Answer:
25/324
Step-by-step explanation:
Make a table of possible products:
![\left[\begin{array}{ccccccc}&1&2&3&4&5&6\\1&1&2&3&4&5&6\\2&2&4&6&8&10&12\\3&3&6&9&12&15&18\\4&4&8&12&16&20&24\\5&5&10&15&20&25&30\\6&6&12&18&24&30&36\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccccc%7D%261%262%263%264%265%266%5C%5C1%261%262%263%264%265%266%5C%5C2%262%264%266%268%2610%2612%5C%5C3%263%266%269%2612%2615%2618%5C%5C4%264%268%2612%2616%2620%2624%5C%5C5%265%2610%2615%2620%2625%2630%5C%5C6%266%2612%2618%2624%2630%2636%5Cend%7Barray%7D%5Cright%5D)
Of the 36 results, 10 are greater than 15.
The probability the product is greater than 15 on a single roll is 10/36 = 5/18.
The probability the product is greater than 15 on two rolls is (5/18)² = 25/324.
Answer:
Doesnt he still have 3? All he did was drop 2
Step-by-step explanation:
Sent a picture of the solution to the problem (s). Used the hint and substituted all the know quantities.
Answer:
40.1%
Step-by-step explanation:
I am assuming that 192 is in 100%.
100% = 192
I then represent the value that we are looking for with
.
x% = 77
By dividing both equations (100% = 192 and x% = 77) and remembering that both left hand sides of BOTH equations have the percentage unit (%).

Now, of course, we take the reciprocal, or inverse, of both sides:

x = 40.1%
Thus making the answer: 40.1% of 192 is 77.
Answer:
A.
Step-by-step explanation: