The problem presented asks us to determine a travel time, $x$, such that 29.46% of the travel times exceed this value, given that Abby's travel times are normally distributed with a mean of 35.6 minutes and a standard deviation of 10.3 minutes.

This is a standard normal distribution problem. We need to find the value of $x$ for which 29.46% of the distribution is greater than $x$. In terms of a z-score, this means we're looking for the z-value where the upper tail of the standard normal distribution is 29.46%.

Steps:

1. Convert the percentage to a probability:
   29.46% translates to a probability of 0.2946. This is the area to the right of $x$, so we are looking for a z-score where the area to the right is 0.2946, or equivalently, the area to the left is $1 - 0.2946 = 0.7054$.

2. Find the z-score corresponding to the cumulative probability of 0.7054 using standard normal tables or a calculator.

3. Use the z-score formula:
   $z = \frac{x - \mu}{\sigma}$ where $\mu = 35.6$ (mean) and $\sigma = 10.3$ (standard deviation). Solve for $x$ once the z-score is known.

Let me calculate the z-score and the corresponding $x$.