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anonymous
 5 years ago
a small school has 100 students who occupy three classrooms. A,B, and C. After the first period of the school day, half the students in room A move to room B, onefifth of the students in room B move to room C, and onethird of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room. Must be put in system and solves on TI 84 calculator.
anonymous
 5 years ago
a small school has 100 students who occupy three classrooms. A,B, and C. After the first period of the school day, half the students in room A move to room B, onefifth of the students in room B move to room C, and onethird of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room. Must be put in system and solves on TI 84 calculator.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's infinite solutions to this problem. Set up the problem like this: (1/2)A = (1/5)B = (1/3)C So, B = (5/2)A C = (3/2)A The value of A must be multiples of the least common multiple of 2, 3, and 5.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using a Matrix Inverse

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure how to help you with the TI 84 part, I'm not familiar with the function involving solving equations. However, I can help you think about the problem to "get the ball rolling". Your biggest hint is that the total number of students in each room is the same for both periods. That means the number of students leaving a particular room must be equal to the number entering the room. So we have: A/2=C/3 B/5=A/2 C/3=B/5 Now, it looks like we have three equations and three unknowns, but really these are all equivalent. We need to add in this fact: A+B+C=100 Let's put B in terms of C and A... B=100(A+C) Then we can use our comparison equations from above with this new substitution. (100(A+C))/5=A/2=C/3 @dcp, there are not infinite solutions because we have the fact that there are 100 students total. Furthermore you cannot have fractions of a student.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, 100 students total. Yep, missed that part. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No worries... I was thinking infinite solutions at first as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok my directions say to write the systems of equations ans set up the matrix equation AX=B and solve it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its supposed to be in a 4x4 matrix
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