If figure 1 has 4 tiles, fig 2 has 10, and fig 3 has 16... What is my rule?

an = 6n -2

Or

an+1 = an +6

Step-by-step explanation:

to go from 4 to 10 we add 6 tiles

to go from 10 to 16 we add 6 tiles

We add 6 tiles each time

an = a1 + d(n-1)  where an is the value of the nth term  a1 is the first term  d is the common factor and n is the term number

an = 4 + 6( n-1)

an = 4 + 6n -6

an = 6n -2

Or

an+1 = an +6

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