Asample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviation of 0.79. calculate a 95% large-sample ci for the true average percentage elongation μ. (give answers accurate to 2 decimal places.)

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!