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NEW QUESTION: 1
Which of the following could cause a browser to display the message below?
"
The security certificate presented by this website was issued for a different website's address."
A. HTTPS://127.0.01 was used instead of HTTPS://localhost.
B. The website is using an expired self-signed certificate.
C. The website is using a wildcard certificate issued for the company's domain.
D. The website certificate was issued by a different CA than what the browser recognizes in its trusted CAs.
Answer: A
Explanation:
Explanation/Reference:
Explanation:
PKI is a two-key, asymmetric system with four main components: certificate authority (CA), registration authority (RA), RSA (the encryption algorithm), and digital certificates. In typical public key infrastructure (PKI) arrangements, a digital signature from a certificate authority (CA) attests that a particular public key certificate is valid (i.e., contains correct information). Users, or their software on their behalf, check that the private key used to sign some certificate matches the public key in the CA's certificate. Since CA certificates are often signed by other, "higher-ranking," CAs, there must necessarily be a highest CA, which provides the ultimate in attestation authority in that particular PKI scheme.
Localhost is a hostname that means this computer and may be used to access the computer's own network services via its loopback network interface. Using the loopback interface bypasses local network interface hardware. In this case the HTTPS://127.0.01 was used and not HTTPS//localhost
NEW QUESTION: 2
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.0
B. Permit 10.0.0.0.255.255.255.254
C. Permit 10.0.0.1.0.0.0.254
D. Permit 10.0.0.0.0.0.0.1
Answer: C
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's 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 odd- numbered hosts in the 10.0.0.0/24 subnet.
NEW QUESTION: 3
Which are 802.1x authentication modes?(Select 2 Answers)
A. Remote mode
B. Proxy mode
C. EAP relay mode
D. EAP termination mode
Answer: C,D
NEW QUESTION: 4
The R75 fw monitor utility is used to troubleshoot which of the following problems?
A. Log Consolidation Engine
B. Phase two key negotiation
C. User data base corruption
D. Traffic issues
Answer: D