To determine the amount of antibiotics left in the dog after 16 hours, we can use the formula for exponential decay. The formula for the amount remaining after time \( t \) is:

Question

\[
A = A_0 \times (1 - r)^n
\]

where:
- \( A \) is the amount remaining,
- \( A_0 \) is the initial amount,
- \( r \) is the decay rate,
- \( n \) is the number of time intervals.

Given:
- \( A_0 = 200 \) mg (initial amount),
- \( r = 0.25 \) (25% decay rate),
- \( t = 16 \) hours.

First, we need to determine \( n \), which is the number of 4-hour intervals in 16 hours:

\[
n = \frac{t}{4} = \frac{16}{4} = 4
\]

Now we can substitute the values into the formula:

\[
A = 200 \times (1 - 0.25)^4
\]

Calculating \( (1 - 0.25) \):

\[
1 - 0.25 = 0.75
\]

Now substituting back into the formula:

\[
A = 200 \times (0.75)^4
\]

Calculating \( (0.75)^4 \):

\[
(0.75)^4 = 0.75 \times 0.75 \times 0.75 \times 0.75 = 0.31640625
\]

Now substitute this value into the equation:

\[
A = 200 \times 0.31640625 \approx 63.28125
\]

Thus, the amount of antibiotic left in the dog after 16 hours is approximately:

\[
\boxed{63.28} \text{ mg}
\]
make this like a 7th grader wrote it

Answer 1

To find out how much antibiotic is left in the dog after 16 hours, we can use a simple formula that helps us understand how the amount decreases over time. The formula is:

\[ A = A_0 \times (1 - r)^n \]

Here's what the parts mean:

• \( A \) is how much antibiotic is left.
• \( A_0 \) is how much antibiotic we started with.
• \( r \) is how quickly it decays (like how much is lost).
• \( n \) is how many time intervals we count.

Now, let's use the numbers given:

• The starting amount \( A_0 \) is 200 mg.
• The decay rate \( r \) is 0.25, which means 25% of the antibiotic is gone each time period.
• The total time \( t \) we are looking at is 16 hours.

First, we need to figure out how many time intervals of 4 hours fit into 16 hours:

\[ n = \frac{t}{4} = \frac{16}{4} = 4 \]

This means there are 4 intervals of 4 hours in 16 hours.

Next, we can put the values into the equation:

\[ A = 200 \times (1 - 0.25)^4 \]

First, let's calculate \( (1 - 0.25) \):

\[ 1 - 0.25 = 0.75 \]

Now, we plug that back into our equation:

\[ A = 200 \times (0.75)^4 \]

Now we need to calculate \( (0.75)^4 \):

\[ (0.75)^4 = 0.75 \times 0.75 \times 0.75 \times 0.75 \]

Doing the math step by step:

1. \( 0.75 \times 0.75 = 0.5625 \)
2. \( 0.5625 \times 0.75 = 0.421875 \)
3. \( 0.421875 \times 0.75 = 0.31640625 \)

Now we can substitute this back into our equation:

\[ A = 200 \times 0.31640625 \]

Calculating that gives us:

\[ A \approx 63.28125 \]

So, the amount of antibiotic left in the dog after 16 hours is about:

\[ \boxed{63.28} \text{ mg} \]

That's how much is still in the dog after all that time!