catmg

Aww, thank you. :))

I really need to work on rereading my answers and being more careful about questions too. 

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Ohh, I do have it backwards. 

First, the probability of drawing a red card is 26/52 = 1/2.

2/52 of them are red kings.  

24/52 of them are red cards. 

If we start by drawing a red king, there are 3 kings left, 3/51. So the chance we draw a red king, then another king is 2/52*3/51. 

If we start by dawing a red card that if not a king, there are 4 kings left, 4/51. So the chance we draw a red card (non king), then a king is 24/42*4/51. 

Adding the chances together, 24/42*4/51 + 2/52*3/51. :))

Please tell me if there are any mistakes or questions. 

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Split this into 2 scenarios, first, the probability of drawing a red king then a red card. Then, the probability of drawing a black king then a red card. Finally, add them together. :))

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Hello, I haven't seen you on in a long time. :))

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replace f(x) with x

x = 4 - 5f(x)

4-x = 5f(x)

f(x) = (4-x)/5

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7x = 21 (mod 14)

x = 3 (mod 14)

2x = -2 (mod 9)

x = -1 (mod 9)

x = 8 (mod 9)

x = 3 (mod 14)

x = 17

x = 17 + 14*9*a, where a is an integer. 

Find the least positive four-digit solution. 

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Draw a diagram and you get a right angled triangle. 

Then, use the pythagorean theorem to solve for sides, and use similar triangles to find the length you want. :))

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If they have an odd number of divisors, they're prime, there are 44 prime numbers less than 2008. sqrt(2008)

2008-44 = 1964

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You're welcome. :))

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log((a^3b^4)/(6cd))

log(a^3b^4) - log(6cd)

3log(a) + 4log(b) - log(6) - log(c) - log(d)

Hopefully that's right. :))

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