How old is Janice J Parziale?
Janice J Parziale is possibly 86 years old and maybe was born in August 1939.
Next Steps: Find a result that matches whom you're looking for the most. Once you find a match, click "Search For Background Reports" to continue. Or use the form below to find the person you are searching for.
By clicking “SEARCH” I agree to the current USSearch TERMS OF USE and PRIVACY POLICY, and agree not use the products or services of USSearch to determine one's eligibility for credit, housing, insurance, employment or any other reason subject to the Fair Credit Reporting Act (“FCRA”). I further understand that USSearch is not a “consumer reporting agency” and does not provide “consumer reports” under the FCRA.
We have 2 records for Janice Parziale with an approximate age of 86 years old. Janice has been found in 2 states including Ohio and Texas. On file we have 4 email addresses. And 2 phone numbers linked to Janice, with area codes such as 513 and 936. The information below may also include images, social media accounts and more.
AKA (Aliases, nicknames, alternate spellings, married and/or maiden names):
landline
consolidated communications of texas company
First Reported June 15, 2004
landline
brightspeed of northwest ohio inc (centurylink)
First Reported May 12, 2020
145 N Hills Blvd
Springboro, Ohio 45066
Warren
Homes, rental properties, businesses, apartments, condos and/or other real estate associated with Janice Parziale.
2236 Windsong Dr
Findlay, Ohio 45840
Hancock
Click here to search for background reports.
Janice J Parziale is possibly 86 years old and maybe was born in August 1939.
Janice J Parziale has been associated with 3 addresses, the last known one is 145 N Hills Blvd, Springboro, Oh 45066-8433.
Janice J Parziale has been known to have 2 phone numbers, including (936) 597-3985.
Janice J Parziale has been known to have 5 email addresses, including xxxxxx@yahoo.com.
Janice J Parziale's possible relatives include Suzan Elizabeth Pardy, Anthony John Parziale, John Vincent Parziale and others.
Search by First Name
Search by Last Name