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.

the answer is su.k.my? ? co.kc

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!