An individual chooses a prime number that is less than 13. What is the probability that the number is odd? Write your answer as an exact fraction which is reduced as much as possible.

4/5

Step-by-step explanation:

Okay, I'm pretty sure about this one! I might be wrong but:

If the number is prime and LESS than 13 (so it can't be 13 either), it's either 2, 3, 5, 7, or 11. Those are 5 numbers in all, and 4 of those are odd. So the fraction would be 4/5!

I might have done this wrong, but I hope it helps!! Have a nice day c:

Step-by-step explanation:

From the picture attached,

Addition of the blocks in first row is 60

a + a + a + 12 = 60

3a + 12 = 60

3a = 60 - 12

3a = 48

a = 16

For second row,

(b + 5) + (b + 5) + (b + 5) + (b + 5) = 60

4(b + 5) = 60

b + 5 = 15

b = 10

For third row,

(a + b) + c = (b + 5) + (b + 5) + (b + 5)

a + b + c = 3(b + 5)

16 + 10 + c = 3(10 + 5) [Since, a = 16 and b = 10}

26 + c = 45

c = 45 - 26

c = 19

For fourth row,

3c + d = 60

3(19) + d = 60

57 + d = 60

d = 60 - 57

d = 3