Question

1odel f(t)=90-15log(t+1),0<=t<=12 where 3. What was the average score on the orig

Answer 1

The average score on the original model f(t)=90-15log(t+1), 0<=t<=12, was approximately 55.84.

Step-by-step explanation:

The average of a function over an interval [a, b] is calculated using the formula:

`\[ \bar{f} = (1)/(b - a) \int_(a)^(b) f(t) \, dt \]`

In this case, the function is
`\( f(t) = 90 - 15 \log(t+1) \)`and the interval is [0, 12]. So, the average score is given by:

`\[ \bar{f} = (1)/(12 - 0) \int_(0)^(12) (90 - 15 \log(t+1)) \, dt \]`

To evaluate the integral, we can use the antiderivative of
`\( f(t) \)`, which is:

`\[ F(t) = 90t - 15(t \log(t+1) - t) \]`

Now, applying the fundamental theorem of calculus, we have:

`\[ \bar{f} = (1)/(12 - 0) [F(12) - F(0)] \]`

After substituting the values, we get:

`\[ \bar{f} = (1)/(12) [(90 \cdot 12 - 15(12 \log(13) - 12)) - (90 \cdot 0 - 15(0 \log(1) - 0))] \]`

Solving this expression yields the final result of approximately 55.84.

In summary, the average score on the original model over the interval [0, 12] is 55.84. This calculation involves finding the antiderivative of the function and applying the fundamental theorem of calculus to evaluate the average value over the given interval.