To find the mean, first sum up all the values.

$\displaystyle \small \sum \left \{ 70,89,67,77,92,83,67,75\right \}=620$

The divide the result by the number of values

$\displaystyle \small 620\div8=77.5$

The mean of six numbers is their sum divided by 6, so the sum is the mean multiplied by 6. This is:

$\displaystyle 77 \times 6 = 462$

Sally's grade is a weighted mean in which her hourly tests have weight 1, her midterm has weight 2, and her final has weight 3. If we call $\displaystyle x$ her score on the final, then her course score will be

$\displaystyle \frac{89 + 85 + 84 + 87 + 2 \cdot 93 + 3x}{1 + 1 + 1 + 1 + 2 + 3}$,

which simplifies to 

$\displaystyle \frac{89 + 85 + 84 + 87 + 186 + 3x}{9} = \frac{3x + 531}{9}$.

Since Sally wants at least a 90 average for the term, we can set up and solve the inequality:

$\displaystyle \frac{3x + 531}{9} \geq 90$ 

$\displaystyle \frac{3x + 531}{9} \cdot 9 \geq 90 \cdot 9$

$\displaystyle 3x + 531 \geq 810$

$\displaystyle 3x + 531-531 \geq 810-531$

$\displaystyle 3x \geq 279$

$\displaystyle 3x \div 3 \geq 279 \div 3$

$\displaystyle x \geq 93$

Sally must score at least 93 on the final.

If Fred does not take the sixth test or gets a 0 on it, he will receive the average of his first five tests. This is

$\displaystyle \frac{78 + 77 + 84+ 89 + 72}{5} =\frac{400}{5} = 80$.

Since he can only improve his class grade by taking the sixth test, Fred is already assured of an average of 80 or better.

Alice's grade is a weighted mean in which her hourly tests have weight 1 and her final has weight 2. If we call $\displaystyle x$ her final, then her term score will be 

$\displaystyle \frac{76 + 85 + 82 + 87 + 2x}{1 + 1 + 1 + 1 + 2}$,

which simplifies to 

$\displaystyle \frac{2x+330 }{6}$.

Since she wants her score to be 90 or better, we solve this inequality:

$\displaystyle \frac{2x+330 }{6}\geq 90$

$\displaystyle \frac{2x+330 }{6} \cdot 6\geq 90\cdot 6$

$\displaystyle 2x+330\geq540$

$\displaystyle 2x+330 -330\geq540-330$

$\displaystyle 2x\geq 210$

$\displaystyle 2x\div 2 \geq 210 \div 2$

$\displaystyle x\geq 105$

Since it is established that 100 is the maximum score, Alice cannot achieve an average of 90 or more for the term.

The first step here, since you're working with the average (or mean), is to determine what is the sum of five test scores that will result in Eric earning at least a mean score of 90.

Start with the fact that he will need to make at least a 90 for his average score. Since you already have the average score, you're working backwards. You need to multiply 90 by 5 for the 5 test scores that will be used to get that average:

$\displaystyle 90\times 5=450$

This tells you that the five scores - the four test scores and the final exam - will need to add up to at least 450 in total.

You know the four original test scores, so you can add those up:

$\displaystyle 94+87+95+89=365$

So you know the four test scores add up to 365. To get an average of at least 90, the fifth score must bring that sum to at least 450. To find the minimum final exam test score, you subtract 365 from 450:

$\displaystyle 450-365=85$

This tells you that Eric must score at least an 85 on the final exam to achieve an overall mean score of 90.

To calculate the average, find the sum of the total hours divided by the number of nights:

$\displaystyle avg=\frac{3.2+1.1+2.7+0.9}{4}=\frac{7.9}{4}=1.975$

The mean is determined by dividing the sum of the values by the total number of values:

$\displaystyle mean=\frac{17+14+22+11+14+16+n}{7}=15$

$\displaystyle 17+14+22+11+14+16+n=105$

$\displaystyle 94+n=105$

$\displaystyle n=11$

The first ten primes form the data set:

$\displaystyle \left \{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29\right \}$

Add them, and divide by :

$\displaystyle \left ( 2 + 3+5+ 7+ 11+13+17+19+ 23+29 \right )\div 10 = 129\div 10= 12.9$