Assessing the data is a vital part of creating an effective marketing strategy. {A; B

A.数据评估就是要制定有效市场营销策略。

B.数据评估和市场策略都非常重要。

C.数据评估是制定有效市场营销策略的重要一环。

A. 数据评估就是要制定有效市场营销策略。

B. 数据评估和市场策略都非常重要。

C. 数据评估是制定有效市场营销策略的重要一环。

One criterion for evaluating the performance of an economic system is to （) changes i

One criterion for evaluating the performance of an economic system is to () changes in productivity.

A. assess .

B. assessing

C. access

S39 ‘Financial Instruments:

Recognition and Measurement’. The directors realise that significant changes may occur in their accounting treatment

of financial instruments and they understand that on initial recognition any financial asset or liability can be

designated as one to be measured at fair value through profit or loss (the fair value option). However, there are certain

issues that they wish to have explained and these are set out below.

Required:

(a) Outline in a report to the directors of Ambush the following information:

(i) how financial assets and liabilities are measured and classified, briefly setting out the accounting

method used for each category. (Hedging relationships can be ignored.) (10 marks)

A.data(O)

B.data(1)

C.data(2)

D.data(3)

A.data（0）

B.data（1）

C.data（2）

D.data（3）

A.data（0）

B. data（1）

C. data（2）

D. data（3）

若有以下程序： #include〈iostream〉 using namespace std； int main() { int data[4],i,j,temp; for (i=O; i＜4; i++) cin＞＞data[i]; for (i=1; i＜4; i++) { j = i-1; temp = data[i]; while (data [j ] ＞temp&&j ＞=0) { data[j+1] = data[j]; j--; } data[j+1] = temp; } for(i=O;i＜4;i++) cout〈〈data[i]〈〈" "; cout〈〈end1; return 0; }

A．2843

B．2348

C．8243

D．8432

A.data=[34,5,-6,12,87,29] data.sort()###SXB###B.data=[34,5,-6,12,87,29] n=len(data) while i＜n: while j＜n-1: if data[j]＞data[j+1]: data[j],data[j+1]= data[j+1],data[j] j+=1 i+=1###SXB###C.data=[34,5,-6,12,87,29] n=len(data) i=0 j=0 while i＜n: while j＜n: if data[j]＞data[j+1]: data[j],data[j+1]= data[j+1],data[j] j+=1 i+=1 print(data)###SXB###D.data=[34,5,-6,12,87,29] n=len(data) i=0 j=0 while i＜n-1: while j＜n-1: if data[i]&g

下列选项中，能填入空白处2的代码是()

A. data[i % cols][i / cols] = Double.parseDouble(dataStr[i]);

B. data[i/cols][i % cols] = Double.parseDouble(dataStr[i]);

C. data[i/ rows][i % rows] = Double.parseDouble(dataStr[i]);

D. data[i % rows][i /rows] = Double.parseDouble(dataStr[i]);

A.data=[34,5,-6,12,87,29] data.sort()###SXB###B.data=[34,5,-6,12,87,29] n=len(data) while i＜n: while j＜n-1: if data[j]＞data[j+1]: data[j],data[j+1]= data[j+1],data[j] j+=1 i+=1###SXB###C.data=[34,5,-6,12,87,29] n=len(data) i=0 j=0 while i＜n: while j＜n: if data[j]＞data[j+1]: data[j],data[j+1]= data[j+1],data[j] j+=1 i+=1 print(data)###SXB###D.data=[34,5,-6,12,87,29] n=len(data) i=0 j=0 while i＜n-1: while j＜n-1: if data[i]&g