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Most Recent HPE6-A85 Exam Dumps

 

Prepare for the HP Aruba Certified Campus Access Associate Exam exam with our extensive collection of questions and answers. These practice Q&A are updated according to the latest syllabus, providing you with the tools needed to review and test your knowledge.

QA4Exam focus on the latest syllabus and exam objectives, our practice Q&A are designed to help you identify key topics and solidify your understanding. By focusing on the core curriculum, These Questions & Answers helps you cover all the essential topics, ensuring you're well-prepared for every section of the exam. Each question comes with a detailed explanation, offering valuable insights and helping you to learn from your mistakes. Whether you're looking to assess your progress or dive deeper into complex topics, our updated Q&A will provide the support you need to confidently approach the HPE6-A85 exam and achieve success.

The questions for HPE6-A85 were last updated on May 3, 2025.
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Question No. 1

You are working with a pair of 6300M switches in a VSF stack. The switch has 48 SmartRate 5G ports, 2 SFP28 ports, and 2 SFP56 ports. Both SFP56 ports are used for stacking.

You need to provide an LACP connection to another identical stack with the maximum available bandwidth possible. What should you configure?

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Correct Answer: A

To provide an LACP connection with the maximum available bandwidth, one should configure a link aggregation group (LAG) using all available ports that can be used for data transfer. Since the SFP56 ports are used for stacking, the next best option is to use the 2 SFP28 ports and as many SmartRate 5G (SR5) ports as possible on each switch, which would allow for a 16-member LAG, with 2 SFP28 and 6 SR5 ports on each switch contributing to the LAG.


Question No. 2

You need to configure wireless access for several classes of loT devices, some of which operate only with 802 11b. Each class must have a unique PSK and will require a different security policy applied as a role There will be 15-20 different classes of devices and performance should be optimized

Which option fulfills these requirements''

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Correct Answer: B

For configuring wireless access for multiple classes of IoT devices with varying security requirements, using a single SSID with Multiple Pre-Shared Keys (MPSK) is an efficient solution. MPSK allows different devices or groups of devices to connect to the same SSID but with unique PSKs, facilitating unique security policies for each class. Given that some IoT devices only support 802.11b, which operates in the 2.4GHz band, it is essential to include the 2.4GHz band in the configuration. The 5GHz band should also be included to support devices capable of operating in that band and to optimize network performance. The 6GHz band (option A) is not suitable since 802.11b devices are not compatible with it. Individual SSIDs for each IoT class (options C and D) would unnecessarily complicate network management and SSID overhead.


Question No. 3

Which commands are used to set a default route to 10.4.5.1 on an Aruba CX switch when ln-band management using an SVl is being used?

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Question No. 4

The noise floor measures 000000001 milliwatts, and the receiver's signal strength is -65dBm. What is the Signal to Noise Ratio?

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Correct Answer: D

The signal to noise ratio (SNR) is a measure that compares the level of a desired signal to the level of background noise. SNR is defined as the ratio of signal power to the noise power, often expressed in decibels (dB).A high SNR means that the signal is clear and easy to detect or interpret, while a low SNR means that the signal is corrupted or obscured by noise and may be difficult to distinguish or recover3. To calculate the SNR in dB, we can use the following formula:

SNR (dB) = Signal power (dBm) - Noise power (dBm)

In this question, we are given that the noise floor measures -90 dBm (0.000000001 milliwatts) and the receiver's signal strength is -65 dBm (0.000316 milliwatts). Therefore, we can plug these values into the formula and get:

SNR (dB) = -65 dBm - (-90 dBm) SNR (dB) = -65 dBm + 90 dBm SNR (dB) = 25 dBm

Therefore, the correct answer is that the SNR is 25 dBm.


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