On Christmas Day 2009, there was a bomb scare aboard a Detroit-bound plane. Soon afterwards, a poll indicated that 42% of Americans were worried that they or a close friend or family member would become a victim of terrorism. The poll had a 5% margin of error. Create an absolute value inequality that describes the ranges of possible poll percentages and rewrite it as an inequality.
I would be scared tbh...
5% of 42 = 2.1
So, 45-2.1 <= x <= 45+2.1
To create the absolute value inequality that describes the ranges of possible poll percentages with a 5% margin of error, we need to consider the maximum and minimum values.
The maximum value would be the poll percentage when it is increased by the margin of error (5% added to 42%):
Maximum value = 42% + 5% = 47%
The minimum value would be the poll percentage when it is decreased by the margin of error (5% subtracted from 42%):
Minimum value = 42% - 5% = 37%
Therefore, the absolute value inequality that describes the ranges of possible poll percentages would be:
|p - 42| ≤ 5
Where "p" represents the poll percentage.
To rewrite this inequality, we can expand the absolute value expression and split it into two separate inequalities:
1. p - 42 ≤ 5
2. -(p - 42) ≤ 5
1. By solving the first inequality:
p - 42 ≤ 5
p ≤ 5 + 42
p ≤ 47
2. By solving the second inequality:
-(p - 42) ≤ 5
-p + 42 ≤ 5
-p ≤ 5 - 42
-p ≤ -37
Multiply both sides by -1 (which reverses the inequality):
p ≥ 37
Therefore, the rewritten inequality is:
37 ≤ p ≤ 47