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Problem 9 (4 points) The following Matlab program exp (1) -1; %3D for i = 1:25 %3D a = i*a-1 end fprintf ('%e ', a);

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Problem 9 (4 points) The following Matlab program exp (1) -1; %3D for i = 1:25 %3D a = i*a-1 end fprintf ('%e ', a); outputs 4.645988e+09. -If I do a similar calculation in Excel produces -2.242373E+09. Explain why such numbers are obtained. If the above code is executed in infinite precision, what would be the output? are th

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