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:
it equals to 1032000
Step-by-step explanation:
because 3000 x 344 = 1032000
6 to the fourth power is 1296 because 6x6x6x6 gives you 1296
Answer:
Step-by-step explanation:
The formula for the area of a circle is 
, r is your radius, A is your area, and π is pi (3.1415...). You plug in your radius, or 5 into r, square it, 25, and multiply it by pi. You may be asked to use a certain number of digits of pi, and in that case you multiply for example 3.14 by 25, however is the exact answer.
Yes because if you answer both you get the same answer
