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To solve this problem, we need to find the average velocity of the arrow over the given time intervals using the height function:

\[
h(t) = 52t - 0.83t^2.
\]

The average velocity \( v_{\text{avg}} \) over an interval \([t_1, t_2]\) is given by:

\[
v_{\text{avg}} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}.
\]

Let's calculate the average velocity for each interval provided:

1. **Interval [3, 4]:**
   \[
   v_{\text{avg}} = \frac{h(4) - h(3)}{4 - 3}.
   \]
   Compute \( h(4) \) and \( h(3) \):
   \[
   h(4) = 52 \times 4 - 0.83 \times 4^2,
   \]
   \[
   h(3) = 52 \times 3 - 0.83 \times 3^2.
   \]

2. **Interval [3, 3.5]:**
   \[
   v_{\text{avg}} = \frac{h(3.5) - h(3)}{3.5 - 3}.
   \]
   Compute \( h(3.5) \) and \( h(3) \):
   \[
   h(3.5) = 52 \times 3.5 - 0.83 \times 3.5^2.
   \]

3. **Interval [3, 3.1]:**
   \[
   v_{\text{avg}} = \frac{h(3.1) - h(3)}{3.1 - 3}.
   \]
   Compute \( h(3.1) \):
   \[
   h(3.1) = 52 \times 3.1 - 0.83 \times 3.1^2.
   \]

4. **Interval [3, 3.01]:**
   \[
   v_{\text{avg}} = \frac{h(3.01) - h(3)}{3.01 - 3}.
   \]
   Compute \( h(3.01) \):
   \[
   h(3.01) = 52 \times 3.01 - 0.83 \times 3.01^2.
   \]

5. **Interval [3, 3.001]:**
   \[
   v_{\text{avg}} = \frac{h(3.001) - h(3)}{3.001 - 3}.
   \]
   Compute \( h(3.001) \):
   \[
   h(3.001) = 52 \times 3.001 - 0.83 \times 3.001^2.
   \]

I will now perform these calculations to provide the values rounded to four decimal places.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the calculations step-by-step if you'd like to do them manually or with a calculator. 

Would you like to go over how to compute these average velocities manually, or do you have any other questions?

Here are some related questions you might find useful:
1. How do you find the instantaneous velocity using a derivative?
2. What is the significance of the average velocity in physics?
3. How would the average velocity change if the arrow were on Earth instead of the moon?
4. How can you estimate the instantaneous velocity using smaller intervals?
5. What does a negative coefficient in the quadratic term indicate about the motion?

**Tip:** To find the average velocity over an interval, always remember to subtract the initial position from the final position, and then divide by the length of the interval.

Suppose an arrow is shot upward on the moon with a velocity of 52 m/s, then its height in meters after t seconds is given by h(t) = 52t - 0.83t^2. Find the average velocity over the given time intervals: [3, 4], [3, 3.5], [3, 3.1], [3, 3.01], [3, 3.001].

Learn how to calculate the average velocity of an arrow shot upward on the moon using the quadratic height function h(t) = 52t - 0.83t^2 over different time intervals. This detailed solution covers concepts in quadratic functions and kinematics, suitable for high school students.

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