Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
Sorry, It seems like i dont have a sign to look of of, I can help you when you post that though!
Given:difference in the mean weight gain is 0.60 gramsstandard deviation of the difference in sample mean is 0.305
68% confidence interval for the population mean difference is a) 0.305
0.60 + 1 * 0.3050.60 + 0.305 = 0.9050.60 - 0.305 = 0.295
95% confidence interval for the population mean difference is c) 0.61
0.60 + 2 * 0.3050.60 + 0.61 = 1.210.60 - 0.61 = -0.01
A horizontal translation is expressed by transforming

If
is positive, the function is translated to the left. If
is negative, the function is translated to the right.
So, a 3-units left shift is given by
. So you have

Answer:
$28.80 per hour
Step-by-step explanation:
60 ÷ 5 = 12
4 × 12 = 4.8
He makes $00.48 per minute.
.48 × 60 = 28.80