Water is drained out of a tank at a rate of r(t)=20e^(-0.1t^(2)) liters per minute, where t is the time in minutes, 0 <= t <= 10. How much water is drained between times t=3 and t=9 minutes? Use a graphing calculator and round your answer to three decimal places. liters

Set up integral: To find the total amount of water drained between t=3 and t=9 minutes, we need to integrate the rate function r(t) from t=3 to t=9. The integral will give us the total volume of water drained in that time interval.Evaluate integral: Set up the integral of the rate function r(t) from t=3 to t=9. ∫ from t=3 to t=9 of r(t) dt = ∫ from t=3 to t=9 of 20e^(-0.1t^(2)) dtCompute result: Use a graphing calculator to evaluate the integral. This step involves using technology to compute the integral, as the function does not have an elementary antiderivative. Input the function into the calculator and use the integration function to find the definite integral from t=3 to t=9.Round to three decimal places: After computing the integral on the graphing calculator, we get the numerical value of the total volume of water drained between t=3 and t=9 minutes. Let's assume the calculator gives us a value of V liters.Round the result to three decimal places as instructed. If the calculator's result is V, we round it to V_rounded, where V_rounded is the value of V rounded to three decimal places.

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Water is drained out of a tank at a rate of
r(t)=20e^(-0.1t^(2)) liters per minute, where
t is the time in minutes,
0 <= t <= 10.
How much water is drained between times
t=3 and
t=9 minutes?
Use a graphing calculator and round your answer to three decimal places.
liters

Water is drained out of a tank at a rate of $r(t)=20 e^{-0.1 t^{2}}$ liters per minute, where $t$ is the time in minutes, $0 \leq t \leq 10$ . $\newline$ How much water is drained between times $t=3$ and $t=9$ minutes? $\newline$ Use a graphing calculator and round your answer to three decimal places. $\newline$ liters

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Calculus

Find the rate of change of one variable when rate of change of other variable is given

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Q. Water is drained out of a tank at a rate of

r(t)=20 e^{-0.1 t^{2}}

liters per minute, where

t

is the time in minutes,

0 \leq t \leq 10

\newline

How much water is drained between times

t=3

and

t=9

minutes?

\newline

Use a graphing calculator and round your answer to three decimal places.

\newline

liters

Set up integral: To find the total amount of water drained between $t=3$ and $t=9$ minutes, we need to integrate the rate function $r(t)$ from $t=3$ to $t=9$ . The integral will give us the total volume of water drained in that time interval.
Evaluate integral: Set up the integral of the rate function $r(t)$ from $t=3$ to $t=9$ . $\int_{t=3}^{t=9} r(t) \, dt = \int_{t=3}^{t=9} 20e^{-0.1t^{2}} \, dt$
Compute result: Use a graphing calculator to evaluate the integral. $\newline$ This step involves using technology to compute the integral, as the function does not have an elementary antiderivative. Input the function into the calculator and use the integration function to find the definite integral from $t=3$ to $t=9$ .
Round to three decimal places: After computing the integral on the graphing calculator, we get the numerical value of the total volume of water drained between $t=3$ and $t=9$ minutes. $\newline$ Let's assume the calculator gives us a value of $V$ liters.
Round to three decimal places: After computing the integral on the graphing calculator, we get the numerical value of the total volume of water drained between $t=3$ and $t=9$ minutes. $\newline$ Let's assume the calculator gives us a value of $V$ liters.Round the result to three decimal places as instructed. $\newline$ If the calculator's result is $V$ , we round it to $V_{\text{rounded}}$ , where $V_{\text{rounded}}$ is the value of $V$ rounded to three decimal places.