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viktelen [127]
2 years ago
10

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
Mathematics
2 answers:
Alecsey [184]2 years ago
7 0
<span>The number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

</span><span> <span><span> Spinner 1 Spinner 2
</span> <span> 1                   1
</span> <span> 1                   2
</span> <span> 1                   3
</span> <span> 1                   4
</span> <span> 2                  1
</span> <span> 2                  2
</span> <span> 2                  3
</span> <span> 2                  4  = NOT 1, AND 4
</span> <span> 3                  1
</span> <span> 3                  2
</span> <span> 3                  3
</span> <span> 3                  4 = NOT 1, AND 4

Only 2 outcomes have passed the given condition.</span></span></span>
dangina [55]2 years ago
6 0

Answer:  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

{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)}

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.

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