Answer:
752.95
Step-by-step explanation:
Data provided in the question
The standard deviation of population = 210
The Margin of error = 15
The confidence level is 75%, so the z value is 1.96
Now the required sample size is
= 752.95
Hence, the number of college students spends on the internet each month is 752.95
Simply we considered the above values so that the n could come
3.24 + 5.70 + 6.17 = 15.11
20.00 - 15.11 = 5.11
He got 5.11$ of change back.
WEEKLY Earning Full time = 40 h x $15 = $600/week
Over time = 3 h 25 min or 3h + 25/60 h
WEEKLY Earning Over time = (3 h)x $30 + (25/60) x $30 = $115
TOTAL WEEKLY INCLUDING OVERTIME: = $715
TOTAL SEMI MONTHLY INCLUDING OVERTIME =$715 X2 = $1,430
BIWEEKLY = SEMI MONTHLY = $1,430
TOTAL MONTHLY INCLUDING OVERTIME =$715 x 4 = $2,860
Answer:
6:1
Step-by-step explanation:
