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Ronald Fast Food randomly selected
20% of locations and asked the manager about the most popular meat. Of the locations surveyed, beef is the most popular meat at 4 locations. Based on the data, what is the most reasonable estimate of the number of Ronald Fast Food locations where the most popular meat is beef?
Choose 1 answer:
(A) 4
(B) 12
(c) 16
(D) 20

Ronald Fast Food randomly selected $20 \%$ of locations and asked the manager about the most popular meat. Of the locations surveyed, beef is the most popular meat at $4$ locations. Based on the data, what is the most reasonable estimate of the number of Ronald Fast Food locations where the most popular meat is beef? $\newline$ Choose $1$ answer: $\newline$ (A) $4$ $\newline$ (B) $12$ $\newline$ (C) $16$ $\newline$ (D) $20$

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Outcomes of compound events

Full solution

Q. Ronald Fast Food randomly selected

20 \%

of locations and asked the manager about the most popular meat. Of the locations surveyed, beef is the most popular meat at

4

locations. Based on the data, what is the most reasonable estimate of the number of Ronald Fast Food locations where the most popular meat is beef?

\newline

Choose

1

answer:

\newline

(A)

4

\newline

(B)

12

\newline

(C)

16

\newline

(D)

20

Determine percentage of beef popularity: Determine the percentage of locations where beef is the most popular meat based on the survey. $\newline$ Since $4$ locations reported beef as the most popular meat and these locations represent $20\%$ of the total locations, we can set up a proportion to find the total number of locations.
Calculate total number of locations: Calculate the total number of locations using the proportion. $\newline$ If $20\%$ of locations correspond to $4$ locations with beef as the most popular meat, then $100\%$ would represent the total number of locations. We can set up the proportion as follows: $\newline$ $\frac{20\%}{4 \text{ locations}} = \frac{100\%}{x \text{ locations}}$
Solve for total number of locations: Solve for $x$ , which represents the total number of locations. $\newline$ Cross-multiply to solve for $x$ : $\newline$ $20\% \times x = 4 \text{ locations} \times 100\%$ $\newline$ $0.20 \times x = 400$ $\newline$ $x = \frac{400}{0.20}$ $\newline$ $x = 2000$

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