**important**
franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.
the probability that the sum of the numbers rolled is either even or a multiple of 5 is [11/18, 13/19, 11/36, 13/36]. the probability that the sum of the numbers rolled is either a multiple of 3 or 4 is [1/12, 1/3, 1/4, 5/9, 7/12].

11/18

5/9

Step-by-step explanation:

we will make punnette square

         dice1

    +

dice2   1       2        3       4        5       6

1            2      3        4       5        6       7

2           3      4         5       6       7        8

3           4      5         6       7       8        9

4           5      6         7       8       9        10

5           6      7         8       9      10       11

6          7       8         9       10     11        12

1)

The probability that the sum of the numbers rolled is either even or a multiple of 5

even number = 18

{(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)},

and

multiple of 5 = 4

{(1,4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}

(we will skip the repeating)

18 + 4 = 22

probability = 22/36

                  = 11/18

2)

The probability that the sum of the numbers rolled is either a multiple of 3 or 4

multiple of 3 = 12

{(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}

and

multiple of 4 = 8

D = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

(we will skip the repeating)

12 + 8 = 20

probability = 20/36

                  = 5/9

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)