The federal government's practice of hiring on the basis of ksa frequently results in the hiring of employees _____ A: based on race,<br/>religion, sex, and so forth. B: who are<br/>unqualified for the job. C: who are qualified<br/>for the job. D: on the basis of<br/>subjective judgment.

戈尔茨坦等将需求评估过程分为五个阶段，即组织支持、组织分析、需求分析、()及人员分析。 A: 组织目标分析 B: 生产计划分析 C: 员工技能分析 D: 任务和KSA分析

  在纯电感正弦交流电路中，当电流[tex=7.643x1.571]WO07BazhtC2fY74UOj8ded35d23A0y1T5ASJAy8iSC5kf91BibRHc0Wo8VXDB7kufwC48eSQ8iFx3tklSSoUKd6Uu5LA+UvI8PM6WPevfC4=[/tex]时, 则电压 (      ) [tex=0.786x1.0]Eur2Tzu+S/KpqWi9c/1ksA==[/tex] 。 未知类型：{'options': ['[tex=9.357x2.214]IoWa75kAqB1hKInaRo8SO2mXcszBvmscBwhoPM2l1gsv0XU6HshYkQ5lH6qcqzYM0que6UxAPQFQzALHNs2H7k43VOrxuTW6iYYkwc2YqUo=[/tex]', '[tex=11.857x2.214]I+h1KWG+ANBj/DKRCQ0nvT8RYMfbIkT5B8CwO2mDLw0k1mD6v8zVOQw98eKdg/OWofMwkZMpfsS4vu6otCw/i6Jw2WHPAGUUKokkCK0wbiY=[/tex][br][/br]', '[tex=10.786x2.214]XWyu9qKEH7hQ+l35y5hX6YQJeLn/09Soiv7Y8JVUYKk+BOJK2txEuCyynN7Uoeuf8hXsD4S2tsKDy3bwfKVaDtRSYaCdb/+a04Tst3eN9FJm1Tv7UlqbRDy+nZlh8LwTlE8/wl6QdTx+eICgVtv3zg==[/tex][br][/br]'], 'type': 102}

常见的金属晶体结构有哪几种？ [tex=2.786x1.143]2xyYMEWCXSosGafUr0ooUw==[/tex]、 [tex=2.786x1.214]NQAL0X+oQZ9Ji2T0z4L2rIIr8FPVtW/39XPcPAJvZFA=[/tex]、[tex=1.071x1.0]8P945eAbdM8rqhp3pIT4GA==[/tex]、[tex=1.286x1.0]o5/ybNFxUFSb0DKV00WnKQ==[/tex] 、[tex=1.071x1.0]mwQR3BG6H2DLYymaty9NgQ==[/tex]、[tex=1.286x1.0]eNFdk/oCZLIU1lqgTeyg6Q==[/tex]、[tex=1.143x1.0]AbXguCEJhaI0OlA0vqy/0w==[/tex]、[tex=0.786x1.0]Eur2Tzu+S/KpqWi9c/1ksA==[/tex] 、[tex=1.5x1.214]MJKnJ5F6g/u5k+B/u17tMw==[/tex]、[tex=1.286x1.0]cLf1+ueR+iiGFyk14tV/sA==[/tex] 各属何种晶体结构?

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。