2-1.3 \( < \) Question \( 9,1.3 .19> \) The selling price of a refrigerator, is \( \$ 924.00 \). If the markup is \( 20 \% \) of the dealer's cost, what is the dealer's cost of the refrigerator?

To find the dealer's cost, let's denote the dealer's cost as \( C \). Since the markup is \( 20\% \) of the dealer's cost, we can express the selling price as: \[ Selling \, Price = C + (0.20 \times C) \] \[ 924 = C(1 + 0.20) \] \[ 924 = C(1.20) \] Now, solve for \( C \): \[ C = \frac{924}{1.20} = 770 \] So, the dealer's cost of the refrigerator is \( \$ 770.00 \). To make calculations easier, remember that percentages can be converted to decimals. For example, \( 20\% \) is the same as \( 0.20 \). When dealing with markups, including them directly in your equations can save you time and reduce mistakes in calculations. Always double-check your arithmetic to avoid simple errors that can throw off your final figure!