Math problems!

Oct 15, 2009 13:34



Hey all you math types!

I have two math problems that are proving to be difficult. If anyone could show me how to solve them I'd appreciate it!

1.     2e^(-2x)   -   3e^(-x)   +   1   =   0

^ = raised to the power of.
I keep getting x = -0.405 which works without the + 1. Not sure where I'm messing up as I am dragging the 1 to the RHS and - ln1 = 0

2. ( Read more... )

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Comments 11

mac_aeda October 15 2009, 21:10:53 UTC
Not sure about #1. It's been too long since I've done that sort of thing, and I never did anything in high school involving e (even though I took Calculus... weird.)

2. The first question is obviously 3/13 chance of drawing a face card.

For the second card, there is a 1/13 chance you drew a queen and a 12/13 chance you didn't. If you drew a queen, theres a 3/51 or 1/17 chance of drawing another. If you didn't draw a queen, there's a 4/51 chance of drawing a queen. Multiply the probabilities, so:
(1/13)*(1/17) + (12/13)*(4/51) = (1/221) + (48/663) = 17/221.

I think their number is wrong.

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mac_aeda October 15 2009, 21:39:49 UTC
Ok, here's what I get for #1:

Assuming n = e^-x, we can rewrite the question as:

2n^2 - 3n + 1 = 0

We can expand this quadratic equation as:

(2n - 1)(n - 1) = 0

In order to have this product be zero either bracket can equal zero, meaning we have two answers:

(2n - 1) = 0
(n - 1) = 0

Therefore, the two answers are:
e^(-x) = 1/2 == x = -ln(1/2)
e^(-x) = 1 == x = -ln1

The two answers are x = 0 and x ≈ 0.693

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dragonflymaria October 15 2009, 22:37:20 UTC
Interesting way to solve that.

x~0.693 doesn't seem to work on the calc. I'll play tho - thanks.

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dragonflymaria October 15 2009, 22:41:47 UTC
could just be the calc. it's one I'm not use to - can't believe I missed x = 0! lol

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dragonflymaria October 15 2009, 22:56:08 UTC
Think I just figured out the probability one - just has taken all week.

If the first is not a queen and the second is a queen that's:
8/52*4/51
If they are both queens:
4/52*3/51

Add them up and reduce you get 44/2652 to 11/663

Big sigh of relief - thanks!

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mac_aeda October 15 2009, 23:39:21 UTC
Ahh, ok. I thought they were asking for 2 seperate probabilities. Those numbers look good.

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dragonflymaria October 16 2009, 20:46:37 UTC
Ahhh!
Thanks for all the help!

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sethstarr October 18 2009, 01:08:12 UTC
Brain Explodes!

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