Math Help Needed, Plzzz!!??

[lil_dcb7]

I royally suck at math. And I have a quiz tomorrow. So If you guys could help me out, I'd appreciate it tons.

There are a few types of examples I have problems with.

1) On one draw of a deck, what's the probability of Getting a Queen or a heart?

2) Find the First four Terms: tn=5n+30

3) Expand the Binomial: (x-y)^3

4) For the Expansion of (a+b)^12, Find the 4th term.

and the final one, this one is KICKING MY ASS!!

5) Find the number of ways a committee can be selected:

A committee of 7 from a group of 7
A committee of 4 from a group of 8
A committee of 9 from a group of 15


you have no idea how much this will help me.

Thanks a ton fellas!

[lil_dcb7]

any body?

[hondajeff]

dude, i dont suck at math---math that real people actually use everyday, that is....

so, uh yeah im not Will Hunting (aka Matt Damon) so uh, how do u like them apples?

sorry i can't help--yer fucked

[ACC0RD22]

1) theres 52 cards in a deck. 1/4 of those are hearts. 4 of them are queens. (52/4 = 18)

18 + 3 (since 1 queen is a heat) = 21. 21/52 or 1 in ≈ 2.5.

2) no clue. first 4 terms of x? I suppose you would set x = to 1, 2, 3, and 4 and then solve for tn.

3) (x-y)^3 = (x-y)(x-y)(x-y)

FOIL it out.

4) no clue.

5) the first one, 7 from 7 people, 1 way. (duh)
second one, 4 from 8 people, assign numbers 1 2 3 4 5 6 7 8 and group accordingly.

1234; 1245; 1256; 1267; 1278;
1235; 1236; 1237; 1238; 1246;
1247; 1248; 1257; 1258; 1268;
1345; 1346; 1347; 1348; 1356;
1357; 1358; 1367; 1368; 1378;
1456; 1457; 1458; 1467; 1468;
1478; 1567; 1568; 1578; 1678;
2345; 2346; 2347; 2348; 2356;
2357; 2358; 2367; 2368; 2378;
2456; 2457; 2458; 2467; 2468;
2478; 2567; 2568; 2578; 2678;
3456; 3457; 3458; 3467; 3468;
3478; 3567; 3568; 3578; 3678;
4567; 4568; 4578; 4678; 5678;

that should be it. 14 * 5 = 70.

I have no clue how to do it any easier way. I must have skipped that class.

[mozzandherb]

What's with all the math threads?...lol...Shit while were at it try this one!

xn + yn = zn

has no non-zero integer solutions for x, y and z when n > 2.

Sorry to jack the thread, but good luck!

[ACC0RD22]

What's with all the math threads?...lol...Shit while were at it try this one!

xn + yn = zn

has no non-zero integer solutions for x, y and z when n > 2.

Sorry to jack the thread, but good luck!

3x + 3y = 3z

where x = 1, y = 2, z = 3.

3(1) + 3(2) = 3(3)

unless Im misunderstanding the problem?

[mozzandherb]

3x + 3y = 3z

where x = 1, y = 2, z = 3.

3(1) + 3(2) = 3(3)

unless Im misunderstanding the problem?

yeah it's not that! I assure you it's not that...lol

[ACC0RD22]

do x y and z have to equal a single number too? Im confused.

[mozzandherb]

do x y and z have to equal a single number too? Im confused.

Do you really want to know the answer?

[blizzard335]

Do you really want to know the answer?

yes please

[ACC0RD22]

yes please

seriously.

[lil_dcb7]

my teacher wants us to use the binomial theorem.

heeeeelp... I looked it up on google, but i am freaking lost.

[ACC0RD22]

http://en.wikipedia.org/wiki/Binomial_theorem

thats the best I can do.

[Accord R33]

#5 is a really elementry probability question.

7 from 7, 1
4 from 8, 8C4 (8 choose 4),

you can do the calcs, but thats how you get the answer, a simple combination.

do the same thing for 15C9, where r = 9, n =15

As for the binomial stuff, im not even going to try. i hate binomial expansion, i can never remember pascals tree!

btw, look up pascals tree, it will show you an easy way to get the answers

[Mista Juice]

back in my day, we would actually learn how to solve this stuff by reading our books and such, not go looking for answers on the internet or message boards.. kids are spoiled these days..