QBD kill waiting for Greenfingers to reset. Being on the Aus server severely limited my killspeed (if you want lag, go to w49). Anyway:-

 

910K Sell price. Should be done with the 100 torstols sometime in the next day or so, then ill use up the excess from SOL, sell up and repeat. I did, however, do a price check on what I've gained so far. I've lost 4 seeds whilst I've been asleep (time to RQ sleep), but I did break my initial required amount today (note: the +5 is the seeds planted currently):-

 

Prices, however, have gone up, rather nicely as well. So now, to break even I only need 738 herbs, delicious 100 herbs more profit. Best run thus far:-

 

Shame that 65 is best, but at least I'm getting a consistent number of herbs

So what is my cut. I'll settle for 35 but I can negotiate.

• RuneScape Status:Retired
  

Clearly I'm going to RWT

Inb4MishyFC

Our friend Mishy doing a scamming FC?

Sounds a bit Fishy.

 

 

teeheehee

GOOD [bleep]ING FIGHT

 

100 Seeds dusted.

 

 

7m profit + 13 seeds saved (2m) which I will use later. Not too bad at all.

- 10.67 herbs/seed planted (233.1k gross, 71.2k profit)

- 13% seeds saved with SOL

 

Numbers would've been higher (closer to 1100 total) had it not been for 4 overnight deaths.

1)

 

f(x) = 2x + 3, and a point (1,3)

 

Which points of f(x) are 5 away of the point (1,3).

 

2) There are 4 black balls and 5 white balls. You pick up 3 balls. What is the probability of having two different coloured balls.

teehee u sed ballz

you smell'd dat word from 4 forums away, only true gaybois can do dat

 

2) There are 4 black balls and 5 white balls. You pick up 3 balls. What is the probability of having two different coloured balls.

 

P(B&W) = 1 - (P(BBB)+P(WWW))

2) There are 4 black balls and 5 white balls. You pick up 3 balls. What is the probability of having two different coloured balls.

 

P(B&W) = 1 - (P(BBB)+P(WWW))

 

And how did you come to this equation. I am curious too now.

The way I read the question is "Probability of getting both black and white in your selection". This, being the complement of P(All black)+P(All white). So you can use the Complement formula P(Complement)=1-P(Event)

[bleep] MATHS

[bleep] MATHS

Well said.

 

And how I sailing plz

 

[bleep] MATHS

Well said.

 

[bleep] you both.

2) There are 4 black balls and 5 white balls. You pick up 3 balls. What is the probability of having two different coloured balls.

 

P(B&W) = 1 - (P(BBB)+P(WWW))

 

And how did you come to this equation. I am curious too now.

 

[spoiler=Potentially Incorrect depending on interpretation of question]

P(BBB) = 5/9 x 4/8 x 3/7

P(BBB) = 5/42

P(WWW) = 4/9 x 3/8 x 2/7

P(WWW) = 2/42 (can be simplified but for convenience)

 

P(2 Colours Drawn) = 1 - (P(BBB)+P(WWW)) ----------- Complementary Events

P(2 Colours Drawn) = 1 - (2/42 + 5/42)

P(2 Colours Drawn) = 5/6

 

Hmm... Thank you very much man. I see you didn't bother to think about the vector question, but it doesn't matter, because vectors are completely gay.

Hmm... Thank you very much man. I see you didn't bother to think about the vector question, but it doesn't matter, because vectors are completely gay.

 

Nah I've thought about it a bit, just haven't figured out yet a [bleep]ing sensible answer. Mind you, I haven't looked at it seriously.

 

 

Ok. It is quite obviously a circle. Don't ask me why I know it, I just [bleep]ing well do. The circle has radius 5, and centre (1,3)

 

[hide=Auror you noob]r² = (x-a)²+(y-b)²

But, you are also given the equation for y (re: f(x))

y = 2x + 3

a = 1

b = 2

 

Sub it all in, you get

 

(x-1)² + (y-3)² = 25

Using y = 2x-3

(x-1)² + (2x+3-3)² = 25

 

Which folds out to

5x² - 2x + 1 = 25

5x² - 2x - 24 = 0

 

Use the quadratic formula to show your working, you get the roots (re: x-values) to be 2.4 and -2

Sub those into f(x) to get your y-values.

 

Answer: (2.4,7.8) and (-2,-1)

 

In summation: [bleep] you[/hide]

"Ok. It is quite obviously a circle. Don't ask me why I know it, I just [bleep]ing well do. The circle has radius 5, and centre (1,3)"

 

holy shit, why the [bleep] couldn't I have thought of that. I tried to use the distance-between-two-points formula... what the [bleep] am I doing in college.

 

But yeah, thank you very much for helping me. I really like the way you sware :) You'll become a great teacher.

Take the square root of the formula for a general circle, you get the distance formula. :wink:

 

That wasn't the problem, it was recognising to begin with it was a circle, AND THEN recognising that you had to solve simultaneously using the substitution method.

i asked my teacher, who gave us these exercises, how to do the probability exercise I asked you here. She told me to write down all possible combinations, but that didn't make sense in my eyes, because there were a total of 9 balls ( 4 black and 5 white),and I was to pick up only three balls, and because a certain ball is not individual, It felt way too abstract for me.

 

Not sure if you understood, but that is what I was thinking.

Writing it down in that sense is a bit of a stupid move, at least for me. I much, much, much prefer the probability trees. Not only is it much more efficient, you can easily figure out probabilities if you do it properly. Here is an example:-

 

3 green marbles, 2 white marbles. When you pick a ball, you note it down then put it back the bag.

 

So you can multiple across vines:-

P(WW) = (2/5)²

 

Your question though, didn't specify to replace the balls, that's why the denominator changes across branches. Apart from that, the process is exactly the same.