第1题:
● 在Excel中,与函数“=AVERAGE(A1:A3)”等价的是 (46) 。
(46)
A.=A1+A2/2
B.=A1+A3/2
C.=(A1+A2+A3)/3
D.=A1+A2+A3/A1*A2*A3
第2题:
计算:
(1)36×(1/2-1/3)2
(2)12.7÷(-8/19) ×0
(3)4×(-3)2+6
(4)(-3/4)×(-8+2/3-1/3)
(5)(-2)3-13÷(-1/2)
(6)0-23÷(-4)3-1/8
(7)(-2)3×0.5-(-1.6)2÷(-2)2
(8)(-3/2)×[(-2/3)2-2]
(9)[(-3)2-(-5)2] ÷(-2)
(10)16÷(-2)3-(-1/8) ×(-4)
(1)1; (2)0; (3)42; (4)5.75; (5)18; (6)0; (7)-4.64; (8)4/3; (9) -8; (10)-2/3
第3题:
先计算,再想一想怎样算得比较快。
1/3-1/4= 1/2+1/5=
1/6-1/7= 1/5+1/6=
1/7-1/8= 1/8+1/9=
1/2-1/3= 1/9+1/10=19/90
1/3-1/4=1/12 1/2+1/5=7/10
1/6-1/7=1/42 1/5+1/6=11/30
1/7-1/8=1/56 1/8+1/9=17/72
1/2-1/3=1/6 1/9+1/10=19/90
第4题:
下列Java表达式 19/3 (int)71.7-2*3 7% 3 (double)1/8+3 计算结果是
A.6.3 64 2 2.5
B.6 63 1 3.125
C.6.4 67 2 2.5
D.6 65 1 3.125
第5题:
A、CYP2C19*2*3
B、CYP2C19*3*3
C、CYP2C19*1*2
D、CYP2C19*1*3
E、CYP2C19*1*1
第6题:
1, 2, 3,8, 21,46, ( )。
A.70
B.87
C.90
D.100
第7题:
计算:
(1)(-8)-(-1) (2)45+(-30)
(3)-1.5-(-11.5) (4)-1/4-(-1/2)
(5)15-[1-(-20-4)] (6)-40-28-(-19)+(-24)
(7)22.54+(-4.4)+(-12.54)+4.4
(8) (2/3-1/2)-(1/3-5/6)
(9)2.4-(-3/5)+(-3.1)+4/5 (10)(-6/13)+(-7/13)-(-2)
(11)3/4-(-11/6)+(-7/3)
(12)11+(-22)-3×(-11)
(13)(-0.1)÷1/2×(-100)
(14)(-3/4)×(-2/3 - 1/3) ×0
(15)(-2)3-32
(16)23÷[(-2)3-(-4)]
(17)(3/4-7/8)÷(-7/8)
(18)(-60)×(3/4+5/6)
(1)(-8)-(-1)=-7
(2)45+(-30)=15
(3)-1.5-(-11.5)=10
(4)-1/4-(-1/2)=1/4
(5)15-[1-(-20-4)]=-10 (6)-40-28-(-19)+(-24)=-73
(7)22.54+(-4.4)+(-12.54)+4.4=10
(8) (2/3-1/2)-(1/3-5/6)=-1
(9)2.4-(-3/5)+(-3.1)+4/5=0.7 (10)(-6/13)+(-7/13)-(-2)=1
(11)3/4-(-11/6)+(-7/3)=1/4
(12)11+(-22)-3×(-11)=22
(13)(-0.1) ÷1/2×(-100)=20
(14)(-3/4) ×(-2/3 - 1/3) ×0=0
(15)(-2)3-32=-17
(16)23÷[(-2)3-(-4)]=-23/4
(17)(3/4-7/8) ÷(-7/8)=1/7
(18)(-60) ×(3/4+5/6)=-95
第8题:
●Let R and S be the relation respection respectively:
Relation R:A B C Relation S:B C D
1 2 3 2 3 4
6 7 8 2 3 5
9 7 8 7 8 1
The number of tuples that the result of the expression R∞S(Here if natural join of relations R and S) is (69) .
(69) A.1
B.4
C.7
D.10
第9题:
1, 2, 3, 8, 21,46, ( )。
A.70
B.87
C.90
D.100
第10题:
-1,1,-8,27,( )
A.-64
B.19
C.46
D.64