Answer:
I think it might be 24.5
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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 )
The area is 32 5/8 ft squared and It would cost$ 195.75
how to slove: you take the to fractions and mulitply them and get 32 5/8 for the area then you multiply 32 5/8 by 6 to find the price of the banner.
(cube root of 5) * sqrt(5)
--------------------------------- = ?
(cube root of 5^5)
This becomes easier if we switch to fractional exponents:
5^(1/3) * 5^(1/2) 5^(1/3 + 1/2) 5^(5/6)
------------------------ = --------------------- = ------------- = 5^[5/6 - 5/3]
[ 5^5 ]^(1/3) 5^(5/3) 5^(5/3)
Note that 5/6 - 5/3 = 5/6 - 10/6 = -5/6.
1
Thus, 5^[5/6 - 5/3] = 5^(-5/6) = --------------
5^(5/6)
That's the correct answer. But if you want to remove the fractional exponent from the denominator, do this:
1 5^(1/6) 5^(1/6)
---------- * ------------- = -------------- (ANSWER)
5^(5/6) 5^(1/6) 5
The correct answer is A, -3 degrees Celsius.
