Basic Math Examples

$38$ , $600 + 6$ , $400 + 4$ , $600 + 3$ , $800 + 3$ , $100$

Step 1

Add $600$ and $6$ .

$38, 606, 400 + 4, 600 + 3, 800 + 3, 100$

Step 2

Add $400$ and $4$ .

$38, 606, 404, 600 + 3, 800 + 3, 100$

Step 3

Add $600$ and $3$ .

$38, 606, 404, 603, 800 + 3, 100$

Step 4

Add $800$ and $3$ .

$38, 606, 404, 603, 803, 100$

Step 5

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

$\frac{38 + 606 + 404 + 603 + 803 + 100}{6}$

Step 6

Simplify the numerator.

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Step 6.1

Add $38$ and $606$ .

$\frac{644 + 404 + 603 + 803 + 100}{6}$

Step 6.2

Add $644$ and $404$ .

$\frac{1048 + 603 + 803 + 100}{6}$

Step 6.3

Add $1048$ and $603$ .

$\frac{1651 + 803 + 100}{6}$

Step 6.4

Add $1651$ and $803$ .

$\frac{2454 + 100}{6}$

Step 6.5

Add $2454$ and $100$ .

$\frac{2554}{6}$

Step 7

Cancel the common factor of $2554$ and $6$ .

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Step 7.1

Factor $2$ out of $2554$ .

$\frac{2 (1277)}{6}$

Step 7.2

Cancel the common factors.

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Step 7.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 1277}{2 \cdot 3}$

Step 7.2.2

Cancel the common factor.

$\frac{2 \cdot 1277}{2 \cdot 3}$

Step 7.2.3

Rewrite the expression.

$\frac{1277}{3}$

Step 8

Divide.

$425.\overline{6}$

Step 9

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.

$425.7$