<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.
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<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.
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<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.
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<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:
6v^5
Step-by-step explanation:
you would leave the 6 alone. since there is 5 of the same variable, you would put 5 as your exponent
Answer:
- <u>You need to estimate: approximately 0.84</u>
Explanation:
Please, find attached a figure with the complete question, including the probability distribution.
<h2>Solution</h2>
The graph does not permit to read exact values, thus you must estimate.
P(X ≥ 8) = P(X=8) + P(X=9) + P(X=10)
- P(X≥8) ≈ 0.62 + 0.14 + 0.08 = 0.84
