There are two spinners. the first spinner has three equal sectors labeled 1, 2 and 3. the second spinner has four equal sectors labeled 3, 4, 5 and 6. the spinners are spun once. what is the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner?
2
6
9
12

 First option is correct.

Step-by-step explanation:

Since we have given that

the first spinner has three equal sectors labelled 1, 2 and 3

The second spinner has four equal sectors labelled 3, 4, 5 and 6.

Sample space will be

So, Number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner :

As we can see that there are 2 such outcomes where we can get the above condition i.e. (2,4) and (3,4).

Hence, First option is correct.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation: