Solve (45+45+48+46+50+46+47+43+47+48)div10=

$$(45+45+48+46+50+46+47+43+47+48) \div 10=$$

Evaluate

$$\frac{93}{2}=46.5$$

Solution Steps

Add 45 and 45 to get 90.

$$\frac{90+48+46+50+46+47+43+47+48}{10}$$

Add 90 and 48 to get 138.

$$\frac{138+46+50+46+47+43+47+48}{10}$$

Add 138 and 46 to get 184.

$$\frac{184+50+46+47+43+47+48}{10}$$

Add 184 and 50 to get 234.

$$\frac{234+46+47+43+47+48}{10}$$

Add 234 and 46 to get 280.

$$\frac{280+47+43+47+48}{10}$$

Add 280 and 47 to get 327.

$$\frac{327+43+47+48}{10}$$

Add 327 and 43 to get 370.

$$\frac{370+47+48}{10}$$

Add 370 and 47 to get 417.

$$\frac{417+48}{10}$$

Add 417 and 48 to get 465.

$$\frac{465}{10}$$

Reduce the fraction \frac{465}{10} to lowest terms by extracting and canceling out 5.

$$\frac{93}{2}$$

Factor

$$\frac{3 \cdot 31}{2} = 46\frac{1}{2} = 46.5$$