To determine the equation that best represents the scenario of the distance a major-league outfielder throws a baseball, we need to look for a relationship between the speed of the throw and the time it is thrown.
Given that:
- The maximum throwing speed is 105 miles per hour, which converts to 154 feet per second.
- Assuming the time of the throw is 2.5 seconds or 90 seconds, we can find the equations based on distance, where distance = speed × time.
If we consider the specific options provided and calculate:
-
For 154 × 2.5, the distance thrown would be:
\( 154 , \text{ft/s} \times 2.5 , \text{s} = 385 , \text{feet} \) -
For 105 × 2.5, it results in:
\( 105 , \text{mph} \) (which needs to be converted to feet per second) is \( 154 , \text{ft/s} \), so this calculation is essentially covering the same speed.
But when multiplying it by 2.5, it suggests a time extension: \( 105 , \text{mph} \times 2.5 , \text{sec (not converted)} \) is not coherent without conversion. -
For 154 × 90:
\( 154 , \text{ft/s} \times 90 , \text{s} = 13,860 , \text{feet} \) -
For 105 × 90:
Again reflecting back to miles per hour but with no direct conversion resulting in
\( 105 , \text{mph} \times 90 , \text{seconds (also needs conversion)} \) is misleading.
The most meaningful and coherent equation based on the scenario given, particularly centered around throwing distance per unit time and reflective of linear motion physics around the provided speed would be:
154 ⋅ 2.5 = 385
Thus the equation that represents this scenario is:
154 ⋅ 2.5 = 385 (154 feet per second for 2.5 seconds results in a distance of 385 feet).
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