Answer:
// program in C++.
// headres
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// array
int temperatures[7];
// count variable
int count=0;
cout<<"Enter the temperature of all days:";
for(int a=0;a<7;a++)
{
// read temperature of 7 days
cin>>temperatures[a];
// find temperature is extreme or not
if(temperatures[a]<-10||temperatures[a]>25)
// count
count++;
}
// print count of extreme temperature
cout<<"number of days of extreme temperature:"<<count<<endl;
return 0;
}
Explanation:
Create an array of size 7 to store the temperature of all days of week.Read the temperature of each day.If the temperature is less than -10 or greater than 25 then increment the count.This will count the number of days of extreme temperature.Print the count.
Output:
Enter the temperature of all days:-20 12 18 30 32 -15 15
number of days of extreme temperature:4
OPTION A would be the answer
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
<h2>Upgrading and retraining are mandatory to move along with the world.</h2>
Explanation:
Let us understand the term deeply,
Upgrading - Updating yourself with the latest
Retraining - learning new skills
Let me give you a real-life example which is nothing but "mobiles". If you are not updated then:
- you will sit with mobile to make calls and
- do money transactions only by stepping into the bank,
- connect with people only through calls or directly visiting them,
- distance break up the relationship,
- booking tickets in classical way, etc.
These could be done in one touch if you have latest mobile with necessary applications.
In a similar way, we need to get retrained to get to learn new skills, technologies so that we can do our job the best, to be on track, be productive, convert your valuable knowledge in terms of money, to be peaceful in day today transactions, etc.
Answer:
#include <iostream>
#include <vector>
using namespace std;
void calGPA();
vector<int> g;
vector<int> h;
int main(){
char pushMore = 'y';
int fg, fh;
for (;;){
if (pushMore == 'n'){
break;
} else{
cout<< "Enter integer for grade: ";
cin>> fg;
cout<< "Enter integer for credit hours: ";
cin>> fh;
g.push_back(fg);
h.push_back(fh);
cout<< "Do you want to add more grade and credit hours? y/n: ";
cin>> pushMore;
}
}
calGPA();
}
void calGPA(){
double total = 0, GPA;
for (int i = 0; i < g.size(); ++i) {
total += g.at(i) * h.at(i) ;
}
cout<< "The GPA is : "<< total/g.size();
}
Explanation:
The C++ source code above defines two vectors 'g' and 'h'. The latter holds the grades of a student while the former holds the credit hours for the subject taken. The void 'calGPA' function calculates and prints the student's GPA.
