F.5 MATHS......有PROBABILIT

yan

2011-01-18 2:30 am

1.If the minimum value of y =4x^2-8x+k is 10,find the value of k
A.10
B.12
C.14
D.16

2.if 2x^2-4x-k^2+7k-8 >= 0 for all real values of x , find the minimum value of k

A 0
B 1
C 2
D 4

3.In how many ways can letters from the word ''VICTORY'' be arranged in a
row if 3 different letters are taken each time ?

A 6
B 42
C 210
D 420

4.all letters are taken from the word ''STUDY'' and then arranged,If ''U'' and
''Y'' are not next to other, and the first letter is neither ''U'' nor ''Y'',how many permutations are there?

A 12
B 36
C 48
D120

5.if nC5 =nP4 /10 , what is the value of n?
A 10
B 12
C 14
D 16

6.if a number is chosen at random from the 5-digit numbers formed by 0,1,2,3 and4 ,the probability that the number chosen is divisible by
2 is

A 5/8
B 1/2
C 5/16
D 4/5

thx !!!

回答 (2)

myisland8132

2011-01-18 2:54 am

✔ 最佳答案

1 y

= 4x^2 - 8x + k

= 4(x - 1)^2 + k - 4Sub. x = 1, we see that k - 4 = 10, k = 14 (C)2 Discriminant <= 016 - 8(-k^2+7k-8) <= 0-k^2+7k-8 >= 2k^2 - 7k + 10 <= 0(k - 2)(k - 5) <= 02 <= k <= 5The minimum value of k is 2 (C)
3 6*5*4 = 210 (C)4 First, only consider that the first letter is neither ''U'' nor ''Y''. If there is no restriction, the no. of permutations = 3*4*3*2*1 = 72. Then consider "U" "Y" as a whole, no. of permutations = 2(3*3*2*1) = 36. So, the required permutations = 72 - 36 = 36. (B)5 nC5 = nP4/1010n!/5!(n-5)! = n(n-1)(n-2)(n-3)10n!(n-4)/(5!n!) = 1n-4 = 12n = 16 (D)6 Total no. of 5-digit number = 5! - 4! = 96The permutations such that the number is divisible by 2= 4*3*2 + 3*3*2 + 3*3*2= 60Probability = 60/96 = 5/8 (A)

👍 0 👎 0

用戶7639

2011-01-18 2:52 am

1)C

2)C...........

👍 0 👎 0