We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
500
7
avatar+11 

Sydney's grades were being averaged. Her grades were 83,63, and 100. What is her final grade that her teacher should put on her report card?  

 Apr 19, 2014
 #1
avatar+158 
0

Sydney's final grade should be 82 if you add all of them up you should get 246 then divide it by 3 you get 82

 Apr 19, 2014
 #2
avatar+103049 
0

Sydney's grades were being averaged. Her grades were 83,63, and 100. What is her final grade that her teacher should put on her report card?  

Add the grades and divide by 3......that will give you your answer   

 Apr 19, 2014
 #3
avatar+11 
0

That's what I thought, but my teacher said 70. Thanks y'all

 Apr 19, 2014
 #4
avatar+890 
0

You should ask your teacher if each piece of work carried the same weighting.

It could be that one or two of the assignments, (if that's what they were), were considered to be minor compared with the other (s) and therefore contributed less to your final grade.

 Apr 19, 2014
 #5
avatar+103049 
0

Bertie is correct.......this would have to be a weighted average problem if "70" is correct.

 Apr 20, 2014
 #6
avatar+103676 
0

I have a different take on this.

I think everyone has assumed that 83,63 and 100 are all percentages, or at least out of the same total.

What if they were not.   Then if they were all equal in weighting they would need to be converted to the same denominator.  Probably converted to a percentage.  Then they could be averaged to get the final score.

 Apr 20, 2014
 #7
avatar+103049 
0

Let's take another look at this one.... (If you want to)

Let's assume that we DO use a "weighted average" system to calculate the average

So we have something like this

[63x + 83y + 100z] / [x + y + z] = 70     where   x, y, z   are the "weights"

Multiplying by [x + y + z] on both sides gives us

63x + 83y + 100z = 70 ( x + y + z)   And after a little manipulation, we have

-7(x) + 13(y) + 30(z) = 0     Now, solving "for z," we get

z = (7x - 13y)/30

Note that, if I let x = 2 and y = 1, I get "1"  as a numerator and "z" = (1/30).......but, I need "z" to be an integer. Therefore, if I just mutiply everything in the nunerator by 30, then we get

z = 30 [7(2) - 13(1)]/ 30 = [7(60) -13(30)]/30 = [(420-390)/30] = (30/30) = 1

Then our "solution set" for (x, y, z) = (60, 30, 1)

And this is just a "3 space" vector. And any positive integer multiple of this vector could be could be a "solution" to the weighted values of x, y, and z.

So, the general soltion to this problem - if "70" is indeed the "answer" - is just given by: n(x, y, z) .....  where (x, y, z) = (60, 30, 1)  and n is a positive integer.

 Apr 20, 2014

30 Online Users

avatar