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網(wǎng)絡工程師每日一練試題(2024/10/1)

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網(wǎng)絡工程師每日一練試題內(nèi)容(2024/10/1)

  • 試題1

    用戶在登錄FTP服務器的過程中,建立TCP連接時使用的默認端口號是(  )。
    A.20
    B.21
    C.22
    D.23

    查看答案

    試題參考答案:B

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題2

    4B/5B編碼先將數(shù)據(jù)按 4 位分組,將每個分組映射到 5 單位的代碼,然后采用()進行編碼。
    A.PCM
    B.Manchester
    C.QAM
    D.NRZ-I

    查看答案

    試題參考答案:D

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題3

    IEEE802.11規(guī)定了多種WLAN通信標準,其中( )與其他標準采用的頻段不同,因而不能兼容。
    A.IEEE802.11a
    B.IEEE802.11b
    C.IEEE802.11g
    D.IEEE802.11n

    查看答案

    試題參考答案:A

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題4

    在RIP路由協(xié)議中,當路由跳數(shù)達到()時,目的地被標記為路由 不可達。
    A.16
    B.14
    C.15
    D.13

    查看答案

    試題參考答案:A

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題5

    在Linux系統(tǒng)中,要查看如下輸出,可使用命令(32)。

    A.[root@localhost]#ifconfig
    B.[root@localhost]#ipconfig eth0
    C.[root@localhost]#ipconfig]
    D.[root@localhost]#ifconfig eth0

    查看答案

    試題參考答案:D

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題6

    在輸入輸出控制方法中,采用?。?) 可以使得設(shè)備與主存間的數(shù)據(jù)塊傳送無需CPU干預。
    A、程序控制輸入輸出
    B、中斷
    C、DMA
    D、總線控制

    查看答案

    試題參考答案:C

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題7

    在瀏覽器的地址欄中輸入 xxxyftp.abc.com.cn,該 URL 中( )是要訪問的主機名。
    A.xxxyfip
    B.a(chǎn)bc
    C.com
    D.Cn

    查看答案

    試題參考答案:A

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題8

    下面關(guān)于幾個網(wǎng)絡管理工具的描述中,錯誤的是  ( )  。
    A、netstat可用于顯示IP、TCP、UDP、ICMP等協(xié)議的統(tǒng)計數(shù)據(jù)
    B、sniffer能夠使網(wǎng)絡接口處于雜收模式,從而可截獲網(wǎng)絡上傳輸?shù)姆纸M
    C、winipcfg采用MS-DOS工作方式顯示網(wǎng)絡適配器和主機的有關(guān)信息
    D、tracert可以發(fā)現(xiàn)數(shù)據(jù)包到達目標主機所經(jīng)過的路由器和到達時間

    查看答案

    試題參考答案:C

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題9

    The TCP protocolis a (1) layer protocol. Each connection connects two TCPs that may be just one physical network apart or located on opposite sides ofthe globe. In other words, each connection creates a (2) witha length that may be totally different from another path created by another connection. This means that TCP cannot use the same retransmission time for all connections. Selecting afixed retransnussion time for all connections can result in serious consequences. Ifthe retransmission time does not allow enough time for a (3) to reach the destination and an acknowledgment to reach the source, it can result in retransmission of segments that are still on the way. Conversely, if the retransnussion time is longer than necessary for a short path, it may result in delay for the application programs.Even for one single connection, the retransmission time should not be fixed.A connection may be able to send segments and receive (4)faster during nontraffic period than during congested periods. TCP uses the dynamic retransmission time,a transmission time is different for each connection and which may be changed during the same connection. Retransmission time can be made(5)  by basing it on the round-trip time (RTT). Several formulas are used for this purpose.
    (1)A、physical
    B、network
    C、transport
    D、application
    (2)A、path
    B、window
    C、response
    D、process
    (3)A、process
    B、segment
    C、program
    D、user
    (4)A、connections
    B、requests
    C、acknowledgments
    D、datagrams
    (5)A、long
    B、short
    C、fixed
    D、dynamic

    查看答案

    試題參考答案:C、A、B、C、D

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

  • 試題10

    計算機系統(tǒng)的主存主要是由()構(gòu)成的。
    A.DRAM
    B.SRAM
    C.Cache
    D.EEPROM

    查看答案

    試題參考答案:A

    試題解析與討論:m.pokkc.com/exam/ExamDay.aspx?t1=8&day=2024/10/1

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