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

Understand and Set Up: Understand the problem and set up the integral. We need to calculate the total amount of water drained from the tank between t=3 and t=9 minutes. The rate of water being drained is given by the function r(t)=20e^(-0.1t^2). To find the total amount of water drained, we need to integrate this rate function from t=3 to t=9.Set Up Integral: Set up the definite integral for the given rate function between the limits of integration. The integral we need to solve is ∫ from t=3 to t=9 of 20e^(-0.1t^2) dt.Evaluate Integral: Use a graphing calculator to evaluate the integral. Since the integral involves an exponential function with a squared term in the exponent, it does not have an elementary antiderivative. Therefore, we will use a graphing calculator to evaluate the integral numerically.Calculate Integral: Enter the function and limits into the graphing calculator and calculate the integral. After entering the function 20e^(-0.1t^2) into the graphing calculator and setting the limits from t=3 to t=9, the calculator will provide the numerical value of the integral.Round Result: Round the result to three decimal places as instructed. Assuming the graphing calculator gave us a result, we would round this result to three decimal places to get our final answer.

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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? Use a graphing calculator and round your answer to three decimal places. $\newline$ $\_\_\_\_\_$ liters

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

\newline

\_\_\_\_\_

liters

Understand and Set Up: Understand the problem and set up the integral. $\newline$ We need to calculate the total amount of water drained from the tank between $t=3$ and $t=9$ minutes. The rate of water being drained is given by the function $r(t)=20e^{-0.1t^2}$ . To find the total amount of water drained, we need to integrate this rate function from $t=3$ to $t=9$ .
Set Up Integral: Set up the definite integral for the given rate function between the limits of integration. The integral we need to solve is $\int_{t=3}^{t=9} 20e^{-0.1t^2} \, dt$ .
Evaluate Integral: Use a graphing calculator to evaluate the integral. $\newline$ Since the integral involves an exponential function with a squared term in the exponent, it does not have an elementary antiderivative. Therefore, we will use a graphing calculator to evaluate the integral numerically.
Calculate Integral: Enter the function and limits into the graphing calculator and calculate the integral. $\newline$ After entering the function $20e^{(-0.1t^2)}$ into the graphing calculator and setting the limits from $t=3$ to $t=9$ , the calculator will provide the numerical value of the integral.
Round Result: Round the result to three decimal places as instructed. $\newline$ Assuming the graphing calculator gave us a result, we would round this result to three decimal places to get our final answer.