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:
D. x = t√y-7
Step-by-step explanation:
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P(t+1) = -2
divide by P
Subtract 1
T equals negative 2 over P minus 1
T = -2/P -1
Answer:
-5 divide by 28
Step-by-step explanation:
so the question supposed be " 15-20 divided by 2(4) + 20
15-20 which is also 15 + (-20)= -5
2(4)=8, 8+20=28
so it can simplify into -5 / 28
(hope it helps :)
Answer:
408
Step-by-step explanation:
You would set up a proportion of 1/12 = 34/x. You would do the butterfly method and get 1x=408. Then divide 408 by 1 to get 408.
