algebra ii & trigonometry (h) 2025 - 2026 mr. gregory (block d) homework: 1.3 hw - models and applications…
algebra ii & trigonometry (h) 2025 - 2026 mr. gregory (block d) homework: 1.3 hw - models and applications question 13, 1.3.19 the selling price of a refrigerator is $811.65. if the markup is 5% of the dealers cost, what is the dealers cost of the refrigerator? points: 0 of 1 save
Answer
Explanation:
Step1: Let the dealer's cost be $x$.
The selling - price is the dealer's cost plus the markup. The markup is 5% of the dealer's cost, so the selling price formula is $Selling\ price=x + 0.05x=1.05x$.
Step2: We know the selling price is $811.65$.
We set up the equation $1.05x = 811.65$.
Step3: Solve for $x$.
Divide both sides of the equation by 1.05: $x=\frac{811.65}{1.05}$. $x = 773$.
Answer:
$773$