A game is played using one die. If the die is rolled and shows 6 , the player wins \( \$ 35 \). If the die shows any number other than 6 , the player wins nothing. Complete parts (a) through (b) below. a. If there is a charge of \( \$ 7 \) to play the game, what is the game's expected value? \( \$ \square \) (Round to the nearest cent.)

To find the expected value of the game, we first need to calculate the probability of winning and losing. There is a 1 in 6 chance (or approximately 16.67%) of rolling a 6 and winning $35, and a 5 in 6 chance (or approximately 83.33%) of winning nothing. Let’s calculate the expected winnings: - Winning scenario: \( \frac{1}{6} \times 35 = 5.8333 \) - Losing scenario: \( \frac{5}{6} \times 0 = 0 \) Now, add these two together: \( 5.8333 + 0 = 5.8333 \) Now, we account for the $7 charge to play: Expected value = Expected winnings - Cost to play Therefore: Expected value = \( 5.8333 - 7 = -1.1667 \) Rounding to the nearest cent, the expected value is \( \$ -1.17 \). So, the answer is: \( \$ -1.17 \)