In how many ways can the numbers 3,4,5,6,7,8,9, and 10 b divided into two sets A and B such that the sum of the numbers in B is three times the sum of the numbers in A?

pls help quick! will mark brainliest!

4 times

Step-by-step explanation:

Given that the sum of all the numbers => 3+4+5+6+7+8+9+10 = 52

And we are to divide it into two sets such that the sum of set B is three times the number of sum of set A => Sum B = 3 * Sum A

Hence Total sum = Sum B + Sum A

Total sum ÷ 4 = 52 ÷ 4 = 13.

Since Sum B = (3 * Sum A) => (3 * 13 ) = 39

Therefore Sum B = 39 and Sum A = 13

To divide them into two sets

Sum A = (10, 3) = 13: Sum B = ( 4, 5, 6, 7, 8, 9) = 39

Sum A = (9, 4) = 13: Sum B = (3, 5, 6, 7, 8, 10) = 39

Sum A = (8, 5) = 13: Sum B = (3, 4, 6, 7, 9, 10) = 39

Sum A = (7, 6) = 13: Sum B = (3, 4, 5, 8, 9, 10) = 39

Hence, the correct answer is 4 times

a. -0.3 > -0.33