Explain how (13 + 10) + 5 is solved differently from 13 + (10 + 5). a) for (13 + 10) + 5, you add 13 + 5 first, then add 10. for 13 + (10 + 5), you add the 10 + 13 first, then add 5. eliminate b) for (13 + 10) + 5, you add 13 + 10 first, then add 5. for 13 + (10 + 5), you add the 10 + 5 first, then add 13. c) for (13 + 10) + 5, you add 13 - 10 first, then add 5. for 13 + (10 + 5), you add the 10 - 5 first, then add 13. d) for (13 + 10) + 5, you add 13 + 10 first, then subtract 5. for 13 + (10 + 5), you add the 10 + 5 first, then subtract 13.

B)For (13 + 10) + 5, you add 13 + 10 first, then add 5. For 13 + (10 + 5), you add the 10 + 5 first, then add 13.

Step-by-step explanation:

Apply the associative property of addition.

Let , then

In this case a=13,b=10,c=5.

When we substitute into the above relation, we get:

For the LHS, you add 13+5 first, then add 5

For the RHS, you add 10+5 first, then add 13

The correct answer is B

b because it contains 40% of the possible results from the random list.  

0,1,2,3 v 4,5,6,7,8,9  

4 integers to 6 integers  

40% success vs 60% fail

step-by-step explanation: