It appears the last column of the output from the problem statement should be 3.5; 0.0; 2.0 instead of 3.5; 0.0; 3.0, but the first test suite does have it correct.
yes true, thank you
Thank you! Modified the problem statement.
Why doesn't it work?
function X = rescale_scores(X)
X = [X(1:end,1:end-1) (X(1:end,end).-60)./4]
end
n = size(X,1);
for c = 1:n
X(c,end) = (X(c,end)-60)/10;
end
A little bit of math work will make your program much simpler. Grade 60-100 for GPA 0 - 4, this is a straight line! The equation for this straight line is GPA = 0.1*(grade-60). Don't forget to take care of negative GPAs.
X(:,end)=mean(rescale(X(:,end-1),-6,4,'InputMin',0,'InputMax',100),2);
Don't forget the negative numbers.
Score GPA
90 - 100 3 - 4
80 - 90 2 - 3
70 - 80 1 - 2
60 - 70 0 - 1
50 - 60 -1 - 0
40 - 50 -2 - -1
30 - 40 -3 - -2
20 - 30 -4 - -3
10 - 20 -5 - -4
0 - 10 -6 - -5
I am getting confused. IIs there a capital and small "x" in use. Getting errors.
Below 60 GPA=0, so why there could be any negative value?
worth to do
good problem
The problem states "Score less than 60: GPA = 0"
But the solution assigns negative values to scores under 60.
So the provided solutions are wrong!
Good question
A couple of if
Nice question. Checks our understanding.
Easy!
Plot the GPA vs score to get the equation.
Please, what is wrong with the following answer
X = (X-60)/10;
thank u
This would operate on the entire matrix X. However we want the GPA to appear only in the last column of X. So try by declaring another variable which would calculate the GPA and then replace the last column of X with this column vector.
We have to modify only last column of matrix X not an entire matrix so find GPA for that column only.
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