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What is Instruction Cycle
A program residing in the memory unit of the computer having of a sequence of instructions. The program is implemented in the computer by going through a cycle for each instruction. Every instruction cycle in turn is subdivided into a sequence of sub cycles or phases. In the basic computer each instruction cycle having of the following phases:
1. Fetch an instruction from memory.
2. Decode the instruction
3. Read the effective address from memory if the instruction has an indirect address.
4. Execute the instruction.
According to the report, network 1 and network 2 are not able to reach network 3. As shown on OTBNetwork Topology above, OTB Inc. has 2 routing protocols running due transition iss
what is jsf
What are wired and wireless transmission systems? Transmission Systems: Modern type long distance transmission systems can be placed in three wide categories: 1. Radio Sy
Proof by Contradiction - Artificial intelligence So, both backward chaining andforward chaining have drawbacks. Another approach is to think regarding proving theorems by contr
tCAS is the number of clock cycles required to access a particular column of data in SDRAM. CAS latency is the column address strobe time, sometimes referred to as tCL.
MAC is the abbreviation for: (A) Multimedia access control (B) Media access control (C) Mobile access control (D) Master access point control Ans: MAC is th
Let Consider a multiple regression model for a response y, with one quantitative idividually variable x1, and one qualitative variable at three levels. a) Write a first-order m
Application layer: It's the topmost layer in OSI model, which allows the user to access network. This layer provides user interface for network applications like remote login, Wor
Q. How to convert Binary to Octal and Hexadecimal? Rules for these conversions are simple. For converting binary to octal binary number is splitted in groups of three, that are
Prove the following Boolean identities using the laws of Boolean algebra (A + B)(A + C) = A + BC Ans. (A+B)(A+C)=A+BC LHS AA+AC+AB+BC=A+AC+AB+BC OR A((C+1)+A(B+1))+BC
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