Answer:
Given the number of platters =5
A. Hence, the number of recording surface =10
B. number of read-write heads =10
C. Number of arms holding the read-write heads = 10
Explanation:
We know that number of recording surface = 2 *( number of platters)
= 2*5=10
And the number of read-write heads = number of recording surface= 10
Also, number of arms holding the read-write head = one for each read-write head= 10
And hence, the above answer.
Answer:
Answer to the following question is as follows;
Explanation:
Windows 7 for 100 users and Windows 8.1 for 25 users are the best options since they both enable networking and active directory, and Windows 8.1 also offers Windows server capabilities.
Winxp would therefore work for $100, but it is unsupported and has no updates.
If you wish to go with open source, you can choose Ubuntu 16 or 18 or Linux.
Answer:
Spiral of silence
Explanation:
The spiral of silence theory states that people remain silent regarding a topic when they feel that their views are in opposition with the rest of the majority.
As stated in the question, the newspaper's stories suggest that majority of the people support cutting taxes, this leads people who supoort raising taxes to not air their opinions because they think that everyone else supports cutting taxes like the papers say.
1 - b - Excel is a computer program, so it'd be Software
2 - a - Process of elimination, haven't heard of the term 'peopleware' before.
3 - d - A joystick accepts input from a user to interact with a computer. ex. Flight Simulators
4 - c - A monitor will display (or output) an image based on what is received from the computer.
Answer:
The function is as follows:
void readAndConvert(){
int n; string symbol,name;
cin>>n;
cin>>symbol;
cin.ignore();
getline (cin,name);
vector<string> trades;
string trade;
for (int inps = 0; inps < n; inps++){
getline (cin,trade);
trades.push_back(trade);}
cout<<name<<" ("<<symbol<<")"<<endl;
for (int itr = 0; itr < n; itr++){
string splittrade[3]; int k = 0;
for(int j=0;j<trades.at(itr).length();j++){
splittrade[k] += trades.at(itr)[j];
if(trades.at(itr)[j] == ' '){
k++; }}
cout<<splittrade[2]<<": "<<floor(stod(splittrade[1]) * stod(splittrade[0]))<<endl; }
}
Explanation:
See attachment for complete program where comments are used to explain each line
