第一部分计划和设计
Design a simple LAN using Cisco Technology
使用思科技术设计一个简单的局域网
1、Which of the following devices can an administrator use to segment their LAN? (Choose all that apply)下列哪个设备能使管理员分段他们的局域网？
A. Hubs
B. Repeaters
C. Switches
D. Bridges
E. Routers
F. Media Converters
G. All of the above
A 集线器
B 中继器
C 交换机
D 网桥
E 路由器
F 媒体转换
G 所有
答案：CDE
解析：Routers, switches, and bridges don’t transmit broadcasts. They segment a large Cumber some network, into multiple efficient networks.
路由器，交换机，和网桥不能传输广播，它们分一个大的网络，到多个网络
错误答案：A. Hubs is incorrect because a hub doesn’t segment a network, it only allows more hosts on one. Hubs operate at layer one, and is used primarily to physically add more stations to the LAN.
B. This also incorrect because the job of a repeater is to repeat a signal so it can exceed distance limitations. It also operates at layer one and provides no means for logical LAN segmentation.
F. This is incorrect because media converters work by converting data from a different media type to work with the media of a LAN. It also operates at layer one and provides no means for logical LAN segmentation.
A 集线器是不正确的因为不能分段，它只允许许多主机在上面，集线器工作在第一层，它用于首先到自身增加许多位置到局域网
B 不正确因为中继器的工作是反复一个信号，因此它能超越远距离限制，它也是工作在一层并且提供无意义的局域网逻辑划分
F 它不正确是因为媒体转换工作通过从一个不同的媒体类型转换数据到局域网的其他媒体，它也是工作在1层并提供无意义的局域网划分。
2、Routers perform which of the following functions? (Select three)
下列哪些是路由器提供的功能？（选3项）
A. Packet switching
B. Collision prevention on a LAN segment.
C. Packet filtering
D. Broadcast domain enlargement
E. Broadcast forwarding
F. Internetwork communication
A 包交换
B 防止局域网分段中的冲突
C 包过滤
D 扩展广播域
E 推进广播
F 网络通讯
答案：ACF
解析：A. Routers work in Layer 3 of the OSI Model. A major function of the router is to route packets between networks.
C. Through the use of access lists, routers can permit and deny traffic using layer 3 and layer 4 packet information.
F. The primary purpose of a router is to route traffic between different networks, allowing for internetworking.
Ａ路由器工作在OSI的3层，主要的功能是在网络中路由数据包Ｃ通过使用访问控制列表，通过使用3层和4层的包信息路由器能允许和拒绝传输数据
F 路由器的主要目的是在不同网络间的路由传送数据，允许网络
错误回答：B. While routers can be used to segment LANs, which will reduce the amount of collisions; it can not prevent all collisions from occurring. As long as there are 2 or more devices on a LAN segment, the possibility of a collision exists, whether a router is used or not.
D. The broadcast domain of a LAN is often segmented through the use of a router. This results in reducing the size of the broadcast domain.
E. Routers do not forward broadcast traffic.
B 路由能够用于局域网的分段，它能减少冲突数量，它不能防止所有的冲突事件，只要在分段中有2个或更多的分段，冲突存在的可能性是是否使用路由D 一个LAN的广播域是通过使用路由器分段，这个结论是减少了广播域的大小 E 路由器不是广播传输数据
3、Which of the following statements most accurately describes the characteristics of the above networks broadcast and collision domains? (Select the two best answer choices)如图所示，销售部门和生产部门通过一个路由器分开，下列哪个陈述正确的描述了关于网络广博域和冲突域的特征？
A. There are two broadcast domains in the network.
B. There are four broadcast domains in the network.
C. There are six broadcast domains in the network.
D. There are four collision domains in the network.
E. There are five collision domains in the network.
F. There are seven collision domains in the network.
A 网络中有2个广播域Ｂ、网络中有４个广播域C、网络中有6个广播域D、网络中有4个冲突域E、网络中有5个冲突域F、网络中有7个冲突域答案：AF
解析：In this network we have a hub being used in the Sales department, and a switch being used in the Production department. Based on this, we have two broadcast domains: one for each network being separated by a router. For the collision domains, we have 5 computers and one port for E1 so we have 6 collision domains total because we use a switch in the Production Department so 5 are created there, plus one collision domain for the entire Sales department because a hub is being used.
在这个网络中我们有一个集线器用于销售部门，和一个交换机用于生产部门。基
于这些，我们有两个广播域：一个通过一个路由器分开。对于这冲突域，我们有5台计算机都在E1中因此一共有6个冲突域，因为我们使用了一个交换机在生产部门这样看来有5个在一起，在加上整个销售部门的一个冲突域，因为是用一个集线器连接的
4、The Testking corporate LAN consists of one large flat network. You decide to segment this LAN into two separate networks with a router. What will be the affect of this change?局域网是由一个大的网络组成。你打算将这个网络通过一个路由器分成两部分，这样改变的影响是什么？
A. The number of broadcast domains will be decreased.
B. It will make the broadcasting of traffic between domains more efficient between segments.
C. It will increase the number of collisions.
D. It will prevent segment 1’s broadcasts from getting to segment 2.
E. It will connect segment 1’s broadcasts to segment 2.
A、广播域的数量减少
B、使在域中的广播段的传输更有效
C、增加冲突域的数目
D、防止段1的广播到段2中
E、连接段1和段2的广播
答案：D
解析：A router does not forward broadcast traffic. It therefore breaks up a broadcast domain, reducing unnecessary network traffic. Broadcasts from one segment will not be seen on the other segment.
一个路由器不能传输广播，它分解广播域，减少不必要的网络传输。一个段中的广播不能被其他段的看见。
错误答案：A. This will actually increase the number of broadcast domains from one to two.
B. All link level traffic from segment one to segment two will now need to be routed between the two interfaces of the router. Although this will reduce the traffic on the LAN links, it does also provide a less efficient transport between the segments.
C. Since the network size is effectively cut into half, the number of collisions should decrease dramatically.
E. Broadcasts from one segment will be completely hidden from the other segment. Ａ、实际上这能增加广播域的数量从１个到２个
Ｂ、从段１到段２的所有的标准连接将通过路由器在两个接口间被路由，在段中它也提供一些有效的小的传输
C、因为网络的大小被有效的打断了一半，冲突域的数目将减少
E、来自一个段的广播将完全隐藏到其他段中
5、Which of the following are benefits of segmenting a network with a router? (Select all that apply)下列哪个是用路由器分段的好处？（选择所有可能的选项）
A. Broadcasts are not forwarded across the router.
B. All broadcasts are completely eliminated.
C. Adding a router to the network decreases latency.
D. Filtering can occur based on Layer 3 information.
E. Routers are more efficient than switches and will process the data more quickly.
F. None of the above.
A、广播不能通过路由器运送
B、所有广播将完全除去
C、增加路由器将减少
延时
D、能过滤基于3层的信息
E、路由器比交换机更快更有效的处理数据
F、都没有
答案：ＡＤ
解析：Routers do not forward broadcast messages and therefore breaks up a broadcast domain. In addition, routers can be used to filter network information with the use of access lists.
路由器不能传送广播信息并因此分割广播域，另外，通过访问控制列表路由器能滤波网络信息
错误答案：B. Broadcasts will still be present on the LAN segments. They will be reduced, because routers will block broadcasts from one network to the other.
C. Adding routers, or hops, to any network will actually increase the latency.
E. The switching process is faster than the routing process. Since routers must do a layer 3 destination based lookup in order to reach destinations, they will process data more slowly than switches.
Ｂ、在局域网段中广播仍然存在，它将减少，因为路由器将妨碍从一个网络到其他网络的广播
Ｃ、在任何网络中增加路由器或跳数实际上都增大延时
Ｅ、交换的过程比路由过程快，为了到达目的地，路由器需要通过３层目的查找，它将比交换机处理数据更慢
6、如图所示：
Of the following choices, which IP address should be assigned to the PC host?
在下列选项中，哪个IP地址能够分配到PC主机上？
A. 192.168.5.5
B. 192.168.5.32
C. 192.168.5.40
D. 192.168.5.6
E. 192.168.5.75
答案：C
解析：The subnet mask used on this Ethernet segment is /27, which translates to 255.255.255.224. Valid hosts on the 192.168.5.33/27 subnet are 192.168.5.33-
192.168.5.62, with 192.168.5.32 used as the network IP address and 192.168.5.63 used as the broadcast IP address. Therefore, only choice C falls within the usable IP range.
这个以太网段的子网掩码是27，转换为二进制为255.255.255.224。在192.168.5.33/27中有效的子网是192.168.5.33-192.168.5.62，习惯使用192.168.5.33作为网络IP地址，192.168.5.62作为广播IP地址。因此，在可用的IP范围中选择C
7、如图所示：
Based on the diagram above, how many collision domains are present in the https://www.doczj.com/doc/2d15699071.html, network??
基于这个图表，在这里有多少个冲突域？
A 、1个 B、2个 C、3个 D、4个 E、5个 F、6个 G、7个
答案：B
解析：Since hubs are being used for both Ethernet segments, there are a total of two collision domains. Routers do not forward broadcast and are used to segment LANs, so TestKingA consists of one collision domain while TestKingB consists of the second collision domain.
因为集线器经常用于以太网段中，有这两个冲突域的总数。路由器不能传送广播和分割局域网，因此TestKingA 是由一个冲突域组成而TestKingB 是第二个冲突域
8、下列的网络拓扑图，注意TestKing1 交换机和TestKing2集线器
Which of the devices shown can transmit simultaneously without causing collisions?
哪些设备能同时传输数据并且不会造成冲突？
A. All hosts
B. Only hosts attached to the switch
C. All hosts attached to the hub and one host attached to the switch
D. All hosts attached to the switch and one host attached to the hub
A、所有主机
B、只有和交换机连接的主机
C、所有和集线器相连的主机和一个同交换机相连的主机
D、所有和交换机相连的主机与一个和集线器相连的主机
答案：B
解析：As we know switch is the device which avoids collisions. When two computers communicate through a switch they make their own collision domain. So, there is no chance of collisions. Whenever a hub is included, it supports on half duplex communication and works on the phenomena of CSMA/CD so, there is always a chance of collision.
正如我们所知道的交换机是避免冲突的设备，当两个计算机通过一个路由器通讯时它们将增大冲突域。因此，没有冲突的机会。无论何时包含一个集线器，它将支持一部分双方的连接并通过CSMA/CD工作，因此也总会有冲突的机会
9、
Study the network topology exhibit carefully, in particular the two switches TestKing1, TestKing2, and the router TestKing3.
Which statements are true in this scenario? Select two.
仔细研究这个网络拓扑图，特别是这两个交换机TestKing1，TestKing2和路由器TestKing3 ，下列哪个是正确的？选择2个
A. All the devices in both networks will receive a broadcast to 255.255.255.255 sent by host TestKingA.
B. Only the devices in network 192.168.1.0 will a broadcast to 255.255.255.255 sent by host TestKingA.
C. All the devices on both networks are members of the same collision domain.
D. The hosts on the 192.168.1.0 network form one collision domain, and the hosts on the 192.168.2.0 network form a second collision domain.
E. Each host is in a separate collision domain.
A、在网络中的所有的设备都能收到一个255.255.255.255的广播，这广播是TestKingA发送的
B、只有在192.168.1.0网络中的设备能够接收到TestKingA 发送的255.255.255.255广播
C、在所有网络中的所有设备在同一个冲突域
D、在192.168.0.1网络中的所有主机是一个冲突域，并且在192.168.2.0网络的主机是另外的一个冲突域
E、所有的主机是单独的冲突域
答案：B E
解析：The switch forms the collision domains. The router divides the broadcast domains and collision domains. The router doesn’t forward the broadcasts. So, hosts in networks 192.168.1.0 and 192.168.2.0 are in two different broadcast domains. Each host is in its own collision domain.
B是正确的，而D不是。如果图表中是使用集线器而不是用的交换机那就是正确的，这里只有2个冲突域，但是这图中有交换机。这作者可能是有意陈述那些正确的广播域 E是正确的，因为整个网络是由交换机组成的
10、Which address represents a unicast address?
哪个地址象征一个单播地址？
A. 224.1.5.2
B. FFFF. FFFF. FFFF.
C. 192.168.24.59/30
D. 255.255.255.255
E. 172.31.128.255/18
答案：E
解析：172.31.128.255 is the only unicast address. It seems to be a broadcast address, because of 255 in the last octett, the broadcast address for this network is
172.31.131.255.
Not A: 224.1.5.2 is a multicast address.
172.31.128.255/18是唯一一个单播地址，它像一个广播地址，因为255 是最后的一位，这个网络的广播地址是172.31.128.255
A 224.1.5.2 是多播地址
11、Wit regard to bridges and switches, which of the following statements are true? (Choose three.)
关于网桥和交换机，下列哪个陈述是正确的？
A. Switches are primarily software based while bridges are hardware based.
B. Both bridges and switches forward Layer 2 broadcasts.
C. Bridges are frequently faster than switches.
D. Switches typically have a higher number of ports than bridges.
E. Bridges define broadcast domain while switches define collision domains.
F. Both bridges and switches make forwarding decisions based on Layer 2 addresses.
A交换机是基于软件的而网桥是基于硬件的B、网桥和交换机都是在2层传送广播的C、网桥比交换机快D、交换机的端口号比网桥高E、网桥定义的是冲突域而交换机是广播域F、网桥和交换机所做的决定都是基于2层地址的
答案：BDF
12、Which Layer 1 devices can be used to enlarge the area covered by a single LAN segment? Select two
在1层设备中，哪个是通过一个单一的局域网段来扩大网络范围的？
A.Switch
B. Router
C. NIC
D. hub
E. Repeater
F. RJ-45 transceiver
A、交换机
B、路由器
C、网卡
D、集线器
E、中继器
F、RJ-45收发器
答案：DE
解析：Both hub, Repeater, Router and Switch repeat the packet. But only hub and Repeater do not segment the network.
集线器，中继器和交换机都能够反复数据包，但是只有集线器和中继器不能分割网络
13、What information is supplied by CDP? Select three.
CDP所提供的信息是什么？
A. Device identifiers
B. Capabilities list
C. Platform
D. Route identifier
E. Neighbour traffic data
A、设备标识符
B、性能列表
C、平台
D、路由标识符
E、相邻间传数据
答案：ABC
解析
CDP是一个思科私有的协议，支持在接口间传送CDP信息，这些接口必须支持SNAP 首部。所有的局域网接口，HDLC，帧中继和ATM都支持CDP。路由器和交换机能够发现相邻路由的详细的3层地址—甚至没有配置3层协议—因为CDP不能支持依靠所有的3层协议
CDP发现的相邻设备的详细信息：
设备标识符—代表主机名
地址列表—网络地址和数据链路地址
端口标识符—标识端口的文本，一个端口的其他
性能列表—关于设备作什么的信息—例如端口，一个路由器或者交换机
平台—在设备中运行的模型和操作系统水平
14、If a host on a network has the address 172.16.45.14/30, what is the address of the subnetwork to which this host belongs?
如果一个网络中的主机地址是172.16.45.14/30，这个主机的子网是哪个地址？
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18
答案：D
解析：The last octet in binary form is 00001110. Only 6 bits of this octet belong to the subnet mask. Hence the subnetwork is 172.16.45.12.
这最后八位字节转换为二进制为00001110，在这八位中只有六个字节属于子网号，因此子网是172.16.45.12
15、How many broadcast domains are shown in the graphic assuming only the default VLAN is configured on the switches?
在下列图里交换机配置的默认VLAMN中显示有多少广播域？
A、一个
B、两个
C、六个
D、十二个
答案：A
解析：There is only one broadcast domain because switches and hubs do not switch the broadcast domains. Only layer 3 devices can segment the broadcast domains.
这里只有一个广播域是因为交换机和集线器都不能交换广播域。只有3层设备能分段广播域
Design an IP addressing scheme to meet design requirements (55 questions)
设计一个IP地址方案用于符合设计要求
1、You have the binary number 10011101. Convert it to its decimal and
hexadecimal equivalents. (Select two answer choices)
你有一个二进制号10011101，把它改变成同等大小的十进制数和十六进制数A.158 B. 0x9D C. 156 D. 157 E. 0x19 F. 0x9F
答案：BD
解析：10011101 = 128+0+0+16+8+4+0+1 = 157
For hexadecimal, we break up the binary number 10011101 into the 2 parts:
1001 = 9 and 1101 = 13, this is D in hexadecimal, so the number is 0x9D. We can further verify by taking the hex number 9D and converting it to decimal by taking 16 times 9, and then adding 13 for D (0x9D = (16x9)+13 = 157).
十进制的转换：10011101 = 128+0+0+16+8+4+0+1 = 157
对于十六进制，我们把二进制数10011101分成两部分：1001=9 1101=13，转为十六进制是D，所以十六进制是OX9D。我们可以验证，通过把十六进制数9D转为十进制：(0x9D = (16x9)+13 = 157)
2、The subnet mask on the serial interface of a router is expressed in binary as 11111000 for the last octet. How do you express the binary number 11111000 in decimal?
在一个路由器串口的子网掩码是用二进制11111000表示的作为最后八位，把它转换为十进制是？
A.210
B. 224
C. 240
D. 248
E. 252
答案：D
解析：128 + 64+32+16+8 = 248. Since this is the last octet of the interface, the subnet mask would be expressed as a /29.
这最后八位组，它的子网可以表示为/29
错误答案：A. The number 210 would be 11010010 in binary.
B. The number 224 would be 11100000 in binary.
C. The number 240 would be 11110000 in binary
E. The number 252 would be 11111100 in binary. This is known as a /30 and is used often in point-point links, since there are only 2 available addresses for use in this subnet.
A、210的二进制数为11010010
B、224的二进制数为11100000
C、240的二进制数为11110000
E、252的二进制数为11111100，它可以看作是/30并用于点对点的连接，因为在这个子网中只有两个可以使用的地址
3、Which one of the binary number ranges shown below corresponds to the value of the first octet in Class B address range?
下列二进制数的范围中哪个显示的是B类前八位地址的范围？
A. 10000000-11101111
B. 11000000-11101111
C. 10000000-10111111
D. 10000000-11111111
E. 11000000-10111111
答案：C
解析：Class B addresses are in the range 128.0.0.0 through 191.255.255.255.
In binary, the first octet (128 through 191) equates to 10000000-10111111
B类地址的范围是128.0.0.0到191.255.255.255，转换为二进制的前八位是10000000-10111111
错误答案：
A. Binary 10000000 does equate to 128 but binary 11101111 equates to 239
B. Binary 11000000 equates to 192 and binary 11101111 equates to 239
D. Binary 10000000 does equate to 128 but binary 11011111 equates to 223
E. Binary 11000000 equates to 192 but binary 10111111 does equate to 191
A、二机制数10000000的等值数是128但是二进制11101111的等值数是239
B、二机制数11000000的等值数是192并且二进制11101111的等值数是239
C、二机制数10000000的等值数是128但是二进制11011111的等值数是223
E、二机制数11000000的等值数是192但是二进制10111111的等值数是191
4、How would the number 231 be expressed as a binary number?
231表示成二进制是？
A. 11011011
B. 11110011
C. 11100111
D. 11111001
E. 11010011
答案：C
解析：Decimal number 231 equates to 11100111 in binary (128+64+32+0+0+4+2+1) 十进制数231等同的二进制数是11100111 (128+64+32+0+0+4+2+1)
错误答案：
A: Binary 11011011 equates to 219 (128+64+0+16+8+0+2+1)
B: Binary 11110011 equates to 243 (128+64+32+16+0+0+2+1)
D: Binary 11101011 equates to 249 (128+64+32+16+8+0+0+1)
E: Binary 11010011 equates to 211 (128+64+0+16+0+0+2+1)
A、11011011的十进制数是219
B、11110011的十进制数是243
Ｄ、11101011的十进制数是249
Ｅ、11010011的十进制数是211
５、How would the number 172 be expressed in binary form?
172如何用二进制表示？
A.10010010
B. 10011001
C. 10101100
D. 10101110
答案：Ｃ
解析：10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172
错误答案：
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174
６、The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of the OUI portion of this NIC card, expressed as a binary number?
你的ＰＣ机网卡ＭＡＣ地址是C9-3F-32-B4-DC-19。这网卡地址的OUI部分用二进制表示是？
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001
答案：Ｄ
解析：The first half of the address identifies the manufacturer of the card. This code, which is assigned to each manufacturer by the IEEE, is called the organizationally unique identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and convert it to decimal form we have:
C9 = (12x16) + 9 = 2013F = (3x16) + 15 = 6332 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 1100100163 = 0011111150 = 00110010
So the correct answer is D: 11001001-00111111-00110010
前半部分的地址标识了厂商号，这个代码都是IEEE分配的，也叫做组织的唯一标识（ＯＵＩ），例如，它的OUI是C9—3F—32。如果我们把它转换为十进制数是：C9 = (12x16) + 9 = 2013F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
因此我们所得到的十进制数是210.63.50．如果把它转换为二进制是：201 = 1100100163 = 0011111150 = 00110010
所以正确的答案是D：11001001-00111111-00110010
７、How do you express the binary number 10110011 in decimal form?
你怎样用十进制数表示二进制数10110011？
A.91
B. 155
C. 179
D. 180
E. 201
F. 227
答案：C
解析：If you arrange the binary number 10110011, against the place value and multiply the values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179
如果你排列二进制，依靠位值并增加值，合计，你就可以得到正确的答案
８、Convert the hex and decimal numbers on the left into binary, and match them to their corresponding slot on the right. (Not all of the hexadecimal and decimal numbers will be used
把左边的十六进制和十进制数转换为二进制，并把它同右边的二进制匹配
答案：
解释：
Explanation:
170 (Decimal) = 10101010
192 (Decimal) = 11000000
F1 (241 = Decimal) = 11110001
9F (159 = Decimal) = 10011111
170（十进制）=10101010
192（十进制）=11000000
F1(241=十进制) =11110001
9f(159=十进制) =10011111
9、The following chart displays all of the possible IP address numbers, expressed in
decimal, hexadecimal, and binary:
下面的表格显示了所有可能的IP地址号，表示为十进制，十六进制和二进制：
Which two of the addresses below are available for host addresses on the subnet 192.168.15.19/28? (Select two answer choices)
下面的哪两个地址可以在192.168.15.19/28子网中作为主机地址？
A．192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
f. 上面的都不是
答案：A，C
解释：
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2). The last bit used to make 240 is the 4th bit (16) therefore the first network will be 192.168.15.16. The network will have 16 addresses (but remember that the first address is the network address and the last address is the broadcast address). In other
words, the networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we are given is 192.168.15.19. Therefore the other host addresses must also be on this network. Valid IP addresses for hosts on this network are:
192.168.15.17-192.168.15.30.
这个网络使用28位子网（255.255.255.240）。也就是说4位用于网络4位用于主机。这允许创建14个网络及每个网络14台主机（2n-2）。第4位是240因此第一个网络是192.168.15.16。这个网络有16个地址（但是记住第一个地址是网络地址，最后一个地址是广播地址）。总之，从192.138.15.16/28开始网络将会增加16个。我们会给192.168.15.19地址，因此其他的主机地址也要在这个网络中。这个网络中有效的主机地址192.168.15.17-192.168.15.30。
错误的答案：
B. This is not a valid address for this particular 28 bit subnet mask. The first network address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.
B.不是这个28位子网掩码的有效地址。第一个网络地址是：192.168.15.16。
D.这是一个网络地址。
E.这是这个子网的广播地址。
10、You have a Class C network, and you need ten subnets. You wish to have as many addresses available for hosts as possible. Which one of the following subnet masks should you use?
你有一个C类地址，你需要10个子网。你希望有尽可能多的主机地址可以使用。你将会选择下面的哪一个子网掩码？
A．255.255.255.192
B. 255.255.255.244
C. 255.255.255.240
D. 255.255.255.248
E. 上面的都不是
答案：C
解释：
Using the 2n-2 formula, we will need to use 4 bits for subnetting, as this will provide for 24-2 = 14 subnets. The subnet mask for 4 bits is then 255.255.255.240.
使用2n-2规则，需要使用4位作为子网，这样就可以提供14个子网，4位作为子网掩码是：255.255.255.240。
错误的答案在：
A. This will give us only 2 bits for the network mask, which will provide only 2 networks.
B. This will give us 3 bits for the network mask, which will provide for only 6 networks.
D. This will use 5 bits for the network mask, providing 30 networks. However, it will provide for only for 6 host addresses in each network, so C is a better choice.
A.用2位作为子网掩码，只能提供2个网络。
B.用3位作为子网掩码，只能提供6个网络。
C.用5位作为子网掩码，只能提供30个网络。然而，每一个子网只提供6个主
机地址，一次C是最好的选择。
11、Which of the following is an example of a valid unicast host IP address?
下面例子中哪一个是有效的单播主机ip地址？
A.172.31.128.255/18
B.255.255.255.255
C.192.168.24.59/30
D.FFFF.FFFF.FFFF
E.244.1.5.2
F.上面的都是
答案：A
解释：
The address 172.32.128.255 /18 is 10101100.00100000.10|000000.11111111 in binary, so this is indeed a valid host address.
地址172.32.128.255/18用二进制表示是10101100.00100000.10000000.11111111，因此这是一个确实有效的主机地址。
错误的答案：
B. This is the all 1’s broadcast address.
C. Although at first glance this answer would appear to be a valid IP address, the /30 means the network mask is 255.255.255.252, and the 192.168.24.59 address is the broadcast address for the 192.168.24.56/30 network.
D. This is the all 1’s broadcast MAC address
E. This is a multicast IP address.
B. 全1是广播地址
C. 尽管乍看这个答案可能出现有效的IP 地址，/30的意思是网络掩码是255.255.255.252，因此192.168.24.59地址是192.168.24.56/30网络的广播地址。
D. 这是一个全1的广播MAK地址。
E. 这是一个多点传送的地址。
12、How many subnetworks and hosts are available per subnet if you apply a /28 mask to the 210.10.2.0 class C network?
如果你申请一个C类地址210.10.2.0并且掩码是28位，每一个子网中有多少个子网和主机可以使用？
A. 30 networks and 6 hosts.
B. 6 networks and 30 hosts.
C. 8 networks and 32 hosts.
D. 32 networks and 18 hosts.
E. 14 networks and 14 hosts.
F. None of the above
A.30个子网和6台主机。
B.6个子网和30台主机。
C.8个子网和32台主机。
D.32个子网和18台主机。
E.14个子网和14台主机。
F.上面的都是
答案：E
解释：
A 28 bit subnet mask (11111111.11111111.11111111.11110000) applied to a class C network uses a 4 bits for networks, and leaves 4 bits for hosts. Using the 2n-2 formula, we have 24-2 (or 2x2x2x2-2) which gives us 14 for both the number of networks, and the number of hosts.
一个28位子网掩码（11111111.11111111.11111111.11110000）被用于一个C 类网络，使用4位作为网络号，剩余的4为作为主机地址。使用2n-2规则，我们有24-2（或者是2*2*2*2-2），网络数和主机数都是14。
错误的答案：
A. This would be the result of a /29 (255.255.255.248) network.
B. This would be the result of a /27 (255.255.255.224) network.
C. This is not possible, as we must subtract two from the subnets and hosts for the network and broadcast addresses.
D. This is not a possible combination of networks and hosts.
A.这是一个/29（255.255.255.248）的网络
B.这是一个/27（255.255.255.224）的网络
C.这是不可能的，因为我们不需从子网和主机地址中减去2个为网络和广播地
址。
D.这是一个不可能的结合对网络和主机。
13、The TestKing network was assigned the Class C network 199.166.131.0 from the ISP. If the administrator at TestKing were to subnet this class C network using the 255.255.255.224 subnet mask, how may hosts will they be able to support on each subnet?
TestKing网络从ISP那分配到一个C类网络199.166.131.0。如果TestKing网络的管理员对这个C 类网络使用255.255.255.224子网掩码，每一个子网中有多少个主机地址可以使用？
A.14
B.16
C.30
D.32
E.62
F.64
答案：C
解释：
The subnet mask 255.255.255.224 is a 27 bit mask
(11111111.11111111.11111111.11100000). It uses 3 bits from the last octet for the network ID, leaving 5 bits for host addresses. We can calculate the number of hosts supported by this subnet by using the 2n-2 formula where n represents the number of host bits. In this case it will be 5. 25-2 gives us 30.
这个子网掩码255.255.255.224是27位（11111111.11111111.11111111.11100000）。它使用最后一个6位位组的前3位作为网络号，剩下的5位作为主机地址。我们可以使用2n-2规则计算支持的主机数，n表示主机的位数。既然这样它就是5 ，25-2就是30。
错误的答案：
A. Subnet mask 255.255.255.240 will give us 14 host addresses.
B. Subnet mask 255.255.255.240 will give us a total of 16 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.
D. Subnet mask 255.255.255.224 will give us a total of 32 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.
E. Subnet mask 255.255.255.192 will give us 62 host addresses.
F. Subnet mask 255.255.255.192 will give us a total of 64 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.
A.子网掩码255.255.255.240 将给我们14个主机地址。
B.子网掩码255.255.255.240 总共有16个主机地址。然而，我们必须要减去两
个地址（网络地址和主机地址）从子网将支持的最大主机数中。
D．子网掩码255.255.255.244将给我们提供总共32个地址。然而，我们必须要减去两个地址（网络地址和主机地址）从子网将支持的最大主机数中。
E.子网掩码255.255.255.192将提供62个主机地址。
F. 子网掩码255.255.255.192将提供62个主机地址。然而，我们必须要减去两个地址（网络地址和主机地址）从子网将支持的最大主机数中。
14、What is the subnet for the host IP address 172.16.210.0/22?
主机IP地址172.16.210.0/22的子网是什么？
A.172.16.42.0
B. 172.16.107.0
C. 172.16.208.0
D. 172.16.252.0
E. 172.16.254.0
F. 上面的都不是
答案 C
解释：
This question is much easier then it appears when you convert it to binary and do the Boolean operation as shown below:
IP address 172.16.210.0 = 10101100.00010000.11010010.00000000
/22 mask = 11111111.11111111.11111100.00000000
AND result = 11111111.11111111.11010000.00000000
AND in decimal= 172 . 16 . 208 . 0
当你把它转换成二进制的时候这个问题就很容易找到答案，下面是用Boolean（布尔书学体系）运算的结果：
IP地址172.16.210.0=10101100.00010000.11010010.00000000
/22子网=11111111.11111111.11111100.00000000
与的结果=11111111.11111111.11010000.00000000
十进制表示与=172.16.208.0
15、What is the subnet for the host IP address 201.100.5.68/28?
主机IP地址201.100.5.68/28的子网是什么？
A. 201.100.5.0
B. 201.100.5.32
C. 201.100.5.64
D. 201.100.5.65
E. 201.100.5.31
F. 201.100.5.1
答案：C
解释：
This question is much easier then it appears when you convert it to binary and do the Boolean operation as shown below:
IP address 201.100.5.68 = 11001001.01100100.00000101.01000100
/28 mask = 11111111.11111111.11111111.11000000
AND result = 11001001.01100100.00000101.01000000
AND in decimal= 200 . 100 . 5 . 64
当你把它转换成二进制的时候这个问题就很容易找到答案，下面是用Boolean（布尔书学体系）运算的结果：
IP 地址201.100.5.68=1101001.01100100.00000101.01000100
/28掩码=11111111.11111111.11111111.11000000
与的结果=11001001.01100100.00000101.01000000
十进制表示与=201.100.5.64
16、3 addresses are shown in binary form below:
A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101
Regarding these three binary addresses in the above exhibit; which statements below are correct? (Select three)
下面用二进制表示的3个地址：
A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101
关于上面列出的3个二进制地址，下面哪些陈述是正确的？（选3个）
A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.
A.地址C是一个公有的C类地址。
B.地址C是一个私有的C类地址。
C.地址B是一个公有的B类地址。
D.地址A是一个公有的A类地址。
E.地址B是一个私有的B类地址。
F.地址A是一个私有的A类地址。
答案：A，D，E
解释：
A. Address C converts to 192.167.178.69 in decimal, which is a public class C address.
D. Address A converts to 100.10.235.39, which is a public class A IP address.
E. Address B converts to 172.18.158.15, which is a private (RFC 1918) IP address. A. 地址C转换为十进制是192.167.178.69, 它是一个公有的C类地址。
D．地址A转换后是100.10.235.39，它是一个公有的A类IP地址
E．地址B转换为172.18.158.15，它是一个私有（RFC 1918）的IP地址17、What is the IP address range for the first octet in a class B address, in binary form?
B类的IP地址的第一个8 位位组用二进制表示的范围是什么？
A. 00000111-10001111
B. 00000011-10011111
C. 10000000-10111111
D. 11000000-11011111
E. 11100000-11101111
F. 上面的都不是
答案：C
解释：
The class B address range is 128.0.0.0-191.255.255.255. When looking at the first octet alone, the range is 128-191. The binary number for 128 is 10000000 and the binary number for 191 is 10111111, so the value rang is 10000000-10111111.
B类地址的范围是128.0.0.0-191.255.255.255。当单独看第一个8位位组时，范围是128-191 。二进制表示128是10000000，二进制数表示191 是10111111，因此值的范围是10000000-10111111。
18、Which one of the binary bit patterns below denotes a Class B address?
下面的表示哪一个是一个B类地址的二进制为的格式？
A. 0xxxxxxx
B. 10xxxxxx
C. 110xxxxx
D. 1110xxxx
E. 11110xxx
答案：B
解释：
Class B addresses start with a binary of 10. The valid class B range is 128.0.0.0- 191.255.255.255.
B类地址的开始二进制是10，B类的有效范围是128.0.0.0-191.255.255.255
错误的答案：
A. Class A addresses start with 0, as they are addresses that are less than 128.
C. Class C addresses start with 110, for a value of 192.0.0.0-223.255.255.255
D. Class D addresses start with 1110. They are reserved for multicast use
E. Class E addresses start with 11110. They are currently reserved for experimental use
A.A类地址的开始是0，因此它的地址少于128。
C．C类地址的开始是11 0，值是192.0.0.0-223.255.255.255
第一章 1关于信息社会，下列说法不正确的是( ) A.在信息社会，信息、知识成为重要的生产力要素。 B.信息社会中所有的工业生产都是自动化的。 C.在信息社会，信息技术催生大批新兴产业，同时，传统产业也普遍实行技术改造。 D.计算机的发明是第三次科技革命的重要标志，是人类文明史上继蒸汽技术革命和电力技术革命之后科技领域的又一次重大飞跃。 2二进制是由下列哪位科学家首先提出来的（） A.布尔 B.巴贝奇 C.莱布尼兹 D.图灵 3在人体器官中，下列说法不正确的是（） A.大脑的思维是对外部事物的直接反应。 B.人体器官可分为直接感觉器官和间接感觉器官。 C.大脑是人体的思维器官。 D.大脑是一种间接感觉器官。 4对于计算思维，下列说法错误的是（） A.计算思维是计算机科学家独有的思维方式。 B.计算思维的产生与信息社会发展的时代背景有关，工具影响我们的思维方式。 C.计算思维是一种借助于计算能力进行问题求解的思维和意识。 D.计算思维的本质是抽象和自动化。 5关于信息技术（Information Technology，IT）下列说法正确的是 A.在信息社会，所有的信息处理中都用到了信息技术。 B.在信息处理的每一个环节，都必须使用信息技术。 C.信息技术无法对工业社会形成的传统设备进行技术改造，成为智能设备。 现代信息技术是指以微电子技术、计算机技术和通信技术为特征的技术。D. 6多选(2分) 关于思维，下列说法正确的是（） A.思维是人类凭借知识、经验对客观事物进行的间接的应。 B.思维是在表象、概念的基础上进行分析、综合、判断、推理等认识活动的过程。 C.思维是人类的大脑对外部客观世界概括的和间接的反应。 D.思维是大脑活动，与工具无关。 7多选 下列内容属于信息素养（Information Literacy）的是（） A.信息意识 B.信息能力 C.信息道德 D.信息知识 8多选(2分)关于推理，下列说法正确的是（） A.推理是由已知判断，根据一定的思维规则导出一个或一组新的判断的过程。 B.归纳推理比演绎推理更具有说服力。 C.三段论是一种演绎推理形式。
基本电路 包括电路模型的元素被称为理想的电路元件。一个理想的电路元件是一个实际的电气元件的数学模型，就像一个电池或一个灯泡。重要的是为理想电路元件在电路模型用来表示实际的电气元件的行为可接受程度的准确性。电路分析，本单位的重点，这些工具，然后应用电路。电路分析基础上的数学方法，是用来预测行为的电路模型和其理想的电路元件。一个所期望的行为之间的比较，从设计规范，和预测的行为，形成电路分析，可能会导致电路模型的改进和理想的电路元件。一旦期望和预测的行为是一致的，可以构建物理原型。 物理原型是一个实际的电气系统，修建从实际电器元件。测量技术是用来确定实际的物理系统，定量的行为。实际的行为相比，从设计规范的行为，从电路分析预测的行为。比较可能会导致在物理样机，电路模型，或两者的改进。最终，这个反复的过程，模型，组件和系统的不断完善，可能会产生较准确地符合设计规范的设计，从而满足需要。 从这样的描述，它是明确的，在设计过程中，电路分析中起着一个非常重要的作用。由于电路分析应用电路模型，执业的工程师尝试使用成熟的电路模型，使设计满足在第一次迭代的设计规范。在这个单元，我们使用20至100年已测试通过机型，你可以认为他们是成熟的。能力模型与实际电力系统理想的电路元件，使电路理论的工程师非常有用的。 说理想电路元件的互连可用于定量预测系统的行为，意味着我们可以用数学方程描述的互连。对于数学方程是有用的，我们必须写他们在衡量的数量方面。在电路的情况下，这些数量是电压和电流。电路分析的研究，包括了解其电压和电流和理解上的电压施加的限制，目前互连的理想元素的每一个理想的电路元件的行为电路分析基础上的电压和电流的变量。电压是每单位电荷，电荷分离所造成的断电和SI单位伏V = DW / DQ。电流是电荷的流动速度和具有的安培SI单位（I= DQ/ DT）。理想的基本电路元件是两个终端组成部分，不能细分，也可以在其终端电压和电流的数学描述。被动签署公约涉及元素，当电流通过元素的参考方向是整个元素的参考电压降的方向端子的电压和电流的表达式使用一个积极的迹象。 功率是单位时间内的能量和平等的端电压和电流的乘积;瓦SI单位。权力的代数符号解释如下： 如果P> 0，电源被传递到电路或电路元件。 如果p<0，权力正在从电路或电路元件中提取。 在这一章中介绍的电路元素是电压源，电流源和电阻器。理想电压源保持一个规定的电压，不论当前的设备。理想电流源保持规定的电流不管了整个设备的电压。电压和电流源是独立的，也就是说，不是任何其他电路的电流或电压的影响;或依赖，就是由一些电路中的电流或电压。一个电阻制约了它的电压和电流成正比彼此。有关的比例常数电压和一个电阻值称为其电阻和欧姆测量。 欧姆定律建立相称的电压和电流的电阻。具体来说，V = IR电阻的电流流动，如果在它两端的电压下降，或V=_IR方向，如果在该电阻的电流流是在它两端的电压上升方向。 通过结合对权力的方程，P = VI，欧姆定律，我们可以判断一个电阻吸收的功率：P = I2R= U2/ R 电路节点和封闭路径。节点是一个点，两个或两个以上的电路元件加入。当只有两个元素连接，形成一个节点，他们表示将在系列。一个闭合的路径是通过连接元件追溯到一个循环，起点和终点在同一节点，只有一次每遇到中间节点。 电路是说，要解决时，两端的电压，并在每个元素的电流已经确定。欧姆定律是一个重要的方程，得出这样的解决方案。 在简单的电路结构，欧姆定律是足以解决两端的电压，目前在每一个元素。然而，对于更复杂的互连，我们需要使用两个更为重要的代数关系，被称为基尔霍夫定律，来解决所有的电压和电流。 基尔霍夫电流定律是： 在电路中的任何一个节点电流的代数和等于零。 基尔霍夫电压定律是： 电路中的任何封闭路径上的电压的代数和等于零。 1.2电路分析技术 到目前为止，我们已经分析应用结合欧姆定律基尔霍夫定律电阻电路相对简单。所有的电路，我们可以使用这种方法，但因为他们而变得结构更为复杂，涉及到越来越多的元素，这种直接的方法很快成为累赘。在这一课中，我们介绍两个电路分析的强大的技术援助：在复杂的电路结构的分析节点电压的方法，并网电流的方
商务英语期末考试参考答案 说明：商务英语期末考试有五种题型： 一、中英短语互译已经全部总结，见下文 二、问答题已经提供了参考答案，仅供参考，见下文 三、案例分析题请根据题目要求自行准备 四、英语段落翻译中文已经注明，请自己翻看商务英语教程 五、写作请根据题目要求自行准备 1、中英短语互译（详见商务英语课件——每个PPT中的短语部分） PPT1： 1. individual proprietorship个体经营，独资企业 2. convertible bonds 可转换债券 3. memorandum of association 公司章程 4. monopolistic competition 垄断竞争 5. business credit 商业信用 1. 财政资源 financial resources 2. 有限责任公司 limited company 3. 销售收入 sales revenue 4. 有形商品 tangible goods 5. 流动资金 working capital PPT2： 1. flexible manufacturing system 弹性生产系统 2. fractional reserve system 部分准备金制度
3. credit instruments 信用工具，信用票据 4. bull market/bear market 牛/熊或多头/空头市场 5. manufacturing process 制造工艺 1. 厂址选择 facility location 2. 库存控制 inventory control 3. 总支出 aggregate expenditures 4. 股权证/权益证 equity instruments 5. 财务管理 financial management PPT3： 1. performance appraisal 绩效/业绩评估 2. promotion-from-within 内部提拔 3. skills inventory 技术库存 4. macroeconomic ramification 宏观经济衍生物/结果 5. closed-end fund 封闭式基金 1. 项目生命周期 project life cycle 2. 软技能和硬技能 soft skills and hard skills 3. 补偿制度 compensation system 4. 开放式基金 open-ended fund 5. 优先股 preferred stock PP4： 1. profit-oriented objective 利润导向的目标 2. marketing intermediary 营销代理商
黄山学院初等教育学院2008－2009学年度第二学期06级英语班《英语翻译》期末考试卷 班级序号姓名成绩 一、选择题。（请将答案写在序号前面）20% 1. Success in many fields depends on getting the latest information. A. 许多领域的成功都取决于能否得到最新消息。 B. 田里的收成在很多方面依靠获得最新的消息。 C. 从很多方面讲，成功主要依靠在最后的时刻还能得到信息。 D. 在众多领域里，成功是获得最晚信息的保障。 2. At least three persons died in the fire and 23 others were injured. A. 至少3人因这次火灾和另外23次火灾造成死亡。 B. 起码有3个人在大火中丧命，23个以外的人受伤。 C. 火灾造成至少3人死亡，23人受伤。 D. 至少有3人死于火海，23人被火海包围。 3. Not only has he a first-class brain, but he is also a tremendously hard worker. A. 在一班，他不仅是头领，而且，也是一个非常严肃的工作者。 B. 他有一流的头脑，是个非常难对付的家伙。 C. 他不仅头脑好用，而且，还喜欢做极端困难的工作。 D. 他不仅头脑非常聪明，而且，工作上也非常勤奋。 4. It is reported that some areas of Australia have had no rain for more than four years. A. 有报道说，澳大利亚地区已经4年多没下雨了。 B. 据报道，澳大利亚的部分地区有4年多没下雨了。 C. 澳大利亚的个别地方被报道说没下雨的时间已经超过了4年。 D. 报纸报道，某些澳大利亚地区已经有4年多没下雨的时间。 5. The village was named after the high mountain standing in front of it. A. 这个村庄是以矗立在它面前的那座高山命名的。 B. 这个村庄的名字是来自于它面前的那座高山。 C. 这个村庄的名字是在前面的那座高山之后起的。 D. 这个村庄的名字是以站立在它面前的那座高山起的。 6. It was officially announced that the birth rate dropped greatly in the past three years. A. 据官方宣布，出生率在过去的三年中有了大幅下降。 B. 正式宣布，过去三年的出生率大幅下降。 C. 正式宣布，出生率大幅下降在过去三年。 D. 据正式声明，过去三年的出生率大幅下降。 7. The introduction of the new technique reduced the cost of the product three times. A. 对新技术的介绍，使产品的成本减少到三分之一。 B. 新技术的引进使产品的成本减少了三倍。 C. 对新技术的介绍使产品的成本减少到原来的三分之二。 D. 新技术的引进使产品成本减少了三分之二。 8. If you are buying a car, you may pay for it out of savings. A. 假使你正在买一辆汽车，你也许会透支购买。 B. 如果你要买一辆汽车，你可以用自己的储蓄存款支付。 C. 假定你打算买一辆汽车，你得花掉你得全部存款。 D. 加入你买一辆汽车，你需要用你的积蓄交此款。 9. Full-service banks can offer services unavailable at smaller financial institutions. A. 全面服务的银行能提供在小财政机构所不能获得的服务。 B. 提供全方位服务的银行能提供较小的金融机构所不能提供的服务。 C. 充满了服务的银行可能提供给较小的财政机构某些无法获得的服务 D. 全面服务的银行可能提供服务，这种服务是较小的金融机构所不能提供的。 10. The following qualifications are required from the candidates for the above position. A. 我们要求应聘高职位的求职者需具备以下资历。 B. 应聘以上职位，需具备下列资格。 C. 候选人为了高职位要求自己具备以下资历。 D. 以下资历是对应聘此职位的候选人的要求。 二、用正反表达法翻译下列句子。20% 1. The teacher found some of the pupils absent. 2. I wonder if he is coming. 3. Your request is beyond my power. 4. You will fail unless you work harder. 5. That lazy boy went to class before he had prepared his lesson. 6. Be sure to lock the door when you leave. 7. Students, with no exception, are to hand in their papers this afternoon. 8. The doubt still unsolved after his repeated explanation. 9. Such flight couldn’t long escape notice. 10. All the articles are untouchable in the museum. 三、用重复法翻译下列句子。20% 1. We should learn how to analyze and solve problems. 2. It's our duty to rebuild and defend our home-land. 3. People forget your face first, then your name. 4. Each had his own business to handle. 5. Big families had their own difficulties.
精心整理计算机能力考试测评.................................................................. 单项选择 计算机系统中存储容量最大的部件是(). A.主存储器 B.高速缓存器 C.硬盘 D.软盘 【评分报告得分：1.0之1.0】 【参考答案】正确 C C A.CPU B.CPU C.CPU D.CPU D B 十进制数 A.10E B.10D C.10C D.10B B A 在Word A.边框 B. C.分页符 D.换行符 C 【学生答案】 C 冯·诺依曼（VonNeumann）在总结ENIAC的研制过程和制订EDVAC计算机方案时，提出两点改进意见，它们是______。 A.机器语言和十六进制 B.采用ASCII编码集和指令系统 C.引入CPU和内存储器的概念 D.采用二进制和存储程序控制的概念 【评分报告得分：0.0之1.0】 【参考答案】错误 D 【学生答案】 B 微机硬件系统中地址总线的宽度<位数>对()影响最大.
A.存储器的字长 B.存储器的稳定性 C.CPU可直接访问的存储器空间大小 D.存储器的访问速度 【评分报告得分：1.0之1.0】 【参考答案】正确 C 【学生答案】 C 1KB表示(). A.1024字节 B.1000字节 C.1000位 D.1024位 A A A. B.CPU C. D.硬盘 B B A.格式 B.编辑 C.工具 D.插入 B B B.RAM C.硬盘 D.ROM 【评分报告得分：0.0之1.0】 【参考答案】错误 B 【学生答案】 C 计算机按处理数据的方式可分为(). A.电子模拟和电子数字计算机 B.科学计算、数据处理和人工智能计算机 C.便携、台式和微型计算机 D.巨型、大型、中型、小型和微型计算机【评分报告得分：0.0之1.0】 【参考答案】错误 A 【学生答案】
电子信息工程专业课程名称中英文翻译对照 （2009级培养计划）
实践环节翻译 高等数学Advanced Mathematics
大学物理College Physics 线性代数Linear Algebra 复变函数与积分变换Functions of Complex Variable and Integral Transforms 概率论与随机过程Probability and Random Process 物理实验Experiments of College Physics 数理方程Equations of Mathematical Physics 电子信息工程概论Introduction to Electronic and Information Engineering 计算机应用基础Fundamentals of Computer Application 电路原理Principles of Circuit 模拟电子技术基础Fundamentals of Analog Electronics 数字电子技术基础Fundamentals of Digital Electronics C语言程序设计The C Programming Language 信息论基础Fundamentals of Information Theory 信号与线性系统Signals and Linear Systems 微机原理与接口技术Microcomputer Principles and Interface Technology 马克思主义基本原理Fundamentals of Marxism 思想、理论和“三个代表” 重要思想概论 Thoughts of Mao and Deng 中国近现代史纲要Modern Chinese History 思想道德修养与法律基 础 Moral Education & Law Basis 形势与政策Situation and Policy 英语College English 体育Physical Education 当代世界经济与政治Modern Global Economy and Politics 卫生健康教育Health Education 心理健康知识讲座Psychological Health Knowledge Lecture 公共艺术课程Public Arts 文献检索Literature Retrieval 军事理论Military Theory 普通话语音常识及训练Mandarin Knowledge and Training 大学生职业生涯策划 （就业指导） Career Planning (Guidance of Employment ) 专题学术讲座Optional Course Lecture 科技文献写作Sci-tech Document Writing 高频电子线路High-Frequency Electronic Circuits 通信原理Communications Theory 数字信号处理Digital Signal Processing 计算机网络Computer Networks 电磁场与微波技术Electromagnetic Field and Microwave Technology 现代通信技术Modern Communications Technology
广东外语外贸大学公开学院辅导资料 商务英语翻译课程试卷 （课程代码：5355） 考生注意：1. 答案必须写在答卷上，写在问卷上无效。 2. 考试时间150分钟。 I. Multiple Choices (20 points, 2 points for each) 第一套试卷 1.It is not surprising, then, that the world saw a return to a floating exchange rate system. Central banks were no longer required to support their own currencies. A.在这种情况下，世界各国又恢复浮动汇率就不足为奇了。各国中央银行也就无须维持本币的汇价了。 B.不足为奇，全世界看到了汇率的回归，因此各国中央银行无需维持本币的汇价了。 C.此时此刻，世界各国又恢复了移动的交换比率，因此各国中央银行无需维持本币汇价。 D.在这种情况下，全世界又恢复了浮动交换率，这已不足为奇了，因此各国中央银行也就无需维持本币价格了。 2．Assuming the laboratory tests go well, and you can quote us a competitive price, we would certainly be able to place more substantial orders on a regular basis. A.假定实验室检验顺利，并且你的报价有竞争力，我们会大量向贵公司订货的。 B.若实验室检验合格，且你们给我们的报价具有竞争力，我们一定会定期大量订货的。 C.若实验室检验良好，且你们给出的报价具有竞争性，我们一定会定期定量订货的。 D.假定实验室发展良好，且你们的报价具有竞争力，我们一定会大量定期订货的。 3．Chinese researchers have made a breakthrough in developing new materials for nickel-hydrogen batteries used in low temperatures, Inhaul reported. A.中国研究者已经在开发新材料用于低温下使用的镍氢电池方面有了突破，据新华社报道。 B.新华社报道，中国科学家在从事新材料制造低温镍氢电池方面有了突破。 C.新华社报道，中国研究人员在开发利用新材料制造在低温下使用的镍氢电池方面已有了突破。 D.中国研究者在开发新材料制造低温镍氢电池有了重大突破，这是新华社报道的。 4. Since the initiation of economic reforms in the late 1970s, China has achieved impressive
Unit1 A new study shows that dogs can detect if someone has cancer just by sniffing the person’s breath. Ordinary household dogs with only a few weeks of basic training learn to accurately distinguish between breath samples of lung- and breast-cancer patients and healthy subjects.One expert, who led the research, said, “Our study provides compelling evidence that cancers hidden beneath the skin can be detected simply by (dogs’) examining the odors of a person’s breath.” Early detection of cancers greatly improves a patient’s survival chances, and researchers hope that man’s best friend, the dog, can become an important tool in early screening. Unit 2 A research team recently replicated a bacterium’s genetic structure entirely from l aboratory chemicals, moving one step closer to creating the world’s first artificial organism. The researchers said, “It’s the second step of a three-step process to create a synthetic organism. But the final step could prove far more difficult.” Nonetheless, the research is pushing forward at a rapid pace. Last June, the team revealed details of an experiment in which researchers inserted the DNA of one species of bacterium into the cells of another. That process almost magically booted up the inserted genome. The research team hopes to use a similar trick to boot up the artificially created genome, so as to create a man-made living organism. Unit 3 In recent years, the psychologists from many countries banded together to do a research which indicated that continued income growth could make people apply themselves to an ongoing consumption race rather than promote overall happiness. Everyone has to spend more and more money in order to maintain a constant level of happiness. This is because our vanity and jealousy are functioning . The way to sort out them out is to cultivate the noble sentiment of being considerate and serving the society. People shouldn’t pin all their hopes on money. Instead , friends, family and work are also playing a very improtant role . Therefore, hapiness is not an inborn trait, but a talent which everyone can learn. Unit 4 I used to feel excited at teaching my students the elegant economic theories that could supposedly cure societal problems of all types. But what is the good of all my complex theories when people were dying of starvation on the sidewalks and porches across from my lecture hall? My lessons were like the American movies where the good guys always win. But when I emerged from the comfort of the classroom, I was faced with the reality of the city streets. Here good guys were mercilessly beaten and trampled. Daily life ws getting worse, and the poor were growing even poorer. Unit 5 No one thought that shy little Einstein would grow up to a prominent scientist. He was slow in learning to speak, and he often paused to consider what he would say during a conversation.In school, Albert Einstein was singled out by his teachers as a troublesome child because he liked to ask difficult strange questions. He did not like to memorize facts and rules, but was interested in what lay below the surface of things. When he was four or five years old, for instance, his father gave him a compass. Little Einstein was curious about the mysterious force that could keep a compass needle always pointing north, which prompted him to read widely in science. His real studies were mostly done at home by reading books on mathematics, physics, and philosophy. Unit 6 Most Americans have great vigor and enthusiasm. They prefer to discipline themselves rather than be disciplined by others. They pride themselves on their independence, their right to make their own minds. They tend to take the initiatives, even when there is a risk in doing so. They have courage and do not give in easily. They are considered sentimental. When they see a flag on ceremonial occasions, or attend parades celebrating Ame rica’s glorious past, tears may come to their eyes. They tend to be emotional at the reunions with family and friends, too. Sometimes, they can laugh at themselves and their country, while they will always remain intensely patriotic.
2、冯·诺依曼式计算机硬件系统的组成部分包括（B）。 A）运算器、外部存储器、控制器和输入输出设备 B）运算器、控制器、存储器和输入输出设备 C）电源、控制器、存储器和输入输出设备 D）运算器、放大器、存储器和输入输出设备 4、（C）设备既是输入设备又是输出设备。 A）键盘B）打印机C）硬盘D）显示器 5、微机中1MB表示的二进制位数是（A）。 A）1024×1024×8B）1024×8 C）1024×1024D）1024 9、在Windows中，剪贴板是程序和文件间用来传递信息的临时存储区，此存储器是（C）。 A）回收站的一部分B）硬盘的一部分C）内存的一部分D）软盘的一部分11、关于WORD保存文档的描述不正确的是（D）。 A）“常用”工具栏中的“保存”按钮与文件菜单中的“保存”命令选项同等功能 B）保存一个新文档，“常用”工具栏中的“保存”按钮与文件菜单中的“另存为”命令选项同等功能 C）保存一个新文档，文件菜单中的“保存”命令选项与文件菜单中的“另存为”命令选项同等功能
D）文件菜单中的“保存”命令选项与文件菜单中的“另存为”命令选项同等功能 13、在WORD中，（D）不能够通过“插入”→“图片”命令插入，以及通过控点调整大小。 A）剪贴画B）艺术字C）组织结构图D）视频 17、在EXCEL单元格中输入正文时以下说法不正确的是（D）。 A）在一个单元格中可以输入多达255个非数字项的字符。 B）在一个单元格中输入字符过长时，可以强制换行。 C）若输入数字过长，EXCEL会将其转换为科学记数形式。 D）输入过长或极小的数时，EXCEL无法表示。 18、下列序列中，不能直接利用自动填充快速输入的是（B） A)星期一、星期二、星期三、…… B)第一类、第二类、第三类、…… C)甲、乙、丙、…… D)Mon、Tue、Wed、…… 19、在PowerPoint中，（B）设置能够应用幻灯片模版改变幻灯片的背景、标题字体格式。 A）幻灯片版式B）幻灯片设计C）幻灯片切换D）幻灯片放映 20在PowerPoint中，通过（A）设置后，点击观看放映后能够自动放映。A）排练计时 B）动画设置 C）自定义动画 D）幻灯片设计
2012-2013学年第一学期2011级《商务英语翻译》毕业清考A 卷答题卡 一，单项选择题（10X2’） 1.__ __D__ 2. ___A_____ 3.__ _C_______ 4. ____C__5. C___ B，（5X3’）6._ _D_____ 7. ______A___ 8._____C_______ 9. ___C____10. _____A___ 二．(10X2’) 11___e___12___g___13____g__14____b___15__j_____ 16___k___17___d____18____m___19___c___20__a___ 三．把下列句子译成汉语，注意翻译技巧（10X4’）黑体为主要评分点 21.他觉得把会议上讨论的事情告诉他妻子是不恰当的。 22. 你必须注意你方出口产品的质量 23. 我的升职多亏你的帮忙；我过去是，现在也是十分感激你的。 24.教师应在教学上有耐心。 25. 他无权签订这样的合同 26.他未遵守安全操作规则导致引擎事故的发生。 27. 显而易见，建一所中学的计划将会被取消 28. 他旅途中的所见所闻给他留下了深刻的印象。 29.他讲得太快了，我们不能听懂他说什么。 30.对亚洲事务感兴趣的大使不禁受宠若惊. 四．请将下面这封信函翻译成汉语，注意翻译技巧（15’） 尊敬的李先生： 4月15日有关付款条件的来函已经收悉。 本公司同意贵公司如下建议： 1. 以见票即付的保兑不可撤销信用证付款，而非见票直接付款。 2. 贵公司的报盘不会有折扣。以上建议获本公司总经理批准，今后将如述执行。现正拟订有关订单，十日内将送达贵公司。 诚望今后两公司间的会谈能促进双方的业务发展。 敬复
2011--2012学年翻译理论与实践期末考试题 班级＿＿＿姓名＿＿＿学号＿＿＿Ⅰ、Translate the following phases into English(2point*5=10point) 1.覆水难收 ___________________________________________________________ 2.指鹿为马 ___________________________________________________________ 3. 爱屋及乌 ___________________________________________________________ 4条条道路通罗马 ___________________________________________________________ 5有钱能使鬼推磨 ___________________________________________________________ ⅡTranslate the following sentences into English(5point*6=30point) 1.我年轻的日子已经一去不返。 ___________________________________________________________ 2.你别狗咬吕洞宾——不识好人心。 ___________________________________________________________ 3.世界上只有海水取之不尽，用之不竭。 ___________________________________________________________ 4.我们是新雇员，得注意自己的一举一动。 ___________________________________________________________ 5.人民军队离不开人民，就像鱼儿离不开水一样。 ___________________________________________________________ 6.虚心使人进步，骄傲使人落后。 ___________________________________________________________ ⅢTranslate the following passages into English(10point*2=20point) 1）当前最重要的任务是发展国民经济，提高人民生活水平。为了实现这一目标，我们必须改革旧的经济体制，进一步解放生产水平；我们应当向世界敞开大门，学习其他国家的先进的科学和技术。只要我们坚持改革开放政策，就一定能把我国建设成强大的社会主义国家。 __________________________________________________________ __________________________________________________________ __________________________________________________________ __________________________________________________________ 2)秋天，无论在什么地方的秋天，总是好的；可是啊，北国的秋，却特别地来得清，来得静，来得悲凉。我的不远千里，要从杭州赶上青岛，更要从青岛赶上北平来的理由，也不过想饱尝一尝这“秋”，这故都的秋味。 __________________________________________________________
1.This electron beam sweeps across each line at a uniform rate，then flies back to scan another line directly below the previous one and so on，until the horizontal lines into which it is desired to break or split the picture have been scanned in the desired sequence． 电子束以均匀的速率扫描每一行，然后飞速返回去扫描下一行，直到把被扫描的图像按所希望的顺序分割成行。 2.The technical possibilities could well exist，therefore，of nation-wide integrated transmission network of high capacity，controlled by computers，interconnected globally by satellite and submarine cable，providing speedy and reliable communications throughout the word 因此，在技术上完全可能实现全国性的集成发送网络。这种网络容量大，由计算机控制，并能通过卫星和海底电缆实现全球互联，提供世界范围的高速、可靠的通信。 3.Transit time is the primary factor which limits the ability of a transistor to operate at high frequency． 渡越时间是限制晶体管高频工作能力的主要因素 4.The intensity of sound is inversely proportional to the square of the distance measured from the source of the sound. 声强与到声源的距离的平方成反比。 5.The attenuation of the filter is nearly constant to within 0.5 dB over the entire frequency band． 该滤波器的衰减近于恒定, 整个频带内的变化在0.5 dB以内。 6.At present, the state of most semiconductor device technology is such that the device design and process technology must be supplemented by screening and inspection procedures, if ultimate device reliability is to be obtained and controlled． 目前, 大多数半导体器件的技术尚未十分完善, 以至若要获得并控制器件最终的可靠性, 就必须辅以筛选和检验, 以弥补设计和工艺技术之不足 7.Bandwidth of transistor amplifiers vary from about 250 MHz in the L band to 1000 MHz in the X band． 晶体管放大器的带宽在L波段约为250 MHz, 在X波段为1000 MHz。 8.The output of the differential amplifier is fed to the circuit’s output stage via an offset-compensation network, which causes the op-amp’s output to center at zero volts. The output stage takes the form of a complementary emitter follower, and provides a low-impedance output. 差动放大级的输出通过一个失调补偿网络与输出级相连, 目的是使运放的输出以0 V为中心。输出级采用互补的射极跟随器的形式以使输出阻抗很低 9.Because of the very high open-loop voltage gain of the op-amp, the output is driven into positive saturation (close to +V) when the sample voltage goes slightly above the reference voltage, and driven into negative saturation (close to-V) when the sample voltage goes slightly below the reference voltage. 由于运放的开环电压增益很高, 当取样电压略高于参考电压时, 输出趋向于正向饱和状态(接近+V)。当取样电压低于参考电压时, 输出趋向于负向饱和状态(接近-V)。 10.If the signal source were direct connected instead of capacitor coupled, there would be a low resistance path from the base to the negative supply line, and this would affect the circuit bias conditions. 如果信号源和电路不是用电容耦合而是直接相连，从基极到负电源线就会一个低阻通路，并且这将影响到电路偏置状态 11.The differential amplifier has a high-impedance (constant-current)“tail”to give it a high input impedance and a high degree of common-mode signal rejection. It also has a high-impedance collector (or drain) load, to give it a large amount of signal-voltage gain (typically about 100 dB). 差动放大极有一个高阻抗的“尾巴”（恒流源）以提供高输入阻抗和对共模信号的深度抑制，同时，它还具有一个高阻抗和集电极或漏极负载以提供高的信号电压增益（典型的数据是100dB). 12.On the other hand, a DC negative-logic system, as in Figure 3.6(b), is one which designates the more negative voltage state of the bit as the 1 level and the more positive as the 0 level. 另一方面, 如图3.6(b)所示, 把比特的较低的电压状态记为1电平, 较高的电压状态记为0电平, 这样的系统称为直流负逻辑系统。 13.For example, to represent the 10 numerals (0, 1, 2, …, 9) and the 26 letters of the English alphabet would require 36 different combinations of 1’s and 0’s. Since 25<36<26, then a minimum of 6 bits per bite are required in order to accommodate all the alphanumeric characters. 例如，要表示0～9十个数字和英文字母表中的26个字母，就需要0和1的36种不同的组合。因为25<36>26,