n a survey of 60 randomly-chosen employees, 15 said they would prefer an hour for lunch with no breaks each day. 45 employees said they would prefer thirty minutes for lunch with two other breaks each day. If there are 500 people employeed by the company, predict the number who would prefer an hour for lunch with no breaks
15/60 or 1/3 chose an hour lunch with no breaks
for in 500 people we would expect
1/3 of 500 or 167 to choose the above option
or, use 15/60 = 1/4 if you want a clser approximation! ...
got me,
( I think 15/60 = 1/3 in non-metric)
anyway...
Thank you Steve
To predict the number of employees who would prefer an hour for lunch with no breaks, we can use proportional reasoning.
First, let's calculate the proportion of employees who prefer thirty minutes for lunch with two other breaks. Out of the 60 surveyed employees, 45 said they preferred this option. We can represent this proportion as:
30-minute lunch with two breaks / Total surveyed employees = 45 / 60
To find the proportion of employees who prefer an hour for lunch with no breaks, we can subtract the above proportion from 1 (since the remaining proportion would represent the hour lunch preference):
1 - 45 / 60 = 15 / 60
To predict the number of employees who would prefer an hour for lunch with no breaks out of the 500 total employees, we can multiply the proportion by the total number of employees:
Number of employees who prefer an hour for lunch with no breaks = (15 / 60) * 500
Calculating this expression:
Number of employees who prefer an hour for lunch with no breaks = 0.25 * 500 = 125
Therefore, we can predict that around 125 employees would prefer an hour for lunch with no breaks.