Carlos has scores 83,74,81,and 85 on his past test.Find the minimum score he can make on the final exam to pass the course with average of 75 or higher, given that the final exam counts as 3 tests.

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- Dec 11th 2005, 07:38 PMPanaboyhelp with this ..
Carlos has scores 83,74,81,and 85 on his past test.Find the minimum score he can make on the final exam to pass the course with average of 75 or higher, given that the final exam counts as 3 tests.

- Dec 11th 2005, 09:23 PMMathGuruQuote:

Originally Posted by**Panaboy**

(83 + 74 + 81 + 85 + x + x + x)/7

so you must solve for x, the equation:

(83 + 74 + 81 + 85 + 3x)/7 = 75 - Jan 13th 2006, 01:34 PMDenMac21Quote:

Originally Posted by**MathGuru**

I think it should be

(83 + 74 + 81 + 85 + 3x) / 5 >= 75 - Jan 13th 2006, 01:50 PMJamesonQuote:

Originally Posted by**DenMac21**

$\displaystyle \frac{(80+2*90)}{n}$, where n is the number of 100 point tests. Now according to your logic I should divide by two, since there were two tests. That would give $\displaystyle \frac{(80+90+90)}{2}=130$ Oops! That can't be possible. Now let's say you divide it by three. $\displaystyle \frac{(80+90+90)}{3}=86.67$.

In the OP's question you should treat the weighted test as three separate ones compared to the max point value of the others. - Jan 13th 2006, 03:19 PMDenMac21
Look at it from this perspective.

Say we don't know what is the maximum points that can be scored on each test. So, you don't know that on fifth (final) test the maximum points is bigger then on other tests.

Would you then divide with 7 or 5? - Jan 13th 2006, 03:54 PMdudWeighted mean, isn't it?
Wikipedia link on weighted mean

http://en.wikipedia.org/math/0/c/4/0...85e9b688b3.png

Following this:

$\displaystyle

\frac{83 \times 1 + 74 \times 1 + 81 \times 1 + 85 \times 1 + x \times 3}{1 + 1 + 1 + 1 + 3}

$

Doing the math...

$\displaystyle

\frac{323 + x}{7}

$

Am I right?

As a side note, does anyone know if LaTeX math has the "curly x symbol with a stroke over it" which should be infront of this answer? - Jan 13th 2006, 11:50 PMCaptainBlack
Do you mean: $\displaystyle \bar x$?

RonL - Jan 14th 2006, 12:26 AMdud
Excellent, thanks!