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:
D. Part of the solution region includes a negative number of erasers purchased; therefore, not all solutions are viable for the given situation.
Step-by-step explanation:
I got it right on the practice
Remember you can do anything to an equation as long as you do it to both sides
5c+16.5=13.5+10c
minus 5c both sides
16.5=13.5+5c
minus 13.5 from both sides
3=5c
divide both sides by 5
3/5=c
Answer:
it is the first option
Step-by-step explanation:
the middle (median)is 31 and where the Q starts is 26
That equation has no solution. There is no number that can be written in place of 'x' that makes the equation true.