Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by
.
a) What was the average score when they initially took the test, t = 0?
S(t) =28 - 20 log (t+1), t>(greater than) 0
You say your equation is not valid for t=0. Why not?
I would assume S(0) = 28 - 20log1 = 28
Note that negative scores are possible for n>26, which seems unlikely
To find the average score when the students initially took the test (t = 0), we can substitute t = 0 into the equation S(t) = 28 - 20 log (t+1).
Let's perform the substitution:
S(0) = 28 - 20 log (0+1)
S(0) = 28 - 20 log (1)
S(0) = 28 - 20 * 0 (since log(1) equals 0)
S(0) = 28 - 0
S(0) = 28
Therefore, the average score when the students initially took the test (t = 0) is 28%.