The mean is the average of all of the numbers:

\(\displaystyle \frac{{1+2+8+9+7+4+1+1+3+2+3}}{11}\)

\(\displaystyle =\frac{{41}}{11}\)

\(\displaystyle =3.72\)

Add the ten elements and divide the sum by 10. The mean is 

\(\displaystyle \left [ 1+ \left ( -2 \right )+3 +\left ( -4 \right )+5+ \left ( -6 \right )+7+ \left ( -8 \right )+9+\left ( -10 \right ) \right ] \div 10\)

\(\displaystyle = -5 \div 10 = -0.5\)

Ellis and Jones gave Sarah the highest and lowest scores, respectively, so her score will be the mean of the other five:

\(\displaystyle \left (9.3+ 9.0+9.2+9.2+8.7 \right )\div 5 =\left (45.4\right )\div 5 = 9.08\)

The formula for the arithmetic mean is:

Mean= \(\displaystyle \frac{x_{1}+x_{2}+...+x_{n}}{n}\)

We can then write:

\(\displaystyle \frac{n+13+27+35}{4}=\frac{20+16+41+23}{4}\)

\(\displaystyle n+13+27+35=20+16+41+23\)

\(\displaystyle n+75=100\)

\(\displaystyle n=25\)

\(\displaystyle \frac{a+b+c+d}{4} = 250\)

\(\displaystyle a+b+c+d = 1,000\)

\(\displaystyle a+b+c+d +e = 1,000 + e\)

\(\displaystyle \frac{a+b+c+d +e }{5}= \frac{1,000 + e}{5} = 200 + \frac{1}{5}e\),

which is the correct choice.

We are looking for the grade Jane must get on her 6th test in order to raise her average of 65 from the 5 previous tests to 72.

The sum of Jane's grades from the previous 5 tests is obtained by using the formula of the arithmetic mean.

mean = sum of values/ number of values

sum of values = mean x number of values

                      \(\displaystyle =65\times5=325\)

Let x be Jane's score on the 6th test:

\(\displaystyle \frac{325+x}{6}=72\)

\(\displaystyle 325+x=432\)

\(\displaystyle x=432-325\)

\(\displaystyle x=107\)

Jane must get a score of 107 on her last test in order to get an average score of 72 in her class.

To find the median, write the numbers in ascending orders. 

\(\displaystyle n-8, n-3, n, n+7, n+9\)

Remember the median is the midpoint value with half of the values below it and the other half of the values above it.

The median is the third number which is n.

The mean is:

\(\displaystyle \frac{(n-8)+(n-3)+n+(n+7)+(n+9)}{5}=\frac{5n-11+16}{5}\)

                                                                                           \(\displaystyle =\frac{5n+5}{5}\)

                                                                                           \(\displaystyle =n+1\)

mean - median = \(\displaystyle n+1-n=1\)

\(\displaystyle \frac{a+b+c+d}{4} = 100\)

\(\displaystyle a+b+c+d = 400\)

\(\displaystyle a+b+c = 400 - d\)

\(\displaystyle \frac{a+b+c }{3}= \frac{400 - d}{3} = \frac{400}{3} - \frac{d}{3} = 133\frac{1}{3} - \frac{1}{3}d\)

The arithmetic mean of \(\displaystyle \left \{ a, b, c \right \}\) is equal to \(\displaystyle 133\frac{1}{3} - \frac{1}{3}d\)

We are told that:

\(\displaystyle \frac{a+b+c+d+e+f+100 }{7} = 250\)

Multiplying each side by \(\displaystyle 7\):

\(\displaystyle \frac{a+b+c+d+e+f+100 }{7} \cdot 7 = 250 \cdot 7\)

\(\displaystyle a+b+c+d+e+f+100 =1,750\)

Subtracting \(\displaystyle 100\) from each side:

\(\displaystyle a+b+c+d+e+f =1,650\)

At this point, we can solve for the arithmetic mean of the set \(\displaystyle \left \{ a, b, c, d, e, f\right \}\):

\(\displaystyle \frac{a+b+c+d+e+f }{6}=\frac{1,650}{6} = 275\), the arithmetic mean of the set \(\displaystyle \left \{ a, b, c, d, e, f\right \}\).

Let \(\displaystyle x\) be the number of croissants sold on Sunday. The daily average of croissants sold in the past week is:

\(\displaystyle \frac{29+15+12+10+34+38+x}{7}=25\)

Solve for \(\displaystyle x\):

\(\displaystyle 29+15+12+10+34+38+x=175\)

\(\displaystyle 138+x=175\)

\(\displaystyle x=175-138=37\)

The number of croissants sold on Sunday is \(\displaystyle 37\).