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Compare American Canyon to Manhattan Beach
This post compares American Canyon, California to Manhattan Beach, California across various dimensions, including population, median income, median age, percentage of housing that is rental, percent of households with kids, and other demographics.
| Dimension | American Canyon (city) | Manhattan Beach (city) |
|---|---|---|
| Population | 20,341 | 35,698 |
| Income (median) | $54,617 | $118,119 |
| Home Value (median) | $418,600 | $1,694,900 |
| White | 38% | 79% |
| Black | 8% | 0% |
| Asian | 35% | 10% |
| With Kids | 50% | 32% |
| Age (median) | 36 | 43 |
| Rentals | 21% | 31% |
Questions and Answers:
Q: Which is more populated, American Canyon or Manhattan Beach?
A: American Canyon has a smaller population count than Manhattan Beach: 20341 compared with 35698 total people.
Q: Do incomes tend to be higher in American Canyon or in Manhattan Beach?
A: Incomes are on average much lower in American Canyon.
Q: Are homes more expensive in American Canyon or Manhattan Beach?
A: Homes tend to be much less expensive in American Canyon.
Q: Which has a higher percentage of housing rental units?
A: Manhattan Beach does. The percentage of rentals differs quite a bit: 21% for American Canyon versus 31% for Manhattan Beach.
Somewhat similar in-state cities, towns, and CDPs to American Canyon may include: Vineyard, Eastvale, and Elk-Grove.
The statistics and analyses on this page are based on data from the U.S. Census American Community Survey.
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