I NEED HELP ASAP!!! Will give brainliest !!!
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.
Answer:
I have attached a picture of me solving the equation.
Now you must be wondering how I got the answer, well first simplify the expression in y = mx + b. Then graph it on demos. Do the same thing for the other expression. Pick any point that lies on the line. That is your solution to the equation.
The green line is the expression of 3x - 4y = 11. I simplify the equation in y = mx + b giving me y = 3/4x - 11/4.
The orange line is the expression of 3x + 2y = 2. I simplify the equation in y = mx + b giving me y = -3/2x + 1.
Hope this helps, thank you !!
