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>
The only reasonable answer I could think for this is "A.<span>The company's expense budget for 2009 was $189,785."
I believe this because since x is the years after 2009 then you wouldn't times that by the expense budget for that year or any year eliminating B and C. When finding the total expense budget for this you would have to take in account the budget from 2009 so that would lead it to add 189,758. Since we are talking about after 2009 then D would be eliminated since 2008 is before 2009.
I hope this helps! Good luck!</span><span>
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Ok, so we know that RST is equal to 6x+12
And RST is also equal to 78 + 3x-12
so we set them equal to each other
6x + 12 = 3x - 12 + 78
And simplify
3x = 54
x = 18
Finally, we solve for the angle with 18 for x
6(18) + 12
108 + 12
120
Hope this helps
The answer is b (I think not really sure)
Look at the ss down below
