Basic Math Examples

$\frac{2}{3}$ , $\frac{4}{9}$ , $\frac{2}{9}$

Step 1

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

$\frac{\frac{2}{3} + \frac{4}{9} + \frac{2}{9}}{3}$

Step 2

Simplify the numerator.

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Step 2.1

To write $\frac{2}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{\frac{2}{3} \cdot \frac{3}{3} + \frac{4}{9} + \frac{2}{9}}{3}$

Step 2.2

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

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Step 2.2.1

Multiply $\frac{2}{3}$ by $\frac{3}{3}$ .

$\frac{\frac{2 \cdot 3}{3 \cdot 3} + \frac{4}{9} + \frac{2}{9}}{3}$

Step 2.2.2

Multiply $3$ by $3$ .

$\frac{\frac{2 \cdot 3}{9} + \frac{4}{9} + \frac{2}{9}}{3}$

Step 2.3

Combine the numerators over the common denominator.

$\frac{\frac{2 \cdot 3 + 4}{9} + \frac{2}{9}}{3}$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Multiply $2$ by $3$ .

$\frac{\frac{6 + 4}{9} + \frac{2}{9}}{3}$

Step 2.4.2

Add $6$ and $4$ .

$\frac{\frac{10}{9} + \frac{2}{9}}{3}$

Step 2.5

Combine the numerators over the common denominator.

$\frac{\frac{10 + 2}{9}}{3}$

Step 2.6

Add $10$ and $2$ .

$\frac{\frac{12}{9}}{3}$

Step 2.7

Cancel the common factor of $12$ and $9$ .

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Step 2.7.1

Factor $3$ out of $12$ .

$\frac{\frac{3 (4)}{9}}{3}$

Step 2.7.2

Cancel the common factors.

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Step 2.7.2.1

Factor $3$ out of $9$ .

$\frac{\frac{3 \cdot 4}{3 \cdot 3}}{3}$

Step 2.7.2.2

Cancel the common factor.

$\frac{\frac{3 \cdot 4}{3 \cdot 3}}{3}$

Step 2.7.2.3

Rewrite the expression.

$\frac{\frac{4}{3}}{3}$

Step 3

Multiply the numerator by the reciprocal of the denominator.

$\frac{4}{3} \cdot \frac{1}{3}$

Step 4

Multiply $\frac{4}{3} \cdot \frac{1}{3}$ .

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Step 4.1

Multiply $\frac{4}{3}$ by $\frac{1}{3}$ .

$\frac{4}{3 \cdot 3}$

Step 4.2

Multiply $3$ by $3$ .

$\frac{4}{9}$

Step 5

Divide.

$0.\overline{4}$

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.

$0.4$