add up the three grades then divide by three
It seems like there might be a misunderstanding in your question, but if you're asking about the average of the grades 83, 86, and 86, you can calculate it by adding the grades together (83 + 86 + 86 = 255) and then dividing by the number of grades (3). This gives you an average grade of 85. If you meant something else, please clarify!
It is not possible to answer the question since there is no information on how many grades there are. Second, the probability of a student getting a particular grade will depend on how "wide" the grade is: some grade boundaries are closer together than others. Furthermore, the grades achieved by the students are not independent events. If the exam is very easy, they might all get a top grade.
On average u should know them in first or second grade !
is level 3. or grades 1-3
You got 75% correct.Reason:5 is 1/4 of 20, therefore meaning that you got 3/4 correct. And of course 3/4 = .75, or 75%
In terms of letter grades, a 3. 48 grade point average falls between an A and a B. A 3. 5 GPA is a definite A.
you add up all the grade totals and divide that number by the number of grades. A=4 B=3 C=2 D=1 F=0 For example if your child received 3 A's and 4 B's you will calculate it like this: 4+4+4+3+3+3+3= 24 then you divide 24 by 7(number of grades) the average is a 3.4 which is the grade point average. =)
There are four main grades of titanium: Grade 1, Grade 2, Grade 3, and Grade 4. These grades vary in terms of their composition, strength, and properties, with Grade 1 being the most pure and Grade 4 the strongest. Each grade is suitable for different applications based on the specific requirements of strength, corrosion resistance, and biocompatibility.
It is around 77-79 percent. The grade would be B+ It's an 85 percent, B+
good grades, keep up GPA(grade point average), come to every practice
level 8 A* level 7 A/A* level 6 B level 5 C level 4 D level 3 E/F level 2 G/U This varies from subject to subject. These are the grades you should get if you reach the written level at the end of KS3.
In the simple case where all the graded courses carry equal weight, assign a numeric value to the letter grades, compute the mathematical average, and then convert the number back to a grade letter if desired. Generally these values are assigned to letter grades: A=4, B=3, C=2, D=1, and F=0. Substitute a number for every grade letter, add up the numbers, and divide by the number of grades/courses to get your numerical average (usally called your GPA, or grade-point average). Use the above scale to convert back to a letter if you wish. For example, if you have 6 courses and earned 2 As, 3Bs, and a C, you would add 2 4s, 3 3s, and a 2, and then divide by 6: (2X4 + 3X3 + 1X2)/6 = (8+9+2)/6 = 19/6 = about 3.2 (rounded), your GPA. 3 is a B, so your average letter grade is a strong B.
The three grades for canned fruits vegetables include: Grade A Grade B Grade C Grade A- desserts, salads (also known as 'Fancy') Grade B- processed (also known as 'Choice') Grade C- puddings, pies (also known as 'Standard')
The following example sets up a two-dimensional array, initialises it with some pseudo-random data, and then prints the table and the averages. #include<iostream> #include<time.h> int main() { const int max_students = 7; const int max_student_grades = 5; const int max_grades = 6; const char grade[max_grades]={'A','B','C','D','E','F'}; srand((unsigned) time(NULL)); // Initialise the array with pseudo-random grades: int table[max_students][max_student_grades]; for(int student=0; student<max_students; ++student) { for(int student_grade=0; student_grade<max_student_grades; ++student_grade) { table[student][student_grade] = rand()%max_grades; } } // Print the table and average the results. int overall=0; for(int student=0; student<max_students; ++student) { int average=0; std::cout<<"Student #"<<student+1; for(int student_grade=0; student_grade<max_student_grades; ++student_grade) { std::cout<<" Grade #"<<student_grade+1<<": "<<grade[table[student][student_grade]]<<", "; average+=table[student][student_grade]; } std::cout<<" Average: "<<grade[average/max_grades]<<std::endl; overall+=average; } std::cout<<"Overall average: "<<grade[overall/max_grades/max_students]<<std::endl; return(0); } Example output: Student #1 Grade #1: A, Grade #2: E, Grade #3: D, Grade #4: E, Grade #5: F, Average: C Student #2 Grade #1: E, Grade #2: D, Grade #3: E, Grade #4: E, Grade #5: E, Average: D Student #3 Grade #1: D, Grade #2: A, Grade #3: D, Grade #4: B, Grade #5: A, Average: B Student #4 Grade #1: C, Grade #2: B, Grade #3: A, Grade #4: A, Grade #5: B, Average: A Student #5 Grade #1: E, Grade #2: D, Grade #3: C, Grade #4: F, Grade #5: E, Average: D Student #6 Grade #1: C, Grade #2: D, Grade #3: A, Grade #4: F, Grade #5: A, Average: B Student #7 Grade #1: B, Grade #2: D, Grade #3: F, Grade #4: B, Grade #5: C, Average: C Overall average: C
Concussions are typically graded on a 3-point scale: Grade 1 = Mild, with temporary confusion and no loss of consciousness Grade 2 = Moderate, with confusion and potential loss of consciousness lasting less than 5 minutes Grade 3 = Severe, with loss of consciousness for more than 5 minutes
A 1953d is common in all grades, the lower grade coins are 3-5 cents, better grades 25 to 50 cents and uncirculated coins are 75 cents to $4.00 depending on the grade
You add the 4 numbers altogether and then divide the sum by 4. Example: 1. there are four numbers: 10, 20, 12, 10 2. Add the numbers: 10+20+12+10= 52 3. Divide the sum by the total amount of numbers which is 4: 52/4 Answer/Average: 13