Answer:
Slope of function B
Step-by-step explanation:
Answer:
p-e< p < p+e
(0.061 - 0.025) < 0.061 < (0.061 + 0.025)
0.036 < 0.061 < 0.086
Step-by-step explanation:
Given;
Confidence interval CI = (a,b) = (0.036, 0.086)
Lower bound a = 0.036
Upper bound b = 0.086
To express in the form;
p-e< p < p+e
Where;
p = mean Proportion
and
e = margin of error
The mean p =( lower bound + higher bound)/2
p = (a+b)/2
Substituting the values;
p = (0.036+0.086)/2
Mean Proportion p = 0.061
The margin of error e = (b-a)/2
Substituting the given values;
e = (0.086-0.036)/2
e = 0.025
Re-writing in the stated form, with p = 0.061 and e = 0.025
p-e< p < p+e
(0.061 - 0.025) < 0.061 < (0.061 + 0.025)
0.036 < 0.061 < 0.086
Answer:
Given definite integral as a limit of Riemann sums is:
![\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Step-by-step explanation:
Given definite integral is:

Substituting (2) in above
![f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=f%28x_%7Bi%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%2B%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%5E%7B3%7D%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%282%2B%5Cfrac%7B3%7D%7B2n%7Di%29%2B%2864%2B%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B3%2816%29%5Cfrac%7B3%7D%7Bn%7Di%2B3%284%29%5Cfrac%7B9%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%29%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B3%7D%7B2n%7Di%2B%5Cfrac%7B144%7D%7Bn%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B291%7D%7B2n%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Riemann sum is:
![= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
2/3 is the answer in lowest terms. i hope this helps!
Lol i dont know the answer but i love your profile pic again sorry