Suppose a fair coin is tossed nine times. Replace the resulting sequence of H’s and 74 Chapter 2 Probability T’s with a binary sequence of 1’s and 0’s (1 for H, 0 for T). For how many sequences of tosses will the decimal corresponding to the observed set of heads and tails exceed 256?

255 sequence

Step-by-step explanation:

Since

Head replaced by 1

Tail replaced by 0

For each toss, there are two possible outcomes: heads 1, or tails 0.

For 9 toss the possible outcome sequence range from

000000000 to 111111111

000000000 in binary = 0 in decimal

111111111 in binary = 511 in decimal

Since we are asked to find decimal corresponding to the observed set of heads and tails that exceed 256?

That is sequence that exceed 256 (100000000)

Which is 257 (100000001) to n

So the outcome sequence will range from 257 to the last possible outcome sequence for 9 tosses which is 511

Therefore between 257 to 511.

Number of outcome sequence is 255

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

i dont know men

step-by-step explanation: