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Moves are \((1,0),~(0,1),~(-1,0),~(0,-1)\)

We have a set of 4 moves and the sum of those must equal (0,0)

There are 36 of these sets of 4 moves that sum to (0,0)

There are a total of \(4^4 = 256\) moves.

P[returning to 0] = \(\dfrac{36}{256} = \dfrac{9}{64}\)

CoolStuffYT is correct.

The set {X, K, M, Z} has 4 possible letters, so we can pick any 4.

The set {W, X, Y, Z} has 4 possible letters, but since we pick 3 out of 4, there are \(4\times 3\times 2=24\).

\(24\times 4 = 96\) crypto codes

You are very welcome!

:P

Let's think about this. Forward, backward, right, and left is 4 possible movements. 

If the ant goes the same direction twice, the only path it can take is back the opposite direction twice.

If the ant goes a direction once and goes back to the origin again, the ant can move any four ways but then must come back.

If the ant goes a direction once and goes a different direction, there are two ways, come back the original way or go in a square-like shape.

Translating this to numbers, we have 

\(4\times 1 \times 1 \times 1=4\)

\(4 \times 1 \times 4 \times 1 = 16\)

\(4\times 2\times 2 \times 1 = 16\)

\(16+16+4=36\)

The total number of pathways are \(4^4=256\)

So, \(\frac{36}{256}=\frac{9}{64}\).

(by the way, can anyone check this, preferably one of the moderators? i'm not 100% sure. thanks!)

You are very welcome!
:P

No problem! Glad to help!

That's what I said. Just trying to list a few squares out, to find the answer.

Also, to cover the \(361\) completely with the square root radical, \(\sqrt{361}\) = \sqrt{361}

Or, you can list the first few squares out. If you know \(20^2=400\), ​\(19^2=381. \)

668/7 = 95.4285714285714286    Number of weeks in a Sol-year

     the  decimal portion will be part of the sol week

         0.4285714285714286 x 7 = 3.0000000000000002 days

                 every 7 years would be 21 days or 3 weeks

                        so in 7 years they would need 3 Leap Years   

                        3/7

Nice work, Guest! I counted the ones that didn't work to get my answer. 

A curve has the equation y= 12 + x^2 (root)x / x    , x > 0

a) express the equation in the form 12x^p + x^q

b) I) hence find dy/dx

ii) find an equation of the normal to the curve at the point on the curve where x=4

iii) the curve has a stationary point P. Show that the x-coordinate of P can be written in the form 2^k, where k is a rational number 

nnnnnn? Why did you delete your question? Other people may have the same or similar questions. 

Actually, tertre is correct. The answer is 5/12. But what are the cases?

Thanks Alan,

I had a super stupid error in mine but I have corrected it and I think our answers are the same now 

To roll either a total of 6 or 12 you have to roll the following:

6+6 = 12

1 + 5 = 6

2 + 4 = 6

3 + 3 = 6

5 + 1 = 6

4 + 2 = 6

I'm assuming that the two dice are distinquishible(such as one is RED and the 2nd GREEN), then you have a total of 6 / 6^2 =6/36 =1/6 probability.

Partitions of (7) =15 as follows:

7 = 7
6 + 1 = 7
5 + 2 = 7
5 + 1 + 1 = 7
4 + 3 = 7
4 + 2 + 1 = 7
4 + 1 + 1 + 1 = 7
3 + 3 + 1 = 7
3 + 2 + 2 = 7
3 + 2 + 1 + 1 = 7
3 + 1 + 1 + 1 + 1 = 7
2 + 2 + 2 + 1 = 7
2 + 2 + 1 + 1 + 1 = 7
2 + 1 + 1 + 1 + 1 + 1 = 7
1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

I count 12 including 7=7, which gives you: 12/15 =4/5 probability.

Look under your question of yesterday where you changed 9.58 seconds to 11.53 ticks. if 11.53 ticks is accurate, then the answer is 1.11 rounded to the nearest hundredth place.

What are the cases? Could you please explain in more detail?

Alan's calculation is exactly 1 1/3. If you have to round it to the 100th decimal place, then it becomes 1.33.