Two dice are tossed together. what is the probability of getting an odd number on the second die and a total of five?
1/18
11/18
7/18
4/9

Probability = (number of ways to succeed) / (total possible outcomes) .

The total possible results of rolling two dice is

   (6 on the first cube) x (6 on the second one) = 36 possibilities.

How many are successful ?  I need you to clarify something first.
You said that the 'second die' shows an odd number.  When a pair
of dice is rolled, the problem usually doesn't distinguish between them. 
And in fact, you said that they're "tossed together" (like a spinach and
arugula salad ?) so I would understand that they would lose their identity
unless they were, say, painted different colors, and we wouldn't know
which one is the second one.

Oh well, I'll just work it both ways:

First way: 
Two identical dice are tossed.
The total is 5 and ONE cube shows an odd number.

How can that happen ?

1 ... 4
4 ... 1
3 ... 2
2 ... 3         

Four possibilities.  Probability = 4/36 = 1/9 = 11.1% .



Second way: 
A black and a white cube are tossed together.
The total is 5 and the white cube shows an odd number.

How can that happen:

B ... W
4 1
2 3

Only two possibilities.  Probability = 2/36 = 1/18 = 5.6% .

yes

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!