Is this Linux? It depends on what thy ask you to do it from.
Answer:The interface is the means by which a user communicates with a system, whether to get it to perform some function or computation directly (e.g., compute a trajectory, change a word in a text file, display a video); to find and deliver information (e.g., getting a paper from the Web or information from a database); or to provide ways of interacting with other people (e.g., participate in a chat group, send e-mail, jointly edit a document). As a communications vehicle, interfaces can be assessed and compared in terms of three key dimensions: (1) the language(s) they use, (2) the ways in which they allow users to say things in the language(s), and (3) the surface(s) or device(s) used to produce output (or register input) expressions of the language. The design and implementation of an interface entail choosing (or designing) the language for communication, specifying the ways in which users may express ''statements" of that language (e.g., by typing words or by pointing at icons), and selecting device(s) that allow communication to be realized-the input/output devices.
Box 3.1 gives some examples of choices at each of these levels. Although the selection and integration of input/output devices will generally involve hardware concerns (e.g., choices among keyboard, mouse, drawing surfaces, sensor-equipped apparel), decisions about the language definition and means of expression affect interpretation processes that are largely treated in software. The rest of this section briefly describes each of the dimensions and then examines how they can be used
Explanation:
Answer:
3.1 ns ; 1.25 ; 3.097
Explanation:
Given :
IF, 3 ns;
ID, 2.5 ns;
EX, 2 ns;
MEM, 3 ns;
WB, 1.5 ns.
Use 0.1 ns for the pipelineregisterdelay
maximum time required for MEM = 3 ns
Pipeline register delay = 0.1 ns.
Clock cycled time of the pipelined machine= maximum time required + delay
3ns+0.1 ns = 3.1 ns
2.) for stall after every 4 instruction :
CPI of new machine :
(1 + (1 /4)) = 1 + 0.25 = 1.25
3.)
The speedup of pipelined machine over the single-cycle machine is given by :
Average time per instruction of single cycle ÷ average time per instruction of pipelined
Clock time of original machine = 12ns
Ideal CP1 = 1
CPI of new machine = 1.25
Clock period = 3.1 ns
(12 * 1) / (1.25 * 3.1) = 12 / 3.875
= 3.097
D. Speed up will equal the number of stages in the machine
Answer:
The correct answer for the given question is " yes".
Explanation:
In the above question the correct answer is yes because, in any social media site. If we like any web page.it will shows the name of person that like the web pages .
For example: if we login in any social side we will see various images,video and story etc.if any user like all. it will show the person name that is liked by them.
So the correct answer for this question is yes.
Answer:
Using C++ to solve the problem as given below
Explanation:
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
string salsa_name[5]={"mild","medium","sweet","hot","zesty"};
int jar_sale[5]={0,0,0,0,0};
int i;
int total=0;
int high=0;
int high_index=0;
int low=0;
int low_index=0;
int temp=0;
for(i=0;i<5;i++)
{
while(temp<=0)
{
cout<<"enter the number of jars sold for "<<salsa_name[i]<<" ";
cin>>temp;
if(temp<=0)
cout<<"invalid data. please try again\n";
}
jar_sale[i]=temp;
temp=0;
}
cout<<"name\t jars sold\n";
cout<<"\n---------------------------\n";
for(i=0;i<5;i++)
{
cout<<" "<<salsa_name[i]<<"\t\t"<<jar_sale[i]<<"\n";
}
low=jar_sale[0];
for(i=0;i<5;i++)
{
total=total+jar_sale[i];
if(jar_sale[i] >= high)
{
high_index=i;
high=jar_sale[i];
}
if(jar_sale[i]<=low)
{
low_index=i;
low=jar_sale[i];
}
}
cout<<"\n total sale : "<<total;
cout<<"\n high seller : "<<salsa_name[high_index];
cout<<"\n low seller : "<<salsa_name[low_index];
}
