Answer:

Step-by-step explanation:
<u>The full question:</u>
<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current positions?"</em>
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The permutation of choosing 3 members from a group of 11 would be:
P(n,r) = 
Where n would be the total [in this case n is 11] & r would be 3
Which is:
P(11,3) = 
So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:
1/990
Answer:
Step-by-step explanation:
sorry do not .know
Answer:
f(-4)= 24
Step-by-step explanation:
-4(-4)+8 We need to multiply -4 and -4 and a negative times a negative is a positive.
16+8 Now it's easy math and we just add them
This gives us 24.
a xuz if u work it out ull get a
Answer:
<em>0 < X < 40</em>
Explanation:
[ Leg A ] 13
[ Leg B ] 27
[ Leg C ] x
Since We Are Missing The Third Side Of The Triangle.
x Has To Be Greater Than 0 But Less Than 40