Answer:
import csv
def Csvreader(filename):
with open("filename", "r") as file:
content = file.csv_reader()
list_content = list(content)
file.close
return list_content
Explanation:
This is a python description of the function that reads the csv file and converts it to a list, from the list, each item can accessed with its index and updated directly. Use this to draw the same conclusion for a java program.
Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
Answer:
C: Security
Explanation:
Communications and information systems principles need to be, among other things, secure. They need to be able to protect sensitive information from those who intentionally not need to know. Some incident information like voice, networks, and data, are very sensitive and thus, should be secure to the right levels and should comply with privacy laws and data protection.
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
The best option that will suite is that there will be no major issues since the offshore leads and the onsite members participated in the demo with the Product Owner/Stakeholders they can cascade the feedback to the offshore members
Explanation:
Iteration demo is the review which is done to gather the immediate feedback from the stakeholders on a regular basis from the regular cadence. This demo will be one mainly to review the progress of the team and the and to cascade and show their working process
They show their working process to the owners and they and the other stakeholders and they get their review from them and so there will be no issues if the members are not able to participate
