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"Quadruple"   = "fourth"  ....so......

⁴√ 242    ≈   3.94415      .....[ assuming the principle root   ]

This is just a forum for maths, not chemestry. 

P.s Put the material in the water and count the bubbles (if any). Record the time (to make it reliable) and compare with the other materials. Discard anomalous results. Repeat 3 times.

1 1/9

I don't see much room for debate here, EP...........LOL!!!!!

Here's my take:

Total identifiable "words"  generated by the arrangements of the letters in"SEVEN"   =

5!  / 2!  =   120  / 2   =   60

Possible identifiable "words"  formed by having "E's"  next to each other  = the "E's"  can occupy any of 4 positions....and for each of these......the oher letters can be arranged in 3!  = 6 ways  =  4 * 6    = 24

So.....the probability =   24/ 60    =    2 / 5

I agree with NinjaAnswer  !!!!

Or...to keep the arguments down to a minimum      .6473 R    ( = 37.09 degrees)         LOL

The square root of  - 1  is DEFINED to be   '   i    '    (as in iMAGINARY number)

Use the inverse cosine  :

arccos ( 0.7977 )  = x  ≈  37.09°

Another way to see this  =  total arrangements possible  less the arrangements where the math books are together

Total arrangements  =  6!   = 720

Total arrangemens where the  math books are together = they can occupy any of 5 positions....and for each of these positions, they can be arranged in two ways......and for each of these, the other books can be arranged in 4! ways  ......so we have     5 * 2! * 4!  =  10 * 24   = 240

So........720 −  240   =    480 ways

As NinjaAnswer found  !!!!

It will take Nash 2.83 months to pay off the entire balance of his Credit Card @$650 per month @ 17.8% APR compounded monthly. Therefore, his total cost will be:2.83 x $650 =$1,840.24, or $51.07 in interest on top his balance of $1,789.17.

Solve for 2017 using these numbers only once: 100, 50, 8, 2, 1, 1

You can only use +, -, x, and / which I forgot to mention in my first post

Total arrangements possible  = 10!

Arrangements where the three prints are together  =

They can occupy any of 8 positions...and for each of these, they can be arranged in 3! ways.....and the other prints can be arranged in 7! ways......so we have

8 * 3!  * 7 !

So....the total probability  is :

[ 8 * 3! * 7! ]  /  10!    =      [8 * 6] / [10 * 9 * 8 ]  = 6  / 90   =  1 / 15

Previous Balance    x    Monthly Interest   +  previous balance    - payment  =

    767                      x         .1625 / 12                + 767                          -    49         = $ 728.55  new balance

Jackson will have have the following amortization table after 1 monthly payment of $49:

Interest =$10.55, Principal =$38.45, Balance of Credit Card =$728.55.

This answer is prob. not even math anymore, so im not sure if this is correct, but 

anyways:

sqrt(100) = 10

10 * 2 = 20

8 - 1 = 7

20 1 7

used numbers:

100, 2, 8, 1, 1

edit:

This is supposed to be a math puzzle right? I hope so, otherwise just forget this answer.

The probability  = 1 − P( 0 heads)  − P(1 head)

P(0 heads)  =  (1/2)^6  =  1/64

P (1 head)  = C(6,1) (1/2)(1/2)^5 = 6 *(1/2)^6  =  6/64

So....the total probability  = 1 − 1/64 − 6/64  = 57/64

If people are to learn from this site NinjaAnswer they would generally appreciate an explanation of how the answer is obtained!

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Note that after choosing one of the 12 black shoes, say, there are only 6 black shoes of the opposite parity left.

p(black pair) = (12/22)*(6/21) → 72/462

p(brown pair) = (6/22)*(3/21) → 18/462

p(grey pair) = (4/22)*(2/21) → 8/462

p(matching pair) = (72 + 18 + 8)/462 → 98/462 → 7/33

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5!/2!3! = .083 or 1/12.

There 120 = 5! ways of aranging the 5 letters in "Seven". You want to know the number of ways out of 120 that you can occupy 2 spaces out of 5, leaving 3 left over.

Source: "Probability, An Introduction" by Samuel Goldberg.

p(at least 1 boy and at least 1 girl) = 1 - p(no boys) - p(no girls)

 → 1 - 2*ncr(20,4)(1/2)²⁰

Its D!

That is incorrect(I think)

The correct answer is 7/33

When you get your final answer, hit the ' = ' sign again.....the answer will display in Sci Notat.

Omg same test!