Hardware
Answer. physical parts of computer which can be seen and touched are called Hardware.
Answer:
Viruses can enter your computer in any number of ways, such as via an email attachment, during file downloads from the Internet or even upon a visit to a contaminated Web site.Jun 2, 2010
Explanation:
Viruses spread when the software or documents they get attached to are transferred from one computer to another using a network, a disk, file sharing methods, or through infected e-mail attachments. Some viruses use different stealth strategies to avoid their detection from anti-virus software.
Answer:
For undirected graph n(n-1) for directed graph 
Explanation:
The maximum number of edges for a undirected graph is given by 
, where n is number of vertices
The maximum number of edges in directed graph is given by , where n is number of vertices
The maximum number of edges is greater in undirected graph than directed graph
Answer:
(a)The CPU B should be selected for the new computer as it has a low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A.
(b) The CPU B is faster because it executes the same number of instruction in a lesser time than the CPU A
.
Explanation:
Solution
(a)With regards to the MIPS performance metric the CPU B should be chosen for the new computer as it low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A and when we look at the amount of process done by the system , the CPU B is faster when compared to other CPU and carries out same number of instruction in time.
The metric of response time for CPU B is lower than the CPU A and it has advantage over the other CPU and it has better amount as compared to CPU A, as CPU B is carrying out more execution is particular amount of time.
(b) The execution can be computed as follows:
Clock cycles taken for a program to finish * increased by the clock cycle time = the Clock cycles for a program * Clock cycle time
Thus
CPU A= 5*10^6 * 60*10^-9 →300*10^-3 →0.3 second (1 nano seconds =10^-9 second)
CPU B= 3 *10^6 * 75*10^-9 → 225*10^-3 → 0.225 second
Therefore,The CPU B is faster as it is executing the same number of instruction in a lesser time than the CPU A
