All Answers 358149 Answers

Veey nice, Alan and Melody......!!!

$${\mathtt{44}}{\mathtt{\,\times\,}}{\mathtt{3}} = {\mathtt{132}}$$     eggs are needed.

$${\frac{{\mathtt{132}}}{{\mathtt{23}}}} = {\mathtt{5.739\: \!130\: \!434\: \!782\: \!608\: \!7}}$$

So you will need 6 packets

$${\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{23}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{6}}$$

The children will get 3 each and Mrs Rodgers will get 6 for herself.  :)

Looks like you and I agree for the "with replacement" case Melody.

.

The anwser is: 4

all you have to do is find the height (h), and the radius (r), and plug it in to a simple formula: V = hπr².

I am going to assume the cards are replaced each time.

Let x be the number of blue and 60-x be the number of red

P(at least 2 red) >=0.5

P(0 or 1 red) <=0.5

P(all blue)+P(1R and 5 blue) <=0.5

$$\\\frac{x^6}{60^6}+6C1\left[\frac{(60-x)}{60}\right]^1\left[\frac{x}{60}\right]^5\le0.5\\\\\\
\frac{x^6}{60^6}+6\left[\frac{(60-x)x^5}{60^6}\right]\le0.5\\\\\\$$

I used Wolfram|Alpha to solve this

Eventually I might get this right!  Here's my latest attempt:

Now with replacement:

.

So, with and without replacement cases require 16 red cards.

I do not understand your formula Nauseated.  Perhaps you could explain how you derived it?  Thanks

Anyway, i have contined this theme on a new post:

here it is   

@@ End of Day Wrap   Sat 28/3/15 Sydney, Australia Time 9:55 am     ♪ ♫

Hi all,

Great answers today from Falkirks, TayJay, Civonumzuk, Alan, Heureka, CPhill, Geno3141 and BrittanyJ.

Thank you  

Interest Posts:

    FTJ means "for the juniors"

use    [2nd]

WOW Heureka !    

What does "only 9 station ways " refer to?

Sun 29/3/15

    FTJ means "for the juniors"

$$nodes:
\begin{array}{ccccccccccc}
A(1) & \rightarrow & 2 & \rightarrow & 3 & \rightarrow & 4\\
\downarrow && \downarrow && \downarrow && \downarrow & \\
5 & \rightarrow & 6 & \rightarrow & 7 & \rightarrow & 8 & \rightarrow & 9 & \rightarrow & 10 \\
\downarrow && \downarrow && \downarrow && \downarrow && \downarrow && \downarrow \\
11 & \rightarrow & 12 & \rightarrow & 13 & \rightarrow & 14 & \rightarrow & 15 & \rightarrow & 16 \\
\downarrow && \downarrow && \downarrow && \downarrow && \downarrow && \downarrow \\
17 & \rightarrow & 18 & \rightarrow & 19 & \rightarrow & 20 & \rightarrow & 21 & \rightarrow & 22 \\
&&&&&& \downarrow && \downarrow && \downarrow\\
&&&& && 23 & \rightarrow & 24 & \rightarrow & B(25)\\
\end{array}$$

adjacency matrix A:

$$Matrix\ A = \bordermatrix{
nodes & A & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & \textcolor[rgb]{1,0,0}{B} \cr
\textcolor[rgb]{1,0,0}{A} &0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&\textcolor[rgb]{1,0,0}{0} \cr
2 &0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
3 &0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
4 &0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
5 &0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
6 &0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
7 &0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0 \cr
8 &0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0 \cr
9 &0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0 \cr
10&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0 \cr
11&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0 \cr
12&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0 \cr
13&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0 \cr
14&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0 \cr
15&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0 \cr
16&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0 \cr
17&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0 \cr
18&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0 \cr
19&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0 \cr
20&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0 \cr
21&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0 \cr
22&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1 \cr
23&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0 \cr
24&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1 \cr
B &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
}$$

 $$\small{\text{
entrys $
\qquad 1 = $ one way $\quad 0 = $no way from node x to node y}}$$

$$\textcolor[rgb]{1,0,0}{Matrix \ element[A][B]}$$

$$Matrix\ A^9 =A\cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A \cdot A = \bordermatrix{
nodes & A & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & \textcolor[rgb]{1,0,0}{B} \cr
\textcolor[rgb]{1,0,0}{A} &0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&\textcolor[rgb]{1,0,0}{106} \cr
2 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
3 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
4 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
5 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
6 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
7 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
8 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
9 &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
10&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
11&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
12&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
13&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
14&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
15&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
16&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
17&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
18&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
19&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
20&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
21&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
22&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
23&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
24&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
B &0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0 \cr
}$$

A -> B

  $$A^1: \text{ Matrix element[A][B] } =0\quad \text{ (1-station-way)} \\
A*A=A^2: \text{ Matrix element[A][B] } =0 \quad\text{ (2-station-way)} \\
A*A*A=A^3 : \text{ Matrix element[A][B] } =0 \quad\text{ (3-station-way)} \\
A*A*A*A=A^4 : \text{ Matrix element[A][B] } =0 \quad\text{ (4-station-way)}\\
A*A*A*A*A=A^5 : \text{ Matrix element[A][B] } =0 \quad\text{ (5-station-way)} \\
A^6 : \text{ Matrix element[A][B] } =0 \quad\text{ (6-station-way)} \\
A^7 : \text{ Matrix element[A][B] } =0 \quad\text{ (7-station-way)} \\
A^8 : \text{ Matrix element[A][B] } =0 \quad\text{ (8-station-way)}\\
A^9 : \text{ Matrix element[A][B] } =106 \quad\text{ (9-station-way)} \\
A^{10} : \text{ Matrix element[A][B] } =0 \quad\text{ (10-station-way)} \\
etc.: \text{ Matrix element[A][B] } = 0$$

A -> B  (only 9-station ways) = 106

Yes you can and it is a little easier but it is a starting point to learn LaTex and with LaTex you can present a lot more things than you can with the calculator. 

Look at the table Heureka added on this thread.  You could not do that with the calculator!

Almost all Heureka's and my good presentations are done using LaTex.

Maybe you could explain your question better or give a proper example because I do not understand what you mean.   

This is often confusing for me too. 

I tend to keep lots of places and only round off at the end.  That is usually safest.

What a great question.  I like it!  

I have a different answer from Geno because have made different assumptions.

I have assumed that steps can be retraced so to get (0,0) for instance you could just keep going one place left then one right, one left, one right etc

I don't think that you can get the ones I have been left out but you can think about it for yourself. 

Let me see,

I have

1 at the origin,

4*5 on the rest of the axes = 20

25 in the 1st quad so that is 100 in all four quads

1+20+100=121

$$\sqrt{130}=\frac{\sqrt{130}}{1}$$

that is it, you cannot simplify sqrt(130)

That is a really good explanation - thanks Geno  

This is Geno's answer.

As you can see it is the same as CPhill's answer.    :)

$$\left({\frac{{\mathtt{15}}}{{\mathtt{60}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{14}}}{{\mathtt{59}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{45}}}{{\mathtt{58}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{44}}}{{\mathtt{57}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{43}}}{{\mathtt{56}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{42}}}{{\mathtt{55}}}}\right){\mathtt{\,\times\,}}{\mathtt{15}} = {\mathtt{0.312\: \!490\: \!387\: \!277\: \!369\: \!3}}$$

enter it into the web2 calc like this

acos(0.765)

Is this what you mean?

$$\frac{(2.34)^2}{68 -3.87}$$

it goes into the web2 calc like this

 (2.34)^2  /( 68 -3.87)

arc tan is the same as inverse tan

so use   atan

Good spotting Chris :)