<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:
58 13/32 ft. cubed
Step-by-step explanation:
Horizontal Shape: 6 x 5/2 x 7/4 = 26 1/4 ft.
Vertical Shape: 7/2 x 21/4 x 7/4 = 32 5/32 ft.
Total Volume: 26 1/4 + 32 5/32 = 58 13/32 ft. cubed
If this helped can I get brainliest :)
You answered all of them if there is only 45 and you answered 810810.
Equalizw the denominators so that the nominators will be equal in this wat sonyou can find the value of x if you want i can solve it
