Answer:
the second one
Step-by-step explanation:
i did it in my head and double checked :)
<span>If set X is made up of the possible ways five students, represented by A, B, C, D, and E, can be formed into groups of three, then the set X consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE} (note that triple ABC is the same as triple ACB, or BCA, or BAC, or CAB, or CBA). The set X totally contains 10 elements (triples). The first statement is true.
</span>
<span>If set
Y is made up of the possible ways five students can be formed into
groups of three if student A must be in all possible groups, then </span><span>the set Y consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE} and contains 6 elements. The second statement is also true.
</span>
<span>If person E must be in each group, then there can be only one group is false statement, because you can see from the set X that triples which contains E are 6.
</span>
<span>There are three ways to form a group if persons A and C must be in it. This statement is true and these groups are ABC, ACD, ACE.</span>
Answer:
(0,0)
Step-by-step explanation:
Okay. You have 15 lbs of concrete, and x bags that can individually hold 2.5 lbs. What you need to do is divide 15 by 2.5, and the quotient will be 6.
Problem: 15 ÷ 2.5 = 6
So the answer to your problem will be: 6 bags.
