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. 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. Which statements about the situation are true? Check all that apply. Set X has 10 possible groupings. X Y Set Y = {ABC, ABD, ABE, ACD, ACE, ADE} If person E must be in each group, then there can be only one group. There are three ways to form a group if persons A and C must be in it.
<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>
Parallel lines will have the same slope but different y-intercepts so again we can use the coordinates to determine the y intercept of the new equation.
So we know that the slope intercept formula is y=mx + b where m is the slope and b is the y-intercept so we will input the x and y coordinates to solve for b.
