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Sagot :
Let's examine the rate at which both Clara's and Heidi's bodies eliminate the medicine.
### Step 1: Clara's Data
We are given the table of how much medicine is left in Clara's body over time:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline t \ (\text{hours}) & 1 & 2 & 3 & 4 & 5 \\ \hline f(t) \ (\text{mg}) & 236.5 & 223.73 & 211.65 & 200.22 & 189.41 \\ \hline \end{array} \][/tex]
### Step 2: Heidi's Function
Heidi's medicine is modeled by the equation:
[tex]\[ f(t) = 300(0.946)^t \][/tex]
### Step 3: Calculate Heidi's Data Points
To compare Clara's and Heidi's rates of elimination, first we need to calculate Heidi's medicine amounts at the same time points given for Clara.
[tex]\[ \begin{array}{|c|c|} \hline t \ (\text{hours}) & f(t) \ (\text{mg}) \\ \hline 1 & 300 \cdot (0.946)^1 \approx 283.8 \\ 2 & 300 \cdot (0.946)^2 \approx 268.57 \\ 3 & 300 \cdot (0.946)^3 \approx 254.16 \\ 4 & 300 \cdot (0.946)^4 \approx 240.52 \\ 5 & 300 \cdot (0.946)^5 \approx 227.57 \\ \hline \end{array} \][/tex]
### Step 4: Rate of Elimination for Clara
Let's compute the rate at which the medicine is eliminated from Clara's body. The rate at each time point is given by the difference in medicine amounts divided by the time interval (which is 1 hour in this case).
[tex]\[ \begin{align*} \text{Rate}_{(1-2)} &= \frac{223.73 - 236.5}{1} = -12.77 \ \text{mg/hour} \\ \text{Rate}_{(2-3)} &= \frac{211.65 - 223.73}{1} = -12.08 \ \text{mg/hour} \\ \text{Rate}_{(3-4)} &= \frac{200.22 - 211.65}{1} = -11.43 \ \text{mg/hour} \\ \text{Rate}_{(4-5)} &= \frac{189.41 - 200.22}{1} = -10.81 \ \text{mg/hour} \\ \end{align*} \][/tex]
Average rate of elimination for Clara:
[tex]\[ \text{Average Rate}_{\text{Clara}} = \frac{-12.77 + -12.08 + -11.43 + -10.81}{4} = -11.7725 \ \text{mg/hour} \][/tex]
### Step 5: Rate of Elimination for Heidi
We perform the same calculations for Heidi.
[tex]\[ \begin{align*} \text{Rate}_{(1-2)} &= \frac{268.57 - 283.8}{1} = -15.23 \ \text{mg/hour} \\ \text{Rate}_{(2-3)} &= \frac{254.16 - 268.57}{1} = -14.41 \ \text{mg/hour} \\ \text{Rate}_{(3-4)} &= \frac{240.52 - 254.16}{1} = -13.64 \ \text{mg/hour} \\ \text{Rate}_{(4-5)} &= \frac{227.57 - 240.52}{1} = -12.95 \ \text{mg/hour} \\ \end{align*} \][/tex]
Average rate of elimination for Heidi:
[tex]\[ \text{Average Rate}_{\text{Heidi}} = \frac{-15.23 + -14.41 + -13.64 + -12.95}{4} = -14.1279 \ \text{mg/hour} \][/tex]
### Step 6: Compare Rates
Finally, we compare the average rates of elimination:
[tex]\[ \text{Average Rate}_{\text{Clara}} = -11.7725 \ \text{mg/hour} \][/tex]
[tex]\[ \text{Average Rate}_{\text{Heidi}} = -14.1279 \ \text{mg/hour} \][/tex]
Since the rate of elimination for Clara’s body is less negative than Heidi’s, it means that Clara’s body eliminates the medicine slower than Heidi’s body.
### Conclusion
Therefore, the best statement to describe the rate at which Clara's and Heidi's bodies eliminated the medicine is:
Clara's body eliminated the antibiotic faster than Heidi's body.
### Step 1: Clara's Data
We are given the table of how much medicine is left in Clara's body over time:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline t \ (\text{hours}) & 1 & 2 & 3 & 4 & 5 \\ \hline f(t) \ (\text{mg}) & 236.5 & 223.73 & 211.65 & 200.22 & 189.41 \\ \hline \end{array} \][/tex]
### Step 2: Heidi's Function
Heidi's medicine is modeled by the equation:
[tex]\[ f(t) = 300(0.946)^t \][/tex]
### Step 3: Calculate Heidi's Data Points
To compare Clara's and Heidi's rates of elimination, first we need to calculate Heidi's medicine amounts at the same time points given for Clara.
[tex]\[ \begin{array}{|c|c|} \hline t \ (\text{hours}) & f(t) \ (\text{mg}) \\ \hline 1 & 300 \cdot (0.946)^1 \approx 283.8 \\ 2 & 300 \cdot (0.946)^2 \approx 268.57 \\ 3 & 300 \cdot (0.946)^3 \approx 254.16 \\ 4 & 300 \cdot (0.946)^4 \approx 240.52 \\ 5 & 300 \cdot (0.946)^5 \approx 227.57 \\ \hline \end{array} \][/tex]
### Step 4: Rate of Elimination for Clara
Let's compute the rate at which the medicine is eliminated from Clara's body. The rate at each time point is given by the difference in medicine amounts divided by the time interval (which is 1 hour in this case).
[tex]\[ \begin{align*} \text{Rate}_{(1-2)} &= \frac{223.73 - 236.5}{1} = -12.77 \ \text{mg/hour} \\ \text{Rate}_{(2-3)} &= \frac{211.65 - 223.73}{1} = -12.08 \ \text{mg/hour} \\ \text{Rate}_{(3-4)} &= \frac{200.22 - 211.65}{1} = -11.43 \ \text{mg/hour} \\ \text{Rate}_{(4-5)} &= \frac{189.41 - 200.22}{1} = -10.81 \ \text{mg/hour} \\ \end{align*} \][/tex]
Average rate of elimination for Clara:
[tex]\[ \text{Average Rate}_{\text{Clara}} = \frac{-12.77 + -12.08 + -11.43 + -10.81}{4} = -11.7725 \ \text{mg/hour} \][/tex]
### Step 5: Rate of Elimination for Heidi
We perform the same calculations for Heidi.
[tex]\[ \begin{align*} \text{Rate}_{(1-2)} &= \frac{268.57 - 283.8}{1} = -15.23 \ \text{mg/hour} \\ \text{Rate}_{(2-3)} &= \frac{254.16 - 268.57}{1} = -14.41 \ \text{mg/hour} \\ \text{Rate}_{(3-4)} &= \frac{240.52 - 254.16}{1} = -13.64 \ \text{mg/hour} \\ \text{Rate}_{(4-5)} &= \frac{227.57 - 240.52}{1} = -12.95 \ \text{mg/hour} \\ \end{align*} \][/tex]
Average rate of elimination for Heidi:
[tex]\[ \text{Average Rate}_{\text{Heidi}} = \frac{-15.23 + -14.41 + -13.64 + -12.95}{4} = -14.1279 \ \text{mg/hour} \][/tex]
### Step 6: Compare Rates
Finally, we compare the average rates of elimination:
[tex]\[ \text{Average Rate}_{\text{Clara}} = -11.7725 \ \text{mg/hour} \][/tex]
[tex]\[ \text{Average Rate}_{\text{Heidi}} = -14.1279 \ \text{mg/hour} \][/tex]
Since the rate of elimination for Clara’s body is less negative than Heidi’s, it means that Clara’s body eliminates the medicine slower than Heidi’s body.
### Conclusion
Therefore, the best statement to describe the rate at which Clara's and Heidi's bodies eliminated the medicine is:
Clara's body eliminated the antibiotic faster than Heidi's body.
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