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Here's an approach to find this

Note that  in base 10

999 - 100 + 1   =  900 three digit integers

So.....we really have

[ 9*10^2 + 9 * 10 + (10 - 1)]  - 10^2 + 1  = 900

So...considering  base 10, we have

 [ (b - 1) * b^2 + (b - 1) * b  + (b - 1) ]  - b^2 + 1  = 900

But....we are looking for the base that only produces 100  three digit integers

So......We can solve this to find the base, b

[ (b - 1) * b^2 + ( b - 1) * b + ( b - 1) ] - b^2 + 1 = 100      simplify

b^3 - b^2 + b^2 - b  + b - 1 - b^2 + 1 = 100

b^3 - b^2  = 100

And note that this is true for b = 5

Because the tangent is > 0   and the secant is < 0 , theta lies in the 3rd quadrant

The secant is -3/1  = 3 /-1  =  r /x

The sine will be negative here  and is given by :

- √ [ r^2 - x^2 ] /  r   =   - √ [ (3)^2 - (-1)^2 ] /   =   -  (√8 )  / 3

We can find P  by finding the unit vector for (3, - 2)

This is given by:

( 3 /  √ [ 3^2 + (-2)^2 ]  ,   -2 /  √  [ 3^2 + (-2)^2 ]  )   =

( 3 / √13  , - 2 /  √13)  =  P 

Having chosen the first vertex at random there are 9 others that can be reached from it with a line along a surface.  Hence there are 19 - 9 = 10 others for which the line must go inside the dodecahedron.  The probability is therefore 10/19.

(b,b,r) or (r,r,b)

\(p = 2 \cdot \dfrac{\dbinom{26}{2}\dbinom{26}{1}}{\dbinom{52}{3}} = \dfrac{13}{17}\)

xy = 144

We can look at the factors of 144

These are 

1 | 2 | 3 | 4 | 6 | 8 | 9 | 12 | 16 | 18 | 24 | 36 | 48 | 72 | 144

So 

The  ordered paIrs  are

(1, 144)

(2, 72)

(3, 48)

(4, 36)

(6, 24)

(8, 18)

(9, 16)

(12, 12)

So.....reversing the first 7 pairs.....we have 7 + 7 + 1 =  15 possible pairs of (x, y)

Prob that they are not the same color =  

1 - prob that they ARE all the same color

P(all the same color)  =

On the first draw, we either get a red or a black  card

The probability  that we match this on the second draw is   25 cards left of the same color / 51 cards left = 25/51

The probability that we match the color on the third draw = 24 cards left of the same color / 50 cards left

So

P( same color)  =   (1) * ( 25/51) * (24/50)  =   25/50 * 24/51  = (1/2) (8/17) =  4/17

So

P (not the same color ) =      1 - 4/17   =   13 / 17

No you might be right.

I found the prob that they are not all of the same suite (not colour) 

CPhill is doing it now anyway.

Does that mean my answer is wrong then? Shoot. I'll have to se what I did wrong.

There are the following palindromes which are multiple of 99: 1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, 9999 = Total of 10
9999 - 1,000 + 1 =9,000 4-digit numbers
10/9,000 =1 / 900 - probability of multiples of 99

Three cards are chosen at random from a standard 52-card deck. What is the probability that they are not all the same suite  (NOT COLOUR)?

Please note that i have answered the wrong question.  

Prob that they are all the same suite  is   \(\frac{52}{52}*\frac{\not{12}^4}{\not{51}^{17}} *\frac{11}{50} =\frac{44}{850}=\frac{22}{425}\\ \)

Prob that they are not all the same suite = \(1-\frac{22}{425}=\frac{403}{425}\approx94.8%\)

This is the correct answer which can be seen by evaluating

\(\sum \limits_{k=0}^4~\dbinom{10}{k}\left(\dfrac 1 2\right)^{10}= \dfrac{193}{512}\)

The  prob of gettint exactly 5 heads is 10C5 (0.5)^10 = 252/1024 = 63/256

So it seems logical that the prob of not getting 5 heads is 1-63/256 = 193/256

and hall of those would have fewer heads than tails

(193/256) /2 = 193/512

The probability that the three cards are not all red (assuming that there are 26 red cards in the deck and each card is not put back in the deck after it has been drawn) is approximately 0.8824 or 88.24%.

Thats cool!!

Thanks, tertre......this is a helpful theorem in many situations......

Assuming that you are meaning cents, we first change \(\$4\) into cents, which is \(400\) cents.

Now, for the division, \(\frac{400}{5}=\boxed{80}\) times/

A quicker way is to use Vieta's, where the sum of the roots are \(\frac{-b}{a}\) and the product of the roots are \(\frac{c}{a}.\)

In the expression \(3x^2-4x-7, \)   we attain \(a=3, b=-4\) , and \(c=-7.\)

Now, by plugging in our values, the sum of the roots is \(\frac{-(-4)}{3}=\frac{4}{3}.\)

Consequently, the product of the roots is \(\frac{-7}{3}.\)

Hope this helped!

We have that either none of them show tails or one does..so....

P(none show tails) =   C(n, 0) (1/2)^n =  (1/2)^n

P( 1 shows a tail) = C(n, 1) (1/2)^n  = n* (1/2)^n

So we have that

(1/2)^n ( n + 1) = 3/16

(n + 1) / 2^n =   3 /  2^4

2^4 / 2^n   =      3 / (n + 1)

2^n / 2^4  =   (n + 1) / 3

2^( n - 4)   =  (n + 1) / 3

Note that the left side is some power of 2

So.....the first positive integer n that makes the right side a power of 2 is when n = 2

But this gives   2^(-2)  not equal to  3/3

The next positive integer that makes the right side a power of 2 is when n  = 5 

So

2^(5 - 4)  =  (5 + 1) /3

2 = 6/3     is true

So.....we need 5 coins

Use quadratic formula to find the roots of x:

x =2 1/3

x = - 1

Their sum = 2 1/3 - 1 =1 1/3

Their product = 2 1/3 x - 1 =- 2 1/3

6! /[3!.3!.2^6] =720/2,304=31.25% - probability of having 3 boys and 3 girls.
So: 1 - 0.3125 =68.75% - probability of having more of one gender than the other(4 boys and 2 girls, 5 boys and 1 girl and 6 boys and no girls, or vice versa).

OK....this one was kinda' interesting....!!!

ohhh i was finding it weird as the equation for asymptotes are +-b/a but as i followed up to the part where u explained how asymptotes can find origin i immediately clicked on. thanks a lot!!!

This is a hyperbola centered at the origin  intersecting the x axis

The equations of the asymptotes are

y = (2/3)x     and    y = (-2/3)x

Notice that  these lines intersect at the hyperbola's center.....if we set these equal, we can verify that the origin is the center

(2/3)x = (-2/3)x

x = -x

2x = 0

x = 0

And putting this value of x  into either linear equation gives y =0

So...the center of C1  is (0,0)

Therefore....set 

(2/3)x - 2  = (-2/3)x + 2       

(4/3)x = 4     Implies that   x = 3

And putting this value back into either equation  gives y  = (2/3)(3) - 2  = 0

So....the center of C2 is (3, 0).....so    C2 is the translation of C1 -  3 units to the right   

So....as you found YEEEEEET.....a =  3.....!!!!

the trick here is that all numbers divisible by 9 are such that their digits add to multiple of 9

so 1089 looks to be the smallest with 2 odds and 2 evens