Answer:
![15 \sqrt[3]{2}](https://tex.z-dn.net/?f=15%20%5Csqrt%5B3%5D%7B2%7D%20)
Step-by-step explanation:
![{(27 \times 250)}^{ \frac{1}{3} } = {(27 \times 125 \times 2)}^{ \frac{1}{3} } \\ = {27}^{ \frac{1}{3} } \times {125}^{ \frac{1}{3} } \times {2}^{ \frac{1}{3} } \\ = \sqrt[ 3]{27} \times \sqrt[3]{125} \times \sqrt[3]{2} \\ = \sqrt[3]{ {3}^{3} } \times \sqrt[3]{ {5}^{3} } \times \sqrt[3]{2} \\ = 3 \times 5 \times \sqrt[3]{2} \\ = 15 \sqrt[3]{2}](https://tex.z-dn.net/?f=%20%7B%2827%20%5Ctimes%20250%29%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%3D%20%20%7B%2827%20%5Ctimes%20125%20%5Ctimes%202%29%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5C%5C%20%20%3D%20%20%7B27%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5Ctimes%20%20%7B125%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5Ctimes%20%20%7B2%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5C%5C%20%20%3D%20%20%5Csqrt%5B%203%5D%7B27%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B125%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B2%7D%20%20%5C%5C%20%20%3D%20%20%5Csqrt%5B3%5D%7B%20%7B3%7D%5E%7B3%7D%20%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B%20%7B5%7D%5E%7B3%7D%20%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B2%7D%20%20%5C%5C%20%20%3D%203%20%5Ctimes%205%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B2%7D%20%20%5C%5C%20%20%3D%2015%20%5Csqrt%5B3%5D%7B2%7D%20)
Answer:
225.78 grams
Step-by-step explanation:
To solve this question, we would be using the formula
P(t) = Po × 2^t/n
Where P(t) = Remaining amount after r hours
Po = Initial amount
t = Time
In the question,
Where P(t) = Remaining amount after r hours = unknown
Po = Initial amount = 537
t = Time = 10 days
P(t) = 537 × 2^(10/)
P(t) = 225.78 grams
Therefore, the amount of iodine-131 left after 10 days = 225.78 grams
The percentage change is 20