7)3000/5000*100=60%
8)7000-6000=1000
1000/7000*100=14.3%
9)33.6/2.1=16
10)201.15/1.25 =160.92 but you only round up in order to get it fill so 161 jugs
Answer:
E
Step-by-step explanation:
The maximum of the data set is the point to the far right.
Point E is the maximum
Point A is the minimum
Point C is the mean
Answer:
C.) p(x), because an increasing quadratic function will eventually exceed an increasing exponential function
HOPE THIS HELPS :3
Answer:
x1 = x 61 = 6
x0 = 1 70 = 1
x-1 = 1/x 4-1 = 1/4
xmxn = xm+n x2x3 = x2+3 = x5
xm/xn = xm-n x6/x2 = x6-2 = x4
(xm)n = xmn (x2)3 = x2×3 = x6
(xy)n = xnyn (xy)3 = x3y3
(x/y)n = xn/yn (x/y)2 = x2 / y2
x-n = 1/xn x-3 = 1/x3
Step-by-step explanation:
i hope this helps if its wrong just delete my answer
Rewrite the sums as
![\displaystyle S_2 = \sum_{k=1}^n \frac{k^2}{2k^2 - 2nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{2k^2}{n^2} - \frac{2k}n + 1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S_2%20%3D%20%5Csum_%7Bk%3D1%7D%5En%20%5Cfrac%7Bk%5E2%7D%7B2k%5E2%20-%202nk%20%2B%20n%5E2%7D%20%3D%20%5Csum_%7Bk%3D1%7D%5En%20%5Cfrac%7B%5Cfrac%7Bk%5E2%7D%7Bn%5E2%7D%7D%7B%5Cfrac%7B2k%5E2%7D%7Bn%5E2%7D%20-%20%5Cfrac%7B2k%7Dn%20%2B%201%7D)
and
![\displaystyle S_3 = \sum_{k=1}^n \frac{k^2}{3k^2 - 3nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{3k^2}{n^2} - \frac{3k}n + 1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S_3%20%3D%20%5Csum_%7Bk%3D1%7D%5En%20%5Cfrac%7Bk%5E2%7D%7B3k%5E2%20-%203nk%20%2B%20n%5E2%7D%20%3D%20%5Csum_%7Bk%3D1%7D%5En%20%5Cfrac%7B%5Cfrac%7Bk%5E2%7D%7Bn%5E2%7D%7D%7B%5Cfrac%7B3k%5E2%7D%7Bn%5E2%7D%20-%20%5Cfrac%7B3k%7Dn%20%2B%201%7D)
Now notice that
![\displaystyle \lim_{n\to\infty} \frac{S_2}n = \int_0^1 \frac{x^2}{2x^2 - 2x + 1} = \frac12](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7BS_2%7Dn%20%3D%20%5Cint_0%5E1%20%5Cfrac%7Bx%5E2%7D%7B2x%5E2%20-%202x%20%2B%201%7D%20%3D%20%5Cfrac12)
and
![\displaystyle \lim_{n\to\infty} \frac{S_3}n = \int_0^1 \frac{x^2}{3x^2 - 3x + 1} = \frac{9 + 2\pi\sqrt3}{27}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7BS_3%7Dn%20%3D%20%5Cint_0%5E1%20%5Cfrac%7Bx%5E2%7D%7B3x%5E2%20-%203x%20%2B%201%7D%20%3D%20%5Cfrac%7B9%20%2B%202%5Cpi%5Csqrt3%7D%7B27%7D)
and the important point here is that
and
converge to constants. For any real constant a, we have
![\displaystyle \lim_{n\to\infty} \frac{\ln(an)}n = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7B%5Cln%28an%29%7Dn%20%3D%200)
Rewrite the limit as
![\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = \lim_{n\to\infty} \exp\left(\ln\left(\sqrt[n]{S_2 \times S_3}\right)\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln(S_2) + \ln(S_3)}n\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln\left(n \times \frac{S_2}n\right) + \ln\left(n \times \frac{S_3}n\right)}n\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csqrt%5Bn%5D%7BS_2%20%5Ctimes%20S_3%7D%20%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cexp%5Cleft%28%5Cln%5Cleft%28%5Csqrt%5Bn%5D%7BS_2%20%5Ctimes%20S_3%7D%5Cright%29%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7B%5Cln%28S_2%29%20%2B%20%5Cln%28S_3%29%7Dn%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7B%5Cln%5Cleft%28n%20%5Ctimes%20%5Cfrac%7BS_2%7Dn%5Cright%29%20%2B%20%5Cln%5Cleft%28n%20%5Ctimes%20%5Cfrac%7BS_3%7Dn%5Cright%29%7Dn%5Cright%29)
Then
![\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = e^0 = \boxed{1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csqrt%5Bn%5D%7BS_2%20%5Ctimes%20S_3%7D%20%3D%20e%5E0%20%3D%20%5Cboxed%7B1%7D)
A plot of the limand for n = first 1000 positive integers suggests the limit is correct, but convergence is slow.