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? Round your answer to a whole percent, if necessary.
S(t)= 68 - 20 log (t+1), t > 0
what's the problem? Just plug in t=0:
S(0) = 68-20log(1) = 68 - 20(0) = 68
Although, as written, S(0) is not defined, since it is only for t>0, not t=0.
Looks like the scores keep going down and down. Is the problem written correctly?
WHAT WOULD THE SCORE BE AFTER 4 MONTHS
To find the average score when they initially took the test (t = 0), we need to substitute t = 0 into the given equation and solve for S(0).
Given the equation:
S(t) = 68 - 20 log(t+1)
Substituting t = 0:
S(0) = 68 - 20 log(0+1)
Since log(0) is undefined, we cannot directly substitute t = 0 into the equation. However, we can use the limit as t approaches 0 to find the answer.
Taking the limit as t approaches 0:
lim (t->0) (68 - 20 log(t+1))
Using the limit properties, we can evaluate the limit separately for each term:
lim (t->0) 68 - lim (t->0) 20 log(t+1)
The limit of 68 is simply 68:
68 - lim (t->0) 20 log(t+1)
Now, let's consider the second term. As t approaches 0, log(t+1) approaches log(1) = 0:
68 - 20 log(1)
68 - 20 * 0
68 - 0
= 68
Therefore, the average score when they initially took the test (t = 0) is 68 percent.