Find the average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$
The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is (Simplify your answer.)

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Final Answer: The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is $\boxed{13}$.

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Step 1 ：The average value of a function $f(t)$ on the interval $[a, b]$ is given by the formula: \[\frac{1}{b-a}\int_{a}^{b}f(t)dt\]

Step 2 ：In this case, $f(t) = (t-5)^2$, $a = 0$, and $b = 12$. So we need to compute the integral of $(t-5)^2$ from $0$ to $12$ and then divide by $12 - 0 = 12$.

Step 3 ：The average value of the function is calculated to be 13.

Step 4 ：Final Answer: The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is $\boxed{13}$.

Find the average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is (Simplify your answer.)

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Find the average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$
The average value of the function $f(t)=(t-5)^{2}$ on $[0,12]$ is (Simplify your answer.)