Question:
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>
<em>I had this same question on my test!</em>
<em>Hope this helped! Good Luck! ~LILZ</em>
Answer:
Try (
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Step-by-step explanation:
OK so, I copied and pasted that so try to put it together lol
The slope of diagonal <em>AB</em> is 0 and its equation is y=2.
Assuming that your question is (x-1)(x+2) = [5(x-1)]/x-1
On the right side, the x-1's will cancel out, leaving you with (x-1)(x+2) = 5
expand the left side, giving you x^2 + x -2 = 5
which goes to x^2 +x -7 = 0
the possible values for x are 2.93 and -3.93. I don't think this was your question, so I'll do the other possible question that you might have been asking.
(x-1)(x+2) = 5(x-1)
divide by x-1 on both sides, leaving you with x+2=5
x+2=5
x=5-3
x=3
