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
<span>t+3/2 = 1/2 // - 1/2
t+3/2-(1/2) = 0
t+3/2-1/2 = 0
t+1 = 0 // - 1
t = -1</span>
Answer: 72 cars.
Step-by-step explanation:
Let be "x" the number of toys that Moby had to start with.
According to the information given in the exercise, Moby sold half his toy car collection. This can be represented with the following expression:

Then he bought 12 more cars, giving a total amount of 48 toy cars.
Therefore, the equation that represents this situation is the equation shown below:

Now you must solve for "x" in order to find its value.
You get that this is:

1.5 x 32 = 48 12 divided 0.03 = 400 12.8 X 3/4 ( 0.75 ) = <span>17.0666667 1.5 x 0.32 = 0 .48 1.2 divided by 0.3 = 4 AND... 102.4 divided 3.2 = 32 Hope this helps! :)
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