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24/ 75    reduces to 8/25 ....  multiplying the top and bottom by 4, we have 32/100  = .32 or 32%

This means we either rolll no fives, one five, two fives, three fives or 4 fives

The probability of rolling no fives is C(12,0)(1/6)^0*(5/6)^12 = 0.1121566547846151

The probability of rolling one five is C(12,1)(1/6)^1*(5/6)^11 = 0.0000000044516784

The probability of rolling two fives is C(12,2)(1/6)^2*(5/6)^10 = 0.2960935686313838

The probability of rolling three fives is C(12,3)(1/6)^3*(5/6)^9 = 0.1973957124209225

The probability of rolling four fives is C(12,4)(1/6)^4*(5/6)^8 = 0.0888280705894151

And summing these we have......about 69.4%  chance of rolling no more than 4 fives

Notice the pattern generated by raising a number ending in 7 to the following powers.....

7^1 = 7      7^2 = 49    7^3 = 343    7^4 = 2401 

And this cyclic pattern repeats every four powers

So    ...   256/4 = 64

And this whole number tells us that 7^256 completes a cycle of 4, so this number ends in a "1"

Why do you think that you are wrong Chris ? I think that yours is the correct solution.

Geno's solution is wrong because it orders the non-Americans twice. The 16×15 is the number of permutations of 2 from 16 and so is already taking order into account. If simply two non-Americans were required, then C (16, 2)=16×15/2 should be used.

Here's an alternative way of arriving at the result

The number of ways in which America wins a single medal, gold = 3×16×15 = 720.

The number of ways in which America wins a single medal, silver = 16×3×15 = 720.

The number of ways in which America wins a single medal, bronze = 16×15×3 = 720.

The total is 720 + 720 + 720 = 2160.

So, how does a troll sincerely wish Alan a happy birthday anniversary, without Melody permutating it into a Scooby snack?

The world may never know.:)

Root’s idea is a good one. How about it, Sorasyn, can you modify DDT to cure CDD? If so, you will be on the short list to win the Nobel Prize –or at least some permutation of it.

I would add a few more snarky comments, but Melody uses an algorithm that does a fair job of permutating snarky comments into Scooby snacks.

Why he get 3 points for wrong answer?

It balances out.  I sometimes get negative points for great answers. We trolls like them. It means we’re doing a good job.

Well it was a shot. Lol.

Since 1° = 60', change one of the 90° into minutes to give  89° 60'.

The problem now is:  89° 60'  -  1° 42'  =  88° 18'.

To begin with, you will need to hire at least 85 employees (to monitor water pollution).

If 27 of these can also monitor air pollution, you will need 73 - 27 = 46 persons who can monitor only air pollution.

So, you will need to hire 85 + 46 = 131 persons.

Another way is to add 85 (water) + 73 (air) to give 158 persons. But since 27 handle both, subtract 27 from 158 to give you 131 persons.

First, factor out -x², giving you -x²(x⁴ - 7x² + 3)

So, one intercept occurs when -x² = 0, so x = 0.

Now, it would be nice if (x⁴ - 7x² + 3) factors, but it doesn't, so use the quadratic formula with the change that the quadratic formula gives you values for x², not x:

x²  =  [7 ± √( (-7)² - 4·1·3)] / [2·1]   --->   x²  =  [7 ± √( 37)] / [2]

You will need to take the ± square roots of that answer.

Now, these x-values will be:  + √( [7 + √( 37)] / 2 )

                                              - √( [7 + √( 37)] / 2 )

                                              + √( [7 - √( 37)] / 2 )

                                              - √( [7 - √( 37)] / 2 )

Funny troll. Maybe Sorasyn formula permutate the cube to termite food instead of mulch. Maybe some DDT will k**l termites and cure CDD. Hahaha

Why he get 3 points for wrong answer?

CPhill -- I think the problem arises by choosing two groups, the groups of Americans and the groups of non-Americans.

By doing the problem this way, there are two arrangements, either choosing the Americans first and then the non-Americans, or vice-versa. Then, there are the six arrangements of the individuals. This would double your answer.

Or, maybe not ... ?

Lollololol

Arent u supposed to fucking combine like terms first?

Algebra is algebra.

5 x 5 - 2(1/2) = 25 - 1 = 24

(7 - 1)(6)(2/3) = 6 * 6 * (2/3)  = 36 * (2/3)  = 24

Let  n  represent the number of chocolates in the box.

Since Shannon ate 1/4 of the chocolates, she ate (1/4)x.

The number remaining is (3/4)x.

Shane ate 1/2 of what was left, so Shane ate (1/2) of (3/4)x, or (3/8)x.

Total number of chocolates = Shannon's + Shane's + 15.

Since the total number is n:  n  =  (1/4)n + (3/8)n + 15

Multiply all the terms by 8:  8n  =  2n + 3n + 120

--->   8n  =  5n + 120

--->  3n  =  120

--->  n = 40.

Shannon's share = (1/4)(40) = 10.

Shane's share = (3/8)(40) = 15.

My share = those that are left = 15.

Total = 40.

There are 1760 yds. in a mile....so

(2 x 1760)/88  = 40 min.

Look in the sticky notes on the right of this page.  ☺

Good answer but don't forget the units.  In this case metres.  :)

-13-(-5)= -8

145 times 0.50= 72.5

Well, Sorasyn it’s a good thing you went to college where you leaned to permutate dimensions. I’ve never heard of it. The closest is permutate wood. This was a program control on a machine used by Purina to make termite food from shipping pallets.

Any way, "Mary, Mary, is quite contrary watching her garden grow. It seems her gardener didn’t buy enough mulch. It’s understandable though, her gardener has the blight of CDD. Permutating this we have DDC and DCD.

Well, until next time, Sorasyn Keep up the permutations.

Using the expansion of  (x + y)⁹  

=  x⁹ + 9x⁸y + 36x⁷y² + 84x⁶y³ + 126x⁵y⁴ + 126x⁴y⁵ + 84x³y⁶ + 36x²y⁷ + 9xy⁸ + y⁹

[Cos(x) + iSin(x)]⁹  =

=  Cos(x)⁹ + 9 Cos(x)⁸ i Sin(x) ⁺ 36 Cos(x)⁷ i² Sin(x)² + 84 Cos(x)⁶ i³ Sin(x)³ 

    + 126 Cos(x)⁵ i⁴ Sin(x)⁴ + 126 Cos(x)⁴ i⁵ Sin(x)⁵ + 84 Cos(x)³ i⁶ Sin(x)⁶ 

    + 36 Cos(x)² i⁷ Sin(x)⁷ + 9 Cos(x) i⁸ Sin(x)⁸ + i⁹ Sin(x)⁹

Since:     i² = -1    i³ = -i     i⁴ = 1    i⁵ = i     i⁶ = -1     i⁷ = -i    i⁸ = 1     i⁹ = i

=  Cos(x)⁹ + 9 i Cos(x)⁸ Sin(x) - 36 Cos(x)⁷ Sin(x)² - 84 i Cos(x)⁶ Sin(x)³ 

    + 126 Cos(x)⁵ Sin(x)⁴ + 126 i Cos(x)⁴ Sin(x)⁵ - 84 Cos(x)³ Sin(x)⁶ 

    - 36 i Cos(x)² Sin(x)⁷ + 9 Cos(x) Sin(x)⁸ + i Sin(x)⁹