A con is tossed n times. How many difference sequences of head and tails can you get?
2 answers:
Answer:
2^n
Step-by-step explanation:
So whenever you flip a coin, you can see it as 2 nodes branching off of each existing node. so for example when you flip a coin once you're going to have 2 sequences initially H and T, now when you flip a coin again for each of those 2 sequences 2 are going to branch off of that, making the total sequences 4, and the next flip 2 sequences are going to branch off each of the 4 sequences and so on. this can generally be described as: 2^n
I attached an image describing this a bit better but the bottom line is that for each 'end node'/sequence you're going to have 2 branch off of it, thus for each coin flip the number of sequences multiplies by 2
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Answer:
see attached
Step-by-step explanation:
Here's your worksheet with the blanks filled.
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Of course, you know these log relations:
log(a^b) = b·log(a) . . . . . power property
log(a/b) = log(a) -log(b) . . . . . quotient property
log(x) = log(y) ⇔ x = y . . . . . . . . . equality property
