X2检验计算中,任一行(R)及一列(C)交叉对应的理论数TRc的计算公式是如下哪一个 A: TRc=(nc+n/n B: TRc=(nc+n/2 C: TRc=(ncn/2 D: TRc=n/(ncn E: TRc=(ncn/n
X2检验计算中,任一行(R)及一列(C)交叉对应的理论数TRc的计算公式是如下哪一个 A: TRc=(nc+n/n B: TRc=(nc+n/2 C: TRc=(ncn/2 D: TRc=n/(ncn E: TRc=(ncn/n
下列哪一个公式是多个样本率x2检验的专用公式 A: X2=∑(A-2/T B: X2=n[∑A2/(ncn-1] C: X2=[n(ad-b2]/[(a+c+(a+(b+] D: P=[(a+!(c+!(a+!(b+!]/(a!b!c!d!n!) E: X2=(1ad-bc|-n/2)2n/2[(a十(c+(a+(b+]
下列哪一个公式是多个样本率x2检验的专用公式 A: X2=∑(A-2/T B: X2=n[∑A2/(ncn-1] C: X2=[n(ad-b2]/[(a+c+(a+(b+] D: P=[(a+!(c+!(a+!(b+!]/(a!b!c!d!n!) E: X2=(1ad-bc|-n/2)2n/2[(a十(c+(a+(b+]
在R×C表的χ2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为 A: TRC=nR+nCR B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCmR
在R×C表的χ2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为 A: TRC=nR+nCR B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCmR
在R×C表的χ2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为 A: TRC=nR+nC2 B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
在R×C表的χ2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为 A: TRC=nR+nC2 B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
在R×C表的X2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为() A: TRC=nR+nC2 B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
在R×C表的X2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为() A: TRC=nR+nC2 B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
20世纪90年代提出的下一代网络NCN的概念,反映了网络融合的发展方向,其主要目标不包括( )。 A: 成本更低、技术更高 B: 支持话音、实时的多媒体业务 C: 缩减服务投向市场的时间 D: 支持种接入方式和多种接入终端
20世纪90年代提出的下一代网络NCN的概念,反映了网络融合的发展方向,其主要目标不包括( )。 A: 成本更低、技术更高 B: 支持话音、实时的多媒体业务 C: 缩减服务投向市场的时间 D: 支持种接入方式和多种接入终端
A1型题 在R×C表的X2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为() A: TRC=nR+nCR B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
A1型题 在R×C表的X2检验中,设nR,nC和n分别为行合计,列合计和总计,则计算每格理论数的公式为() A: TRC=nR+nCR B: TRC=nR+nCn C: TRC=nR×nCn D: TRC=nR+nCnC E: TRC=nR×nCnR
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
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
【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
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