某工厂生产滚珠,从某日生产的产品中随机抽取 9 个,测得直径[tex=2.571x1.357]WjseTLDPuNQ7B0t1S0jb6A==[/tex]如下:[tex=25.286x1.357]tOuhNvizKyRZPn+XCdGxL3lFsJTa+uWTErIATE4sWU5wsFBch4Ml4hy4GoZjSD9WWNnrsIT99nPFZlWPoEkFnQ7VIWM06cnleRcu9Qujjl/6u9IfrP2zpJACrL2AnJQe[/tex]设滚珠直径服从正态分布[tex=3.929x1.571]gORhU/2TMGLbpqNyfyzmZ6k+pBGdBkdAl/WqEq1GQaA=[/tex],求直径均值[tex=0.643x1.0]R8j6nFNrQJBYHOT5c6hCaw==[/tex]的置信水平为[tex=1.786x1.0]L31gvkuUIdeYrFQQ/WabHQ==[/tex]的置信区间,如果未知[tex=0.571x0.786]KMF8QHqVjNLkn7nK5uaSag==[/tex].

某工厂生产滚珠,从某日生产的产品中随机抽取 9 个,测得直径[tex=2.571x1.357]WjseTLDPuNQ7B0t1S0jb6A==[/tex]如下:[tex=25.286x1.357]tOuhNvizKyRZPn+XCdGxL3lFsJTa+uWTErIATE4sWU5wsFBch4Ml4hy4GoZjSD9WWNnrsIT99nPFZlWPoEkFnQ7VIWM06cnleRcu9Qujjl/6u9IfrP2zpJACrL2AnJQe[/tex]设滚珠直径服从正态分布[tex=3.929x1.571]gORhU/2TMGLbpqNyfyzmZ6k+pBGdBkdAl/WqEq1GQaA=[/tex],求直径均值[tex=0.643x1.0]R8j6nFNrQJBYHOT5c6hCaw==[/tex]的置信水平为[tex=1.786x1.0]L31gvkuUIdeYrFQQ/WabHQ==[/tex]的置信区间,如果已知直径标准差[tex=5.714x1.357]xCiUKJOkTkXvb9WzG0a3ue3rD7UHQ2iqA+bJsksEoZ4=[/tex].

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）

共有四个站进行码分多址CDMA通信。四个站的码片序列为：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) , 现在收到这样的码片序列：（-1 -1 +1 +1 +3 +3 -1 -1 ）