We are given that the times are normally distributed with mean \(\mu = 125\) and standard deviation \(\sigma = 24\). We need to find the value of \(t\) such that 90% of the time, Karli spends more than \(t\) minutes on social media.

This implies that 10% of the time, Karli spends less than \(t\) minutes. We need to find the z-score corresponding to the 10th percentile of the standard normal distribution, which is \(z = -1.282\).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(-1.282 = \frac{t - 125}{24}\)

Solving for \(t\):

\(t - 125 = -1.282 \times 24\)

\(t - 125 = -30.168\)

\(t = 125 - 30.168\)

\(t = 94.832\)

Rounding to one decimal place, \(t \approx 94.2\).