Ajournal article reports that a sample of size 5 was used as a basis for calculating a 95% ci for the true average natural frequency (hz) of delaminated beams of a certain type. the resulting interval was (230.061, 233.807). you decide that a confidence level of 99% is more appropriate than the 95% level used. what are the limits of the 99% interval?

(229.5450, 234. 3230)

Step-by-step explanation:

Given that a  journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type.

95% confidence interval was

From confidence interval we find mean as the average of lower and upper bound.

Mean =

Margin of error = upper bound -mean

=

i.e. 1.96 * std error = 1.873

For 99% interval margin of error would be 2.58*std error

so margin of error for 99% =

Confidence interval 99% lower bound = Mean - 2.3890 =229.5450

Upper bound = Mean +2.3890 = 234.3230

step-by-step explanation:

i need to answer questions so i dont know but i think its the first one because it goes up but a consent rate

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2