Question

1. $g = \frac{9}{5}d + 30$ if d equals 27, solve for g round to 2 decimal places to the right of the decimal 18. cost + markup = selling price. the cost is $5 and the selling price is $25. solve for markup? round to cents 19. for the preceding problem what is the markup based on selling price as a percentage? round to two places to the right of the decimal and than write this answer as a percentage. 20. cost + markup = selling price. the cost is $5 and the selling price is $25. what is the cost as a percentage of selling price? round to two places to the right of the decimal and than write this answer as a percentage.

Explanation:

Question 17

Step1: Substitute D = 27 into the formula

We have the formula \( G=\frac{9}{5}D + 30 \), and we know that \( D = 27 \). So we substitute \( D \) with 27 in the formula.
\( G=\frac{9}{5}\times27 + 30 \)

Step2: Calculate \(\frac{9}{5}\times27\)

\(\frac{9}{5}\times27=\frac{9\times27}{5}=\frac{243}{5} = 48.6\)

Step3: Add 30 to the result

Now we add 30 to 48.6. \( 48.6+30 = 78.6 \)

Step1: Recall the formula \( \text{Cost}+\text{Markup}=\text{Selling Price} \)

We know that Cost = $5 and Selling Price = $25. We need to find Markup. Let's denote Markup as \( M \). So the formula is \( 5+M = 25 \)

Step2: Solve for M

Subtract 5 from both sides of the equation. \( M=25 - 5=20 \)
Since we need to round to cents (two decimal places), \( M = 20.00 \) dollars.

Step1: Recall the formula \( \text{Cost}+\text{Markup}=\text{Selling Price} \)

We know Cost = $5, Selling Price = $25, so Markup (from question 18) is $20. We need to find the markup as a percentage of the selling price. The formula for percentage is \( \text{Percentage}=\frac{\text{Markup}}{\text{Selling Price}}\times100\% \)

Step2: Substitute the values

Substitute Markup = 20 and Selling Price = 25 into the formula. \( \text{Percentage}=\frac{20}{25}\times100\% \)

Step3: Calculate the fraction

\(\frac{20}{25}=0.8\)

Step4: Multiply by 100%

\( 0.8\times100\% = 80.00\% \)

Answer:

\( G = 78.60 \) (rounded to two decimal places)

Question 18