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 )
5×4=20 , 12×4=48
53÷4 = 13 then
x=13
Answer:
9
Step-by-step explanation:
7r+18=r²
r²-7r-18=0
factorizing
r²+2r-9r-18=0
r(r+2)-9(r+2=0
(r-9)(r+2)=0
r-9=0
r=9
check
(r²)=(9*9)=81
7r+18=7(9) + 8=81
Answer:
The answer is high inflation
Step-by-step explanation:
Answer:
120/6=20
Step-by-step explanation:
