A standard fair dice is rolled twice. What is the probability of getting an odd number on the first roll and any factor of 12 on the other?

Step-by-step explanation:

It is too bad that the singular of the word dice is die.

Anyway, if your die is a 6-sided, unloaded cube with from 1 to 6 dots on a side, the numbers represented by the dots are 1 , 2 , 3 , 4 , 5 , and 6.

The odd numbers are 1 , 3 , and 5 . So three of the six numbers are odd. Rolling any one of these is a favorable outcome.

There are six outcomes in all.

P(rolling an odd number) = number of favorable outcomes ÷ number of all outcomes

=36=12, or .50 in decimal.

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3