Assignment 6.7: Exponential and Logarithmic Models Progress save Score: $1 / 10$ $1 / 10$ answered Question 4 At the beginning of an experiment, a scientist has 200 grams of radioactive goo. After 255 minutes, her sample has decayed to 12.5 grams. What is the half-life of the goo in minutes? Find a formula for $G(t)$, the amount of goo remaining at time $t$. \[ G(t)= \] How many grams of goo will remain after 65 minutes? You may enter the exact value or round to 2 decimal places. Question Help: Video Submit Question

Step 1 ：Given that the initial amount of the substance, \(G_0\), is 200 grams and the amount of the substance after 255 minutes, \(G(255)\), is 12.5 grams, we can substitute these values into the exponential decay formula \(G(t) = G_0 * e^{kt}\) to find the decay constant \(k\).

Step 2 ：Substituting the given values into the formula gives us \(12.5 = 200 * e^{255k}\).

Step 3 ：To solve for \(k\), we first divide both sides by 200 to get \(\frac{12.5}{200} = e^{255k}\).

Step 4 ：Then we take the natural logarithm of both sides to get \(\ln(\frac{12.5}{200}) = 255k\).

Step 5 ：Finally, we solve for \(k\) to get \(k = \frac{\ln(\frac{12.5}{200})}{255} \approx -0.0116\).

Step 6 ：Now that we have the decay constant, we can find the half-life of the substance. The half-life is the time it takes for half of the substance to decay, so we set \(G(t) = \frac{G_0}{2}\) and solve for \(t\).

Step 7 ：Substituting the values into the formula gives us \(\frac{200}{2} = 200 * e^{-0.0116t}\).

Step 8 ：Solving for \(t\) gives us \(t = \frac{\ln(\frac{1}{2})}{-0.0116} \approx 59.7\) minutes. So, the half-life of the goo is approximately 59.7 minutes.

Step 9 ：The formula for \(G(t)\), the amount of goo remaining at time \(t\), is \(G(t) = 200 * e^{-0.0116t}\).

Step 10 ：To find out how many grams of goo will remain after 65 minutes, we substitute \(t = 65\) into the formula to get \(G(65) = 200 * e^{-0.0116 * 65} \approx 94.12\) grams.

Step 11 ：\(\boxed{94.12}\) grams of goo will remain after 65 minutes.