L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
<h3>How do you know if a language is decidable?</h3>
A language is said to be decidable only when there seems to exists a Turing machine that is said to accepts it,
Here, it tends to halts on all inputs, and then it answers "Yes" on words that is seen in the language and says "No" on words that are not found in the language. The same scenario applies to recognizable language.
So, L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
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Hi! Did you forget to add something to this?
In this program, I am using the school-based grading system and the program should accept the subject and the number of students.
Program approach:-
- Using the necessary header file.
- Using the standard I/O namespace function.
- Define the main function.
- Declare the variable.
- Display enter obtain marks in 5 subjects.
- Return the value.
Program:-
//header file
#include<iostream>
//using namespace
using namespace std;
//main method
int main()
{
//declare variable
int j;
float mark, sum=0, a;
//display enter obtain marks in 5 subjects
cout<<"Enter Marks obtained in 5 Subjects: ";
for(j=0; j<5; j++)
{
cin>>mark;
sum = sum+mark;
}
a = sum/5;
//display grade
cout<<"\nGrade = ";
if(a>=91 && a<=100)
//display a1
cout<<"a1";
else if(a>=81 && a<91)
//display a2
cout<<"a2";
else if(a>=71 && a<81)
cout<<"b1";
else if(a>=61 && a<71)
cout<<"b2";
else if(a>=51 && a<61)
//display c1
cout<<"c1";
else if(a>=41 && a<51)
//display c2
cout<<"c2";
else if(a>=33 && a<41)
//display d
cout<<"d";
else if(a>=21 && a<33)
//display e1
cout<<"e1";
else if(a>=0 && a<21)
//display e2
cout<<"e2";
else
//display invalid
cout<<"Invalid!";
cout<<endl;
//return the value
return 0;
}
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The answer is : variables
When viewing data entered into a spreadsheet, the columns identify Variable. The variables later can be used on a formula to help you process any sort of data that is implemented within excels' formula system
Answer: A. Usability testing
Explanation:
Usability testing would be the correct answer because if the user interface isn't "frendly" then people won't be able to use it properly. Or if you are refering to the word friendly as looking safe, then even more so would be usability testing. Because if people don't like the fonts and styles of the user interface, they might not use it at all.
