susan 發問時間： 社會與文化語言 · 1 0 年前

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?

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

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

0.6*0.6*0.6*0.6=0.1296→這是4次都正面。

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?

(A) 3

(B) 5

(C) 8

(D) 13

(E) 15

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 ?

Q4

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

Q5

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

2 個解答

• 1 0 年前
最佳解答

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！

• 1 0 年前

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

參考資料： 剛申請到2008秋季MBA的自己