Let the households with cars be represented by = c
As given 52% households have 2 or more cars and margin of error is +- 4%
The equation becomes
[tex]|c-52|\leq4[/tex]
Solving this we get,
[tex]-4\leq c-52\leq4[/tex]
[tex]-4+52\leq c\leq 4+52[/tex]
[tex]48\leq c\leq56[/tex]
Hence, the least percent is 48% and greatest percent is 56%
Answer:
[tex]48\leq x\leq 56[/tex].
Step-by-step explanation:
Let x represent percent of households with two or more cars.
We will use margin of error formula to solve our given problem.
[tex]|\text{Actual}-\text{Ideal}|\leq\text{Tolerance}[/tex]
Substitute the given values:
[tex]|x-52|\leq4[/tex]
Using absolute value property [tex]|u|\leq a[/tex], then [tex]-a\leq u\leq a[/tex], we will get:
[tex]-4\leq x-52\leq4[/tex]
[tex]-4+52\leq x-52+52\leq4+52[/tex]
[tex]48\leq x\leq 56[/tex]
Therefore, our required inequality would be [tex]48\leq x\leq 56[/tex] and least percent is 48% and greatest percent is 56%.
