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Leto [7]
3 years ago
7

List 2 negative impacts of cybersecurity.Answer and I will give you brainiliest ​

Computers and Technology
1 answer:
Nitella [24]3 years ago
5 0

Answer:

Cyber security can be used for good but nevertheless some people will find ways to do bad things. Basically the negative impacts of cyber security is the opposite to the good type of cyber security they take advantage of it and exploit computer system to do what they want. An example to this is planting a malware on the victim's computer or using a DDOS attack on a WiFi router the possibility are endless.

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What is the value of count after this nested FOR loop executes fully.
Alisiya [41]

Answer:

168 (although the =< must be corrected to <=)

Explanation:

int count = 0;

for (int row = 4; row <= 15; row++)

for (int col = 0; col < 13; col = col +2)

count+=2;

The inner for loop runs 7 times (for col = 0,2,4,6,8,10,12). Anything higher is not less than 13. Therefore the inner loop increments count by 2 seven times, i.e. it increments count by 14.

The outer for loop runs 12 times (for row = 4,5,6,7,8,9,10,11,12,13,14,15).

If the count is incremented by 14 twelve times, you are incrementing it by 14*12 = 168.

Therefore the count goes from 0 to 168 after the nested loops.

5 0
3 years ago
Each symbol of an octal number corresponds to 3 bits of a binary number. is it true or false​
grin007 [14]

Answer: True.

Explanation: It uses only the 3 bits to represent any digit in binary and easy to convert from octal to binary and then to vice-versa. Hope that helps

5 0
3 years ago
How does VIDEO CONFERENCE change the way we work today???
ArbitrLikvidat [17]
Well theirs less travel. Its alot easier. To work from home.

But this is what it said on a website
Most of them figure they can use conference call time to get 'real work' done, chat, read the newspaper.
7 0
3 years ago
Read 2 more answers
Write an algorithm to find perimeter of circle<br>​
Anna11 [10]

Answer:

Here’s one!

Given [math]R[/math], the radius of the circle.

Let [math]N,D\leftarrow 0[/math]

Repeat until [math]D[/math] is large enough (about 1,000,000)

[math]x,y\leftarrow U[0,1][/math]

If [math]x^2 + y^2\le 1[/math] then [math]N\leftarrow N+1[/math]

[math]D\leftarrow D+1[/math]

[math]P\leftarrow\frac{8NR}{D}[/math]

Return [math]P[/math]

[math]U[0,1][/math] is a uniform random number in the range [math][0,1][/math].

Explanation:

6 0
2 years ago
Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
zalisa [80]

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;

}

}

8 0
3 years ago
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