Two dice are rolled. Find the probability of getting a. A sum of 5, 6, or 7 b. Doubles or a sum of 6 or 8 c. A sum greater than 8 or less than 3 d. Based on the answers to parts a, b, and c, which is least likely to occur

The answer is below

Step-by-step explanation:

The possible outcome from rolling two dice is:

1,1      1,2      1,3     1,4      1,5     1,6

2,1     2,2     2,3    2,4     2,5    2,6

3,1     3,2     3,3    3,4      3,5    3,6

4,1     4,2     4,3     4,4     4,5     4,6

5,1     5,2     5,3     5,4     5,5     5,6

6,1     6,2     6,3     6,4     6,5     6,6

The total number of outcomes = 36

a)

P(sum of 5) = 4/36, P(sum of 6) = 5/36, P(sum of 7) = 6/36

Hence:

P(sum of 5 or sum of 6, or sum of 7) = P(sum of 5) + P(sum of 6) + P(sum of 7) = 4/36 + 5/36 + 6/36

P(sum of 5, 6, or 7) = 15 / 36

b)

P(doubles) = 6/36, P(sum of 6) = 5/36, P(sum of 8) = 5/36

Hence:

P(doubles or sum of 6 or sum of 8) = P(doubles) + P(sum of 6) + P(sum of 8) = 6/36 + 5/36 + 5/36

P(doubles or sum of 6 or sum of 8) = 16 / 36

c)

P(sum greater than 8) = 10/36, P(sum lesser than 3) = 1/36

P(sum greater than 8 or less than 3) = P(sum greater than 8) + P(sum lesser than 3) = 10/36 + 1/36

P(sum greater than 8 or less than 3) = 11 / 36

Part c has the least probability of occurrence, hence c is least likely to occur.

answer: the two figures are congruent.

step-by-step explanation: they're congruent because figure efgh is reflected across the x and y axis but it was also rotated.

hope it !

1,407.5 hours

step-by-step explanation:

we first convert all the time to hours and then sum them up.

there are 24 hours in 1 day, in 58 days, there are 24 x 58 = 1,392 hours.

there are 60 minutes in 1 hour, thus, 30 minutes will be equivalent to 30 / 60 = 0.5 hours.

therefore, the number of hours it will take mercury to complete a full rotation is 1,392 + 15 + 0.5 = 1,407.5 hours