and 6.....or......they would have to end in 8 and 2 or 2 and 8.......so, there are 4 possibilities, here

And there would be 9 of each, since the first digit could be 1 - 9....so......9 x 4 = 36

wait, not sorry. I think there was another question similar to this one, but got deleted. They were very similar it's just that one part was different. 

Here it is if someone would like to help me please:

In how many ways can the 10 kids in my class be seated in a circle if John and Sam insist on being seated diametrically opposite each other? (As usual, two seatings which are rotations of each other are considered the same.)

I am so sorry, but I put this answer in and it said it was wrong. Sorry again because I still don't know so I won't be able to explain it.

 -6*sec(sin(5x^2+3x+2)).....we will apply the -6 back at the end....using the chain rule several times we have

sec(sin(5x^2 + 3x + 2))*tan(sin(5x^2 + 3x + 2)) *cos(5x^2 + 3x + 2) * (10x + 3)

And applying the -6 back, we have

(-60x - 18)*sec(sin(5x^2 + 3x + 2))*tan(sin(5x^2 + 3x + 2)) *cos(5x^2 + 3x + 2)

Thanks for that well-presented answer......and......welcome to the forum.....!!!

$${\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{p}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}} = {\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{p}}$$

$${\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{p}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}} = {\mathtt{15}}$$

$${\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{p}} = {\mathtt{3}}$$

$${\mathtt{p}} = {\frac{{\mathtt{3}}}{{\mathtt{14}}}}$$ or 0.2142857142857143...

$${\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

$${\frac{{\mathtt{4}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{18}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

$${\frac{{\mathtt{5}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{18}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

$${\frac{{\mathtt{5}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{19}}}{{\mathtt{3}}}}$$

$${\frac{{\mathtt{15}}}{{\mathtt{6}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{38}}}{{\mathtt{6}}}}$$

$${\frac{{\mathtt{53}}}{{\mathtt{6}}}}$$ or $${\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{6}}}}$$ or 8.8333333333333333...

I don't understand.    

Maybe it is just too late for me.

When this music finishes I am off to bed.  :)

This refers to distinct hands....."Distinct Hands" is the number of different ways to draw the hand, not counting different suits

For instance.....there is 1 "distinct" royal flush......but there are  4 suits of them......

Clearly....there are WAY more than 858 two-pair hands......

One thing about probabiliity is that it can be really difficult to interprete what you are being asked for.

The other thing about prob is that although you may be able to understand the correct answer when it is presented to you, you often still cannot understand why your own original answer is incorrect!

It is an interesting and a frustrating feild of methematics. 

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I'm listening to this

so what is the 858 for?

Check the probability.....  123552/ 2598560   = about 4.75%

This says out of 2,598,560 possible hands, there are 123,552 hands consisting of two pairs....

doesn't that snip say that there are 858 possibilities?

Sorry, Melody...I believe this question  asks how many two-pair hands are possible in a draw of 5 cards....if I'm interepreting this correctly.....I think my answer is the only correct one....

Look at the number of possible royal flushes in the table......4....!!! 

@@ End of Day Wrap   Sun 15/3/15     Sydney, Australia Time   11:55 pm     ♪ ♫

Hi everyone,

We were not swamped with questions today - it was sat/sun afterall - but there were a lot of difficult probability questions that took a lot of the mathematicians' time.  As always there were some great answers provided by Nauseated, CPhill, Alan and Kitty(๑‵●‿●‵๑)

Interest Posts:

mON 16/3/15

Please can someone else discuss this one.   :)

I know my CDD is probably causing problems here but I always seem to get to it late at night and it has confused me :(

You would need more than just the tangent.

Do you  have the centre or something else?

Thanks Chris - that would be a good site for us to study from I expect.

So there are 858 distinct possibilities.

Maybe this is interpreted differently from how I interpreted it. 

I interpreted there being 4*3=12 ways of choosing 2 of any individual rank.

I wonder if I was suppose to think of all those as the same?

I just did an investigative problem here

3!*4!   but I think this should be divided by 2 because each clockwise possibilty has an anticlockwise possiblility tht would be classed as the same.

so I think

$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$

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this is something we have been debating today so I am going to try and look at an easier question and see what happens.

I am going to have just 4 children sitting in a circle   they will be ABC and D   

no restrictions except that clockwise is considered the same as anticlockwise so, fo instance,

ABCD = ADCB

I think there should be 3!/2  ways = 3 ways  Mmm

ABCD = BCDA = CDAB=DABC   now counter clockwise   = DCBA=ADCB=BADC=CBAD   8 all the same

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

ADBC       8 the same

ADCB       8 the same

That is 6 choices starting with A, there will be another 6 starting with B, 6 with C and 6 with D

that is 6*4=24 permutations.

BUT how many of these are really the same?

24 divided by 8 = 3

I have colour coded to show the 3 distinct possibilities         

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Notice that "a" could come first and be matched with 25 letters  {the rest of them} for a total of 25 words

'b' could come first and be matched with 24 letters for a total of 24 words

"c" could come first and be matched with 23 lettters for a total of 23 words

etc........

So....this appears to just be the sum of the first 25 integers = [26 * 25] / 2  = 325

Hey....Melody and I agree on something!!!