Choose the correct option that completes the sentence.

A news website uses 32-bit integers to count the number of times an article has been viewed. The website is becoming more popula

Irina-Kira [14]

Answer:

2^32 times as many values can be represented

Explanation:

32-bit. This means that the number is represented by 32 separate one’s and zero’s. 32 bits of 2 possible states = 2^32=4,294,967,296 possible values.

Integer meaning that only whole multiples of one are accepted.

Signed meaning that negative values are accepted. This halves the number of possible positive values (roughly), so the largest number you can represent is 2^31–1=2,147,483,647, but instead of 0, the smallest number you can represent is -2,147,483,648. An unsigned 32-bit integer, by contrast, can represent anything from 0 to 4,294,967,295.

7 0

2 years ago

What software is my teacher using to do this and how can i stop her from doing this

aliina [53]

Answer:

I’m not sure if you can only teachers can acess that

Explanation:

7 0

3 years ago

Read 2 more answers

Matching parentheses. An math expression may have a number of parentheses like (, ), [, ], { and }. Each openning parenthesis (,

Solnce55 [7]

C++ Code

#include
using namespace std;

bool match(const char exp[],const int s)
{
// declare a character array to perform stack operations
char stack[s];

// declare top and initialize to -1 and flag to 0
int top=-1,i,flag=0;

// visit all characters in the expression string
for(i=0;i {
// if the character is [ or ( or { then push it into stack
if(exp[i]=='[' || exp[i]=='(' || exp[i]=='{')
{
top++;
stack[top]=exp[i];
}
// if the character is ] or ) or } then check conditions
else if(exp[i]==']' || exp[i]==')' || exp[i]=='}')
{
// check stack is empty or not
if(top!=-1)
{
// check all possible failure conditions
if(exp[i]==')' && (stack[top] == '{' || stack[top]=='['))
{
flag = 1;
break;
}
else if(exp[i]==']' && (stack[top] == '{' || stack[top]=='('))
{
flag = 1;
break;
}
else if(exp[i]=='}' && (stack[top] == '(' || stack[top]=='['))
{
flag = 1;
break;
}
top--;
}
else
{
flag=1;
break;
}
}
}
// after visiting all characters of expression string check if stack is not empty and flag is 1. if any one of the condition is true return false. otherwise return true

if(top>=0 || flag==1)
return false;
else
return true;
}

int main()
{

// declare character array to store expression
char exp[10000];
cout<<"Enter an Expression"<
// read expression from user
cin.getline(exp, 10000);
int s=0;

// find the length of the expression string
for(int i=0;exp[i]!='\0';i++)
{
s++;
}

// call the match function
bool status = match(exp,s);

// print the result based on value returned by match() function
if(status == 1)
cout<<"true"< else
cout<<"false"<
}

Sample Input/Output is attached