Mon-Fri 8 hrs a day.
Moderate stress.
2006-11-03 12:16:47 · 12 answers · asked by Anonymous in Business & Finance ➔ Careers & Employment
Only of it offered really great benefits upon hire. $42 K isn't much these days, so it would really depend on what kind of experience and background you can bring to the table. If you're just starting out, that's a good rate.
2006-11-03 12:20:53 · answer #1 · answered by pknutson_sws 5 · 0⤊ 0⤋
Well, would depend on what my other options were. If my education, skills, and experience qualified me for a job with a salary much higher than that, then no. Otherwise, it's a pretty decent salary.
2006-11-03 12:19:24 · answer #2 · answered by Judy 7 · 0⤊ 0⤋
I make $8000 a year..your dang right. Can't be more stress than sitting home and wondering how you are going to live on $600 a month
2006-11-03 12:53:45 · answer #3 · answered by chilover 7 · 0⤊ 0⤋
Well, I've been at my job nearly 20 years, and I don't make that much!
I am hourly, though, so I get overtime.
Depends where you live--in the midwest, that's decent. In Washington DC, that's a JANITOR!
2006-11-03 12:25:44 · answer #4 · answered by Anonymous · 0⤊ 0⤋
AS LONG AS I GET SUMMERS OFF (and all school holidays) AND CAN SPEND TIME AWAY FROM WORK NOT THINKING ABOUT WORK SO I CAN: -TAKE CARE OF MY HUSBAND -TAKE CARE OF MY BABIES and lastly, as long as my salary goes up a thousand bucks or more a year (free dental and medical)
2016-03-28 06:04:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Without a doubt.
2006-11-03 12:29:52 · answer #6 · answered by Mariposa 7 · 0⤊ 0⤋
That isn't very much now a days.
2006-11-03 12:24:10 · answer #7 · answered by mamatohaley+1 4 · 0⤊ 0⤋
yes,if it fit my requirements for my life schedule,and I'm sure theres more questions it would have to satisfy
2006-11-03 12:26:20 · answer #8 · answered by crazy b 3 · 0⤊ 0⤋
heck yah! That is good pay rate.
2006-11-03 12:21:22 · answer #9 · answered by dakotanmisty 4 · 0⤊ 0⤋
if it pays better than my current job, why not?
2006-11-03 12:19:29 · answer #10 · answered by mgd1k 3 · 0⤊ 0⤋