某联立方程计量经济学模型有3个方程，3个内生变量[tex=4.929x1.357]vkCJYkBVQKvLYWNv133Jj2sL2BC6uiaFwYwr+/AYb1zFZJcZ3Y5OsnU9qZ4ngw0m[/tex]3个外生变量[tex=5.214x1.357]CVnIHALKpREOmNiXwIbJmiVLQiRvqZtxZaoo41O9aX8WQTuMNPn8dl7bwS/ho8P/[/tex]和样本观测值始终为1的虚变量c,样本容量为n。其中第2个方程:[tex=13.071x1.214]w/VzgK+sVZy1SqdUBvmh/wFJ22OjFLzd62vp7Eikuz2LpPt9dIQ4/yIVLiC5Sczs4Tc2/rabTc8vZ9aNIHRMrameWZK4ECCFfJzsgnFpV90=[/tex]为恰好识别的结构方程。写出用工具变量法估计该方程参数的正规方程组; 

 某联立方程计量经济学模型有3个方程，3个内生变量[tex=4.929x1.357]vkCJYkBVQKvLYWNv133Jj2sL2BC6uiaFwYwr+/AYb1zFZJcZ3Y5OsnU9qZ4ngw0m[/tex]3个外生变量[tex=5.214x1.357]CVnIHALKpREOmNiXwIbJmiVLQiRvqZtxZaoo41O9aX8WQTuMNPn8dl7bwS/ho8P/[/tex]和样本观测值始终为1的虚变量c,样本容量为n。其中第2个方程:[tex=13.071x1.214]w/VzgK+sVZy1SqdUBvmh/wFJ22OjFLzd62vp7Eikuz2LpPt9dIQ4/yIVLiC5Sczs4Tc2/rabTc8vZ9aNIHRMrameWZK4ECCFfJzsgnFpV90=[/tex]为恰好识别的结构方程。用间接最小二乘法估计该方程参数，也可以看成一 种工具变量法，指出工具变量是如何选取的，并写出参数估计量的矩阵表达式;

 某联立方程计量经济学模型有3个方程，3个内生变量[tex=4.929x1.357]vkCJYkBVQKvLYWNv133Jj2sL2BC6uiaFwYwr+/AYb1zFZJcZ3Y5OsnU9qZ4ngw0m[/tex]3个外生变量[tex=5.214x1.357]CVnIHALKpREOmNiXwIbJmiVLQiRvqZtxZaoo41O9aX8WQTuMNPn8dl7bwS/ho8P/[/tex]和样本观测值始终为1的虚变量c,样本容量为n。其中第2个方程:[tex=13.071x1.214]w/VzgK+sVZy1SqdUBvmh/wFJ22OjFLzd62vp7Eikuz2LpPt9dIQ4/yIVLiC5Sczs4Tc2/rabTc8vZ9aNIHRMrameWZK4ECCFfJzsgnFpV90=[/tex]为恰好识别的结构方程。 用二阶段最小二乘法估计该方程参数，也可以看成一种工具变 量法，指出[tex=1.0x1.214]wmS4Og53BRUWJ9qIqm1VdQ==[/tex]的工具变量是什么，并写出参数估计量的矩阵表达式。

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4

共有4个站进行码分多址通信。4个站的码片序列为：a：（-1 -1 -1 +1 +1 -1 +1 +1） b：（-1 -1 +1 -1 +1 +1 +1 -1） c：（-1 +1 -1 +1 +1 +1 -1 -1） d：（-1 +1 -1 -1 -1 -1 +1 -1） 现收到这样的码片序列：（-1 +1 -3 +1 -1 -3 +1 +1），则（ ）发送1。

共有四个站进行CDMA通信，现收到这样的码片序列（-3 +1 -1 -1 +1 +1 +1 -3），则下列站点发送了数据1的是（ ） A: （-1 -1 -1 +1 +1 -1 +1 +1） B: （-1 -1 +1 -1 +1 +1 +1 -1） C: （-1 +1 -1 +1 +1 +1 -1 -1） D: （-1 +1 -1 -1 -1 -1 +1 -1）