math 求解

liliannechang

In a certain board game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down.  If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

(A)  1/8
(B)  1/6
(C)  1/5
(D)  3/23
(E)  4/23


Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume.  If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z ?

(A) y + 3z
(B) (y + z) /4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z

Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

(A)  X4 ≧ 1
(B)  X3 ≦ 27
(C)  X2 ≧ 16
(D)  2 ≦ |X| ≦ 5
(E)  2 ≦ 3X + 4 ≦ 6

If x ≠0, then  sqrt(x^2)/x

(A)  -1
(B)  0
(C)  1
(D) x
(E) |x| / x

melodyflying

E, 8/46
A, 1*X+2*Y+3*Z=1.5(X+Y+Z)
E, ABC都是无数多解，D有两条线段
E, x的绝对值作用等同sqrt(x^2)

liliannechang

谢谢， 你能说说你做题的思路吗

saintaa

You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23

melodyflying

谢谢， 你能说说你做题的思路吗
liliannechang 发表于 2013-1-21 19:25

第二题初中高中化学最简单的浓度题.A牛奶浓度为1，B为2,C为3，那混合牛奶的溶质就是1x+2y+3z,混合牛奶共有x+y+z加仑..牛奶浓度就是(1x+2y+3z)/(x+y+x)  题目说混合后浓度为1.5,则(1x+2y+3z)/(x+y+x)=1.5。将x用y和z表示下就好了。
第三题ABC解集无限多可以理解吧？...D的话解集是-5<=x<=-2和2<=x<=5所以有两条线段不满足题目要求。
第四题真的就像我之前那么解释就对了..

liliannechang

不好意思，我现在才看明白