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Me

lol

If you are talking about separately.....the chances of winning the first is .03 and the chance of winning the second  is about  5/150 = 1/30  = about  .033

If you are talking about winning both, then the answer is   .03 * .033  = about .0099  = about .01 rounded =  1% chance

And I am also single....................a master answerer.........

me :/

Cross multiply, so 10b=8b-64 subtract 8b from both sides, so 2b=-64, divide both sides by 2 and you have b=-32

\(36 : 546 = x : 100\)

\(x = \frac{36*100}{546} = 6,59%\)

.03*.033=.099% chance

Nah, Bruh

6.59%

2x^2-7x+5=0   factor

(2x - 5) (x -1 )  =  0

Set each factor to 0

2x - 5  = 0                                     and                  x  - 1  = 0

add 5 to both sides                                              add 1 to both sides

2x = 5                                                                   x = 1

divide both sides b 2

x = 5/2

So.....the two answers are   x =5/2      and     x =  1

2x^2 - 7x + 5 = 0.

2x^2 - 2x - 5x + 5 = 0. (In any formula ax^2+bx+c, find two numbers, m and n, where m+n = b and m*n = a*c).

2x(x-1) - 5(x-1) = 0. (Step 2: Separate the equation in two, and find the Highest Common Factor (in these cases, 2x and 5, respectively. Make sure the expressions in brackets match!)

(x-1)(2x-5) = 0. (Finally, remove the in-brackets expression from both halves, and place what remains in it's own bracket.)

And there you have it.

8.91*10⁹

P(X)=F+V(X)

Vans.

Better than those!

I'm not entirely sure how to calculate it, but there is a way to visualise the solution. Picture 6 dice, each with a different number on it. These collectively represent the first die. Then draw 6 lines from each of those, at the end of which are 6 dice per original die, each with their own unique number from 1-6. They represent the second thrown die. Do this 9 more times, and you will have graphically represented each and every possible outcome of rolling 11 dice in a row (exactly 6^11 outcomes). Now, all you need to do is look at each one, and determine x/(6^11), where x is how many of thoise outcomes fit the original conditions. I definitely don't have the time to do all that, but it's the only way I can think of to calculate those conditions.

Sorry if this doesn't help...

Infinity times 2

Me 2! Except you havent been messageing me sooo....

???

LOL, ;}

Yeah you have asked that so many times now

oh, ok

its not me ;) lolol

24.7

I messaged someone.They didn't message me back, so.