My son heard me read this question out loud and immediately piped up with six. Then he said five and went back to say six again. He is not 100% certain, but would love to hear from others on what they think is the correct answer. OK, Crystal. My pipes are warmed up and ready to go. The answer would be 6 and here is one scenario: First batter hits a triple and is thrown out at the plate trying for an inside the park home run. No runners on and one out. Second batter hits a triple and is thrown out at the plate trying for an inside the park home run. No runners on and two outs. Third batter hits a single. Runner on first and two outs. Fourth batter hits a single. Runners on first and second and two outs. Fifth batter hits a single. Bases loaded and two outs. Sixth batter hits a ball that strikes one of the runners. The runner is declared out and the batter is credited with a single. Six hits, no runs.
5
i believe it is 27 because the most hits you can score in one inning without scoring is 3. 3 multiplied by 9 is 27 so 27
13 times, a team has gotten shut out on 13 hits
3 hits to load bases followed by 2 pick off outs. Then 2 more hits to re-load bases equals five total. The 6th hit comes when a hit ball then hits a runner. The runner is called out but the batter gets credited with a hit.
on April 18th, 2009 the Cleveland Indians set a major league record scoring 14 runs in one inning.[2nd inning]
I don't know what the actual answer is, but I say 6 hits without scoring a run is theoretically possible: three hits to load the bases, two fielder's choice outs of the lead runners, then two more singles to load the bases again. Then the 6th hit would be a batted ball into a baserunner. The baserunner would be the 3rd out and the batter is credited with a hit. Or the 6th hit could be, although unlikely, an infield hit where there's an attempted putout at 1st, then the batter rounds 1st in attempt to go to 2nd and is immediately tagged out before the run crosses the plate.
This must be from the guy who asks how may ways a batter can score from third without a hit. 45 is the answer. Example. First batter singles. Second batter hits into a double play. Next three batters all have infield singles to load the bases. Sixth batter's ground ball strikes the base runner going from 1st-2nd. The batter is out, the ball is dead runners may not advance and the batter gets credited with a base hit. So ... five hits in the inning. The team plays eight more that go the same way. 9x5=45.
An attacker is the one who hits or spikes the ball with the intention of scoring a point for their team.
If it hits the umpire behind home plate, it is a foul ball. If it hits a field umpire who is in bounds, then it is a live ball, meaning that the batter might be thrown out at first, or he might reach first base safely (and then this would be scored as a hit without an error),
I depends. If it hits the half closer to the mount, it's safe. If it hits the half toward the stands, it's out.
The answer is six. Load the bases with 3 singles (3) single up the middle, runner thrown out at home 3 times in a row. (6) The answer is six, but not in the manner listed above, which are actually 3 fielders choices. The real answer is: 2 singles to start the inning. Runners at first and second. 2 more singles with the lead runner thrown out at home. First and second, two out. 1 more single to load the bases. Bases loaded, two out. The sixth batter hits a ball that hits one of the runners. The runner is the third out, and the batter is credited with a single.
poo