Finite Math Examples

$9$ , $3$ , $1$ , $\frac{1}{3}$ , $\frac{1}{9}$

Step 1

The mean of a set of numbers is the sum divided by the number of terms.

$\frac{9 + 3 + 1 + \frac{1}{3} + \frac{1}{9}}{5}$

Step 2

Simplify the numerator.

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Step 2.1

Add $9$ and $3$ .

$\frac{12 + 1 + \frac{1}{3} + \frac{1}{9}}{5}$

Step 2.2

Add $12$ and $1$ .

$\frac{13 + \frac{1}{3} + \frac{1}{9}}{5}$

Step 2.3

To write $13$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{13 \cdot \frac{3}{3} + \frac{1}{3} + \frac{1}{9}}{5}$

Step 2.4

Combine $13$ and $\frac{3}{3}$ .

$\frac{\frac{13 \cdot 3}{3} + \frac{1}{3} + \frac{1}{9}}{5}$

Step 2.5

Combine the numerators over the common denominator.

$\frac{\frac{13 \cdot 3 + 1}{3} + \frac{1}{9}}{5}$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $13$ by $3$ .

$\frac{\frac{39 + 1}{3} + \frac{1}{9}}{5}$

Step 2.6.2

Add $39$ and $1$ .

$\frac{\frac{40}{3} + \frac{1}{9}}{5}$

Step 2.7

To write $\frac{40}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{\frac{40}{3} \cdot \frac{3}{3} + \frac{1}{9}}{5}$

Step 2.8

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.8.1

Multiply $\frac{40}{3}$ by $\frac{3}{3}$ .

$\frac{\frac{40 \cdot 3}{3 \cdot 3} + \frac{1}{9}}{5}$

Step 2.8.2

Multiply $3$ by $3$ .

$\frac{\frac{40 \cdot 3}{9} + \frac{1}{9}}{5}$

Step 2.9

Combine the numerators over the common denominator.

$\frac{\frac{40 \cdot 3 + 1}{9}}{5}$

Step 2.10

Simplify the numerator.

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Step 2.10.1

Multiply $40$ by $3$ .

$\frac{\frac{120 + 1}{9}}{5}$

Step 2.10.2

Add $120$ and $1$ .

$\frac{\frac{121}{9}}{5}$

Step 3

Multiply the numerator by the reciprocal of the denominator.

$\frac{121}{9} \cdot \frac{1}{5}$

Step 4

Multiply $\frac{121}{9} \cdot \frac{1}{5}$ .

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Step 4.1

Multiply $\frac{121}{9}$ by $\frac{1}{5}$ .

$\frac{121}{9 \cdot 5}$

Step 4.2

Multiply $9$ by $5$ .

$\frac{121}{45}$

Step 5

Divide.

$2.6\overline{8}$

Step 6

The mean should be rounded to one more decimal place than the original data. If the original data were mixed, round to one decimal place more than the least precise.

$2.7$