Pregunta
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(a). Kisiwaa puts \( \frac{1}{4} \) of her salary in savings and uses \( \frac{1}{2} \) of the remaining money to pay back her sister. If the amount left is GHC \( 2,400.00 \), how much did she have as salary? (b). A dance academy charges GHC 20.00 to use the facility and GHC 25 per hour of instruction. (i) Write an equation that gives the total cost to learn at the dance academy as a function of hours of instruction. (ii) Find the total cost of 2 hours of dance instruction. Find the perimeter and area of the shape below. (c).

Ask by Herrera Schneider. in Ghana
Jan 29,2025

Solución de inteligencia artificial de Upstudy

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**Problem (a):** Kisiwaa's total salary is **GHC 6,400.00**. **Problem (b):** - **(i) Total cost equation:** \( C(h) = 20 + 25h \) - **(ii) Cost for 2 hours:** GHC 70.00 **Problem (c):** Cannot determine the perimeter and area without information about the shape. Please provide details or an illustration.

Solución

Sure, let's tackle each part of your question step by step. --- ### **Problem (a)** **Kisiwaa** puts \( \frac{1}{4} \) of her salary in savings and uses \( \frac{1}{2} \) of the remaining money to pay back her sister. The amount left is GHC \( 2,400.00 \). We need to find her total salary. #### **Solution:** 1. **Let’s denote Kisiwaa's total salary as \( S \).** 2. **Savings:** - She saves \( \frac{1}{4} \) of her salary. \[ \text{Savings} = \frac{1}{4}S \] 3. **Remaining after Savings:** \[ \text{Remaining} = S - \frac{1}{4}S = \frac{3}{4}S \] 4. **Paying Back Her Sister:** - She uses \( \frac{1}{2} \) of the remaining amount to pay back her sister. \[ \text{Amount paid back} = \frac{1}{2} \times \frac{3}{4}S = \frac{3}{8}S \] 5. **Amount Left:** - After paying back, the amount she has left is GHC \( 2,400.00 \). \[ \text{Amount left} = \frac{3}{4}S - \frac{3}{8}S = \frac{3}{8}S = 2,400 \] 6. **Solving for \( S \):** \[ \frac{3}{8}S = 2,400 \\ S = 2,400 \times \frac{8}{3} \\ S = 2,400 \times \frac{8}{3} = 6,400 \] #### **Answer:** Kisiwaa's total salary is **GHC 6,400.00**. --- ### **Problem (b)** A dance academy charges GHC 20.00 to use the facility and GHC 25 per hour of instruction. #### **(i) Write an equation that gives the total cost to learn at the dance academy as a function of hours of instruction.** #### **Solution:** Let \( h \) represent the number of hours of instruction. - **Fixed Cost (Facility Use):** GHC 20.00 - **Variable Cost (Per Hour Instruction):** GHC 25 per hour The total cost \( C \) can be expressed as: \[ C = 20 + 25h \] #### **Answer:** \[ C(h) = 20 + 25h \] where \( C \) is the total cost and \( h \) is the number of hours of instruction. --- #### **(ii) Find the total cost of 2 hours of dance instruction.** #### **Solution:** Using the equation from part (i): \[ C(h) = 20 + 25h \] Plug in \( h = 2 \): \[ C(2) = 20 + 25 \times 2 = 20 + 50 = 70 \] #### **Answer:** The total cost for 2 hours of dance instruction is **GHC 70.00**. --- ### **Problem (c)** **Find the perimeter and area of the shape below.** *Unfortunately, I can't see the shape you're referring to. To help you calculate the perimeter and area, please provide a description of the shape, including its type (e.g., rectangle, triangle, circle) and relevant measurements (e.g., lengths of sides, radius, etc.).* --- If you provide more details or an illustration of **Problem (c)**, I'd be happy to assist you further!

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Beyond the Answer

Kisiwaa’s salary puzzle is a classic example of problem-solving with fractions! To find her salary, let’s denote her salary as \( S \). After saving \( \frac{1}{4}S \), she has \( \frac{3}{4}S \) left. If she pays back her sister \( \frac{1}{2} \) of this remaining amount, she uses \( \frac{1}{2} \times \frac{3}{4}S = \frac{3}{8}S \). After these transactions, she’s left with GHC \( 2,400.00 \), leading to the equation \( \frac{3}{8}S = 2400 \). Solving this gives her salary as GHC \( 6,400.00 \). At the dance academy, the total cost can be calculated easily! If you use \( H \) to represent hours of instruction, the total cost \( C \) can be expressed as: \[ C = 20 + 25H \]. For 2 hours of dance instruction, plug in \( H \): \[ C = 20 + 25(2) = 20 + 50 = GHC 70.00 \]. So, the total cost for 2 hours is GHC 70.00 — time to dance!

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