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