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There are 6 green balls and 4 red balls, making 10 total balls. The probability of picking a green ball first is \(\frac{6}{10}\) .

Since there are replacements, there will be still 10 balls in the bag. Now, for the second draw, you pick green again; making the probability, \(\frac{5}{10}\) . So, the probability of picking a green ball both times is \(\frac{6}{10}*\frac{5}{10}=\frac{30}{100}=\frac{3}{10}\) .

Next, the probability of picking a red ball is \(\frac{4}{10}\) . Doing the same as before, we get: \(\frac{4}{10}*\frac{3}{10}=\frac{3}{25}\) .

I'm leaving the next step up to you, to solve... What can we do after this?

I understand and b, just not c

You do not want help, you just want someone else to do all your homework for you. 

sorry I'm new to this

I will admit that I am tired but you lost me on the third line.

If it takes 68 meters per minute for the ship's ascent, the question is asking us to find how many minutes it will take reach 631 meters.

First, the ship is at a depth of 1583 meters, so to reach 631 meters, it has to go: \(1583-631=952\) meters.

We also have our proportion: 68 meters every 1 minute, so 952 meters in \(\frac{952}{68}=\boxed{14}\) minutes.

Aren't all these terms in your textbook?

1-Circumcenter
2-The incenter
3-Sides
4-Equilateral
5- 5 3/5. The only explanation you need in number 5 is to subtract the first term(1/5) from the second term(2). So: 2 - 1/5 = 1 4/5 - This is the common difference you ADD to each term.

Here is a video to watch that explains it.

xyz =2[x+y+z], solve for x

x = (2 (y + z))/(y z - 2) and yz cannot=2

Therefore x cannot be > 1, because it will violate x ≤ y ≤ z

I can only see 2 solutions as follows:

x = 1,  y = 3   z = 8, and

x = 1,  y = 4,  z = 5

Sum of z =5 + 8 = 13

Looks interesting :)

Yes i am in agreement.

You have asked this same question at least 2 times before!

If you still need an answer to an older question then put a link to your original question in your new post and aske people to answer on the original post please.

If you do not like someone's answer. If you do not understand it or if you have reason to think it is incorrect then say so (politely) on the original post!

you already asked this question, and a guest asked you to clarify because your question makes no sense...

edit: jesus christ, this is the third time you post this question, you don't even check the answers, stop reposting questions

[4/52]*[13/52]*[4/52]

=1/676

if with replacement

if without replacement ,

[4/52]*[12/48]*[3/36}

=1872

Here's a graph that conforms to the specified vertices:

I know little or nothing about graph theory, so I wrote a brute force and ignorance program to calculate the number of paths of length 2 here and came up with 77.  However, it wouldn't surprise me if this were wrong!

The initial volume of tank: \(40*25*30=30,000\) inches^3

The depth of 24 inches: \(40*25*24=24,000\) inches^3

If the rock, with a volume of 500cm^3, is put in, the total volume will be:\(24,000+500=24,500\) icnhes^3

We know that our aquarium is 40 inches long, and 25 inches wide, so we have to find the height, that will equal our volume of 24,500 icnhes^3.

So, \(40*25=1000\), and inches\(\frac{24500}{1000}=24.5 \). We know that our initial height was 24 inches, thus \(24.5-24=\boxed{0.5}\) inches increase.

Here's an example: Dan has 7 2/3 meters of yarn and his friend, Evan, has 4 1/15 meter of yarn. What is the difference between these two amounts of yarn? 

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.
What is the 23rd term?

AP Formula:

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_5 &=& 9 \qquad & i = 5 \\ a_{32} &=& -84 \qquad & j = 32 \\ a_{23} &=& \ ? \qquad & k = 23 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ a_5(32-23) + a_{32}(23-5) +a_{23}(5-32) &=& 0 \\ 9\cdot 9 - 84 \cdot 18 - a_{23}\cdot 27 &=& 0 \\ a_{23}\cdot 27 &=& 9\cdot 9 - 84 \cdot 18 \\ a_{23}\cdot 27 &=& 81 - 1512 \\ a_{23}\cdot 27 &=& - 1431 \\\\ a_{23} &=& -\dfrac{ 1431 } {27} \\\\ \mathbf{ a_{23} } & \mathbf{=} & \mathbf{ -53 } \\ \hline \end{array}\)

The 23rd term is -53

For some integers that are not palindromes, like 91, a person can create a palindrome by repeatedly reversing
the number and adding the original number to its reverse. For example, .
Then , which is a palindrome, so 91 takes two steps to become a palindrome. Of all positive integers

between 10 and 100,
what is the sum of the non-palindrome integers that take exactly six steps to become palindromes?

Why would a difference in the dice (assuming it does not affect the fairness) make the probability outcomes different?  

However, the most irritating of all, are guests who hide behind their anonymity and spew BS that absolutely does not help the growth of the questioner. 

Saying "I hope this helped" is just simply a positive signature of mine, concluding my solution, much better than starting your answer off with "this is easy". I think this, we can both agree on. 

So, Guest, instead of taking your time to write these futile responses, please go learn some math, or how to use the comma, or something else that will contribute to the overall mentality of the user of this website. 

I hope this helped,

Gavin. 

Thanks Guest :)

Are the dice all the same or all different, colours or sizes I mean?

It is complicated either way but it does make a difference.

It would be 1 - the prob of 2 not being the same.

= 1- the prob that they are all different.

If the dice are all different then there are 6! way that they can be all different.

If the dice are all different then there are 6^6 variations on what can be thrown.

So prob that they are all different = \(\frac{6!}{6^6}\)

and the probability that they ar not all different = \(1-\frac{6!}{6^6}\)

1-6!/(6^6) = 0.9845679012345679 

so the prob that at least 2 die show the same number is about 98.5%

Answer to the first question:a path of length 2 consists of 3 ordered vetrices, where the first vetrice is connected to the second, and the second vertice is connected to the third (the first vetrice and the third vertice can't be the same vetrice).

In a path of length 2, the second vertice has to be connected to at least 2 other vertices, meaning that the second vetrice's degree has to be at least 2. suppose we have a vetrice v, deg(v)=r>1, to find the number of paths of length 2 where the second vertice is v, we have to find the number of vertices that are connected to v (that's r, by definition) and multiply it by r-1 (so, if we have a vertice with degree r, the number of paths of length 2 where the second vertice is v is r(r-1)).

why? we already know that the second vertice is v, all we have to do is find other 2 vertices, that are connected to the second one. the order of the vertices matters, because one has to be the first vertice (where the path begins) and one has to be the last vertice (where the path ends). we have r vertices to choose from, so we have a total number of r(r-1) combinations.

to find the total number of paths with length 2, we have to find deg(v)*(deg(v)-1) for every vertice v with deg(v)>1, and find the sum of the results.

so the answer should be 2*(2-1)+3*(3-1)+3*(3-1)+3*(3-1)+4*(4-1)+4*(4-1)+5*(5-1)+8*(8-1)=2+6+6+6+12+12+20+56=120

i hope im right