Answer:
a. drop the attack packet(s)
Explanation:
An IPS not only detects and alerts system administrators but it also stops attacks. It is built to consistently monitor a network, report malicious incidents and take preventative actions. So, how does an IPS work exactly? An IPS does a deep packet inspection and either blocks the IP of whoever sent the malicious packet or removes the malicious packet's content and repackages its payloads. Thus, this means that the malicious packet is completely dropped by stripping off of its header information and any infected items in the packet.
Answer:
We will be using Average function of Excel Calculation sheet.
Its Syntax in Calc sheet is : =AVERAGE (number1, [number2], ...)
or you can use =Average and then click on the first cell and drag the whole row till end and close the brackets. It will calculate the average of that row.
Once you have applied average function on first row, you can drag this cell down the column and it will replicate the same function for respective rows.
Here you are, its that button.
Answer:
Open an application (Word, PowerPoint, etc.) and create a new file like you normally would. ...
Click File.
Click Save as.
Select Box as the location where you'd like to save your file. If you have a particular folder that you'd like to save it to, select it.
Name your file.
Click Save.
Explanation:
Answer:
a) 500 Kbps b) 64 sec c) 320 sec
Explanation:
a) We define the throughput of a network, as the actual maximum transmission rate that the network is able to deliver, which in this case is equal to the lowest transmission rate of any of the links that the traffic must go through:
R1 =500 kbps
b) If the file size is given in bytes, and we have the throughput in bps, we need to convert to bits first, as follows:
4*10⁶ bytes * (8 bits/byte) = 32*10⁶ bits.
The time needed to transfer the file, will be given by the quotient between the file size and the throughput, as follows:
t = 32*10⁶ bits / 500*10³ bits/sec = 64 sec
c) If the transmission rate R2 is reduced to 100 kbps, R2 becomes the lowest transmission rate in the network, so it becomes the new throughput.
So, the time needed for the same file to be transferred to host B is as follows:
t= 32*10⁶ bits / 100*10³ bits/sec = 320 sec
