Answer:
The 802.11ac wireless standard takes channel bonding to a higher level because it can support 20MHz, 40MHz, and 80MHz channels, with an optional use of 160MHz channels.
Explanation:
The 802.11ac is a standardized wireless protocol established and accepted by the institute of electrical and electronics engineers (IEEE). 802.11ac as a wireless local area network (WLAN) protocol, has multiple amplitude and bandwidth, thus making it to be the first standard wireless protocol to have the ability to operate on a Gigabit (Gb) network.
Generally, the 802.11ac wireless standard provides an advantage over 802.11n by incorporating increased channel bonding capabilities. The 802.11ac wireless standard takes channel bonding to a higher level because it can support 20MHz, 40MHz, and 80MHz channels, with an optional use of 160MHz channels.
<em>On the other hand, 802.11n is a standardized wireless protocol that can support either a 20MHz or 40MHz channel. </em>
Answer:
I'm guessing C
Explanation:
It just makes the most sense to me, but please don't be really confident with it, I'm in 6th grade. I'm really sorry if I'm incorrect.
Answer:
spanning-tree portfast bpduguard
Explanation:
spanning- tree protocol (STP) is a layer 2 protocol in the OSI model. It is automatically configured in a switch to prevent continual looping of BPDUs, to avoid traffic congestion. The fastport bpduguard is only applicable in non-trunking access in a switch. It is more secure to configure the fastport mode in switch port connected directly to a node, because there are still bpdus transfer in a switch to switch connection.
BPDUs Guard ensures that inferior bpdus are blocked, allowing STP to shut an access port in this regard.
Answer:
hi your question lacks the necessary matrices attached to the answer is the complete question
1024 bytes
Explanation:
A) The minimum size of the cache to take advantage of blocked execution
The minimum size of the cache is approximately 1 kilo bytes
There are 128 elements( 64 * 2 ) in the preceding simple implementation and this because there are two matrices and every matrix contains 64 elements .
note: 8 bytes is been occupied by every element therefore the minimum size of the cache to take advantage of blocked execution
= number of elements * number of bytes
= 128 * 8 = 1024 bytes ≈ 1 kilobytes
