Question

problem: a batter’s on - base average (seedi) is a way to rate a hitters success getting on base. it is calculated using the formula: seedi = (h + bb + hbp - cs) / ab, where h is total hits, bb is walks, hbp is hit - by - pitch, cs is caught stealing, and ab is at - bat. the table below lists 5 or fewer batters. calculate the on - base average (seedi) for each batter.

player	t	h	bb	hbp	cs	ab
murphy	883	185	17	2	3	1095
other	927	194	45	3	9	1249
ramon	735	90	325	0	5	1258

find which player’s (seedi, round to the nearest hundredth if necessary)
alison: blank
murphy: blank
cory: blank

Explanation:

Step1: Recall the formula for \(\text{SecA}\)

The formula for \(\text{SecA}\) is given by \(\text{SecA}=\frac{f + b+ s + p - C}{A}\), where \(f\) is total bases, \(b\) is runs, \(s\) is home runs, \(p\) is RBIs (Runs Batted In), \(C\) is strikeouts, and \(A\) is at - bats.

For Player "Allison":

• \(f = 396\), \(b = 195\), \(s = 20\), \(p = 120\), \(C = 8\), \(A = 1320\)
• First, calculate the numerator: \(f + b + s + p - C=396 + 195+20 + 120-8\)
• \(396+195 = 591\), \(591+20 = 611\), \(611 + 120=731\), \(731-8 = 723\)
• Then, calculate \(\text{SecA}=\frac{723}{1320}\approx0.548\)

For Player "Maggie":

• \(f = 385\), \(b = 185\), \(s = 17\), \(p = 105\), \(C = 2\), \(A = 1295\)
• Numerator: \(f + b + s + p - C=385+185 + 17+105-2\)
• \(385 + 185=570\), \(570+17 = 587\), \(587+105 = 692\), \(692-2 = 690\)
• \(\text{SecA}=\frac{690}{1295}\approx0.533\)

For Player "Cora":

• \(f = 387\), \(b = 194\), \(s = 42\), \(p = 219\), \(C = 3\), \(A = 1249\)
• Numerator: \(f + b + s + p - C=387+194 + 42+219-3\)
• \(387+194 = 581\), \(581+42 = 623\), \(623+219 = 842\), \(842-3 = 839\)
• \(\text{SecA}=\frac{839}{1249}\approx0.672\) (Wait, maybe there is a mis - reading of the table. Let's re - check the table values. If the "Other" row for Cora has \(f = 387\), \(b = 194\), \(s = 42\), \(p = 219\), \(C = 3\), \(A = 1249\). But let's check the "Raven" row as well. For "Raven": \(f = 135\), \(b = 90\), \(s = 325\) (this seems too high, maybe a typo, but we will work with the given). But according to the problem, we need to find for Allison, Maggie, and Cora.

Wait, maybe the table columns are: Player (Allison, Maggie, Cora, Raven, and two others), and rows: At Bats (A), Hits (f), Runs (b), Home Runs (s), RBIs (p), Strikeouts (C)? Wait, the original formula is \(\text{SecA}=\frac{f + b + s + p - C}{A}\), so we need to identify \(f\) (hits), \(b\) (runs), \(s\) (home runs), \(p\) (RBIs), \(C\) (strikeouts), \(A\) (at - bats) for each player.

Let's re - interpret the table:

Assume the columns are Player: Allison, Maggie, Cora, (others), and rows:

• At Bats (A): 1320 (Allison), 1295 (Maggie), 1249 (Cora)
• Hits (f): 396 (Allison), 385 (Maggie), 387 (Cora)
• Runs (b): 195 (Allison), 185 (Maggie), 194 (Cora)
• Home Runs (s): 20 (Allison), 17 (Maggie), 42 (Cora)
• RBIs (p): 120 (Allison), 105 (Maggie), 219 (Cora)
• Strikeouts (C): 8 (Allison), 2 (Maggie), 3 (Cora)

Recalculating \(\text{SecA}\) for each:

Allison:

\(f + b + s + p - C=396 + 195+20 + 120 - 8=396+195 = 591; 591+20 = 611; 611+120 = 731; 731 - 8=723\)

\(\text{SecA}=\frac{723}{1320}\approx0.548\)

Maggie:

\(f + b + s + p - C=385+185 + 17+105 - 2=385 + 185=570; 570+17 = 587; 587+105 = 692; 692 - 2 = 690\)

\(\text{SecA}=\frac{690}{1295}\approx0.533\)

Cora:

\(f + b + s + p - C=387+194 + 42+219 - 3=387+194 = 581; 581+42 = 623; 623+219 = 842; 842 - 3=839\)

\(\text{SecA}=\frac{839}{1249}\approx0.672\) (But this seems high, maybe the home runs for Cora is 4 (a typo, 42 seems too high for home runs). If we assume \(s = 4\) instead of 42:

\(f + b + s + p - C=387+194 + 4+219 - 3=387+194 = 581; 581+4 = 585; 585+219 = 804; 804 - 3 = 801\)

\(\text{SecA}=\frac{801}{1249}\approx0.641\)

But since the problem is about calculating \(\text{SecA}\) for each player, we will proceed with the given values.

Answer:

• Allison: \(\frac{723}{1320}\approx0.548\)
• Maggie: \(\frac{690}{1295}\approx0.533\)
• Cora: \(\frac{839}{1249}\approx0.672\) (or adjusted if there is a typo in the table)