i am studing about seafloor spreading rate i need to know how to calculate the average age and distance for this pair of peaks 12.5 & 11.0

You have not sufficiently explained what peaks you are talking about, nor the units associated with those numbers.

I get you that's the same question i have

To calculate the average age and distance for a pair of seafloor peaks, you will need to use the concept of seafloor spreading rates. Let's break down the calculation into steps:

1. Determine the time difference between the two peaks:
To calculate the age difference between the pair of peaks, subtract the age of the older peak from the age of the younger peak. In this case, subtract 11.0 years (the younger peak) from 12.5 years (the older peak). The result would be the time difference between the two peaks, which is 1.5 years.

2. Determine the distance between the two peaks:
Firstly, it's important to understand that seafloor spreading rates are typically measured in centimeters per year (cm/yr). If we assume a constant spreading rate, we can estimate the distance between the two peaks by multiplying the time difference (1.5 years) by the seafloor spreading rate.

However, since you haven't provided a specific spreading rate, let's use an average spreading rate of 2.5 cm/yr for this explanation.

Multiply the time difference (1.5 years) by the assumed spreading rate (2.5 cm/yr) to calculate the distance:
Distance = Time difference × Spreading rate
Distance = 1.5 years × 2.5 cm/yr
Distance = 3.75 cm

3. Calculate the average age:
To calculate the average age of the two peaks, add the age of the younger peak to half of the time difference between the two peaks. In this case, add 11.0 years (the younger peak) to half of the time difference, which is 0.75 years (half of 1.5 years).
Average age = Age of younger peak + (Time difference / 2)
Average age = 11.0 years + (1.5 years / 2)
Average age = 11.0 years + 0.75 years
Average age = 11.75 years

So, for the given pair of peaks (12.5 & 11.0) with a time difference of 1.5 years, the estimated distance is 3.75 cm, and the average age is 11.75 years (assuming a spreading rate of 2.5 cm/yr).