Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
Performance monitor tool can you use to display hardware utilization statistics.
<h3>what is a performance Monitor?</h3>
- Performance Monitor is a system monitoring program introduced in Windows NT 3.1. It monitors various activities on a computer such as CPU or memory usage.
To learn more about performance monitoring, refer
to brainly.com/question/12960090
#SPJ4
HTML is a very basic markup language and requires memorization of a few dozen HTML commands that structure the look and layout of a web page. Before writing <span>any HTML code or designing your first web page, you must decide on an HTML editor or text editor, such as Notepad or Word Pad.</span>
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
I do not believe you answer is right. I believe it would be B the average montly sales for the big toy company. Only documenting the last month is not enough information to make a histogram, and a histogram asks for information based on one subject not multiple subjects. Asking for the number of each type of candy sold last month would make a normal graph comparing the difference in how much each candy sold, so your answer should be B.
