公認されたMTCNA勉強ガイド & 資格試験のリーダー & 便利なMTCNA試験勉強書あなたは短い時間でMTCNA試験に合格できるために、我々は多くの時間と労力を投資してあなたにMikroTikのMTCNA試験を開発しますから、我々の提供する商品はIT認定試験という分野で大好評を得ています。だからこそ、我々はJPNTestの問題集に自信があります。自信があるから、我々は失敗返金ということを承諾します。 MikroTik Certified Network Associate Exam 認定 MTCNA 試験問題 (Q75-Q80):質問 # 75
What layer in the TCP/IP stack is equivalent to the Transport layer of the OSI model?
A. Internet
B. Network Access
C. Application
D. Host-to-Host
正解:D
質問 # 76
What does this simple queue do (check the image)?
A. Queue limits host 192.168.1.10 upload data rate to one megabit per second.
B. Queue guarantees upload data rate of one megabit per second for host 192.168.1.10
C. Queue limits host 192.168.1.10 download data rate to one megabit per second.
D. Queue guarantees download data rate of one megabit per second for host 192.168.1.10
正解:A
質問 # 77
Select valid MAC-address:
A. 00:00:5E:80:EE:B0
B. AEC8:21F1:AA44:54FF:1111DAE:0212:1201
C. 192.168.0.0/16
D. G2:60:CF:21:99:H0
正解:A
解説:
A valid MAC address must meet the following criteria:
* Be 6 bytes (48 bits) in length
* Consist of only hexadecimal digits (0-9, A-F)
* Written in six groups separated by colons or hyphens (e.g., 00:1A:2B:3C:4D:5E) MTCNA Course Material - RouterOS MAC Address Basics:
"MAC addresses are 48-bit identifiers written as six pairs of hexadecimal digits. Invalid characters or incorrect length disqualifies an address." Rene Meneses MTCNA Guide - MAC Addressing Section:
"Each MAC is made up of 12 hexadecimal characters (6 octets). If a character like 'G' appears, or if it's longer than 6 bytes, it is invalid." MikroTik Wiki - MAC Addressing Rules:
"Valid MAC format: XX:XX:XX:XX:XX:XX using only 0-9 and A-F. 192.168.0.0/16 is an IP subnet, not a MAC."
* Option A: Invalid - "G" and "H" are not hexadecimal characters
* Option B: Valid - proper format and hex content
* Option C: Invalid - Too long (appears to be IPv6 or malformed)
* Option D: Invalid - this is an IP network (CIDR notation), not a MAC
Only Option B is correct.
質問 # 78
Which of the following is the valid host range for the subnet on which the IP address 192.168.168.188
255.255.255.192 resides?
A. 192.168.168.128-192
B. 192.168.168.129-191
C. 192.168.168.129-190
D. 192.168.168.128-190
正解:B
解説:
IP address: 192.168.168.188
Subnet mask: 255.255.255.192 # /26 # Block size = 64
Subnets:
* 192.168.168.0/26 # 192.168.168.0 - 63
* 192.168.168.64/26 # 192.168.168.64 - 127
* 192.168.168.128/26 # 192.168.168.128 - 191 # Contains 192.168.168.188
* 192.168.168.192/26 # 192.168.168.192 - 255
Valid host range = 192.168.168.129 - 190
(Broadcast = 191, Network address = 128)
MTCNA Course Material - Subnetting Practice:
"To find valid hosts, exclude the subnet and broadcast address. In /26, each block is 64 addresses." Rene Meneses MTCNA Study Guide - IP Addressing:
"For /26 subnetting, calculate block size as 2
さらに、JPNTest MTCNAダンプの一部が現在無料で提供されています:https://drive.google.com/open?id=1dJHun2n5da31jzIT8xo8qzz9cif6WJ4B
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