Help (4/d)^3 anyway help
1 answer:
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First, find what percentage of students had 3 or more by adding up your known percents:
45% + 23 % + 21% + x% = 100%
x = 11%
Since you're given that 96 students had 2 or more, you add up the percentages of 2 and 3 or more:
11 + 21 = 32%
Now set up a proportion that relates it to the whole:
This will allow you to find the total number of students at the school.
Cross multiplying and solving for x results in 300 total students.
Question 1:
45% had one or more absences. 45% of 300 students is 135 students.
Question 2:
As we found before, 11% of students had three or more. 11% of 300 is 33 students.
45% + 23 % + 21% + x% = 100%
x = 11%
Since you're given that 96 students had 2 or more, you add up the percentages of 2 and 3 or more:
11 + 21 = 32%
Now set up a proportion that relates it to the whole:
This will allow you to find the total number of students at the school.
Cross multiplying and solving for x results in 300 total students.
Question 1:
45% had one or more absences. 45% of 300 students is 135 students.
Question 2:
As we found before, 11% of students had three or more. 11% of 300 is 33 students.