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.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6