Question

the table below shows the water level (in inches) of a reservoir for three months compared to the yearly average. is the water level for the three month period greater than, or less than the yearly average? explain.

june	july	august

Explanation:

Step1: Find the sum of the three - month levels

We need to add the water - level values for the three months. The values are \(1\frac{1}{4}\), \(- 2\frac{1}{2}\), and \(- 1\frac{1}{8}\).
First, convert the mixed numbers to improper fractions:
\(1\frac{1}{4}=\frac{1\times4 + 1}{4}=\frac{5}{4}\), \(-2\frac{1}{2}=-\frac{2\times2+1}{2}=-\frac{5}{2}\), \(-1\frac{1}{8}=-\frac{1\times8 + 1}{8}=-\frac{9}{8}\).
The common denominator of 4, 2, and 8 is 8.
\(\frac{5}{4}=\frac{5\times2}{4\times2}=\frac{10}{8}\), \(-\frac{5}{2}=-\frac{5\times4}{2\times4}=-\frac{20}{8}\).
Now, add the fractions: \(\frac{10}{8}-\frac{20}{8}-\frac{9}{8}=\frac{10-(20 + 9)}{8}=\frac{10 - 29}{8}=-\frac{19}{8}=-2\frac{3}{8}\).

Step2: Calculate the three - month average

The formula for the average of \(n\) numbers \(x_1,x_2,\cdots,x_n\) is \(\bar{x}=\frac{x_1 + x_2+\cdots+x_n}{n}\). Here \(n = 3\).
The sum of the three - month levels is \(-\frac{19}{8}\), so the average is \(\frac{-\frac{19}{8}}{3}=-\frac{19}{8}\times\frac{1}{3}=-\frac{19}{24}\).

Step3: Compare with the yearly average

Let the yearly average be \(y\). Assume the problem is to compare the three - month average with the yearly average. If the yearly average is \(0\) (a common reference point if not given otherwise).
Since \(-\frac{19}{24}\approx - 0.79\) and \(0>-\frac{19}{24}\).

Answer:

The three - month average is less than the yearly average.