Texexp-314 is an isotope that is decaying $$ 17 \% $$ each day. Chemists initially have a 289-gram sample. Which of the following equations represents the amount of the isotope remaining after $$ d $$ days? $$ y=289(0.17)^{d} $$ $$ y=289(0.83)^{d} $$ $$ y=289(1.17)^{d} $$ $$ y=314(0.83)^{d} $$

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To determine the correct equation representing the amount of the isotope **Texexp-314** remaining after $$ d $$ days, let's analyze the problem step by step.

### Given:
- **Initial amount** ($$ y_0 $$) = 289 grams
- **Daily decay rate** = 17%

### Understanding Decay:
A 17% daily decay means that each day, **83%** of the isotope remains (since $$ 100\% - 17\% = 83\% $$).

### Exponential Decay Formula:
The general formula for exponential decay is:
$$
y = y_0 \times (1 - \text{decay rate})^d
$$
Plugging in the given values:
$$
y = 289 \times (0.83)^d
$$
This equation accurately represents the remaining amount of the isotope after $$ d $$ days.

### Evaluating the Options:

1. **$$ y = 289(0.17)^d $$**
   - Incorrect. This represents only the decayed portion, not the remaining amount.

2. **$$ y = 289(0.83)^d $$**
   - **Correct.** This correctly models the remaining isotope after $$ d $$ days.

3. **$$ y = 289(1.17)^d $$**
   - Incorrect. This implies an increase, which contradicts the decay process.

4. **$$ y = 314(0.83)^d $$**
   - Incorrect. The initial amount should be 289 grams, not 314 grams.

### **Answer:**
$$
y = 289(0.83)^{d}
$$
The correct equation is $$ y = 289(0.83)^{d} $$.

Texexp-314 is an isotope that is decaying \( 17 \% \) each day. Chemists initially have a 289-gram sample. Which of the following equations represents the amount of the isotope remaining after \( d \) days? \( y=289(0.17)^{d} \) \( y=289(0.83)^{d} \) \( y=289(1.17)^{d} \) \( y=314(0.83)^{d} \)

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