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:
its porpotional
Step-by-step explanation:
hope this helps
Answer:
The value of x would be 
Step-by-step explanation:
Given,
The dimension of the cardboard = 10 ft by 10 ft,
∵ After removing four equal squares of size x ( in ft ) from the corners,
The dimension of the resultant box would be,
Length = ( 10 - 2x ) ft,
Width = ( 10 - 2x ) ft,
Height = x ft,
The volume of box,
Differentiating with respect to x,
Again differentiating with respect to x,
For maxima or minima,
By quadratic formula,
For x = 5/3, V'' = negative,
While for x = 5, V'' = Positive,
Hence, the value of x would be 5/3 ft for maximising the volume.
Answer:
First one is DC
Second one is AB.
Step-by-step explanation:
Just did them on edge.
Answer:
66 67/100
Step-by-step explanation:
