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NEW QUESTION: 1
Click the Exhibit button.
Given the routing table shown in the exhibit, what is the next-hop to reach the host
192.168.1.129?
A. 10.126.131.248
B. 10 126.131 250
C. 10.126.131.252
D. 10.126.131.251
Answer: C
NEW QUESTION: 2
Which statement about enrollment in the IP telephony PKI is true? (Source. Understanding Cisco IP Telephony Authentication and Encryption Fundamentals)
A. The CAPF itself has to enroll with the Cisco CTL client.
B. CAPF enrollment supports the use of authentication strings.
C. MICs are issued by the CAPF itself or by an external CA.
D. LSCs are issued by the Cisco CTL client or by the CAPF.
Answer: B
Explanation:
Explanation/Reference:
Incorrect answer: B, C, D
Explanation: The CAPF enrollment process is as follows:
1. The IP phone generates its public and private key pairs.
2. The IP phone downloads the certificate of the CAPF and uses it to establish a TLS session with the CAPF.
3. The IP phone enrolls with the CAPF, sending its identity, its public key, and an optional authentication string.
4. The CAPF issues a certificate for the IP phone signed with its private key.
5. The CAPF sends the signed certificate to the IP phone.
Link: http://my.safaribooksonline.com/book/certification/cipt/9781587052613/understanding-cisco-ip- telephony-authentication-and-encryption-fundamentals/584.
NEW QUESTION: 3
質問をドラッグアンドドロップ
IPv4ネットワークサブネットを左側から右側の選択して配置できる正しい使用可能なホスト範囲にドラッグアンドドロップします。
Answer:
Explanation:
Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet
172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.