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slava [35]
3 years ago
12

The gap between those who have easy access to the internet and technology and those who do not is known as the

Computers and Technology
1 answer:
kirill [66]3 years ago
7 0
Computer concepts... hope it helps
You might be interested in
Help me please and thanks
Yuri [45]

Answer:

b

Explanation:

6 0
4 years ago
Computer are most wonderful creation of 21st century how ?​
son4ous [18]

Answer:

It can do all the functions at a speedy rate and also helps us to search and progress in our homes and businesses. A computer can therefore be called a calculator with a twist for not only does it perform fast calculations, but it also has other special characteristics.

Explanation:

hope it helps you

6 0
2 years ago
Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the pro
AnnZ [28]

Answer:

a. (210^6)((210^7)/(2.510^8)) = 1.610^5 bits or 160,000 bits

b. 1.6*10^5 bits or 160,000 bits

c. Bandwidth delay product of link is maximum number of bits that can be in the link

d. Width of bit = Length of link / bandwidth-delay product so 1 bit is 125 meters long. Yes, this is longer than a football field.

e. Width of bit = s/R

Hope this helps :)

3 0
3 years ago
What are files that are placed on a computer by some websites when the user visits the site?
Otrada [13]

Answer:

A: Cookies.

Explanation:

Usually you will get a message about the site using "cookies" to help user experience.

3 0
3 years ago
Suppose you are given a string containing only the characters ( and ). In this problem, you will write a function to determine w
Whitepunk [10]

Answer:

Code is completed below

Explanation:

Source Code in Java:

class Parenthesis

{

boolean hasBalancedParentheses(String eq) //method to check and return if a set of parenthesis is balanced or not

{

int count=0; //to store count of current unclosed opening brackets

for(int i=0;i<eq.length();i++)

{

if(eq.charAt(i)=='(')

count++;

else if(eq.charAt(i)==')')

count--;

if(count<0) //if count falls below zero, there were more closing brackets than opening brackets till this point

return false;

}

if(count>0) //if count is still more than zero, there were less closing brackets than opening brackets

return false;

return true; //if program reaches this point, it has passed all the tests

}

public static void main(String args[])

{

//testing the function

Parenthesis ob=new Parenthesis();

System.out.println(ob.hasBalancedParentheses("()(()())((())())"));

System.out.println(ob.hasBalancedParentheses(")((()())(()))())"));

}

}

8 0
3 years ago
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