Answer:
A≈94.25in²
Step-by-step explanation:
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:hehdjjdjfhrjrg
Step-by-step explanation:jejejdjfhbwbqkdjjcifirocicufjr
Answer:
27 years
Step-by-step explanation:
P is measured in thosands therefore, P will have a value of 120
so it would be...
120=80e^0.015t
80e^0.015t=120
e^0.015t=120/80=3/2=1.5
e^0.015t=1.5
0.015t= In(1.5)
t=In(1.5)/0.015
∴ t≈27.03 round it to 27 years
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