Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
i know b only Sorry hope this helps tho
Step-by-step explanation:
its 4 because each space is 25% no matter how big or small it is. so if you land on -9 to -5 thats 25% and to -2 thats another 25% total 50% its asking for 75% so you go one more which lands on 4 making it 75%
Answer:
p = 2 and q = 2
Step-by-step explanation:
hope this helps please like and mark as brainliest
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