https://file.aiquickdraw.com/mathbot/file/1726710261996l0hmo1cm.jpg

Let's break down the questions one by one based on the table provided in the image.

### Given Table of Temperatures:
- **Monday:**
  - 8:00 AM: 100.5°F
  - 8:00 PM: 102.3°F
- **Tuesday:**
  - 8:00 AM: 103.1°F
  - 8:00 PM: 100.7°F
- **Wednesday:**
  - 8:00 AM: 99.9°F
  - 8:00 PM: 98.6°F

### Question (a):
**What was the average rate of change in Lisa’s temperature from 8:00 AM on Monday to 8:00 PM on Monday?**

To calculate the **average rate of change** of Lisa's temperature, we use the formula:

\[
\text{Average Rate of Change} = \frac{\text{Change in Temperature}}{\text{Change in Time}}
\]

For Monday (from 8:00 AM to 8:00 PM):
- Temperature at 8:00 AM = 100.5°F
- Temperature at 8:00 PM = 102.3°F
- Time difference: 12 hours

\[
\text{Average Rate of Change} = \frac{102.3 - 100.5}{12} = \frac{1.8}{12} = 0.15 \, \text{°F per hour}
\]

**Answer to (a):** The average rate of change was 0.15°F per hour.

---

### Question (b):
**What was the average rate of change in Lisa’s temperature from 8:00 AM on Tuesday to 8:00 PM on Wednesday? Is your answer reasonable? What does the sign of the rate mean?**

For Tuesday (8:00 AM) to Wednesday (8:00 PM):
- Temperature at 8:00 AM Tuesday = 103.1°F
- Temperature at 8:00 PM Wednesday = 98.6°F
- Time difference: 36 hours (8:00 AM Tuesday to 8:00 PM Wednesday)

\[
\text{Average Rate of Change} = \frac{98.6 - 103.1}{36} = \frac{-4.5}{36} = -0.125 \, \text{°F per hour}
\]

The negative sign indicates that Lisa's temperature was **decreasing** over this period.

**Answer to (b):** The average rate of change was \(-0.125 \, \text{°F per hour}\). The negative sign indicates that her temperature was dropping. This is reasonable because the data shows a decreasing trend in temperature over this period.

---

### Question (c):
**During which 12-hour period was the average rate of change in Lisa’s temperature the greatest?**

We will calculate the average rate of change for each 12-hour period and compare.

1. **Monday (8:00 AM to 8:00 PM):**
   - Already calculated: **0.15°F per hour**

2. **Tuesday (8:00 AM to 8:00 PM):**
   - Temperature at 8:00 AM = 103.1°F
   - Temperature at 8:00 PM = 100.7°F
   - Time difference = 12 hours

\[
\text{Average Rate of Change} = \frac{100.7 - 103.1}{12} = \frac{-2.4}{12} = -0.2 \, \text{°F per hour}
\]

3. **Wednesday (8:00 AM to 8:00 PM):**
   - Temperature at 8:00 AM = 99.9°F
   - Temperature at 8:00 PM = 98.6°F
   - Time difference = 12 hours

\[
\text{Average Rate of Change} = \frac{98.6 - 99.9}{12} = \frac{-1.3}{12} = -0.1083 \, \text{°F per hour}
\]

**Conclusion for (c):** The greatest average rate of change was on **Tuesday (8:00 AM to 8:00 PM)**, where the rate was \(-0.2°F per hour\).

---

Would you like more details or have any further questions?

Here are 5 related questions you could explore:
1. What would the total temperature change be from Monday 8:00 AM to Wednesday 8:00 PM?
2. How can we interpret the rates of change in terms of Lisa's health?
3. What would the average rate of change be over the entire 3-day period?
4. How would the average rate of change look if we only consider day-to-day differences?
5. Can we model the temperature changes using a linear equation?

**Tip:** When calculating rates of change, a positive rate indicates an increase, while a negative rate indicates a decrease over the interval.

The table below shows Lisa's temperature during an illness over a 3-day period. The user is asked to determine the average rate of change of Lisa's temperature and the period with the greatest change.

Explore how to calculate the average rate of change in temperature over a 3-day period for Lisa. The step-by-step solution covers the formula for average rate of change and compares temperature changes over different time intervals. Suitable for Grades 6-8.

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