Melody

A stack contains 7 green marbles and 3 red marbles. 2 marbles are taken from the sack w/out replacement. What is the probability of taking one green marble and one red marble (in any order)?

P(Green followed by Red)  =  \(\frac{7}{10}*\frac{3}{9}=\frac{21}{90}\)

Now do something very similar to get the prob of a red followed by a green.

Then add the answers together and simplify.

you use the formula     Tn=a+(n-1)d

\(a+22d=\frac{2}{3}\qquad (1)\\ a+52d=\frac{3}{2}\qquad (2)\\~\\ (2)-(1)\\ 30d=\frac{3}{2}-\frac{2}{3}\\ 30d=\frac{5}{6}\\ d=\frac{1}{36}\\ \)

Check what i have done, then sub d in to find a

Then find  a₃₅

_{Any questions? , then i am happy to discuss further.}

Each of the digits​ 0-9 is written on a slip of​ paper, and the slips are placed in a hat. If 5 slips of paper are selected at​ random, determine the probability that the 5 numbers selected are greater than 3.

4,5,6,7,8,9    there are 6 favourable numbers out of 10

6C5  /  10C5   =  6 / 252 =      1/42

I did a similar question yesterday.  ...

what have you done towards solving this question.

Originally there are  B bags in shop B   and  B+180 bage in shop A

then after transfer

 there will be 

  \(B+\frac{B+180}{5}\;\;bags\;\ in\; shop\;\;B\)

and

\(\frac{4}{5}(B+180)\)    bags in shop A

You can take it from there.

(2z - 1/sqrt(z) + z^3)^9.

\((2z - \frac{1}{\sqrt z}+ z^3)^9\)

Is this your intended question?

C to Y is 37 and  C to L is 51  so the minimum distance possible between Y and L is 14

Now M to Y is 6 and  and M to L is 8 .   

 I already know that YL is a minimum of 14

So that means that   MYL must be collinear points  ( all on one line.

this means that C, Y, M, L are all  collinear.

You can finish it.

Here is a pic.

\(ar^{22}=16\\ ar^{27}=24\)

solve simultaneously to find a and r

etc

You may see this properly depending on what extentions you have on your computer but we do not.

Get rid of the dollar signs.

I do not know if guests answer is right or wrong, I have not checked.

Thanks for answering guest :)

Let

\(f(x)=x^3(ax^2+bx+c)\quad so\\ f(x)+2=x^3(ax^2+bx+c)+2\qquad (1)\\ f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But also}\\~\\ f(x)+2=(x+1)^3(dx^2+ex+f)\qquad (2)\\ \text{By inspection I can see that }f=2\\ f(x)+2=(x+1)^3(dx^2+ex+2)\\ f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\ \text{The x term will be }6x+ex=(6+e)x\\ \text{ but the coefficient of x must be 0 so e=-6}\\~\\ \text{The x squared term will be }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\ \text{ but the coefficient of x squared must be 0 so d=12}\\ so\\ f(x)+2=(x+1)^3(dx^2+ex+f)\\ f(x)+2=(x+1)^3(12x^2-6x+2)\\ \text{expand etc} \)

The logic is good but you need to check for careless errors.

LaTex:

f(x)=x^3(ax^2+bx+c)\quad so\\
f(x)+2=x^3(ax^2+bx+c)+2\qquad (1)\\
f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But  also}\\~\\
f(x)+2=(x+1)^3(dx^2+ex+f)\qquad (2)\\
\text{By inspection I can see that   }f=2\\
f(x)+2=(x+1)^3(dx^2+ex+2)\\
f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\
\text{The x term will be   }6x+ex=(6+e)x\\
\text{ but the coefficient of x must be 0 so e=-6}\\~\\
\text{The x squared term will be   }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\
\text{ but the coefficient of x squared must be 0 so d=12}\\
so\\
f(x)+2=(x+1)^3(dx^2+ex+f)\\
f(x)+2=(x+1)^3(12x^2-6x+2)\\
\text{expand etc}