Answer:
Both compiled and interpreted languages are high-level languages and translate code for a computer to understand.
Explanation:
The one similarity between compiled and interpreted languages is that they are both high-level languages.
A high-level language is a computer language written in easy to understand human language which is then converted to machine code for the computer to understand.
A high-level language can either be interpreted or compiled.
An interpreted is a language in which the code is translated line by line before execution while a compiled language is one in which the source code is converted directly into machine language before execution.
So, <u>the similarity between both languages is that they are high level languages and translate code for a computer to understand. </u>
The pseudocode to calculate the average of the test scores until the user enters a negative input serves as a prototype of the actual program
<h3>The errors in the pseudocode</h3>
The errors in the pseudocode include:
- Inclusion of unusable segments
- Incorrect variables
- Incorrect loops
<h3>The correct pseudocode</h3>
The correct pseudocode where all errors are corrected and the unusable segments are removed is as follows:
start
Declarations
num test1
num test2
num test3
num average
output "Enter score for test 1 or a negative number to quit"
input test1
while test1 >= 0
output "Enter score for test 2"
input test2
output "Enter score for test 3"
input test3
average = (test1 + test2 + test3) / 3
output "Average is ", average
output "Enter score for test 1 or a negative number to quit"
input test1
endwhile
output "End of program"
stop
Read more about pseudocodes at:
brainly.com/question/11623795
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
True. At least that's how it is for camera's that print photos. Not digital cameras
Answer:most of the computer campanies use java script, phyton, c++ or c sharp .
Explanation:
