6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320
plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n
Answer:
0.1111
Step-by-step explanation:
Given that you roll two dice.
the average of the high and low roll is exactly 3,
Since die can show only 1 to 6 we can say average can be 3 in each of the following case.
(1,5) (2,4) (3,3) (4,2) (5,1)
There cannot be any other combination to get average of 3.
Thus favourable events = 4
Sample space will have
(1,1)...(1,6)
(2,1)....
(6,1)...(6,6) i.e. 36
So probability that the average of the high and low roll is exactly 3
=
Answer:
<em>The answer to your question is</em> <em>C.n ≥ 7</em>
Step-by-step explanation:
<u><em>I hope this helps and have a good day!</em></u>
Answer:
k=9
Step-by-step explanation:
4/3 = 8/6 = 12/9