The average is found by calculating the total payroll and then dividing by the total number of employees. \(\displaystyle \frac{(1\cdot 25,000)+(4\cdot 40,000)+(2\cdot 50,000)+(5\cdot 75,000)}{1+4+2+5}\)

\(\displaystyle \frac{25,000+160,000+100,000+375,000}{12} = \frac{660,000}{12}= $55,000\)

For the first 5 games the bowler has averaged 215. The equation to calculate the answer is

\(\displaystyle \frac{(215\cdot 5)+x}{6}=220\)

where \(\displaystyle \dpi{100} \small x\) is the score for the sixth game. Next, to solve for the score for the 6th game \(\displaystyle \dpi{100} \small (x)\) multiply both sides by 6:

\(\displaystyle (215\cdot 5)+x =1,320\)

which simplifies to:

\(\displaystyle 1,075+x =1,320\)

After subtracting 1,075 from each side we reach the answer:

\(\displaystyle x =1,320 - 1, 075 = 245\)

We can't just average 87 and 93! This will give the wrong answer! The average formula is \(\displaystyle \dpi{100} \small average = \frac{sum}{number\ of\ terms}\).

For the first 5 tests, \(\displaystyle \dpi{100} \small 87=\frac{sum}{5}\). Then \(\displaystyle \dpi{100} \small sum=87\times 5=435\).

Now combine that with the 6th test to find the overall average.

\(\displaystyle \dpi{100} \small average = \frac{435+93}{6}=88\)

\(\displaystyle \dpi{100} \small average = \frac{3\times 3000 + 4000 + 2\times 5200}{6} = \$ 3900\)

\(\displaystyle average = \frac{39+18+24+51+40+15+23}{7} = 30\)

The average of 10, 25, and 70 is 35: \(\displaystyle \frac{10+25+70}{3}=35\)

So the average of 15, 30, and the unknown number is 25 or, 10 less than the average of 10, 25, and 70 (= 35)

so \(\displaystyle \frac{15+30+x}{3}=25\)

\(\displaystyle 15+30+x=75\)

\(\displaystyle 45+x=75\)

\(\displaystyle x=30\)

\(\displaystyle average = \frac{sum}{terms} = \frac{2x + 3x + 2 + 7x + 4}{3} = \frac{12x + 6}{3} = 4x +2\)

\(\displaystyle average = \frac{sum}{days}\)

\(\displaystyle 85=\frac{89+76+92+90+80+84+x}{7}=\frac{511+x}{7}\)

\(\displaystyle 595=511+x\)

\(\displaystyle x=84\)

\(\displaystyle avg=\frac{98+64+82+90+70+88}{6}=\frac{492}{6}=82\)

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 \(\displaystyle N\):

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

\(\displaystyle \small 572 + 3N = 800\)

\(\displaystyle \small 3N = 228\)

\(\displaystyle \small N = 76\)