Question

1. dwaine tomlinson dirige un programa de baloncesto en california. el primer día de la temporada se presentaron 60 jóvenes y fueron clasificados por nivel de edad y por su preferencia en la posición de juego, como se muestra en la siguiente tabla. utilizando el conjunto de etiquetas (letras) en la tabla, encuentre el número de jugadores en cada uno de los siguientes conjuntos. (a) j∩g (b) s∩n (c) n∪(s∩f) (d) s′∩(g∪n) (e) (s∩n′)∪(c∩g′) (f) n′∩(s′∩c′)

posición

	guardia (g)	delantero (f)	centro (n)	totales
preparatoria (s)	12	5	9	26
universidad (c)	5	8	2	15
totales	26	19	15	60

Explanation:

Step1: Recall set - intersection and union definitions

Intersection ($\cap$) gives the elements common to both sets, and union ($\cup$) gives all the elements in either set.

Step2: Solve (a) $J\cap G$

We look for the number of elements that are in both the "Secundaria ($J$)" row and the "Guardia ($G$)" column. From the table, the value is 9.

Step3: Solve (b) $S\cap N$

We find the number of elements in the "Preparatoria ($S$)" row and the "Centro ($N$)" column. From the table, the value is 9.

Step4: Solve (c) $N\cup(S\cap F)$

First, find $S\cap F$. The number of elements in the "Preparatoria ($S$)" row and the "Delantero ($F$)" column is 5. Then, for $N\cup(S\cap F)$, we use the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Here, $n(N) = 15$ and $n(S\cap F)=5$, and $n(N\cap(S\cap F)) = 0$. So $n(N\cup(S\cap F))=15 + 5-0=20$.

Step5: Solve (d) $S'\cap(G\cup N)$

$S'$ is the complement of $S$, so it is the set of non - preparatoria students (i.e., $J$ and $C$). $G\cup N$ is the set of guards or centers. First, find the number of elements in $J\cap(G\cup N)$ and $C\cap(G\cup N)$. For $J\cap(G\cup N)$, we have $9 + 4=13$. For $C\cap(G\cup N)$, we have $5+2 = 7$. So $n(S'\cap(G\cup N))=13 + 7=20$.

Step6: Solve (e) $(S\cap N')\cup(C\cap G')$

For $S\cap N'$, we look at the "Preparatoria ($S$)" row and non - "Centro ($N$)" columns. So $n(S\cap N')=12 + 5=17$. For $C\cap G'$, we look at the "Universidad ($C$)" row and non - "Guardia ($G$)" columns. So $n(C\cap G')=8 + 2=10$. Then $n((S\cap N')\cup(C\cap G'))=17+10 - 0=27$.

Step7: Solve (f) $N'\cap(S'\cap C')$

$S'\cap C'$ is the set of non - preparatoria and non - university students, which is just the "Secundaria ($J$)" students. $N'$ is the set of non - centers. For $J\cap N'$, we have $9+6 = 15$.

Answer:

(a) 9
(b) 9
(c) 20
(d) 20
(e) 27
(f) 15