Let's break down the **Rick and Morty Learning Curve Challenge** step by step. ## 1. **Finding the Labor Time for the First Unit (\(T_1\))** You are given: - **Total time for first 4 units**: 1,800 hours - **Learning curve**: 85% (which means each time cumulative production doubles, average hours per unit drops to 85% of the previous average) - **Units produced**: 4 We'll use the **unit time formula** for the learning curve: \[ T_n = T_1 \cdot n^{\log_2(b)} \] where: - \(T_n\) = time for the \(n\)th unit - \(T_1\) = time for the 1st unit - \(b\) = learning curve rate (as a decimal, so 0.85) - \(n\) = unit number ### **Step 1: Calculate \(\log_2(b)\)** \[ \log_2(0.85) = \frac{\ln(0.85)}{\ln(2)} \approx \frac{-0.1625}{0.6931} \approx -0.2344 \] ### **Step 2: Express total time for first 4 units** Total time for first 4 units: \[ \text{Total} = T_1 + T_2 + T_3 + T_4 \] Write each term: \[ T_2 = T_1 \cdot 2^{\log_2(0.85)} = T_1 \cdot 0.85 \] \[ T_3 = T_1 \cdot 3^{\log_2(0.85)} = T_1 \cdot 3^{-0.2344} \] \[ T_4 = T_1 \cdot 4^{\log_2(0.85)} = T_1 \cdot 4^{-0.2344} \] Calculate: - \(3^{-0.2344} = e^{-0.2344 \cdot \ln(3)} \approx e^{-0.2344 \cdot 1.0986} \approx e^{-0.2573} \approx 0.7732\) - \(4^{-0.2344} = e^{-0.2344 \cdot \ln(4)} \approx e^{-0.2344 \cdot 1.3863} \approx e^{-0.3248} \approx 0.7227\) So, \[ \text{Total} = T_1[1 + 0.85 + 0.7732 + 0.7227] = T_1 \times 3.3459 \] Set equal to 1,800: \[ T_1 \times 3.3459 = 1800 \] \[ T_1 = \frac{1800}{3.3459} \approx \boxed{538.1 \text{ hours}} \] --- ## 2. **Labor Hours for the 6th, 12th, and 18th Units** Use the formula: \[ T_n = T_1 \cdot n^{\log_2(0.85)} \] Recall: - \(T_1 \approx 538.1\) - \(\log_2(0.85) \approx -0.2344\) ### **6th Unit** \[ T_6 = 538.1 \cdot 6^{-0.2344} \] Calculate \(6^{-0.2344}\): \[ \ln(6) = 1.7918 \] \[ -0.2344 \times 1.7918 = -0.420 \] \[ e^{-0.420} = 0.6573 \] So, \[ T_6 = 538.1 \cdot 0.6573 \approx \boxed{353.7 \text{ hours}} \] --- ### **12th Unit** \[ T_{12} = 538.1 \cdot 12^{-0.2344} \] \[ \ln(12) = 2.4849 \] \[ -0.2344 \times 2.4849 = -0.5824 \] \[ e^{-0.5824} = 0.5587 \] \[ T_{12} = 538.1 \cdot 0.5587 \approx \boxed{300.7 \text{ hours}} \] --- ### **18th Unit** \[ T_{18} = 538.1 \cdot 18^{-0.2344} \] \[ \ln(18) = 2.8904 \] \[ -0.2344 \times 2.8904 = -0.6775 \] \[ e^{-0.6775} = 0.5077 \] \[ T_{18} = 538.1 \cdot 0.5077 \approx \boxed{273.4 \text{ hours}} \] --- ## **Summary Table** | Unit | Labor Hours (rounded) | |-------|----------------------| | 1 | 538.1 | | 6 | 353.7 | | 12 | 300.7 | | 18 | 273.4 | --- ## **Final Answers** 1. **Labor time for the first unit (\(T_1\))**: \(\boxed{538.1 \text{ hours}}\) 2. **Labor times for specific units:** - 6th unit: \(\boxed{353.7}\) hours - 12th unit: \(\boxed{300.7}\) hours - 18th unit: \(\boxed{273.4}\) hours Let me know if you want the labor hours for any other units!

The learning curve describes how production efficiency improves over time, resulting in decreased labor time per unit as cumulative output increases.