An urn contains four fair dice. Two have faces numbered 1, 2, 3, 4, 5, and 6; one has faces numbered 2, 2, 4, 4, 6, and 6; and one has all six faces numbered 6. One of the dice is randomly selected from the urn and rolled. The same die is rolled a second time. Calculate the probability that a 6 is rolled both times.

7/24 = 0.2917 = 29.17%

Step-by-step explanation:

We have four dice, and they are different, so we need to calculate the probability for each one, and multiply it by 1/4 (the chance of that die be picked)

First and second die:

The chance of rolling 6 is 1/6

So prob = (2/4) * (1/6) * (1/6) = 1/72

Third die:

The chance of rolling 6 is 2/6 = 1/3

So prob = (1/4) * (1/3) * (1/3) = 1/36

Fourth die:

The chance of rolling 6 is 1

So prob = (1/4) * 1 * 1= 1/4

Summing all the probabilities: we have 1/72 + 1/36 + 1/4 = (1+2+18)/72 = 21/72 = 7/24 = 0.2917 = 29.17%

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points