The price of Mid-Range Hotel (Room/Night) in Cleveland, United States is 199.10 USD
Mid-Range Hotel in Cleveland, United States
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| Currency | Code | Symbol | Value |
|---|---|---|---|
| Gold Ounce | XAU | g | 0.1599 XAU |
| Sierra Leonean Leone | SLL | Le | 861,505.7000 SLL |
| Indian Rupee | INR | Rs | 12,403.2132 INR |
| Polish Zloty | PLN | zł | 616.3758 PLN |
| Tajikistani Somoni | TJS | SM | 950.2087 TJS |
| Burundian Franc | BIF | Fr | 308,632.8740 BIF |
| Other cities | Price | COMPARED WITH Cleveland |
|---|---|---|
| Zuni Pueblo | 181.00 USD | -9.09% |
| Zephyrhills | 181.00 USD | -9.09% |
| Zellwood | 170.14 USD | -14.55% |
| Zelienople | 166.52 USD | -16.36% |
| Zeeland | 181.00 USD | -9.09% |
| Zapata | 161.09 USD | -19.09% |
Mid-Range Hotel price comparison chart
The Mid-Range Hotel price comparison chart shows that price for Mid-Range Hotel (Room/Night) in Cleveland is % higher than the same Mid-Range Hotel price in , United States.