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Right back at ya

A bag contains five white balls and four black balls. Your goal is to draw two black balls.

You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

Winning draws

BB                        (4/9)*(3/8)   =  12/72

BW   B                 (4/9)*(5/8)*(3/8)    =  60/576

WB    B                 (5/9)*(4/8)*(3/8)    =  60/576

WW   BB               (5/9)*(4/8)*(4/9)*(3/8)  =  240/5184

ADDED = 12/72+120/576+240/5184 = 91/216

Non winning draws

WW   WB             (5/9)*(4/8)*(5/9)*(4/8)    =   400/5184

WW   BW             (5/9)*(4/8)*(4/9)*(5/8)      =   400/5184

WW   WW             (5/9)*(4/8)*(5/9)*(4/8)    =    400/5184

WB     W                (5/9)*(4/8)*(5/8)            =    100/576

BW      W                (4/9)*(5/8)*(5/8)            =   100/576

ADDED= 1200/5184+200/576 =125/216

125+91=216 great.

I draw a probability tree and listed all the probabilities on it.

I ended up with 9 possible scenarios and 4 of them were favourable.

If I work out the probabilities of each of the favourable ones and add them together then hopefully that will be the answer.

Although I would check it by adding all possible outcomes probabilities together (They should add to 1)

Sorry but that answer is incorrect!

Let  x = 0.51010101010....

10x = 5.1010101010....

1000x = 510.1010101010....

1000x - 10 x = 505

990x = 505

x = 505/990 = 101/198

It is everything BELOW and INCLUDING the line...so it is a   <=   which rules out the first two choices

    it includes the line so the last choice is rules out ....leaving only the third choice 

THANK YOU FOR THE HEADS UP! SORRY AGAIN. Hope you can understand my apology. 

Sorry Guest, your answer is very well thought out, but I think there could be some misunderstanding in the problem. Also, I have thumbs up your answer. And THANK YOU so much for you effort as well as time. SORRY, I won't do it again. :(

It's about the likes you have

I am assuming you mean "no two girls". Anyways, we see, that the only way for no girl to be together is if they sit BGBGBG or GBGBGB.

Therefore, we can do a premutation to see how many different ways this is possible. 

So:

2 * 3! * 3! = 72.

The unit's digit must be either a 4 or a 6 for the square to end in 6; all other possibilites can be excluded by trial.

If you are squaring a two-digit number, you can represent this number as  10a + b  (knowing that b must be

either 4 or 6).

(10a + b)²  =  100a² + 20ab + b²

The  100a²  can be ignored for it does not affect the ten's digit.

20ab  must be even; an even number, 20, times either an even number or an odd number results in an even number.

To this term (20ab) must be added the carry of the b² term.

If  b = 4,  the carry is 1.

If  b = 6,  the carry is 3.

Adding an odd number (either 1 or 3) to an even number (20ab) results in an odd number.

By trying various possibiliteis, you will discover that the ten's digit can be any of the odd digits.

okay Melody, I am sorry, and will not do that again, I just am really stuck on this problem

The formula for the surface area of a sphere is:  Area  =  4·pi·r².

If the basketball has a diameter of 24 cm, its radius is 12 cm.

Area  =  4·pi·r²   --->   Area  =  4·pi·12²  =  1809.56 sq cm.

If the cost is  $71.62  per sq m, the cost, per square cm is  $71.62 / 100²  =   $ 0.007162.

Final cost:  $ 0.007162 x 1809.56  =  $12.96.

For this, we can make an inequality. So, here is what we have, according to the question:

2 sacks of sugar + 3 sacks of flour \(\leq\) 40 lbs

flour \(\leq\) 2 sacks of sugar + 6lbs

Using variable form, this is:

2s + 3f \leq 40

f \leq 2s+6

Since the question is asking for the largest possible weight, we can change the /leq to equal signs

2s + 3f = 40

f = 2s+6

Now, we can isolate the two stacks of sugar. So:

f-6 = 2s

40-3f = 2s

Now, our equation is:

f-6=40-3f

4f=46

f = 11.5 lbs

Therefore, the flour weighs 11.5 lbs.

Thanks for the answer! Just wondering, (this may be a glitch) do you see more than one post on this? It says that there are 3 replys, but only one post has shown up.

To do this, we first need to expand the equation. This will be : 

8a+12=20

Subtracting 12 from both sides, we get:

8a=8

Divide both sides by 8, and we get:

a=1

Hello Guest!

To do this problem, we can use a version of the \(d=rt\) formula as a justification of division. 

We have \(\frac{4000}{25} = 160\). In essence, this is really a division problem.

So, 160 days.

Hope this helps Guest!

Okay, by how did you get that? I was squaring some of the numbers, and I noticed that they were odd, but do you know how to prove that?

Here's a ittle more help:

Now differentiate k with respect to \(\theta\) , set to zero and solve for \(\theta\)  Select the solution that gives a value for k leading to the maximal area.

As with Melody above to,

\(\displaystyle 5x^{4}-8x^{3}+2x^{2}+4x+7\equiv a(x+2)^{4}+b(x+2)^{3}+c(x+2)^{2}+d(x+2)+e,\)

but now substitute x = -1.

Consider theta to be given and calculate the time of the flight, ( solve y = 0), and from that, the horizontal distance covered.

This will be the upper limit for the integral that follows.

The parameter is t, so eliminate that between the equations x = and y = to arrive at an equation y = ' a function of x and theta '.

The area under the flight path will be the integral of y wrt x between x = 0 and x = the limit calculated earlier.

The result of that is a function involving v, g and theta. You now have to calculate the value of theta for maximum area, 

(differentiate wrt theta and put that equal to zero etc.).

Post again if you need further help.

Attn: Iamhappy

I do not like that you have given your guest answerer thumbs down.

If someone answers you, you should thank them for their interest and their time.

If you do not think the answer is correct then you open a polite conversation with your reasons.

To give thumbs down to someone who has tried to help you is rude.

Note: I have not looked at the answer as yet so I have no idea whether or not it has any merit.

But thank you guest answerer for your time and interest.

Given that \(x+y=\dfrac7{10}\) and \(x-y=\dfrac5{14}\),

find the value of \(x^2-y^2\), expressed as a common fraction.

\(\begin{array}{|rcll|} \hline x^2-y^2 &=& (x+y)(x-y) \\\\ x^2-y^2 &=& \dfrac7{10}*\dfrac5{14} \\\\ \mathbf{x^2-y^2} &=& \mathbf{\dfrac14} \\ \hline \end{array}\)