—May I use my credit card here? —A. You can’t.B. No one uses it.C. am afraid you can’t.D. Who told you?

(b) The marketing director of CTC has suggested the introduction of a new toy ‘Nellie the Elephant’ for which the

following estimated information is available:

1. Sales volumes and selling prices per unit

Year ending, 31 May 2009 2010 2011

Sales units (000) 80 180 100

Selling price per unit ($) 50 50 50

2. Nellie will generate a contribution to sales ratio of 50% throughout the three year period.

3. Product specific fixed overheads during the year ending 31 May 2009 are estimated to be $1·6 million. It

is anticipated that these fixed overheads would decrease by 10% per annum during each of the years ending

31 May 2010 and 31 May 2011.

4. Capital investment amounting to $3·9 million would be required in June 2008. The investment would have

no residual value at 31 May 2011.

5. Additional working capital of $500,000 would be required in June 2008. A further $200,000 would be

required on 31 May 2009. These amounts would be recovered in full at the end of the three year period.

6. The cost of capital is expected to be 12% per annum.

Assume all cash flows (other than where stated) arise at the end of the year.

Required:

(i) Determine whether the new product is viable purely on financial grounds. (4 marks)

I'd never achieve my dreams ______ here.

A: working

B: worked

C: work

D: to work

A.X

B.y

C.j

D.i

A、three times as much

B、as much three times

C、much as three times

D、as three times much

A. buy

B. spend

C. bill

D. use

阅读以下函数说明和C语言函数，将应填入(n)处的字句写在对应栏内。

[说明]

某银行共发出M张储蓄卡，每张储蓄卡拥有唯一的卡号，每天每张储蓄卡至多支持储蓄卡持有者的N笔“存款”或“取款”业务。程序中用数组card[M][N+3]中的每一行存放一张储蓄卡的有关信息，其中：

card[i][0]存放第i张卡的卡号；

card[i][1]存放第i张卡的余额；

card[i][2]存放第i张卡的当日业务实际发生笔数；

card[i][3]～card[i][N+2]存放第i张卡的当日存取款金额，正值代表存款，负值代表取款。

当持卡者输入正确的卡号、存款或取款金额后，程序进行相应的处理；若输入不正确的数据，程序会提示持卡者重新输入；若输入的卡号为负数时，银行终止该卡的当日业务。

[C程序]

include＜stdio.H＞

define M 6

define N 5

long card[M][N+3]={{9801,2000,0,},{9812,2000,2,},{9753,3000,1,},

{8750,500,0,},{9604,2800,3,),(8901,5000,5,}};

int locate(long card[][N+3],int m,long no)

{ int i;

for(i=0;i＜m;i++)

if((1)==no) return i;

(2);

}

main()

{long cardNo,money;

int k;

while(1){

printf("请输入卡号:\n");

scanf("%1d",&cardNo);

if(cardNo＜0) break;

k=locate(card,M,cardNo);

if(k==-1){

printf("不存在%id号的储蓄卡\n",cardNo);

continue;

}

printf("请输入金额(正值代表存款，负值代表取款):\n");

scanf("%id",&money);

if(card[k][1]+money＜0){

printf("存款余额不足，不能完成本次的取款业务\n");

continue;

}

if(card[k][2]==N){

printf("已完成本卡的当日业务\n");

continue;

}

/*处理一笔业务的数据*/

card[k] (3)=money;

(4);

(5);

}

}