When two fair 6-sided dice are tossed, the numbers $a$ and $b$ are obtained. What is the probability that the two digit number $ab$ (where $a$ and $b$ are digits) and $a$ and $b$ are all divisible by 3?

Explanation:

If either a or b (or both) are multiples of 3, then so is a*b

Consider 'a' being a multiple of 3. This means a = 3k for some integer k. Multiply both sides by b to get ab = 3k*b, and we see that 3 is a factor of ab, making ab to be a multiple of 3.

There are six sides to the dice and those sides are {1,2,3,4,5,6} of which only {3,6} are multiples of 3.

If we want both dice to be multiples of 3, then the possible outcomes are

There are four outcomes. This is out of 6*6 = 36 total outcomes overall.

Dividing the values gives 4/36 = 1/9 as the probability we want

Side note: for a roll like (a,b) = (6,3) we see that a*b = 6*3 = 18, which is a multiple of 3. The same happens with the other three (a,b) pairings mentioned above.

21

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2