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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 \( x \).

Suppose Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. Assume the travel times are normally distributed with a standard deviation of 10.3 min. Determine the travel time x such that 29.46% of the 60 days have a travel time that is greater than x.

Solve a normal distribution problem to determine the travel time such that 29.46% of values exceed this number. Given a mean of 35.6 minutes and a standard deviation of 10.3, we use Z-scores to find the solution.

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