Understanding Ticker Tape: Solving for Period and Frequency

[davie08]

Homework Statement

A recording timer is used to produce a section of ticker tape. 165 dots were
produced over a 2.5 s time interval.
a) Determine the period of the timer. 0.015 seconds
b) Determine the frequency of the timer. 66 hertz
c) What is the change in time between the ninth and thirteenth dot?
Im taking this online and for this quiz it didn't have the answers to check your work, I'm just unsure of how to do C.

Homework Equations

T=(change in time)/N(vibrations) f(frequency)=N/(change in time)

The Attempt at a Solution

T=(change in time)/N 0.015=(change in time)/4 = 0.06 seconds

 

[CompuChip]

In a, you calculated that there is one dot every 0.015 seconds, right?
In other words, 0.015 seconds is the time between any dot and the next.
How long is there between any dot and the second next dot (e.g. between 1 and 3, or between 9 and 11)?

(By the way, a and b are correct, although I would prefer to write 2.5/165, 5/330 or 1/66 second rather than the rounded 0.015).