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Pass Guaranteed 2025 The SecOps Group CNSP: Certified Network Security Practitioner Accurate Braindumps Downloads
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The SecOps Group Certified Network Security Practitioner Sample Questions (Q22-Q27):
NEW QUESTION # 22
A system encrypts data prior to transmitting it over a network, and the system on the other end of the transmission media decrypts it. If the systems are using a symmetric encryption algorithm for encryption and decryption, which of the following statements is true?
Answer: A
Explanation:
Symmetric encryption is a cryptographic technique where the same key is used for both encryption and decryption processes. In the context of network security, when data is encrypted prior to transmission and decrypted at the receiving end using a symmetric encryption algorithm (e.g., AES or Triple-DES), both the sender and receiver must share and utilize an identical secret key. This key is applied by the sender to transform plaintext into ciphertext and by the receiver to reverse the process, recovering the original plaintext. The efficiency of symmetric encryption makes it ideal for securing large volumes of data transmitted over networks, provided the key is securely distributed and managed.
Why A is correct: Option A accurately describes the fundamental property of symmetric encryption-using a single shared key for both encryption and decryption. This aligns with CNSP documentation, which emphasizes symmetric encryption's role in securing data in transit (e.g., via VPNs or secure file transfers).
Why other options are incorrect:
B: This describes asymmetric encryption (e.g., RSA), where different keys (public and private) are used for encryption and decryption, not symmetric encryption.
C: Symmetric encryption inherently relies on keys; the absence of keys contradicts its definition and operational mechanism.
D: Symmetric encryption is not inherently insecure; its security depends on key strength and management practices, not the algorithm itself. CNSP highlights that algorithms like AES are widely regarded as secure when implemented correctly.
NEW QUESTION # 23
What is the response from an open TCP port which is not behind a firewall?
Answer: B
Explanation:
TCP's three-way handshake, per RFC 793, establishes a connection:
Client → Server: SYN (Synchronize) packet (e.g., port 80).
Server → Client: SYN-ACK (Synchronize-Acknowledge) packet if the port is open and listening.
Client → Server: ACK (Acknowledge) completes the connection.
Scenario: An open TCP port (e.g., 80 for HTTP) with no firewall. When a client sends a SYN to an open port (e.g., via telnet 192.168.1.1 80), the server responds with a SYN-ACK packet, indicating willingness to connect. No firewall means no filtering alters this standard response.
Packet Details:
SYN-ACK: Sets SYN and ACK flags in the TCP header, with a sequence number and acknowledgment number.
Example: Client SYN (Seq=100), Server SYN-ACK (Seq=200, Ack=101).
Security Implications: Open ports responding with SYN-ACK are easily detected (e.g., Nmap "open" state), inviting exploits if unneeded (e.g., Telnet on 23). CNSP likely stresses port minimization and monitoring.
Why other options are incorrect:
A . A FIN and an ACK packet: FIN-ACK closes an established connection, not a response to a new SYN.
B . A SYN packet: SYN initiates a connection from the client, not a server response.
D . A RST and an ACK packet: RST-ACK rejects a connection (e.g., closed port), not an open one.
Real-World Context: SYN-ACK from SSH (22/TCP) confirms a server's presence during reconnaissance.
NEW QUESTION # 24
What is the response from an open UDP port which is not behind a firewall?
Answer: C
Explanation:
UDP's connectionless nature means it lacks inherent acknowledgment mechanisms, affecting its port response behavior.
Why B is correct: An open UDP port does not respond unless an application explicitly sends a reply. Without a firewall or application response, the sender receives no feedback, per CNSP scanning guidelines.
Why other options are incorrect:
A: ICMP Port Unreachable indicates a closed port, not an open one.
C: SYN packets are TCP-specific, not UDP.
D: FIN packets are also TCP-specific.
NEW QUESTION # 25
The application is showing a TLS error message as a result of a website administrator failing to timely renew the TLS certificate. But upon deeper analysis, it appears that the problem is brought on by the expiration of the TLS certificate. Which of the following statements is correct?
Answer: B
Explanation:
TLS (Transport Layer Security) secures communication (e.g., HTTPS) using certificates, per RFC 8446. A certificate includes:
Validity Period: Start and end dates (e.g., "Not After: March 8, 2025").
Purpose: Authenticates the server and encrypts the session.
Scenario: An expired TLS certificate (e.g., past "Not After" date). Modern browsers (e.g., Chrome, Firefox) validate certificates during the handshake:
ClientHello: Browser initiates TLS.
ServerHello: Server sends its certificate.
Validation: Browser checks expiration, CA trust, etc.
If expired, browsers reject the handshake, displaying errors (e.g., "NET::ERR_CERT_DATE_INVALID"). No session key is negotiated, and communication doesn't proceed over TLS. Users may bypass warnings (e.g., "Advanced > Proceed"), but this is unencrypted or uses a fallback (not standard TLS), breaking security guarantees.
Security Implications: Expired certificates expose sites to MITM attacks, as trust is lost. CNSP likely emphasizes certificate management (e.g., automation with Let's Encrypt) to avoid this.
Why other options are incorrect:
B . The communication is still over TLS: False; an expired certificate halts the TLS handshake in compliant browsers. Legacy systems might negotiate insecurely, but this isn't "TLS" per standards.
Real-World Context: The 2019 Equifax breach partially stemmed from expired certificates missing vulnerabilities.
NEW QUESTION # 26
What will be the subnet mask for 192.168.0.1/18?
Answer: C
Explanation:
An IP address with a /18 prefix (CIDR notation) indicates 18 network bits in the subnet mask, leaving 14 host bits (32 total bits - 18). For IPv4 (e.g., 192.168.0.1):
Binary Mask: First 18 bits are 1s, rest 0s.
1st octet: 11111111 (255)
2nd octet: 11111111 (255)
3rd octet: 11000000 (192)
4th octet: 00000000 (0)
Decimal: 255.255.192.0
Calculation:
Bits: /18 = 2