A computer is programmed to generate a sequence of three digits, where each digit is either 0 or 1, and each of these is equally likely to occur. Construct a sample space that shows all possible three-digit sequences of 0s and 1s and then find the probability that a sequence will contain at least one 0. A. 000, 001, 010, 011, 100, 101, 110, 111; P(at least one 0) = Equals StartFraction 7 Over 8 EndFraction b. 000, 001, 010, 100, 101, 110 ; P(at least one 0) = Equals StartFraction 6 Over 6 EndFraction c. 100, 101, 110, 111; P(at least one 0) = Equals three-fourths d. 000, 001, 010, 011; P(at least one 0) = Equals StartFraction 4 Over 8 EndFraction = one-half

hi there!

the answer to this question is d: 35

step-by-step explanation:

5 to the power of 2 is 25

2 multiplied by 5 is 10

then you add the 2 numbers and you get 35

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt