The output needs to look like the box (picture below) Thanks 1. In this problem

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The output needs to look like the box (picture below) Thanks 1. In this problem we explore a Monte Carlo approach to approximating the value of pi. The area of a circle of radius r 1 is Adrie-Tr2- whereas the area of the square inscribing it is Aaquare-( 2-4. We then have Acirde Anquare4 Now, consider the following process: * Generate a random x-coordinate between-1 and i Generate a random y-coordinate between -1 and 1 Calculate the length of vector from (0,0) to the random point (,) Use a Boolean expresion to create a logical variable that stores whether or not the random point is inside the circle of radius 1 centered at the origin . Add the resulting logical variable to a running sum variable A Monte Carlo approach would perform the above process for thousands of trials. We can approximate Aralio by mumber of times random point lies inside the circle number of trials Write a script that performs this Monte Carlo approach to approximating pi but with only 5 trials. Feel free to write some code and just repeat it 5 times in your script. Sample output:

Explanation / Answer

% Matlab code

% Comment if u have any doubt

% like the answer,if u liked it ;-)

min = -1;

max = 1;

Inside_Circle = 0;

for i = 0:4

x = min + ((max-min).*rand(1));

y = min + ((max-min).*rand(1));

d = sqrt((x - 0)^2 + (y - 0)^2);

if((x < 1 && x > -1) && (y < 1 && y > -1))

inside = 1;

else

inside = 0;

end

Inside_Circle = Inside_Circle + inside;

fprintf("Trial %d : x = %.2f, y = %.2f, length = %.2f, This point in circle ? %d, Total in circle = %d ",(i+1),x,y,d,inside,Inside_Circle);

end

ratio = Inside_Circle/(i+1);

pi_approx = 4 * ratio;

fprintf("An approximation of pi is %.4f ",pi_approx);

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