Answer:
Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds
Step-by-step explanation:
State the null hypothesis
H0: u = 60against the claim
Ha u≠ 60 (this is a two tailed test)
Sample size n= 36
Sample mean=X`= 55
Population mean = u= 60
The significance level α = 0.05
Standard deviation= Sd = 22 seconds
Z= X`- u / Sd /√n
Z= 55- 60 / 22/√6
z= - 5/22/6
Z= -1.3635
The value of z from the table is Z∝/2= ±1.96
The critical region is less than - 1.96 and greater than 1.96.
Since the calculated value of Z does not fall in the critical region we accept our null hypothesis that mean = 60 seconds
Answer:
48 mph
Step-by-step explanation:
First we need to find the distance from Elkhart to Chicago. Toledo to Elkhart is 136 miles and Toledo to Chicago 244 miles.
So the distance from Elkhart to Chicago can be calculated, since Chicago is farther from Toledo than Elkhart, as: distance(Toledo to Chicago) - distance(Toledo to Elkhart). These distances are given in the problem, so the distance from Elkhat to Chicago is: 244 miles - 136 miles = 108 miles.
This problem basically wants to know the slowest you can be yet still ariving on time. If you are the minimum speed, you will arrive in Chicago exactly at 10:30 A.M. So you have 2 hours and 15 minutes(10:30 A.M - 8.15A.M.) to drive 108 miles.
15 minutes is a fourth of a hour, so you have 2.25hours to go through 108 miles.
The minimum speed you must maintain is 108mph/2.25h = 48mph.
Answer:
-24
Step-by-step explanation:
(-6)(4)
The indicated operation is multiplication
-24
