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I think it would be 4/52 for the first ace, which could be any of the four aces

and then              3/52 for an undrawn ace, one of the remaining three  

and then              2/52 for one of the two remaining undrawn aces  

and then              1/52 for the last ace that hadn't been previously drawn  

for the total probability multiply them together    

                             4        3        2      1           24                  3   

                            ––  x  ––  x  ––  x  ––  =  ––––––––  =  ––––––––    

                            52      52     52     52       7,311,616        913,952   

a - bi = 2 - 7i.

There are 60 different possible rational roots.

You can isolate x, y, z to get that the possible values of xyz are -2, -1, 1, and 2.

There are 11 triangles, so the area of the 11-gon is 11*15 = 165.

z + 1/z = -1

(z + 1/z)^3 = -1

z^3 + 3z + 3/z + 1/z^3 = -1

z^3 + 3(z + 1/z) + 1/z^3 = -1

z^3 + 3(-1) + 1/z^3 = -1

z^3 + 1/z^3 = 2

can you take it from here?

=^._.^=

Anybody got the answer?

Take the average of the two bases   (12+18)/2 = 15

Area =  15 * 11 = 165 mm²

From wolframalpha:

Sin law

19 / sin25  = DE / sin 65

sin 65^o * 19 / sin 25^o   = DE   

The maximum numbr of handshakes is 104.

The perimeter is just the circumference of two circles with diameter = 2/pi

circumference of circle = pi * d 

   two of these   would be    2 * pi * 2/pi = 4 units

n must be between 81 and 10000, so there are 10000 - 81 = 9919 possible values of n.

x(1) = -10.

The perimeter is 8.

maybe explain if possible since i'm still confused on how you got to the answer

can you explain if possible?

The smallest possible value of f(1) is 120.

The smallest possible area of triangle OAB is 80.

Good point.....questions are often unclear and subject to interpretation by the answerer.

If you want to get EACH of the aces it would be a little more difficult :

(1/52)⁴      

mhm, just realized melody and calthegreat found the error before me

People frequently compain about this trivial typo.....you are not the first ....let it go.  Maybe someday it will be fixed.

   I worry more about the language and syntax  skills of people asking questions.....really poor.  Their errors are FAR more egregious than the " I'am  vs I'm"  in the CAPTCHA.

.....what is the probability that I will end up with all four Aces?      

By this, do you mean  

...probability that I will have pulled an ace each time?    

OR  

...probability that I will have pulled the ace of each suit?   

I think you can do that much on your own.....

What is the percentage?