Answer:
The correct answer is letter "A": commit with fallback.
Explanation:
American Professor Alfred A. Marcus (born 1950) in his book "<em>The Future of Technology Management and the Business</em>" (2015) describes that hedging may be a strategy to shield businesses from the rapidly evolving world they face as a result of the continuous implementation of technology in the market. According to Marcus, there are 5 hedge approaches that firms should implement:
- Gamble on the most probable:<em> work on the product with the highest success rate.
</em>
- Take the robust route: <em>invest in as many products as possible.
</em>
- Delay until further clarity emerges:<em> waiting for a proper moment to react in front of market changes.
</em>
- <u>Commit with a fallback</u>:<em> adapt according to the market.
</em>
- Try to shape the future:<em> innovate.</em>
Answer:
Following are the statement is given below
if(updateDirection ==1) // check condition
{
++numUsers; // increments the value
}
else
{
--numUsers; // decrement the value
}
Explanation:
Following are the description of statement
- Check the condition in the if block .If the "updateDirection" variable is 1 then then control moves to the if block otherwise control moves to the else block statement.
- In the if block the statement is "++numUsers" it means it increment the value by 1 .
- In the else block the statement is "--numUsers" it means it decrement the value by 1 .
Session Initiation Protocol, brainliest ?
The correct answer is a. effective communication
- - -
Ineffective and barriers to communication are problems that make communication unclear. Workplace communication is at work or at a job. This is not a job newsletter for workers, but for people at home.
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
