An element with mass 650 grams decays by 29.5% per minute. How much of the element is remaining after 20 minutes, to the nearest 10th of a gram? Answer:

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An element with mass 650 grams decays by  29.5% per minute. How much of the element is remaining after 20 minutes, to the nearest 10th of a gram? Answer:

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Identify: Identify the initial amount, decay rate per minute, and total time. Initial amount (a) = 650 grams Decay rate per minute (r) = 29.5% Total time (t) = 20 minutes We need to calculate the remaining amount of the element after 20 minutes.Convert Rate: Convert the decay rate from a percentage to a decimal. To convert a percentage to a decimal, divide by 100. r = 29.5% / 100 = 0.295Calculate Percentage: Calculate the remaining percentage after each minute. Since the element decays by 29.5% each minute, it retains 100% - 29.5% = 70.5% of its mass each minute. Remaining percentage per minute = 100% - 29.5% = 70.5% Convert this to a decimal: 70.5% / 100 = 0.705Use Exponential Decay: Use the exponential decay formula to calculate the remaining amount after 20 minutes. The formula for exponential decay is: remaining amount = initial amount * (1 - decay rate)^time Substitute the values into the formula: remaining amount = 650 * (0.705)^20Calculate Remaining Amount: Calculate the remaining amount using the values from Step 4. remaining amount = 650 * (0.705)^20 Use a calculator to find (0.705)^20. (0.705)^20 ≈ 0.016258 Now, multiply this by the initial amount: remaining amount ≈ 650 * 0.016258 remaining amount ≈ 10.5677Round to Nearest: Round the remaining amount to the nearest 10th of a gram. The remaining amount rounded to the nearest 10th of a gram is approximately 10.6 grams.

An element with mass $650$ grams decays by $29.5 \%$ per minute. How much of the element is remaining after $20$ minutes, to the nearest $10$ th of a gram? $\newline$ Answer:

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An element with mass $650$ grams decays by $29.5 \%$ per minute. How much of the element is remaining after $20$ minutes, to the nearest $10$ th of a gram? $\newline$ Answer: