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A radioactive compound with mass 280 grams decays at a rate of
18.3% per hour. Which equation represents how many grams of the compound will remain after 2 hours?

C=280(1.183)^(2)

C=280(1-0.183)(1-0.183)

C=280(1+0.183)^(2)

C=280(1-0.183)(1-0.183)(1-0.183)(1-0.183)

A radioactive compound with mass $280$ grams decays at a rate of $18.3 \%$ per hour. Which equation represents how many grams of the compound will remain after $2$ hours? $\newline$ $C=280(1.183)^{2}$ $\newline$ $C=280(1-0.183)(1-0.183)$ $\newline$ $C=280(1+0.183)^{2}$ $\newline$ $C=280(1-0.183)(1-0.183)(1-0.183)(1-0.183)$

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Write exponential functions: word problems

Full solution

Q. A radioactive compound with mass

280

grams decays at a rate of

18.3 \%

per hour. Which equation represents how many grams of the compound will remain after

2

hours?

\newline

C=280(1.183)^{2}

\newline

C=280(1-0.183)(1-0.183)

\newline

C=280(1+0.183)^{2}

\newline

C=280(1-0.183)(1-0.183)(1-0.183)(1-0.183)

Decay Factor Calculation: Understand the decay rate and convert it to a decay factor. $\newline$ The decay rate is given as $18.3\%$ per hour, which means that each hour, the compound loses $18.3\%$ of its mass. To find the decay factor, we subtract the decay rate from $1$ (since $1$ represents the whole mass). $\newline$ Decay factor $= 1 - \text{decay rate}$ $\newline$ Decay factor $= 1 - 0.183$ $\newline$ Decay factor $= 0.817$
Exponential Decay Equation Formulation: Write the exponential decay equation using the decay factor. $\newline$ The general form of the exponential decay equation is $C = \text{initial\_mass} \times (\text{decay\_factor})^{\text{time}}$ . $\newline$ Here, the initial mass ( $C_0$ ) is $280$ grams, the decay factor we found in Step $1$ is $0.817$ , and the time ( $t$ ) is $2$ hours. $\newline$ $C = 280 \times (0.817)^2$
Options Comparison: Check the given options to see which one matches the equation we derived. $\newline$ Option A: $C=280(1.183)^{2}$ - This option is incorrect because it suggests an increase by $18.3\%$ instead of a decrease. $\newline$ Option B: $C=280(1-0.183)(1-0.183)$ - This option is correct because it represents the mass after two hours, taking into account the decay rate twice. $\newline$ Option C: $C=280(1+0.183)^{2}$ - This option is incorrect because it suggests an increase by $18.3\%$ instead of a decrease. $\newline$ Option D: $C=280(1-0.183)(1-0.183)(1-0.183)(1-0.183)$ - This option is incorrect because it suggests the decay happens over four hours instead of two.

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