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:
Response 1
Step-by-step explanation:
Response 1 is the only wording that is self-consistent. Response 1 is the appropriate response.
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Response 2 argues about angles, then concludes sides are congruent. This makes no sense.
Response 3 identifies congruent segments, but they are not opposite sides. The conclusion is unsupported.
Answer: 3
Step-by-step explanation: Hope I helped
Answer:
H=21ft
Step-by-step explanation:
Height=2•Area/Base
Area=126ft^2
Base=12ft.
H=2•126/12=21ft
H=21ft
Hope this helps :)
Answer:
P = 0.55 or 55 %
Step-by-step explanation:
First step: Fred has probability of 0,5 when chossing jar 1 or jar 2
Second step : The probability of chossing one chocolate chp cookie in jar 1 is 3/5 and from the jar 2 is 1/2
Then the probability of Fred to get a chocolate chip cookie is
P ( get a chocolate chip cookie ) =( 0.5 * 3/5) +( 0.5* 1/2)
P = 0.3 + 0.25
P = 0.55 or 55 %
