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 )
I’m here to help what’s the problem?
Answer:
Equate the divisor to 0
2x-1=0
2×=1
×=1/2
Putting onto the polynomial
f(1/2) = (1/2)³-1/2)²+3(1/2)-2
=-5/8
Answer:
35.6 m
Step-by-step explanation:
Number of revolutions = 40
Length of thread = 0.89 m
Total Distance = Number of revolutions * Length of Thread
Total Distance = 40 * 0.89 = 35.6 m
<8 * (-7), 8 * (-2)>
<-56, -16>
Hope it helps, good work!
