Ametronome uses a pendulum to keep time. the table gives the angular displacement of a pendulum as it swings back and forth. what is the period of its oscillations? time angle 0.0 15.0 0.1 2.55 0.2 -14.4 0.3 -7.35 0.4 11.6 0.5 11.3

With the given information, it appears that the period of the pendulum is about 0.45 seconds.

The period is the length of one complete cycle of the pendulum. That mean that is will start positive go negative and return to positive. 

If we assume that time 0 is the start with an angle of 15  degree, then we are looking for the point in the graph where it returns to about 15 degrees.

If you create a scatter plot of the data, it would appear that it would reach 15 between 0.4 and 0.5 seconds. At 0.4 it is at 11.6, then it moves up. At 0.5, it is on it's way down at 11.3.

It is safe to assume that it reached 15 between these points to complete one period.

Lol english

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.