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NEW QUESTION: 1
CORRECT TEXT
An administrator is configuring a secured webserver, however connecting to https://127.0.0.1 is not working. The administrator runs netstat -ntl, which returns the following output: tcp 0 0
0.0.0.0:80 0.0.0.0:* LISTEN What port should be listening before a successful connection is possible? (Provide only the numerical value of the port).
Answer:
Explanation:
443
NEW QUESTION: 2
Which type of computing and storage environments does EMC NetWorker protect?
A. LAN, SAN, DAS, and NIS
B. SAN, NAS, DNS, and LAN
C. NAS, SAN, DAS, and WAN
D. DAS, DNS, SAN, and WAN
Answer: C
NEW QUESTION: 3
How are "Rule-based related products" added to the shopping cart when you check their checkbox together with the checkbox for the basic product?
A. As the basic product's custom options
B. Separately from the basic product
C. As the basic product's configuration options
D. As a single grouped product
Answer: B
NEW QUESTION: 4
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.0.0.0.1
B. 許可10.0.0.0.255.255.255.254
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.1.0.0.0.254
Answer: D
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).
The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.