數學不懂做..T_T

44.
The answer : C. 336

Case 1 :
Arrange 1 person from C and D to take the 1st position (2P1),1 person from of A, B, E and F to take the 2nd position (4P1),and the rest 4 people take the rest positions (4P4).

Case 2 :
Arrange 1 person from E and F to take the 1st position (2P1),1 person from A, B and the one from E to F who has not been arranged to takethe 2nd position (3P1), and the rest 4 people take therest positions (4P4).

Number of arrangement
= 2P1 × 4P1 × 4P4+ 2P1 × 3P1 × 4P4
= 2 × 4 × 24 + 2 × 3 × 24
= 336


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45.
The answer : C. 4.8

Variance of the group of the 6 numbers :
[(x1 - 8)² + (x2 - 8)² + ...... + (x6- 8)²] / 6 = 6
(x1 - 8)² + (x2 - 8)² + ...... + (x6- 8)² = 36

Variance of the group of the 4 numbers :
[(x7 - 8)² + (x8 - 8)² + ...... + (x10- 8)²] / 4 = 3
(x7 - 8)² + (x8 - 8)² + ...... + (x10- 8)² = 12

Variance of the group of these 10 numbers
= [(x1 - 8)² +(x2 - 8)² + ...... + (x6- 8)²] / 10
= [(x1 - 8)² +(x2 - 8)² + ...... + (x6- 8)² + (x7 - 8)² + ...... + (x10- 8)²] / 10
= (36 + 12) / 10
= 4.8


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17.
The answer : B. 1 : 3

Join BE.

ΔABE and ΔBDC have the same altitude, and with the bases AB : BC = 1 : 2respectively.
(Area of ΔABE) : (Area of ΔBCD) = 1 : 2
Let (Area of ΔABE) = k and (Area of ΔABC) = 2k

ΔABF and ΔABE have the same altitude, and with the bases AF : AE = 2 : (2 + 1)= 2 : 3
Area of ΔABF
= (Area of ΔABF) × (2/3)
= (2/3)k

(Area of ΔABF) : (Area of ΔBCD)
= (2/3)k : 2k
= 1 : 3

2015-03-10 00:22:32 補充：
我用的方法是必須兩組數的平均數相同，而兩組數合成一組時，新的平均數也與原來的兩個平均數相同。

若兩組數的平均數不同，則要用 知足常樂 師兄所提供的方法了。

Q45 有Ｄ技巧。

(1) μ = ∑x / n
⇒ ∑x = nμ

(2) σ² = ∑(x - μ)² / n
⇒ nσ² = ∑(x - μ)² = ∑x² - nμ²
⇒ ∑x² = nσ² + nμ² = n(σ² + μ²)

Therefore, the combined sum is
∑z = ∑x + ∑y = (6)(8) + (4)(8) = 80
∑z² = ∑x² + ∑y² = (6)(6 + 8²) + (4)(3 + 8²) = 688

2015-03-05 17:17:37 補充：
Combined mean
= 80/10
= 8

Combined variance
= (∑z² - Nμ²)/N
= (688 - 10*8²)/10
= 48/10
= 4.8

(C)