There are 3636 chocolates in a box, all identically shaped. there 1212 are filled with nuts, 99 with caramel, and 1515 are solid chocolate. someone randomly selects one piece, eats it, and then selects a second piece. find the probability of selecting 2 nut candies. express your answer as a simplified fraction

1211/10905

Step-by-step explanation:

In the beginning, the probability to pick a nut piece is 1212/3636. Once you eat it you have 3635 candies left, of which 1211 are nuts, and therefore the probability to pick nut again is 1211/3635.

So, the combined probability is 1212/3636 * 1211/3635 = 1/3 * 1211/3635 = 1211/(3*3635) = 1211/10905

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752