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>
1.208 ounce because 30.2 divided by 25 equals 1.208 ounces
Answer: hello your question is incomplete attached below is the complete question
answer :
a) 
( option D )
b) A = 1/2 (6)(3) ( option B )
c) Area of shaded region = 9
Step-by-step explanation:
<u>a) Using integration with respect to x </u>
Area = 
( note a = 10 )
= y^2/2 - 4y |⁷₄ + 10y - y^2/2 |¹⁰₇
= 33/2 - 12 + 30 - 51/2 = 9
hence the best integral from the options attached is option D
= [ 10x - x^2 /2 - x^2/2 - 9x ] ³₀
= 30 - 9/2 - 9/2 - 12 = 9
<u>b) Using Geometry </u>
Area = 1/2 * base * height
= 1/2 * 6 * 3
= 9
4:1 means for every four soccer player there is 1 volleyball player
that means there are four times the number of soccer players than volleyball players
