In a random sample of 90 students in first grade, the mean hours of sleep per night was 9.2 with a standard deviation of 0.37 in a random sample of 70 students in the 12th grade, the mean hours of sleep per night was 7.3 with a standard deviation of 0.58. construct a 99% confidence interval for the difference between the mean hours of sleep. a. (1.82 h, 1.98 h)b. (1.79 h, 2.01 h)c. (1.74 h, 2.06 h)d. (1.69 h, 2.11 h)

Option C is right

Step-by-step explanation:

Given are two samples

I group has sample size 90, mean = 9.2 and std dev =0,.37

II group has sample size 70, mean = 7.3 and std dev = 0.58

we find that mean difference i.e. difference between mean hours of sleep between two groups

= 9.2-7.3 = 1.9

Std error for difference = 0.0620

Since sample size is sufficiently large we can use Z critical for 99%

i.e. 2.58

Margin of error =±2.58(0.062)

=0.16

Hence confidence interval 99% for difference in hours of sleep

= (1.9-0.16,1.9+0.16)

=(1.74, 2.06)

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

step-by-step explanation: