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Compare Jacksonville Beach to Sanford
This post compares Jacksonville Beach, Florida to Sanford, Florida across various dimensions, including population, median income, median age, percentage of housing that is rental, percent of households with kids, and other demographics.
| Dimension | Jacksonville Beach (city) | Sanford (city) |
|---|---|---|
| Population | 22,833 | 57,979 |
| Income (median) | $57,776 | $36,585 |
| Home Value (median) | $317,000 | $113,100 |
| White | 91% | 60% |
| Black | 2% | 26% |
| Asian | 2% | 3% |
| With Kids | 19% | 31% |
| Age (median) | 42 | 33 |
| Rentals | 33% | 47% |
Questions and Answers:
Q: Which is more populated, Jacksonville Beach or Sanford?
A: Jacksonville Beach has a much smaller population count than Sanford: 22833 compared with 57979 total people.
Q: Do incomes tend to be higher in Jacksonville Beach or in Sanford?
A: Incomes are on average much higher in Jacksonville Beach.
Q: Are homes more expensive in Jacksonville Beach or Sanford?
A: Homes tend to be much more expensive in Jacksonville Beach.
Q: Which has a higher percentage of housing rental units?
A: Sanford does. The percentage of rentals differs quite a bit: 33% for Jacksonville Beach versus 47% for Sanford.
Somewhat similar in-state cities, towns, and CDPs to Jacksonville Beach may include: Winter-Park, Jupiter, and Wekiwa-Springs.
The statistics and analyses on this page are based on data from the U.S. Census American Community Survey.
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