Feedback on the changes made to "college algebra" 8/19

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I'm trying to understand what exactly is the proctor fee... Are they really asking students to pay $120 to be able to sit 3 exams!!!! And why should "I"pay someone to watch me take a ###% exam. I am so over this school...the school should pay for it!

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The proctoring service is to make sure there is no cheating on an exam. The fee for this varies by person and/or location. If the fee is too steep for you, look around for the next nearest location until you find an acceptable option. In my opinion, $120 is a bit much but when I think about the alternative, which is giving up a major portion of my life to sit in a classroom for 15 weeks per course, I find a way to work out the fee. The people who proctor the exam have to be paid by someone and in an online course, it would be kind of difficult for a college or university to arrange and provide the varying compensation for proctors all over the world.

I just recently took a proctored exam and would have loved to have taken it at my job since they offer that service to any employee in need of it for free. But the problem is, they are only partnered with local universities and are not registered nationally, and the course exam that I needed was from a university in another state. So, I had to go with the testing center in the university closest to my home, that happened to be registered nationally and verifiable by the university in Utah, whom I chose for the particular course. It cost me $30 but that was ok because I would have had to have made sure I was working and not being cancelled in order for my job to do it because that's a 3 1/2 hour drive in one direction....far more expensive than $30. Even with the additional fee, which I was aware of before I registered for the course, I still saved more than $700 since I didn't have to take the class that was offered by the college that I'm currently enrolled in.

Keep looking. Surely there are other places that proctor exams for far less than $120. If this is the total amount for the 3 exams that you mentioned, I'd say that sounds about right.

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