求助几个GRE的数学题
1,fromthesetof6letters,A,B,C,D,EandF,thereare20different3-lettersubsetsthatcouldbesel...
1, from the set of 6 letters, A, B, C, D, E and F, there are 20 different 3-letter subsets that could be selected.
A: the number of 3-letter subsets that include F
B: 10
2, 10, 15, 25, 30, x
if d is the standard deviation of the numbers in the list above, for which of the following values of x would the value of d be least?
0
5
20
30
80
3,
A: the least positive integer with four different prime factors, each greater than 2
B: 1,155
4,which of the following can't be a factor of 2n,3k(2的n次方乘以3的k次方), where n and k are positive integers?
6
8
27
42
54
5, if 72.42=k(24+n/100), where k and n are positive integers and n<100, then k+n=?
17
16
15
14
13 展开
A: the number of 3-letter subsets that include F
B: 10
2, 10, 15, 25, 30, x
if d is the standard deviation of the numbers in the list above, for which of the following values of x would the value of d be least?
0
5
20
30
80
3,
A: the least positive integer with four different prime factors, each greater than 2
B: 1,155
4,which of the following can't be a factor of 2n,3k(2的n次方乘以3的k次方), where n and k are positive integers?
6
8
27
42
54
5, if 72.42=k(24+n/100), where k and n are positive integers and n<100, then k+n=?
17
16
15
14
13 展开
展开全部
1 六个字母里选出三个字母,其中包括F的有几组。 问题等价于从剩余五个字母中选两个 答案为 c(5,2) = 10组
2 标准差要小,所以x要尽量接近前面数字的平均值,平均值为20 故选20
3 求有四个质因数的最小整数:3*5*7*11=1155
4 如果一个数可以成为2^n3^k的因数,那么它必须能表示成2^x3^y的形式(x y 可以为0)因此42不行 (6 = 2^1 3^1, 8 = 2^3 3^0, 27 = 2^0 3^3, 54=2^1 3^3)
5 要满足等式 k=3(其他值都不能满足0<n<100)于是n=14 故答案为17
希望满意~~~
2 标准差要小,所以x要尽量接近前面数字的平均值,平均值为20 故选20
3 求有四个质因数的最小整数:3*5*7*11=1155
4 如果一个数可以成为2^n3^k的因数,那么它必须能表示成2^x3^y的形式(x y 可以为0)因此42不行 (6 = 2^1 3^1, 8 = 2^3 3^0, 27 = 2^0 3^3, 54=2^1 3^3)
5 要满足等式 k=3(其他值都不能满足0<n<100)于是n=14 故答案为17
希望满意~~~
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