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If the ideal is 50 and the max variation is 7/32in, then that's 7/32 under 50 and 7/32 over 50.
So, 50-7/32= 49 25/32, and 50+7/32 = 50 7/32.
So the acceptable width is 49 25/32 to 50 7/32.
If the ideal is 50 and the max variation is 7/32in, then that's 7/32 under 50 and 7/32 over 50.
So, 50-7/32= 49 25/32, and 50+7/32 = 50 7/32.
So the acceptable width is 49 25/32 to 50 7/32.
Answer;
The acceptable width is 49 25/32 and 50 7/32
That is; 49 25/32 ≤ x ≥ 50 7/32
Explanation;
-If the ideal is 50 and the max variation is 7/32 inches, then that's 7/32 under 50 and 7/32 over 50.
Therefore; 50-7/32= 49 25/32, and 50+7/32 = 50 7/32.
x ≤ 50 7/32 and x ≥ 49 25/32
Thus; the acceptable width is 49 25/32 ≤ x ≥ 50 7/32.
