Answer:
we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
Step-by-step explanation:
Given data
n=29
mean of x = 49.98 mm
S = 0.14 mm
μ = 50.00 mm
Cl = 95%
to find out
Can we be 95% confident that machine calibrated properly
solution
we know from t table
t at 95% and n -1 = 29-1 = 28 is 2.048
so now
Now for 95% CI for mean is
(x - 2.048 × S/√n , x + 2.048 × S/√n )
(49.98 - 2.048 × 0.14/√29 , 49.98 + 2.048 × 0.14/√29 )
( 49.926757 , 50.033243 )
hence we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
Answer:
560,000L
Step-by-step explanation:
Given data
L=14m
W=8m
H=5m
Volume= L*W*H
Volume= 14*8*5
Volume= 560 m^3
1m^3 = 1000L
560m^3 = x
cross multiply
x= 560*1000
x= 560,000L
Hence the volume is 560,000L
Answer:
2^3×3^2
and yh
Step-by-step explanation:
2×2×2×3×3
Answer:
Step-by-step explanation:
<u><em>System A and System B are </em></u><u><em>not equivalent</em></u> !!!
