[brainliest! ] on mina's journey to mexico the plane flies the 2000 km distance at 1600 km/h.
on the way back there is a head wind and the plane only flies at a speed of 1000 km/h.
find the average speed for the two journeys. give your answer to the nearest
km/h.
give and explanation with your answer! ! you will get brainliest. : )

  1231 km/h

Step-by-step explanation:

The average speed is given by ...

  average speed = (total distance)/(total time)

The total time will be the sum of times down and back, each of which is given by ...

  time = distance/speed

Going to Mexico, the time required was ...

  time down = (2000 km)/(1600 km/h) = 1.25 h

Coming back, the time required was ...

  time back = (2000 km)/(1000 km/h) = 2.00 h

Then the average speed is ...

  average speed = (2000 km + 2000 km)/(1.25 h +2.00 h) = 4000/3.25 km/h

  ≈ 1231 km/h

Comment on average speed

The value we computed was ...

  2/(1/s1 + 1/s2) . . . . where s1 and s2 are the speeds over the same distance

This is the harmonic mean of two numbers (the two speeds). The harmonic mean of n numbers is ...

  harmonic mean = n/(1/a1 +1/a2 +1/a3 + ... + 1/an)

This mean generally finds less use than the typical arithmetic mean or geometric mean.

  arithmetic mean = (a1 + a2 + a3 + ... + an)/n

  geometric mean = (a1·a2·a3·...·an)^(1/n)

x

step-by-step explanation:

mxb67

step-by-step explanation: