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【多选题】
下列有关乘数分析法,正确的是A.
阐述了国际贸易与国民收入的关系
B.
探析了各国经济通过进出口途径相互影响的机制
C.
其分析考虑了国际间的资本流动
D.
理论建立在凯恩斯乘数原理下
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【单选题】电路如下图所示,是一种光电报警电路的一部分。三极管 T 的放大倍数β =100 ,基极和发射极的正向压降为 0.7V ;电阻 R 1 =6kΩ , R 2 =1kΩ ;蜂鸣器是阻性元件,且电阻为 100 Ω;当流过发光二极管的电流大于等于2mA时,发光二极管就会发光;当流过蜂鸣器的电流大于等于 60mA 时,蜂鸣器就会鸣叫;电压 U CC =12V 。在正常情况下, B 点电位为 0V 。若 B ...
A.
发光二极管不发光 蜂鸣器不鸣叫
B.
发光二极管发光 蜂鸣器不鸣叫
C.
发光二极管不发光 蜂鸣器鸣叫
D.
发光二极管发光 蜂鸣器鸣叫
【单选题】程序填空 while(low<=high) { ______________ if(date[mid]==x) {printf("%d 是数组中的第%d个数\n",x,mid);break;} else if(date[mid]
A.
mid=(low+high)/2;
B.
mid=high/2;
C.
low=mid+high;
D.
mid=low%2;
【单选题】代码填空【使用递归实现二分查找】 int binarySearch(int a[], int key, int low, int high) { if (low > high) return -1; int mid; mid = (low + high) / 2; if (key == a[mid]) return mid; else if (key < a[mid]) return _____...
A.
(1) binarySearch(a, key, low, mid+1) (2) binarySearch(a, key, mid - 1, high)
B.
(1) binarySearch(a, key, low, mid+1) (2) binarySearch(a, key, mid + 1, high)
C.
(1) binarySearch(a, key, low, mid - 1) (2) binarySearch(a, key, mid + 1, high)
D.
(1) binarySearch(a, key, low, mid - 1) (2) binarySearch(a, key, mid - 1, high)
【单选题】________ among the “Angry Young Men” novelistin the mid 1950s and early 1960s was the first to start the attack on middleclass privileges and power in his novel Lucky Jim .
A.
Kinsley Amis
B.
John Wain
C.
John Osborne
D.
John Brain
【简答题】In the 1950s, classroom violence ________. A) was something unheard of B) was by no means a rare occurrence C) attracted a lot of pubic attention D) began to appear in ysts’ data
【单选题】It was in the 1950s that the American government finally took action to build a national highway system.
A.
Y
B.
N
C.
NG
【单选题】A psychological issue that began to be discussed in the 1950s was the question of the most appropriate age for second language learning. The ability of young children to learn language 'easily' had, f...
A.
interest.
B.
controversy.
C.
compliments.
D.
encouragement.
相关题目：
【单选题】程序填空 while(low<=high) { ______________ if(date[mid]==x) {printf("%d 是数组中的第%d个数\n",x,mid);break;} else if(date[mid]
A.
mid=(low+high)/2;
B.
mid=high/2;
C.
low=mid+high;
D.
mid=low%2;