Question

1. two players have the same number of at-bats (250). player a has 35 singles, 15 doubles, 5 triples, and 20 home runs. player b has 50 singles, 10 doubles, 3 triples, and 10 home runs. who has the higher slugging percentage?

Explanation:

Step1: Recall Slugging Percentage Formula

Slugging percentage (\(SLG\)) is calculated as \(\frac{(1\times \text{singles}) + (2\times \text{doubles}) + (3\times \text{triples}) + (4\times \text{home runs})}{\text{at - bats}}\)

Step2: Calculate Slugging Percentage for Player A

For Player A:
Singles = 35, Doubles = 15, Triples = 5, Home Runs = 20, At - bats = 250
First, calculate the numerator:
\(1\times35 + 2\times15+3\times5 + 4\times20=35 + 30+15 + 80\)
\(35+30 = 65\), \(65 + 15=80\), \(80+80 = 160\)
Then, \(SLG_A=\frac{160}{250}=0.64\)

Step3: Calculate Slugging Percentage for Player B

For Player B:
Singles = 50, Doubles = 10, Triples = 3, Home Runs = 10, At - bats = 250
First, calculate the numerator:
\(1\times50+2\times10 + 3\times3+4\times10=50 + 20+9 + 40\)
\(50+20 = 70\), \(70 + 9=79\), \(79+40 = 119\)
Then, \(SLG_B=\frac{119}{250}=0.476\)

Step4: Compare the two slugging percentages

Since \(0.64>0.476\), Player A has a higher slugging percentage.

Answer:

Player A