Question

The ideal width of a certain conveyor belt for a manufacturing plant is 50 in. Conveyor belts can vary from the ideal width by at most 7/32in. Which of the following represents the acceptable widths for the conveyor belt? Check all that apply.
x>=49 25/32
x<=50 7/32
49 25/32<=x<=50 7/32
x+7/32<=50
|x-50|< 7/32

Answer 1

Answer:

The answer is:

$|x-50|\leq \dfrac{7}{32}$

$49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$

Step-by-step explanation:

The ideal  width of a certain conveyor belt for a manufacturing plant is 50 in.

Conveyor belts can vary from the ideal width by at most 7/32 in.

i.e. the absolute change in the width has to be less than or equal to 7/32 in. and it can't exceed more than that.

That is if it will decrease than the  the acceptable width for the conveyor belt is represented by the expression;

$|x-50|\leq \dfrac{7}{32}$

Also on solving it we get:

$\dfrac{-7}{32}\leq x-50\leq \dfrac{7}{32}\\\\\\50-\dfrac{7}{2}\leq x\leq 50+\dfrac{7}{2}\\\\49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$

Hence, the correct choices are:

$|x-50|\leq \dfrac{7}{32}$

$49\dfrac{25}{32}\leq x\leq 50\dfrac{7}{32}$