In American roulette, the wheel has the 38 numbers, \( 00,0,1,2, \ldots, 34,35 \), and 36 , marked on equally spaced slots. If a player bets \( \$ 4 \) on a number and wins, then the player keeps \( \$ 4 \) and receives an additional \( \$ 140 \). Otherwise, the player is awarded nothing, and the casino takes the player's \( \$ 4 \). Find the expected value \( E(x) \) to the player for one play of the game. If \( x \) is the gain to a player in a game of chance, then \( E(x) \) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is \( \$ \square \). (Round to the nearest cent as needed.)

To find the expected value \( E(x) \) in American roulette when betting on a single number, let's consider the two outcomes: 1. **Player wins**: The probability of this happening is \( \frac{1}{38} \) because there are 38 total slots. If the player wins, they gain \( \$140 \) plus their initial \( \$4 \), so their total gain is \( \$144 \). 2. **Player loses**: The probability of losing is \( \frac{37}{38} \). If the player loses, they lose their \( \$4 \). Now we can calculate the expected value: \[ E(x) = P(\text{win}) \cdot \text{Gain when win} + P(\text{lose}) \cdot \text{Loss when lose} \] \[ E(x) = \left(\frac{1}{38}\right) \cdot 144 + \left(\frac{37}{38}\right) \cdot (-4) \] \[ E(x) = \frac{144}{38} - \frac{148}{38} \] \[ E(x) = \frac{144 - 148}{38} = \frac{-4}{38} \approx -0.1053 \] This results in an expected value of approximately \( -0.11 \) when rounded to the nearest cent. Thus, the expected value is \( \$ -0.11 \).