45 divided by 3 gives you 15 which is the amount of money he spent on the gas and 45 divided by 5 is 9 which is the amount of money he spent on the flowers which means he spent $24 so he has $21 left for the big date.
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
The probability that the train will be there when Alex arrives is 5/18
If Alex arrives at any time after 1.20pm the chances that train will be there is 1/3.
However if alex arrives at 1.00pm exactly there is no chance the train will be arrive there.
The probability that the train will be there increase linearly to 1/3 as alex's arrival time moves from 1.00pm to 1.20pm.
By arranging the probabilities over the first 20 minutes to get a 1/6 chance the train will be there if alex arrives between 1.00pm to 1.20pm
we get the final answer by
=1/3( 1/6 + 1/3 + 1/3)
=5/18
So, the probability that the train will be there when Alex arrives is 5/18
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x^2 + 4x + 3 + 1 = (x + 2)^2
So 1 need to be added.
Answer:
Is this high school work?
Step-by-step explanation:
