We are given the learning curve for the number of units $N$ produced per day after a new employee has worked $t$ days, modeled by:

$N = 30(1 - e^{kt})$

We also know that after 20 days, the employee produces 16 units. We can use this information to find the value of $k$.

Part (a) - Find the value of $k$

We can substitute $N = 16$ and $t = 20$ into the equation:

$16 = 30(1 - e^{20k})$

Now, let's solve for $k$.

1. Divide both sides of the equation by 30: $\frac{16}{30} = 1 - e^{20k}$

$\frac{16}{30} = 0.5333$

$0.5333 = 1 - e^{20k}$

2. Subtract 0.5333 from 1: $e^{20k} = 0.4667$

3. Take the natural logarithm of both sides: $20k = \ln(0.4667)$

4. Solve for $k$: $k = \frac{\ln(0.4667)}{20}$

Let's calculate $k$.The value of $k$ is approximately $-0.038$ (rounded to three decimal places).

Thus, the learning curve equation becomes:

$N = 30(1 - e^{-0.038t})$

Part (b) - How many days should pass before this employee produces 23 units per day?

We need to find $t$ when $N = 23$. Using the equation $N = 30(1 - e^{-0.038t})$, substitute $N = 23$:

$23 = 30(1 - e^{-0.038t})$

Let's solve for $t$.It appears that the result for $t$ is negative, which is not physically meaningful in this context. This suggests a mistake in the interpretation of the model. Let me correct and adjust for the proper solution approach for part (b).It will take approximately 38 days for the employee to produce 23 units per day.

Summary:

• $k \approx -0.038$
• Learning curve equation: $N = 30(1 - e^{-0.038t})$
• The employee will produce 23 units per day after about 38 days.

Would you like more details on any step or have any further questions?

Here are 5 related questions to consider:

1. How would the learning curve change if the maximum number of units was higher?
2. What does the parameter $k$ represent in the learning curve equation?
3. How would the model change if the rate of learning was slower?
4. What is the significance of the constant 30 in this equation?
5. How can we modify this model to account for workers who may regress in productivity?

Tip: In learning curve models, $k$ represents the rate of improvement. A more negative $k$ indicates faster learning.