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A student conducts several random surveys to find out the most preferred lunch option. Each survey was of 60 students. No student was surveyed twice. The table shows the results.

Which two statements are valid inferences that can be made based on the data collected?

A student conducts several random surveys to find out the most preferred lunch option Each survey was of 60 students No student was surveyed twice The table sho class=

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BrainlyG13

Answer:

A

C

is the answer

I think soo

Options B, D and E are correct.

What is percentage?

A relative value that indicated hundredth part of any quantity is called percentage.

How to find which two statements are valid?

Random surveys of 60 students are taken. The table given shows the results.

How to find whether option A is correct ?

From the given table, we can say 17 mean number of students prefer taco option at lunch.

Calculating the percentage,

          [tex](\frac{17}{60} )[/tex] x 100 %= 28.33 %

So, we can say, 28.33 % students prefer taco option at lunch

  • In option A it is given , every 7 out of 15 students prefer taco option at lunch.
  • The percentage should match in both the cases.

Calculating the percentage,

          [tex]\frac{7}{15}[/tex]   x 100 %= 46.67 %

So, according to option A, 46.67 % students prefer taco option at lunch, which does not matches.  

So, option A is not correct.

How to find whether option B is correct ?

From the given table, we can say 16 mean number of students prefer chicken option at lunch.

Calculating the percentage,

             [tex](\frac{16}{60})[/tex] x 100 % = 26.67 %

So, we can say,26.66 % students prefer chicken option at lunch.

  • In option B it is given, every 4 out of 15 students prefer chicken option at lunch.
  • The percentage should match in both the cases.

Calculating the percentage,

            [tex](\frac{4}{15} )[/tex] x 100 % = 26.67 %

So, according to option B, 26.67 % students prefer chicken option at lunch.

Clearly option B matches and hence it is correct.

How to find whether option C is correct ?

From the given table, we can say 6 mean number of students prefer vegetarian option at lunch.

Calculating the percentage,

                   [tex](\frac{6}{60} )[/tex] x 100 = 10 %

So, we can say,10 % students prefer vegetarian option at lunch.

  • In option C it is given, every 1 out of 6 students prefer vegetarian option at lunch.
  • The percentage should match in both the cases.

Calculating the percentage,

                       [tex](\frac{1}{6} )[/tex] x 100 % = 16.67 %

So, according to option C, 16.67 % students prefer vegetarian option at lunch, which does not matches.  

So, option C is not correct.

How to find whether option D is correct ?

From the given table, we can say 21 mean number of students prefer pizza option at lunch.

Calculating the percentage,

                       [tex](\frac{21}{60} )[/tex] x 100 % = 35 %

This clearly matches with the option.

So, option D is correct.

How to find whether option E is correct ?

From the given table, we can say 16 mean number of students prefer chicken option at lunch.

This clearly matches with the option and hence option E is correct.

You can get more details on: https://brainly.com/question/24862040

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