2006-12-01 01:14:20 · 9 answers · asked by pnatt89 1 in Computers & Internet ➔ Software
I use AVAST!, and it's free...It updates regularly, and if you get a virus, etc., it blares a horn to let you know, and you can put it into your virus chest...
2006-12-01 01:25:04 · answer #1 · answered by Anonymous · 0⤊ 0⤋
NOD 32 hands down...its a proven fact.
Qutoe from this Article: http://www.pcmech.com/show/reviews/914/
"Effectiveness
NOD32's on-demand scanner module is the program's bragging point. There are many 'types' of scans that you can do: Virus Signature, Heuristics, Advanced Heuristics, Adware/ Spyware/ Riskware, and Potentially dangerous applications. On AV-Comparative, NOD32 earned the Advanced + certification on both the New "Zoo" samples in proactive and retrospective testing and the on-demand scans, recording a staggering 62% and 98% respectively. In a test conducted by CheckVir, NOD32 earned top honors, detecting and cleaning all 420 virus samples, earning ESET's product the CheckVir Advanced Certification."
2006-12-01 01:21:40 · answer #2 · answered by Austin 2 · 0⤊ 0⤋
I agree, AVG Antivirus.
2006-12-01 01:18:46 · answer #3 · answered by mrfoxhorn 5 · 0⤊ 0⤋
I perfer AVG antivirus.
2006-12-01 01:16:39 · answer #4 · answered by brmwk 3 · 0⤊ 0⤋
Your pref, If you pay, you get a little bit more like, support, faster updates, usually free versions are not has high on the priority list on new viruses....etc,
I prefer free Check out www.thefreecountry.com
I use AVG myself
2006-12-01 01:50:45 · answer #5 · answered by Arkane Steelblade 4 · 0⤊ 0⤋
norton is a good option or mcafee, both are great in there protection and removal.
unlike avg or avast that dont't detect as much or remove
2006-12-01 01:22:57 · answer #6 · answered by great one 6 · 0⤊ 0⤋
I prefer Norton.
2006-12-01 01:18:38 · answer #7 · answered by samthecatrocks 3 · 0⤊ 0⤋
That is up to you to decide. I use AVG Free.
2006-12-01 01:28:10 · answer #8 · answered by ? 4 · 0⤊ 0⤋
NORTON ALWAYS works for us.
2006-12-01 01:36:07 · answer #9 · answered by wicked 2 · 0⤊ 0⤋