Which is a recursive formula for the sequence 99.4, 0, –99.4, –198.8, where f(1) = 99.4?
f(n + 1) = f(n) + 99.4, n ≥ 1 f(n + 1) = f(n) – 99.4, n ≥ 1 f(n + 1) = 99.4f(n), n ≥ 1
f(n + 1) = –99.4f(n), n ≥ 1

the answer is B: f(n + 1) = f(n) - 99.4,n > 1

Step-by-step explanation:

answer:

what is your question that you ask

step-by-step explanation:

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560