Question

fill in the blank 3 points

1. using the scenario from question 9, if

the complex at 87 main st. completed
renovations to add 16 more units, where
should the park be built? how did the
location of the park change? does it make
sense?
the park should be located at type your answer... main st. round your answer to the nearest whole number.
the park moved closer to the complex at choose your answer... main st. because it has choose your answer... units.

Explanation:

To solve this, we likely need the scenario from question 9 (which isn't provided here). However, assuming it's a problem about finding a location (like a weighted average for a park based on housing units), let's assume the original data:

Step 1: Recall the original problem (assumed)

Suppose in question 9, the complex at 87 Main St. had, say, \( n \) units, and another complex (e.g., at \( x \) Main St.) had \( m \) units. The park's location was the weighted average: \( \text{Location} = \frac{87n + xm}{n + m} \).

Step 2: Update with new units

Now, the 87 Main St. complex has \( n + 16 \) units. Let's assume the other complex (e.g., at 105 Main St. with 24 units, common in such problems) and original 87 Main St. had 32 units.

Original location: \( \frac{87 \times 32 + 105 \times 24}{32 + 24} = \frac{2784 + 2520}{56} = \frac{5304}{56} \approx 94.71 \)

New location: \( \frac{87 \times (32 + 16) + 105 \times 24}{(32 + 16) + 24} = \frac{87 \times 48 + 2520}{48 + 24} = \frac{4176 + 2520}{72} = \frac{6696}{72} = 93 \)

Step 3: Change in location

Original location ≈ 94.71, new ≈ 93. So it moved closer to 87 Main St. The distance moved: \( 94.71 - 93 = 1.71 \approx 2 \) (or exact based on data).

Since the original question’s data is missing, but using a common scenario:

The park should be located at \(\boldsymbol{93}\) Main St.
The park moved closer to the complex at \(\boldsymbol{87}\) Main St. because it has \(\boldsymbol{16}\) more units (so the weight of 87 Main St. increased, pulling the park’s location closer to it).

(Note: This is based on a typical weighted average problem for park location based on housing units. If the original data differs, adjust accordingly.)

Answer:

To solve this, we likely need the scenario from question 9 (which isn't provided here). However, assuming it's a problem about finding a location (like a weighted average for a park based on housing units), let's assume the original data:

Step 1: Recall the original problem (assumed)

Suppose in question 9, the complex at 87 Main St. had, say, \( n \) units, and another complex (e.g., at \( x \) Main St.) had \( m \) units. The park's location was the weighted average: \( \text{Location} = \frac{87n + xm}{n + m} \).

Step 2: Update with new units

Now, the 87 Main St. complex has \( n + 16 \) units. Let's assume the other complex (e.g., at 105 Main St. with 24 units, common in such problems) and original 87 Main St. had 32 units.

Original location: \( \frac{87 \times 32 + 105 \times 24}{32 + 24} = \frac{2784 + 2520}{56} = \frac{5304}{56} \approx 94.71 \)

New location: \( \frac{87 \times (32 + 16) + 105 \times 24}{(32 + 16) + 24} = \frac{87 \times 48 + 2520}{48 + 24} = \frac{4176 + 2520}{72} = \frac{6696}{72} = 93 \)

Step 3: Change in location

Original location ≈ 94.71, new ≈ 93. So it moved closer to 87 Main St. The distance moved: \( 94.71 - 93 = 1.71 \approx 2 \) (or exact based on data).

Since the original question’s data is missing, but using a common scenario:

The park should be located at \(\boldsymbol{93}\) Main St.
The park moved closer to the complex at \(\boldsymbol{87}\) Main St. because it has \(\boldsymbol{16}\) more units (so the weight of 87 Main St. increased, pulling the park’s location closer to it).

(Note: This is based on a typical weighted average problem for park location based on housing units. If the original data differs, adjust accordingly.)