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6 .某企业购入设备一台,买价1 0 0 0 0 0 元,增值税1 7 0 0 0 元,运杂费1 3 0 0 元,款项以银 行存款支付,该笔经济业务编制的会计分录是( ) 。
A .一借一贷
B .一借多贷
C .一贷多借
D .多借多贷
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报告期比基期销售额增加了4200元,为分析因销售量变动使销售额增长的绝对额,应选用的算式及结果为( )。A.∑P1Q1-∑P1Q0=2350元B.∑P1Q1-∑P0Q1=2300元C.∑P1Q0-∑P0Q0=1580元D.∑P0Q1-∑P0Q0=1900元
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Router R1 currently has no configuration related to IPv6 or IPv4. The following configuration exists in a planning document, intended to be used to copy/paste into Router R1 to enable OSPFv3 on interfaces F0/0 and S0/0/0. No other related configuration exists. Assuming F0/0 and S0/0/0 reach an up/up state, which of the following is true about OSPFv3 on R1 after this configuration has been pasted into R1?()ipv6 router ospf 1ipv6 unicast- routinginterface f0/0ipv6 address 2000 ::1/64ipv6 ospf 1 area 1interface s0/0/0ipv6 address 2001::/64 eui - 64ipv6 ospf 1 area 0A. OSPF works on F0/0 and S0/0/0 without further configurationB. OSPF works with the addition of one command: a no shutdown command in OSPF router configur ation modEC. OSPF works with the addition of one command: an router - id command in OSPF router configuration modED. OSPFv3 needs at least two more configuration commands before it works on R1
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8选1数据选择器74LS151的输出Y与输入数据端D0……D7及输入地址选择输入端A2A1A0的对应表达式为()。A.Y=(D0+D1+D2+D3+D4+D5+D6+D7)A2A1A0#B.Y=D0A2'A1'A0'+D1A2'A1'A0+D2A2'A1A0'+D3A2'A1A0+D4A2A1'A0'+D5A2A1'A0+D6A2A1A0'+D7A2A1A0#C.Y=D0A2A1A0+D1A2'A1'A0+D2A2'A1A0'+D3A2'A1A0+D4A2A1'A0'+D5A2A1'A0+D6A2A1A0'+D7A2'A1'A0'#D.Y=(D0+D1+D2+D3+D4+D5+D6+D7)A2'A1'A0'
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12、下列向量组中是线性无关的向量组是().A.(1,2),(-3,0),(5,1).B.(1,1,0),(0,0,3),(2,2,0).C.(2,6,0),(3,9,0),(0,0,2).D.(1,1,0),(0,2,0),(0,0,3).
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下列向量组中是线性无关的向量组是().A.(1,2),(-3,0),(5,1).B.(1,1,0),(0,0,3),(2,2,0).C.(2,6,0),(3,9,0),(0,0,2).D.(1,1,0),(0,2,0),(0,0,3).
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序列e(k)={-1,2,1;k= -1,0,1}与 h(k)={1,2,-1;k=0,1,2} 的卷积和序列为A.{-1,0,6,0,-1;k=-1,0,1,2,3}B.{-1,0,6,0,-1;k=0,1,2,3,4}C.{0,6,0,-1;k=0,1,2,3}D.{-1,0,6,0;k=-1,0,1,2}
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2、下列向量组中,()是线性无关向量组A.(1,2),(-3,0),(5,1)B.(1,1,0),(0,0,3),(2,2,0)C.(2,6,0),(3,9,0),(0,0,2)D.(1,1,0),(0,2,0),(0,0,3)
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以下给出了一个代码逻辑段,请问下列哪一组测试用例满足多条件覆盖要求 if (a>0 c==1 ) { x = x+1; } if (b==3 || d<0 ) { y = 0; }A.a>0,c=1, b!=3, d<0;a<=0, c=1, b=3,d>=0;a>0, c!=1, b!=3,d<0;a<=0, c!=1, b!=3, d>=0B.a>=0,c=1, b=3, d<0;a<0, c=1, b=3,d>=0;a>0, c!=1, b!=3, d<0;a<=0, c!=1, b!=3, d>=0C.a>0,c=1, b=3, d<0;a<=0, c=1,b=3,d>=0;a>0, c!=1,b!=3, d<0;a<=0, c!=1, b!=3, d>=0D.a>0,c=1, b!=3, d<=0;a<=0, c=1, b=3,d>=0;a>0, c!=1, b!=3,d<0;a<=0, c!=1, b!=3, dg
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