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>
Your question is so blur i cant see
plzz send a nice pics
Answer:
<h2>
The eleventh term of the sequence is 64</h2>
Step-by-step explanation:
The sequence given is an arithmetic sequence
14, 19, 24, …………., 264
The nth term of an arithmetic sequence is given as;
Tn = a+(n-1)d where;
a is the first term = 14
d is the common difference = 19-14=24-19 = 5
n is the number of terms = 11(since we are to look for the eleventh term of the sequence)
substituting the given values in the formula given;
T11 = 14+(11-1)*5
T11 = 14+10(5)
T11 = 14+50
T11 = 64
The eleventh term of the sequence is 64
Answer:
b
Step-by-step explanation:
just took it i got 100%
Answer
D $48.00 u need to multiply 216 and 12 to get 48
