Answer:
True.
Explanation:
An investigator would like to do an internet-based college student survey to gather data regarding their behavior as well as drug use. Even so, specific identifications are still not gathered and IP addresses might be available in the set of data. Damage threat must be assessed both by the magnitude (or severity) and the likelihood (or likelihood) of the hurt.
Answer:
Number of fragments is 3
Explanation:
The maximum size of data field in each fragment = 4468 - 20(IP Header)
= 4448 bytes
Hence, the number of required fragment = (10000 - 20)/4448
= 3
Fragment 1
Id = 218
offset = 0
total length = 4468 bytes
flag = 1
Fragment 2
Id = 218
offset = 556
total length = 4468 bytes
flag = 1
Fragment 3
Id = 218
offset = 1112
total length = 1144 bytes
flag = 0
Answer:
It is the ALU or the Arithmetic Logic Unit.
Explanation:
It is the ALU. However, keep in mind that registers and buses do a very important task. The number of registers we have, faster is the processing, and the opposite is true as well. And there is a reason behind this if we have different channels for sending and receiving the data from the memory, and several registers for storing the data, and we can formulate the requirement seeing the requirements for full adder and half adders. Remember we need to store several variables in case of the full adder, and which is the carry, and if we have separate registers for each of them, our task becomes easier. Remember its the CU that tells the ALU what operation is required to be performed. Also remember we have the same channel for input and output in the case of Van Neumann architecture, as we have a single bus. and we also have a single shared memory. And Harvard architecture is an advanced version of it.
