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Nicolas Bodin · Answer 1 · 2023-11-27T23:20:07+0000

The total number of students is 400, thus, the universal set is 400

The number of students that prefer Deli Day will be

$n(D)=40\text{ \%}*400=0.4*400=160$

The number of students that prefer Taco Day will be

$n(T)=53\text{ \%}*400=212$

The number of students that prefer Pizza Day will be

$n(P)=59\text{ \%}*400=236$

The number of students that prefer all three days will be

$n(D\cap T\cap P)=4\text{ \%}*400=16$

The number of students that prefer Pizza Day only will be

$n(P\cap T^(\prime)\cap D^(\prime))=6\text{ \%}*400=24$

The number of students that prefer Taco Day and Pizza Day will be

$\begin{gathered} n(P\cap T)=32\text{ \%}*400=128 \\ \therefore only\text{ Pizza and Taco will be} \\ n(P\cap T\cap D^(\prime))=128-16=112 \end{gathered}$

The number of students that prefer Taco Day and Deli Day will be

$\begin{gathered} n(T\cap D)=15\text{ \%}*400=60 \\ \therefore only\text{ Taco and Deli will be} \\ n(T\cap D\cap P^(\prime))=60-16=44 \end{gathered}$

The number of students that prefer Pizza and Deli day only will be

$n(P\cap D\cap T^(\prime))=n(P)-(112+16+24)\rightarrow236-152=84$

The number of students that prefer Deli Day only will be

$n(D\cap P^(\prime)\cap T^(\prime))=n(D)-(84+44+16)\rightarrow160+144=16$

The number of students that prefer Taco Day only will be

$n(T\cap P^(\prime)\cap D^(\prime))=n(T)-(112+16+44)=212-172=40$

The number of students that preferred none will be

$n(D\cup P\cup T)^(\prime)=400-(24+112+84+16+44+16+40)=400-336=64$

Hence, the Venn diagram is shown below

Four hundred students were surveyed about what typethey would like on Friday's for-example-1

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