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During the 2012-2013 NBA season, Jamal Crawford playing for the Los Angeles Clippers, had a free throw percentage of 0.871. Using complete sentences, describe the likelihood that he would make the free throw each time he stepped up to the line that season. Explain your reasoning.

Respuesta :

igoroleshko156

Answer:

When Jamal Crawford comes to the free throw line, in the 2012-2013 NBA season, he would make 871 free throws out of 1000.  Therefore, his percentage of making a free throw would by %87.1

Jamal Crawford percentage of making a free throw would by 87.1%

What is percentage?

Percentage is defined as ratio expressed as a fraction of 100.

When Jamal Crawford comes to the free throw line, in the 2012-2013 NBA season,

He had a free throw percentage = 0.871 = 871/1000

So he would make 871 free throws out of 1000.  

Therefore, his percentage of making a free throw would by 87.1%

Learn more about percentage here:

brainly.com/question/24159063

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