All Answers 358411 Answers

Let's look at the squared number list. 

1^2 =1

2^2 =4

3^2 =9

4^2 =16

5^2 =25

6^2 =36

7^2 =49

8^2 =64

9^2 =81

10^2 =100

11^2 =121

108 doesn't fit into any of these, so this is going to be done using a calculator. Or an elaborae formula, so called, that I don't want to go through. 

To find this number's square root, punch into the calculator 

$$square root$$ 108

TO check your answer, look back at the list. The number 108 falls between 100 and 121, so the final answer should be between 10 and 11. Make sense?

Hope it helps! 

8 people are sitting at a round table. 3 of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?

the number of ways that 8 people can sit around a table is   $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$

the number of ways 3 distinct people can sit together around the table is  $$(\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{1}}\right)){!}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{720}}$$

so I think the prob will be    $${\frac{{\mathtt{1}}}{{\mathtt{7}}}}$$

$${\frac{{\mathtt{720}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}} = {\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$

the meaning is not clear - brackets are needed BUT

inverse tan is the same as arc tan

enter this into the web2 calc.    put the brackets in the correct place of course.  

atan(4.7/3.5)

Please help me quick. I have atime limit

C(6,3) * C(11,5)  = 9240 ways

6 people are sitting around a table. Let x be the number of people sitting next to at least one woman and y be the number of people sitting next to at least one man. How many possible values of the ordered pair (x,y) are there? (For example, (6,0) is the pair if all 6 people are women, since all 6 people are sitting next to a woman, and 0 people are sitting next to a man.)

6 people

x be the number of people sitting next to at least one woman

y be the number of people sitting next to at least one man

I tried to do some of this with combinations but it didn't work.

I ended up drawing lots of hexagons, and just looking at what happens.

I think it is correct. 

(6,0) if all are women

(5,2)   if 1 man

(4,4)   if 2 men sitting together

(6,3)   if the 2 men and they are separated by 1 seat

(6,4)   if the 2 men are opposite each other

This pattern must be symmetrical

(0,6) if all are men

(2,5) if 1 woman

(4,4) if 2 woman and they sit together

(3,6) if the 2 mwomen and they are separated by 1 seat

(4,6)   if the 2 women are opposite each other

This one was harder.  But after drawing pics it seems to me that there are only 3 posibilities for 3 men and 3 women.

(5,5)  The men all sit together

(5,5)  A pair of men, a pair of women, a man, a woman

(6,6)  Alternating

so what do we have

(6,0), (0,6)

(5,2)(2,5)

(4,4)

(6,3)(3,6)

(6,4)(4,6)

(5,5)

(6,6)

11 pairs

"At most" connotes 1, 2 or 3 groups

For one group = one way....all of them together

For two groups.....2 + 2  or 3 + 1  = two ways   ...{the oranges are indistinguishable, so 1 + 3 is the same grouping as 3 + 1 }

For 3 groups = 1 + 1 + 2  = one way   {again, they are indistinguishable}

So ..... four ways.....

C(15,2) * C(10,3) = 12,600 ways

Sorry Chris, I misread your first answer - but I am still really dubious.

I like the working of my answer but I don't understand why  the number would be so much bigger (or bigger at all) than if everyone could just sit where they wanted.  

Both your answer and mine have a bigger value than if everyone was individuals and to me this seems a little crazy.

There has to be doubling up happening.  !!

Well...if I can use my secret decoder ring, I think you're saying that for every 2 votes Billy got, Haley got 7.

So...the ratio of votes for Haley to Billy = 7 : 2

Here's the justification for my answer....

Consider a "birdseye" view of the table....

One person either occupies the left seat on one side of the table or the right seat on that side.

And for each of these arrangements, the other 7 seats can be filled in all possible ways = 7! ways.

So, 2 x 7!   = 10,080 ways

"Rotations" of the table don't matter......one rotation would look like any other for a particular arrangement.

How many ways are there to put 5 b***s in 2 boxes if the b***s are distinguishable but the boxes are not?

B***s different,  boxes are the same.

5,0          1ways

4,1           5ways

3,2           5C2 =  $${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{10}}$$      ways

16 ways maybe

I believe that it is just a lie. There is no way that if you add 1+2 you get 4. what you would get is 3. no matter what you do.

Yes that looks right Chris ,

The 5! comes from (6-1)! you always take of 1 for circles because the first person marks the beginning and then end. 

(There are 5 individual people plus the siamese triplets)

(2x^3)(4x^1/3) =

8x^3 * x^(1/3) =

(8x^3)³√x

This post  is all nonsense -

I am just leaving it here so that Alan's and Bertie's comments make sense.

I think you misread the question Chris,

And I think it might be  (table rotations are considered the same)

8C2*6C2*4C2*2C2*(4-1)!

$${\left({\frac{{\mathtt{8}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{15\,120}}$$

This seems much to big

Chris's answer - where no one was pair up - was    $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$

And i thought my answer should be smaller than this. 

-5d -4 (+4) = 21 (+4)

-5d (/5) =25 (/5)

d=-5

1 million times 4billion equals 4.0e15

Answer me CPHIL

My FAVORITE IS NEON ORANGE!