Question

The earthquake monitoring station in Albuquerque, NM, is at an approximate elevation of 1849 meters. The station in Hockley, TX, is approximately 2264 metersbelow the elevation of the Albuquerque station. Which expression gives the elevation of the Hockley station?O A. -2264 - 1849|OB. 1849 – 2264OC. 2264 - 1849O D. 1849] + [2264

Answer 1

We know the elevation of the station in Albuquerque, which is 1849 meters, and we know that this station is about 2264 meters above the station located in Hockley, then the elevation of the last station can be calculated by subtracting the absolute value of the vertical distance between the two stations (2264) from the elevation of the Alburqueue station (1849), like this:

Elevation = |1849| - |2264|

When we solve this expression, we get a negative result, which is -415, since we are calculating the elevation with respect to the sea level, this means that the station in Hockley is 415 meters below the sea level.

Then, the correct answer is option B. |1849| - |2264|