susan asked in 社會與文化語言 · 1 decade ago

不須翻譯的-英文數學問題

Q1:

For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

答案是 E

(D) 4(0.6)^4(0.4) + (0.6)^5

(E) 5(0.6)^4(0.4) + (0.6)^5

正面機率是0.6 反面是機率是0.4

題目要求正面最少要 4次

我的算法是0.6*0.6*0.6*0.6*0.6=0.07776→這是5次都是正面

0.6*0.6*0.6*0.6=0.1296→這是4次都正面。

兩個相加為0.20736

我的問題是:我看不懂 這個式子→5(0.6)^4(0.4) + (0.6)^5 請幫忙解說謝謝 !!

Q2:

In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average (arithmetic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?

答覆是 D

(A) 3

(B) 5

(C) 8

(D) 13

(E) 15

題目要求學生可以借最大數量為多少本?

我的算法:從題目得知總共有60本書

60-(12+20+18)=10 →兩個學生借書的最大值也等於其他學生借書的最大值。 請解說謝謝!

Q3

If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

條件1: n = 2

條件2: m = 1

答案是條件2成立。

我覺得要條件1跟條件2成立,才能的知餘數吧!! 請幫忙解說

謝謝!!

Q4

If x is a negative number, what is the value of x ?

條件1:x2 = 1 是X的平方=1

條件2: x2 + 3x + 2 = 0

答案是條件1成立

條件一:(-1)的平方=1 →X 值 成立

我的問題是:我覺得條件2 也能成立(-1)的平方+3(-1)+2=0→ 1-3+2=0 成立 請解釋謝謝!!

Q5

If R = 1 + 2xy + (x^2)(y^2), what is the value of xy ?

條件1:R = 0

條件2: x > 0

答案是條件1成立

我的問題是條件1:R=0 可以得知x跟y值不會是零,但如何得知xy相乘的值呢?

請解說謝謝。

2 Answers

Rating
  • 1 decade ago
    Favorite Answer

    Q1: For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

    (E) 5(0.6)^4(0.4) + (0.6)^5

    正面機率是0.6 反面是機率是0.4,題目要求正面最少要 4次。

    為什麼5(0.6)^4(0.4) + (0.6)^5是對的呢?

    0.6*0.6*0.6*0.6*0.6=0.07776→這是5次都是正面,YES!

    0.6*0.6*0.6*0.6=0.1296→這是4次都正面,NO!

    4次正面+1次反面的機率是:5(0.6)^4(0.4)。因為可能HHHHT, HHHTH, HHTHH, HTHHH, THHHH。有C(5,1)種可能,每種可能發生的機率是0.6*0.6*0.6*0.6*0.4。所以是5*0.6*0.6*0.6*0.6*0.4。

    Q2: In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average (arithmetic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?

    (D) 13

    題目要求某一個學生借了最大數量為多少本?

    我的算法:從題目得知總共有60本書

    60-(12+20+5*3)=13 →剩下6個人中,有5個人借了3本,最後一個人就是借了最大數量書的人:13本。

    Q3: If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

    條件1: n = 2

    條件2: m = 1

    答案是條件2成立。

    3的n次方除以10餘數只有四種可能:3, 9, 7, 1, 3, 9, 7, 1……如果條件1成立,3^(10 + m) = (3^10)*(3^m) = 59409*3^m,餘數可能有很多種。如果條件2成立,3^(4n+3),除以10,餘數必為7。

    Q4: If x is a negative number, what is the value of x ?

    條件1:x^2 = 1 是X的平方=1

    條件2: x^2 + 3x + 2 = 0

    答案是條件1成立。

    條件一:(-1)的平方=1 →X 值 成立

    如果條件2成立,(x+2)*(x+1)=0,X可能是-2或-1,無法充分回答X等於幾的問題。

    Q5: If R = 1 + 2xy + (x^2)(y^2), what is the value of xy ?

    條件1:R = 0

    條件2: x > 0

    答案是條件1成立。

    如果條件1成立,R=0=(xy+1)*(xy+1),則xy=-1。如果條件2成立,x > 0,不可能知道y的數值,以及xy的數值!

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  • 1 decade ago

    你要考700分以上可能還要加點油喔.

    Q1.

    你的算法沒有考慮到排列, 而且4次都正面的時候你沒有計算到反面的機率(第五次你乘1的話就變成正反都可以了).

    解析:

    the probability that the outcome will be heads at least 4 times = P(head 4 times) P(head 5 times)

    P(head 4 times)= C(5,4) x (0.6)^4 x (0.4)^1 = 5x(0.6)^4x(0.4)

    P(head 5 times)= C(5,5) x (0.6)^5 x (0.4)^0 = 1x(0.6)^5

    兩個相加就是你的答案了. 這裡的C(5,4)代表C5取4

    選E

    Q2.

    你60-(12 20 18)=10 的18是題目看錯或是想錯了.

    解析:

    平均2本, 共30人 ---> 總共60本

    2人拿0本 ---> 共0本

    12人拿1本 ---> 共12本

    10人拿2本 ---> 共20本

    所以60-(12-20)=28本 (這些給其他6人分)

    要有一個人拿最多的情況之下, 一人拿最多, 其他人拿最少,

    所以那六個人應該是3,3,3,3,3,13本這樣的分配情況.

    因此the maximum number of books that any single student could have borrowed = 13

    選D

    Q3.

    解析:

    3^1= 3 除10餘3

    3^2= 9 除10餘9

    3^3= 27 除10餘7

    3^4= 81 除10餘1

    3^5= 243 除10餘3 (每4次開始循環了)

    題目已知 n and m are positive integers, 所以4n代表4或8或12或.....

    因為都一直循環所以對餘數無影響.

    故, 條件2充份, 選B.

    Q4.

    解析:

    條件1充分你已知, 略.

    X^2 3X 2= 0 可以因式分解(X 1)(X 2)=0

    又題目說x is a negative number, 所以條件2的解X=-1或-2都滿足,

    故不充份, 選A

    Q5.

    解析:

    請先假設xy=z

    則R= 1 2z z^2

    條件1為R=0, 所以原式為 0= 1 2z z^2

    可以因式分解(z 1)(z 1)=0 (有z=-1的重根),

    故可知z = xy = -1, 即條件1滿足.

    x>0仍未之xy之值, 不充份.

    選A

    Source(s): 剛申請到2008秋季MBA的自己
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