geno3141

I did miss the √5 --- thanks for the correction!

When you (corectly!) got step 2 to equal 1.107  -- because it's cosine, you get two answers both + and - 1.107

So, after you get your first answer, use -1.107 and finish to get your second answer.  

For the first player:

The total number of possible hands of 13 cards from a deck of 52 is ₅₂C₁₃.

The total number of possible hands for this player that contains one ace and twelve other cards:

₄C₁·₄₈C₁₂  (which is choosing 1 ace from 4 aces and 12 other cards from the 48 remaining cards)

So, the probability for the first person getting one ace is:   ( ₄C₁·₄₈C₁₂ ) / ₅₂C₁₃ 

For the second player:

Probability  =  ( ₃C₁·₃₆C₁₂ ) / ₃₉C₁₃ 

because:  there are only 3 aces left, only 36 other cards left, and only a total of 39 cards left.

For the third player:

Probability  =  ( ₂C₁·₂₄C₁₂ ) / ₂₆C₁₃ 

because:  there are only 2 aces left, only 24 other cards left, and only a total of 26 cards left.

For the fourth player:

Probability  =  ( ₁C₁·₁₂C₁₂ ) / ₁₃C₁₃ 

because:  there is only 1 ace left, only 12 other cards left, and only a total of 13 cards left.

Multiply these four probabilities together to get the total probability. ...

The smallest prime number is 2.

The reciprocal of 2 is ½.

The square of ½ is 1/4.

cos(2x + 0.464)  =  1

2x + 0.464  =  cos^-1(1)

But:  cos^-1(1)  =  0   ---   and only 0, so there will  be only one answer within that domain.

And:  2x + 0.464  =  0     --->     2x  =  -0.464     --->     x  =  -0.232

To get within the domain, add 3.14159   --->   x  =  2.909.

A negative exponent basically means that the term is in the wrong place: if it is in the numerator, put it into the denominator and make the exponent positive; if it is in the denominator, put it into the numerator and make the exponent positive.

For  c⁶ / c^-7   --->   rewrite it as:  c⁶ · c⁷  =  c¹³.

(Since the exponent is negative and the term is in the denominator, write the term in the numerator with a positive exponent.)

The probability of rolling at least one 4 plus the probability of rolling no 4s is 1.000.

It is easier to calculate the probability of rolling no 4s and subtracting that value from 1.000 to get the probability of rolling at least one 4 than it is to calculate the probability of rolling at leawst one 4 directly.

Probability of rolling no 4's:   --->   the probability of rolling anything but a 4 is 5/6. This probability holds for each roll, so if you roll a die 7 times, the probability of never getting a 4 is:  5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6. Now, subtract this answer from 1.000 and the result is the probability of rolling at least one 4.

You can change the minutes into decimal hours by dividing by 60, like this:  8.4/60  =  0.14,

so  1:08.4  =  1.14, etc.

If, at the end, you wish to turn the decimal hours back into minutes, multiply by 60.

Yes, it is. The only non-real solutions are those that have a negative number under the square roots sign. 

Since the last digit is a zero, let's focus on the first two digits -- they will go from 10 through 99

First:  count how many numbers there are from 10 through 99. (Answer to part A)

For the whole number to be divisible by 4 and since the last number is a zero, it will already be divisible by 2.

The problem now becomes:  how many of the numbers from 10 through 99 are divisible by 2; or, in other words, how many of the number from 10 through 99 are even? (Answer to part B)

Divide the answer to part B by the answer to part A to get the probability ...

C(10,10) =  ₁₀C₁₀  =  10! / ( 10! · 0! )

But:  0! = 1           <----------------------------------

--->   10! / ( 10! · 1! )   =   10! / 10!  =  1