(b) Calculate the value of the closing stocks of finished goods at the end of the three-mo

题目

(b) Calculate the value of the closing stocks of finished goods at the end of the three-month period, and the value

of cost of sales for the period. (3 marks)

如果没有搜索结果或未解决您的问题,请直接 联系老师 获取答案。
相似问题和答案

第1题:

When there is no indication of damage to the ______ , a bill of lading is said to be ______.

A.documents.., clear

B.goods.., clear

C.documents.., clean

D.goods.., clean


正确答案:D
解析:a bill of lading提单,提单是货物由运输公司到托运人的收据,也是准可议付单据(quasi negotiable document)。

第2题:

It took me three hours _______ the task.

A、finishing

B、to finish

C、finish

D、finished


参考答案:B

第3题:

The students ______ four hours doing the reading assignments.

A.spent

B.cost

C.took

D.finished


正确答案:A

第4题:

The boy lay in the street,his eyes_______and his hands______.

A.closing;trembling
B.closed;trembled
C.closing;trembled
D.closed;trembling

答案:D
解析:

第5题:

下面程序的运行结果为【】。 include void main() {unsigned char value=127;int tota

下面程序的运行结果为【 】。

include <iostream.h>

void main()

{

unsigned char value=127;

int total=100;

value++;

total+=value;

cout<<total<<end1;

}

A) 227 B) 100 C) 127 D) 27


正确答案:4 10
4 10 解析:本题主要考察C++类中静态数据成员的使用。程序首先定义了类 count,其内部含有private类型数据成员static int n;同时含有public类型构造函数 count()和静态成员函数static inttest(),这两个函数的功能分别是为对象申请系统资源并将静态数据成员n加1和将静态数据成员n加4。主函数前,程序将静态数据成员n初始化为0,该数据成员为所有类count的对象所共有的数据成员;主函数中程序首先执行静态成员函数test()(由于test声明为static,因此其调用时无需通过具体对象),其执行过程中,静态数据成员n应该加4变成n=4,因此此处输出为4;此后程序创建对象c1和c2,由于在每次创建过程中都要调用构造函数count(),而每次调用count()函数后,静态数据成员n值都会加1。因此,创建两个对象之后,n值变为n=6;再次执行test()函数后,n的值再次加4,因此变为n=6+4=10。故程序全部执行后,变量n值变为10,而中间程序输出为“410”。

第6题:

1.publicclassSimpleCalc{

2.publicintvalue;

3.publicvoidcalculate(){value+=7;}

4.}

And:

1.publicclassMultiCalcextendsSimpleCalc{

2.publicvoidcalculate(){value-=3;}

3.publicvoidcalculate(intmultiplier){

4.calculate();

5.super.calculate();

6.value*=multiplier;

7.}

8.publicstaticvoidmain(String[]args){

9.MultiCalccalculator=newMultiCalc();

10.calculator.calculate(2);

11.System.out.println(”Valueis:+calculator.value);

12.}

13.}

Whatistheresult?()


参考答案:A

第7题:

Youexecutedthefollowingcode:BEGINDBMS_SCHEDULER.SET_ATTRIBUTE(NAME=>’JOB_A’,ATTRIBUTE=>’JOB_PRIORITY’,VALUE=>7);END;/Afteranalyzingtheabovecode,whatconclusionwillyoudraw?()

A.Thecodewillbeexecutedsuccessfully.

B.ThecodewillnotbeexecutedsuccessfullybecausethevalueoftheVALUEparametermustbe1,2,or3.

C.ThecodewillnotbeexecutedsuccessfullybecausethevalueoftheVALUEparametermustrangebetween1and5.

D.ThecodewillnotbeexecutedsuccessfullybecausenoSET_ATTRIBUTEprocedureexistsintheDBMS_SCHDULERpackage.


参考答案:C

第8题:

The test ______, we began our holiday.

A. was finished

B. finishing

C. was finishing

D. finished


正确答案:D
答案为D。考完试我们开始放假了。过去分词表示已经被动、已完成的动作,相当于 when the test was finished...。

第9题:

以下程序的执行结果为()。includeclass myClass{Dublic:myClass(intdata) ;myClass(

以下程序的执行结果为( )。 #include<iostream.h> class myClass { Dublic: myClass(int data) ; myClass(){value=0; }; myClass(myClass &obj); void print(){ cout<<value<<end1; } ~myClass(){} private: int value; }; myClass::myClass(int data) {value=data; } myClass::myClass(myClass &Obj) { value=obj.value; } void main() { myClass a(100); myClass b(a) ; b.print(); }

A.0

B.100

C.10

D.程序编译时出错


正确答案:B
解析:题中类myClass定义了一个拷贝初始化构造函数myClass(myClass&obj)。在主函数中,系统首先自动调用相应的构造函数建立对象,然后调用拷贝构造函数对对象b进行初始化。所以程序最后的输出为100。

第10题:

The boy lay in the street,his eyes_______and his hands______.

A.closing;trembling
B.closed;trembled
C.closing;trembled
D.closed;tremblin

答案:D
解析: