Question

an element with mass 210 grams decays by 8.3% per minute. how much of the element is remaining after 15 minutes, to the nearest 10th of a gram?

Explanation:

Step1: Identify the decay formula

The formula for exponential decay is \( A = P(1 - r)^t \), where \( A \) is the final amount, \( P \) is the initial amount, \( r \) is the rate of decay (in decimal), and \( t \) is the time.
Here, \( P = 210 \) grams, \( r = 8.3\% = 0.083 \), and \( t = 15 \) minutes.

Step2: Substitute values into the formula

Substitute \( P = 210 \), \( r = 0.083 \), and \( t = 15 \) into the formula:
\( A = 210(1 - 0.083)^{15} \)
First, calculate \( 1 - 0.083 = 0.917 \).
Then, calculate \( 0.917^{15} \). Using a calculator, \( 0.917^{15} \approx 0.260 \) (more precise calculation: \( 0.917^{15} \approx e^{15\ln(0.917)} \approx e^{15\times(-0.0863)} \approx e^{-1.2945} \approx 0.273 \) (wait, let's do it directly: \( 0.917^1 = 0.917 \), \( 0.917^2 = 0.917\times0.917 \approx 0.840 \), \( 0.917^3 \approx 0.840\times0.917 \approx 0.770 \), \( 0.917^4 \approx 0.770\times0.917 \approx 0.706 \), \( 0.917^5 \approx 0.706\times0.917 \approx 0.648 \), \( 0.917^6 \approx 0.648\times0.917 \approx 0.594 \), \( 0.917^7 \approx 0.594\times0.917 \approx 0.545 \), \( 0.917^8 \approx 0.545\times0.917 \approx 0.500 \), \( 0.917^9 \approx 0.500\times0.917 \approx 0.458 \), \( 0.917^{10} \approx 0.458\times0.917 \approx 0.420 \), \( 0.917^{11} \approx 0.420\times0.917 \approx 0.385 \), \( 0.917^{12} \approx 0.385\times0.917 \approx 0.353 \), \( 0.917^{13} \approx 0.353\times0.917 \approx 0.324 \), \( 0.917^{14} \approx 0.324\times0.917 \approx 0.297 \), \( 0.917^{15} \approx 0.297\times0.917 \approx 0.272 \)).

Then, \( A = 210\times0.272 \approx 57.12 \). Wait, let's use a calculator for \( 0.917^{15} \):

Using a calculator, \( 0.917^{15} \approx e^{15\ln(0.917)} \). \( \ln(0.917) \approx -0.0863 \), so \( 15\times(-0.0863) = -1.2945 \), \( e^{-1.2945} \approx 0.273 \). Then \( 210\times0.273 = 57.33 \). Wait, maybe my step-by-step multiplication was wrong. Let's use a calculator for \( 0.917^{15} \):

\( 0.917^{15} \approx 0.273 \) (more accurately, using a calculator: \( 0.917^{15} \approx 0.273 \)). Then \( A = 210\times0.273 = 57.33 \), which rounds to 57.3 grams (to the nearest tenth).

Wait, let's do it more accurately. Let's use a calculator for \( 0.917^{15} \):

\( 0.917^{15} = e^{15\ln(0.917)} \)

\( \ln(0.917) \approx -0.086303 \)

\( 15\times(-0.086303) = -1.294545 \)

\( e^{-1.294545} \approx 0.273 \)

Then \( 210\times0.273 = 57.33 \), which is 57.3 when rounded to the nearest tenth.

Wait, maybe I made a mistake in the initial step. Let's check with a calculator:

\( 0.917^{15} \approx 0.273 \), so \( 210 \times 0.273 = 57.33 \), which is 57.3 grams.

Answer:

57.3