|By Jens (Jens) on Sunday, April 04, 2004 - 08:09 pm: Edit|
1. A right circular cone has height h and radius r. If the cone is cut in two pieces by a plane parallel to the base and midway to the vertex, the volume of the larger of the two resulting figures is?
Answer: 7(pi)(r squared)h/ 24
I tried plugging into the calc, but i never got the right answer!
Thanks a ton.
|By Dmscramjet (Dmscramjet) on Sunday, April 04, 2004 - 09:10 pm: Edit|
= 2.8353 or 2.84
|By Jens (Jens) on Sunday, April 04, 2004 - 09:25 pm: Edit|
any takers for the first one?
|By Beero1000 (Beero1000) on Sunday, April 04, 2004 - 10:23 pm: Edit|
the bottom part is bigger
bottom = cone-top=
1/3(pi)r^2h - 1/3(pi)(.5r)^2.5h=8/24(pi)r^2h - 1/24pir^2h
bottom part = 7/24(pi)r^2h
|By Jens (Jens) on Sunday, April 04, 2004 - 11:02 pm: Edit|
1. 7 blue marbles & 6 red marbles in jar. Marbles are randomly selected one at a time & not put back. If first 2 selected are both blue, what is probability that AT LEAST 2 red marbles will be chosen in the next 3 selections?
2. A menu lists 10 items in list A & 20 items in list B. A family of 5 plans to share 5 items from A and 5 items from B. How many diff combos of items could the family choose?
3. 2 Trucks A & B start from same point X at same time & travel along two diff roads, which both begin at point X and from an 100 degree angle. If truck A is travelling twice as fast as B, & the trucks are separated by 334 mi after 4 hrs, how fast is truck B travelling?
Ans: 35 mph
Thats all. Thanks!
|By Jens (Jens) on Monday, April 05, 2004 - 12:28 am: Edit|
|By Lildude_Ravi (Lildude_Ravi) on Monday, April 05, 2004 - 02:56 am: Edit|
1. After 2 blue marbles are selected , 5 blue and 6 red are left.
Now for choosing at least 2 red in the next three sections, there can be 4 ways :
6/11*5/10*5/9 + 6/11*5/10*5/9 + 5/11*6/10*5/9 + 6/11*5/10*4/9
2. This is a simple combination prob
10C5 * 20C5 = 3907008 (nCr can be calculated with help of calculator)
I hope u find this useful!!!!!!!!!!
|By Beero1000 (Beero1000) on Monday, April 05, 2004 - 03:21 am: Edit|
3. law of cosines
c^2 = a^2 + b^2 - 2abcosC
x^2 = 334^2/80-64cos100=1224
|By Conker (Conker) on Monday, April 05, 2004 - 05:55 am: Edit|
You might want to consider taking the Math Ic exam. None of these problems are particularly difficult, and you will have a lot of trouble with the IIc exam if you are struggling with these problems.
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