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)
Answer:
m>14 or m< -2
Step-by-step explanation:
Answer:
Time taken for the ball to hit the ground back = 3.08 s
Step-by-step explanation:
h(t)= -16t² + 48t + 4
when will rhe object come back to hit rhe ground?
When the ball is at the level.of the ground, h(t) = 0.
0 = -16t² + 48t + 4
-16t² + 48t + 4 = 0
Solving the quadratic equation
t = 3.08 s or t = -0.08 s
Since the time cannot be negative,
Time taken for the ball to hit the ground back = 3.08 s
Hope this Helps!!!
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hope it helps you!!
#IndianMurga(. ❛ ᴗ ❛.)
