In a test for the breaking strength of concrete slabs for freeway construction, 3 of the 200 slabs tested did not meet the safety requirements. What percent of the slabs tested met the safety requirements?
2007-06-03 18:01:05 · 13 answers · asked by Transformers20 2 in Science & Mathematics ➔ Mathematics
98.5%
2007-06-03 18:03:07 · answer #1 · answered by hornedphrog 2 · 1⤊ 1⤋
3 were faultly so the # of slabs that were not faulty is(200-3)=197
so the percentage that passes the safety requirements is:
197\200*100%= 98.5%
2007-06-03 18:05:27 · answer #2 · answered by Dr. Eddie 6 · 1⤊ 1⤋
there are 200 - 3 = 197 labs that meet the safety requirements.
197/200 = .985 or 98.5%
2007-06-03 18:04:47 · answer #3 · answered by 7 · 1⤊ 1⤋
i.e., 197 out of 200 met the safety requirements.
(197/200)*100%=98.5%
2007-06-03 18:04:44 · answer #4 · answered by Jain 4 · 1⤊ 1⤋
Just how were they tested, anyway?
Your method could be flawed, skewing the results in an unintended direction!!!
Are your lawyers well versed in engineering liability?
2007-06-03 18:12:58 · answer #5 · answered by Revenant Hamster 4 · 1⤊ 0⤋
If it is the same batch of concrete than 0%. Any failure tosses the entire batch.
2007-06-03 18:45:01 · answer #6 · answered by jd 3 · 1⤊ 0⤋
The Answer is 98.5%
2007-06-03 18:07:07 · answer #7 · answered by Prince O 2 · 1⤊ 1⤋
I think you do not have questions to ask thats y u r asking such foolish questions
2007-06-03 18:05:47 · answer #8 · answered by SHAIKH 1 · 1⤊ 2⤋
98.5
2007-06-03 18:07:49 · answer #9 · answered by ★Greed★ 7 · 1⤊ 1⤋
98.5% i think
2007-06-03 18:04:11 · answer #10 · answered by Anonymous · 1⤊ 1⤋