Astudy of three hundred teenagers found that the number of hours they spend on instant messaging sites each week is normally distributed with a mean of 13 hours. the population standard deviation is 5 hours. what is the margin of error for a 99% confidence interval?
0.743
0.334
0.320
0.413

OPtion A

Step-by-step explanation:

Given that a study of three hundred teenagers found that the number of hours they spend on instant messaging sites each week is normally distributed with a mean of 13 hours. The population standard deviation is 5 hours.

Std deviation = 5 hours

Sample size = 300

Hence std error of sample = 5/sqrt 300

=0.289

Margin of error = z Critical x std error

Since confidence level = 99%

Z critical = 2.58

Margin of error = 2.58 (0.289)

=0.743

Option A is right

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a

10 b/c 5*2 =10

step-by-step explanation: