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【单选题】
某产品有F1、F2、F3、F4四项功能，采用环比评分法得出相邻两项功能的重要性系数为：F1/F2=1.75，F2/F3，=2.20，F3/F4=3.10。则功能F2的重要性系数是（ ）。
A.
0.136
B.
0.179
C.
0.224
D.
0.298
相关题库：造价工程师《建设工程造价管理》
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参考答案：
举一反三
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】甲组变异系数为3.59%,乙组变异系数为4.89%说明( )
A.
说明甲组比乙组变异程度大
B.
此时不能说明哪一组变异程度大,要经过显著性检验
C.
说明乙组比甲组变异程度大
D.
说明甲组的均数比乙组小
E.
不能说明任何问题
相关题目：
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
【单选题】甲组变异系数为3.59%,乙组变异系数为4.89%说明( )
A.
说明甲组比乙组变异程度大
B.
此时不能说明哪一组变异程度大,要经过显著性检验
C.
说明乙组比甲组变异程度大
D.
说明甲组的均数比乙组小
E.
不能说明任何问题
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