Question

Write a script that prompts the user to enter a value n (n > 6) that will generate an n element vector of random integers between 0 and 100 that will represent quiz grades. Then have the script calculate the average of the quizzes where the smallest grade is dropped. Then script will display the average of the quizzes along with the highest quiz grade. Given the following system of equations: 8s - 11t + 7u + 5v - w + 2x + 3y + 4z = 115 3s - 8t + 4u - 9v + 2w + 9x + y - z = 20 3t - 6u - 13w - 2x - 5z = - 38 w + 14x + y + z = - 20 19s-6t - w + 3y = 75 -12s - 5t - 3u + 17v - 5w + 7z = 44 23u - 15v + 7w + 14y + z = 108 2s - 3t - 4w + 5y = 40 Solve these equations using both left division, right division, and multiplying matrices using the inverse of a matrix by using both the inverse function and raising the matrix to the -1 power. Use MATLAB to show that the sum of the infinite series sigma^infinity_n = 0 1/(2n + 1)(2n + 2) converges to In 2. Do this for computing the sum for a) n = 50 b) n = 500 c) n = 5000 For each case compute the percent error using the equation:

Answer 1

1)% for quiz grades.

clear all;
clc

prompt = 'enter n value ';
n = input(prompt);
quiz_grades = randi([0,100],1,n)    %random value of number b/w 0 -100
minimum = min(quiz_grades);         % finding minimum of values

average = mean(quiz_grades(quiz_grades~=minimum))       %finding mean value leaving minimum value
maximum = max(quiz_grades)             %finding maximum value
   

enter n value 5 quiz_grades 76 27 68 66 16 average 59.2500 maximum 76

%2) to find left right division

A = [8,-11,7,5,-1,2,3,4;
    3,-8,4,-9,2,9,1,-1;
    0,3,-6,0,-13,-2,0,-5;
    0,0,0,0,1,14,1,1;
    19,-6,0,0,-1,0,3,0;
    -12,-5,-3,17,-5,0,0,7;
    0,0,23,-15,7,0,14,1;
    2,-3,0,0,-4,0,5,0]
B = [115; 20;-38;-20; 75; 44; 108; 40]
left= A\B
linv = inv(A)*B          %left inverse
lraised = A^-1 *B      %raised to power -1
right = B/A
rinv = A*inv(B)
rraised = A *B^-1

3.0000 1.0000 -1, 0000 -2.0000 4.0000 5.0000 2.0000 -4.0000 3.0000 1.0000 -1, 0000 -2.0000 4.0000 5.0000 2.0000 -4.0000

%3) to find the summation of infinite series

clear all;
clc

prompt = 'enter n value ';
n = input(prompt);

L =log(2)

for i=1:n
    P =(1/((2*n+1)*(2*n+2)));
   
end

S= sum(P)

per_err = ((L - S)/L)*100

enter n value 50 0.6931 9.7069e-005 pererr = - 99.9860

enter n value 500 0.6931 9.9701e-007 pererr = - 99.9999 fx >>

enter n value 5000 0.6931 9.9970e-009 pererr = - 100.0000