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:
Step-by-step explanation:
KT because triangles KIT and KET are congruent ( mirror images).
Tanα=h/x
h=xtanα, we are told that x=6.5ft and α=74° so
h=6.5tan74 ft
h≈22.67 ft (to nearest hundredth of a foot)
Answer:
4th option - over the interval (4,7) the local minimum is -7
Step-by-step explanation:
There's only one local minimum in this graph and it's the one between (4,7), so this is the only plausible answer.
