We know that
with
scale factor 1=1 in/4 m
<span>the longer side of the ballroom-------> 44 m
scale factor= measure model/measure actual
measure model=scale factor*measure actual
</span>measure model=(1/4)*44------> 11 in
with
scale factor 2=1 in/7 m
measure actual=measure model/scale factor
measure actual=11/(1/7)-----> 77 m
the answer is
77 m
X=5 is a definite answer and one solution. While X greater than OR equal to 5 means that it could be 5 AND anything above 5, which gives multiple of choices for solutions. (I believe this write but it’s been awhile since I did algebra stuff)
27,491,739 times 93,792 is 2,578,505,184,288.
<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>