I currently have AT&T Yahoo! Broadband connection but I want to switch service because they suspended my connection without notice and I want to try something new. I live in Southern California so I was thinking about getting Time Warner Roadrunner Basic interntet cable (the cheapest one). My question is will I get better speeds with Roadrunner?
2007-01-22 10:31:33 · 6 answers · asked by j2dalo87 1 in Computers & Internet ➔ Other - Computers
I don't know.... but one thing is for sure... if COX is in LA too like here in NV... It is the fastest Internet.... cable Internet that doesn't need the telephone line and which stays online forever... no need to connect or anything.....
2007-01-22 10:41:09 · answer #1 · answered by CRA 3 · 0⤊ 0⤋
Just to have a laugh at cost to someone who said " cable internet that does not need the phone line and doesn't even need to connect" LMAO THAT IS THE FUNNIEST DUMBEST THING EVER SAID cable internet doesn't need the phone line...and even DSL that uses the phone line don't need to be connected unless you set it that way!!
2007-01-22 11:22:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋
My experience is that the RR service are not as fast as they proclaim.. but I would bet you wont see TOO MUCH of a difference. It is generally a littel slower though
2007-01-22 10:45:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Well it depends on which city you live in. You probably will though. However, I suggest that you try Verizon DSL...it is very fast for wherever you live and they have step by step guides for you. It is very easy to install.
2007-01-22 11:25:10 · answer #4 · answered by GT Star_Gyal 3 · 0⤊ 0⤋
They have different packages, some aren't even advertised. What you should do is call them for the exact rates and ask them if there are ANY other packages that they didn't tell you about
2007-01-22 11:27:38 · answer #5 · answered by windfishfighter 3 · 0⤊ 0⤋
most isps' have several levels of service and u get the speed that u pay for ..
2007-01-22 10:35:02 · answer #6 · answered by Anonymous · 0⤊ 0⤋