$p(t)=100\left(\frac{1}{2}\right)^{\frac{t}{8.02}}$ $\newline$ Radioactive isotopes are unstable atoms that decay into other atoms. lodine $-131$ is a radioactive isotope. The function models $p$ , the percentage of iodine $-131$ atoms remaining in a sample after $t$ days. To the nearest tenth of a percent, what percent of iodine $-131$ atoms remain in the sample after $10$ days? $\newline$ (Ignore the $\%$ symbol when entering your answer. For example, if the answer is $11.2 \%$ , enter $11$ . $2$ .)

Full solution

p(t)=100\left(\frac{1}{2}\right)^{\frac{t}{8.02}}

\newline

Radioactive isotopes are unstable atoms that decay into other atoms. lodine

-131

is a radioactive isotope. The function models

p

, the percentage of iodine

-131

atoms remaining in a sample after

t

days. To the nearest tenth of a percent, what percent of iodine

-131

atoms remain in the sample after

10

days?

\newline

(Ignore the

\%

symbol when entering your answer. For example, if the answer is

11.2 \%

, enter

11

2

Identify Function and Value: Identify the given function and the value of $t$ for which we need to find $p(t)$ . The function given is $p(t) = 100 \times \left(\frac{1}{2}\right)^{\frac{t}{8.02}}$ , which models the percentage of iodine $-131$ atoms remaining after $t$ days. We need to find the percentage remaining after $t = 10$ days.
Substitute Value into Function: Substitute the value of $t$ into the function. $\newline$ We substitute $t = 10$ into the function to find $p(10)$ . $\newline$ $p(10) = 100 \times \left(\frac{1}{2}\right)^{\frac{10}{8.02}}$
Calculate Exponent: Calculate the exponent. $\newline$ We need to calculate $(\frac{1}{2})^{(\frac{10}{8.02})}$ . $\newline$ First, we calculate the fraction $\frac{10}{8.02}$ . $\newline$ $\frac{10}{8.02} \approx 1.247$ $\newline$ Now we raise $(\frac{1}{2})$ to the power of $1.247$ . $\newline$ $(\frac{1}{2})^{1.247}$
Use Calculator for Calculation: Use a calculator to find the value of $(\frac{1}{2})^{1.247}$ . $\newline$ Using a calculator, we find that $(\frac{1}{2})^{1.247} \approx 0.421$ .
Find Percentage by Multiplication: Multiply the result by $100$ to find the percentage. $\newline$ Now we multiply the result from the previous step by $100$ to find the percentage of iodine- $131$ remaining after $10$ days. $\newline$ $p(10) = 100 \times 0.421$ $\newline$ $p(10) \approx 42.1$