This is a weighted mean, with the hourly tests assigned a weight of 1, the midterm assigned a weight of 2, and the final assigned a weight of 3. The total of the weights will be

 $\displaystyle 5 (1) + 2 + 3 = 10$.

If we let $\displaystyle N$ be Sandra's final exam score, Sandra's final weighted average will be

$\displaystyle \small \frac{84 + 86 + 76+ 89+ 93 + 72 \cdot 2 + N \cdot 3}{10}$

$\displaystyle \small = \frac{84 + 86 + 76+ 89+ 93 + 144 + 3N}{10}$

$\displaystyle \small = \frac{572 + 3N}{10}$

For Sandra to get a final average of 80, then we set the above equal to 80 and calculate $\displaystyle N$:

$\displaystyle \small \frac{572 + 3N}{10} = 80$

$\displaystyle \small 572 + 3N = 800$

$\displaystyle \small 3N = 228$

$\displaystyle \small N = 76$

$\displaystyle mean=\frac{85+87+87+82+89}{5}=86$

Reorder the values in numerical order:82, 85, 87, 87, 89

The median is the center number, 87.

The mean of a sample data set is the sum of all of the values divided by the total number of values. In this case:

$\displaystyle \frac{28+30+31+35+41}{5}=\frac{165}{5}=33$

$\displaystyle (75)(6)=450$

Therefore, the sum of all 6 digits must equal 450.

$\displaystyle 450=80+78+78+70+71+x$

$\displaystyle 450=377+x$

Subtract 377 from both sides.

$\displaystyle x=73$

The worst-case scenario is that she will score 65 or less, in which case her score will be the mean of the scores she has already achieved.

$\displaystyle \frac{76+ 84+80+65+ 91}{5} = 79.2$

The best-case scenario is that she will score 100, in which case the 65 will be dropped and her score will be the mean of the other five scores.

$\displaystyle \frac{76+ 84+80+ 91+100}{5} = 86.2$

Add the numbers and divide by 6:

$\displaystyle \left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$\displaystyle = 1.11111\div 6 = 0.185185$

The mean of $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, $\displaystyle d$, and $\displaystyle e$ is $\displaystyle \frac{1}{5} \left (a + b + c + d + e \right )$

If you add both sides of each equation:

$\displaystyle a + 2b + 3c + 4d + 5e = 3,400$

$\displaystyle \underline{5a + 4b + 3c + 2d + e = 5,300}$

$\displaystyle 6a + 6b + 6c + 6d + 6e = 8,700$

or 

$\displaystyle 6\left (a + b + c + d + e \right )= 8,700$

Equivalently,

$\displaystyle 6\left (a + b + c + d + e \right ) \div 6= 8,700 \div 6$

$\displaystyle a + b + c + d + e = 1,450$

$\displaystyle \frac{1}{5} \left (a + b + c + d + e \right )= \frac{1}{5} \cdot 1,450 = 290$,

making 290 the mean.

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\displaystyle \left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

$\displaystyle = (a + a + a + a + 2a + 2a + 2a + 2a) + (-x +x+2x-x +x+2x)$

$\displaystyle = 12a + 4x$

Divide this by 8:

$\displaystyle \left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

Let $\displaystyle X$ be her final grade. Julie's final score is calculated as a weighted mean, so we can set up the following inequality:

$\displaystyle \frac{85 + 84 + 74 + 0.5 \cdot90 + 1.5 \cdot 72 + 2 \cdot X}{1 + 1 + 1 + 0.5 + 1.5 + 2} \geq 80$

Simplify and solve for $\displaystyle X$:

$\displaystyle \frac{85 + 84 + 74 + 45+ 108 + 2 \cdot X}{7} \geq 80$

$\displaystyle \frac{396 + 2 X}{7} \geq 80$

$\displaystyle \frac{396 + 2 X}{7} \cdot 7 \geq 80 \cdot 7$

$\displaystyle 396 + 2 X \geq 560$

$\displaystyle 396 + 2 X -396 \geq 560 -396$

$\displaystyle 2 X \geq 164$

$\displaystyle 2 X \div 2 \geq 164 \div 2$

$\displaystyle X \geq 82$

Julie must make a minimum of 82% on the final to meet her goal.