Question

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

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!