• Dec 11th 2005, 08:38 PM
Panaboy
help 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, 10:23 PM
MathGuru
Quote:

Originally Posted by Panaboy
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.

Carlos' final score will be calculated as follows:

(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, 02:34 PM
DenMac21
Quote:

Originally Posted by MathGuru
Carlos' final score will be calculated as follows:

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

so you must solve for x, the equation:

(83 + 74 + 81 + 85 + 3x)/7 = 75

Why dividing with 7 when there are 5 tests totally?

I think it should be
(83 + 74 + 81 + 85 + 3x) / 5 >= 75
• Jan 13th 2006, 02:50 PM
Jameson
Quote:

Originally Posted by DenMac21
Why dividing with 7 when there are 5 tests totally?

I think it should be
(83 + 74 + 81 + 85 + 3x) / 5 >= 75

I see your point, but you do need to divide by 7. Look at this easy example. Let's say you have two tests and you made an 80 on one and a 90 on the other. Let's say the second test, the one you made a 90 on, though is worth two test grades. Here's how you would calculate your average.

$\frac{(80+2*90)}{n}$, 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 $\frac{(80+90+90)}{2}=130$ Oops! That can't be possible. Now let's say you divide it by three. $\frac{(80+90+90)}{3}=86.67$.

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, 04:19 PM
DenMac21
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, 04:54 PM
dud
Weighted mean, isn't it?
Wikipedia link on weighted mean
http://en.wikipedia.org/math/0/c/4/0...85e9b688b3.png

Following this:
$
\frac{83 \times 1 + 74 \times 1 + 81 \times 1 + 85 \times 1 + x \times 3}{1 + 1 + 1 + 1 + 3}
$

Doing the math...
$
\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 14th 2006, 12:50 AM
CaptainBlack
Do you mean: $\bar x$?

RonL
• Jan 14th 2006, 01:26 AM
dud
Excellent, thanks!