Morgan is playing a board game that requires three standard dice to be thrown at one time. Each die has six sides, with one of the numbers 1 through 6 on each side. She has one throw of the dice left, and she needs a 17 to win the game. What is the probability that Morgan wins the game (order matters)?
Answer:
1/72
Step-by-step explanation:
<em>Morgan can roll a 17 in 3 different ways. The first way is if the first die comes up 5, the second die comes up 6, and the third die comes up 6. The second way is if the first die comes up 6, the second die comes up 5, and the third die comes up 6. The third way is if the first die comes up 6, the second die comes up 6, and the third die comes up 5. For each way, the probability of it occurring is 1/6 x 1/6 x 1/6 = 1/216. Therefore, since there are 3 different ways to roll a 17, the probability that Morgan rolls a 17 and wins the game is 1/216 + 1/216 + 1/216 = 3/216 = 1/72</em>
The first equation given is y = 3 - 1/2x In other words, y is the same as 3 - 1/2x. We can replace y in the second equation with 3 - 1/2x This is known as substitution (think of a substitute teacher who is a temporary replacement for your teacher)
Doing this leads to... 3x+4y = 1 3x+4*y = 1 3x+4*( y ) = 1 3x+4*( 3 - 1/2x ) = 1 <<--- y has been replaced with 3-1/2x 3x+4*(3) +4*(-1/2x) = 1 3x+12-2x = 1 3x-2x+12 = 1 x+12 = 1 x+12-12 = 1-12 <<-- subtracting 12 from both sides x = -11 Which is why the answer is choice C) -11
