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Thanks for sharing this :)

Thank you so much! You were right!!! (Just checked)

Since the expected loss is 50 cents, we know it is more likely to lose than to win. It is also impossible to get -50, and the total difference between the 2 possibilities is 2, 50/200=1/4, so the chance of winning is 1/4 and the chance of losing is 3/4. Since (1/4)*3=3/4,  the amount of black balls are 3 times the amount of white balls, so k=15. 

Hope this helps!! :)

Nevermind the answer is 3/16.

If someone asked me to turn my back while they rolled two dice onto a table top, and then asked me how many 1's I would expect to see when I turned round to look at them, I would say that I expected to see none.

The likelyhood of there being either one or two 1's is roughly 30.56% so it's more than twice as likely that there will be no 1's as there being either one or two.

It will not matter how many times we repeat this, my answer will always be the same, none.

If though I was told that we were going to do this say 100 times, and on how many times would I expect my expectation to be wrong, well that's a different question.

Since 30 = 2 x 3 x 5, therefore it has common factors with the following numbers:
2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30[22 numbers]
It is relatively prime with following numbers:
1, 7, 11, 13, 17, 19, 23, 29[8 numbers]
Therefore, the probability that a number =< 30 is relatively prime to 30 is:
8 / 30 = 4 / 15

And then, each of the six tiles can be either blue or red (2 choices for each one of a set of 6 elements)

Solve for t:
56 = 60 t - 16 t^2

56 = 60 t - 16 t^2 is equivalent to 60 t - 16 t^2 = 56:
60 t - 16 t^2 = 56

Divide both sides by -16:
t^2 - (15 t)/4 = -7/2

Add 225/64 to both sides:
t^2 - (15 t)/4 + 225/64 = 1/64

Write the left hand side as a square:
(t - 15/8)^2 = 1/64

Take the square root of both sides:
t - 15/8 = 1/8 or t - 15/8 = -1/8

Add 15/8 to both sides:
t = 2 or t - 15/8 = -1/8

Add 15/8 to both sides:
t = 7/4 = 1.75 seconds. But after 1.75 seconds it continues to go up for about 1/4 foot. And coming down to 56 feet again at exactly t= 2 seconds.

Note: somebody should check this.

No I don't think so...

Hint: How many permutations of three numbers{-1, 0, 1} can you arrange so that their SUM is always zero?

3! = {-1, 0, 1} | {-1, 1, 0} | {0, -1, 1} | {0, 1, -1} | {1, -1, 0} | {1, 0, -1} (total: 6). But since you have unlimited supply of these tiles, then {0, 0, 0} would also qualify?? Makes sense?

One very simple but "foolproof" method of converting rational decimals to proper fractions is the "continued fraction" method, which goes as follows:
Take the reciprocal of the fraction: 0.12515622210319585788252097839671, and you should get:
7.9900142653352353780313837375178. Subtract the integer part, or 7, and take the reciprocal of the fraction and you should get:
1.0100864553314121037463976945245. Subtract the integer part, or 1, and take the reciprocal of the fraction and you should get:
99.142857142857142857142857142857. Subtract the integer part, or 99, and take the reciprocal of the fraction and you should get:
7. Once it converges to a whole number, then do this:
Multiply: 7 x 99.142857142857142857142857142857 =694. Then multiply this: 694 x 1.0100864553314121037463976945245 = 701. Since this is the second last multiplication, then this must the NUMERATOR.
Finally, multiply this NUMERATOR x the first number that we listed and you should get:
701 x  7.9900142653352353780313837375178=5,601 - This is the DENOMINATOR. So, the proper fraction of this rational decimal is:
701 / 5,601.
Note: This decimal fraction does not repeat its digits for 933 decimal places. Or, Its "period" is 933 digits long.

What happened to your picture?

Thanks so much!! :D

3)

Solve for n:
sqrt(sqrt(sqrt(n) + 2) + 1) = 2

Raise both sides to the power of two:
sqrt(sqrt(n) + 2) + 1 = 4

Subtract 1 from both sides:
sqrt(sqrt(n) + 2) = 3

Raise both sides to the power of two:
sqrt(n) + 2 = 9

Subtract 2 from both sides:
sqrt(n) = 7

Raise both sides to the power of two:
n = 49

The expected value is the probability-weighted average.

There are two dice, each with a probability of (1/6) of rolling a one, so multiply the probability by two (2).  (1/6) *(2) = (1/3)

The expected number of 1's is (1/3).

GA

I don't know the difference between "the expected value of 1's obtained" and the probability of obtaining at least one 1 on a 2 dice roll. However, I shall attempt to answer the latter:

The probability of NOT getting a 1 on the first die is: 5/6, and:

The probability of NOT getting a 1 on the second die is also 5/6. Then:

The probability of getting at least one 1 would be:

1 - [5/6 x 5/6] =1 - [25/36] = 11/36 =~30.56%

https://web2.0calc.com/questions/what-is-the-sum-of-all-positive-integers-v-for-which-lcm-v-20-60#r1

Let the size of Tom's field = F, then:
The size of Brad's field =1.75F, and:
The size of Karl's field =2.5 x 1.75F = 4.375F
Tom plows: (1/35)F in 1 hour, and:
Brad plows:(1.75/50)F in 1 hour, and:
Karl plows:(4.375/100)F in 1 hour, then:
All 3 men working together can plow the 3 fields in H hours:
[F + 1.75F + 4.375F] / [(1/35)F +(1.75/50)F + (4.375/100)F] =H, solve for H
     7.125F / 0.107321428571....F = H
H = 66.38935 hours =~66 hours and 23.361 minutes.

You already posted this question...

Oh wauw haha! Dat maakt mijn leven heel wat makkelijker :D Super bedankt btw

it helped! thx! NOW IMA GLEAN YA(joke. just read sythe. it's awesome.!)

I believe we  need  to include 60, as well

So...the sum  is   66 + 60   =   126