I need help simplifying 8^3
2 answers:
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49. From 3 coin tosses, there are 8 possible outcomes:
... TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
All but the first have at least one head, so 7/8 of the possibilities have at least one head. That's 87.5% (not among your choices).
Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.
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Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.
50. See for yourself: the calculator says 66.82%. Your best choice is selection d.
Answer:
Option (C)
Step-by-step explanation:
System of the equations have been given as,
--------(1)
-------(2)
Substitute the value of y from equation (1) to equation (2),
Substitute the value of 'x' in equation (1),
By using a graphing utility, graph of the equations that will intersect each other at a point (2, 3).
Therefore, solution of the system of equations will be .
Option (C) will be the answer.
