Answer:
6
Step-by-step explanation:
To find the median you simply look for the number in the middle of the list.
In this particular case 1 and 11 are both the fifth number in the series of numbers. When this happens you have to add the two terms together (1+11) and divide by two (12/2)=6
Answer:
![a^{6}](https://tex.z-dn.net/?f=a%5E%7B6%7D)
Step-by-step explanation:
when you have the same base and its times you can add the exponents
a² + a³ + a have exponents of 2+ 3 + 1 = 6
so its ![a^{6}](https://tex.z-dn.net/?f=a%5E%7B6%7D)
multiply total weight by unit cost
8.8 x 2.09 = 17.765 = $17.77
Answer:
I'm gonna go with C
Step-by-step explanation:
Don't take my word for it
Answer:
Diverges
General Formulas and Concepts:
<u>Algebra I</u>
- Exponential Rule [Dividing]:
![\displaystyle \frac{b^m}{b^n} = b^{m - n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bb%5Em%7D%7Bb%5En%7D%20%3D%20b%5E%7Bm%20-%20n%7D)
<u>Calculus</u>
Limits
- Limit Rule [Variable Direct Substitution]:
![\displaystyle \lim_{x \to c} x = c](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20c%7D%20x%20%3D%20c)
Series Convergence Tests
- P-Series:
![\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^p}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum%5E%7B%5Cinfty%7D_%7Bn%20%3D%201%7D%20%5Cfrac%7B1%7D%7Bn%5Ep%7D)
- Direct Comparison Test (DCT)
- Limit Comparison Test (LCT):
![\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Ba_n%7D%7Bb_n%7D)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
![\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum%5E%7B%5Cinfty%7D_%7Bn%20%3D%2019%7D%20%5Cfrac%7B8n%5E3%20-%202n%5E2%20%2B%2019%7D%7B6%20%2B%203n%5E4%7D)
<u>Step 2: Apply DCT</u>
- Define Comparison:
![\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{n^3}{n^4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum%5E%7B%5Cinfty%7D_%7Bn%20%3D%2019%7D%20%5Cfrac%7Bn%5E3%7D%7Bn%5E4%7D)
- [Comparison Sum] Simplify:
![\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum%5E%7B%5Cinfty%7D_%7Bn%20%3D%2019%7D%20%5Cfrac%7B1%7D%7Bn%7D)
- [Comparison Sum] Determine convergence:
![\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n} = \infty , \ \text{div by P-Series}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum%5E%7B%5Cinfty%7D_%7Bn%20%3D%2019%7D%20%5Cfrac%7B1%7D%7Bn%7D%20%3D%20%5Cinfty%20%2C%20%5C%20%5Ctext%7Bdiv%20by%20P-Series%7D)
- Set up inequality comparison:
![\displaystyle\frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \geq \frac{1}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B8n%5E3%20-%202n%5E2%20%2B%2019%7D%7B6%20%2B%203n%5E4%7D%20%5Cgeq%20%5Cfrac%7B1%7D%7Bn%7D)
- [Inequality Comparison] Rewrite:
![\displaystyle n(8n^3 - 2n^2 + 19) \geq 6 + 3n^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20n%288n%5E3%20-%202n%5E2%20%2B%2019%29%20%5Cgeq%206%20%2B%203n%5E4)
- [Inequality Comparison] Simplify:
![\displaystyle 8n^4 - 2n^3 + 19n \geq 6 + 3n^4 \ \checkmark \text{true}](https://tex.z-dn.net/?f=%5Cdisplaystyle%208n%5E4%20-%202n%5E3%20%2B%2019n%20%5Cgeq%206%20%2B%203n%5E4%20%5C%20%5Ccheckmark%20%5Ctext%7Btrue%7D)
∴ the sum
is divergent by DCT.
<u>Step 3: Apply LCT</u>
- Define:
![\displaystyle a_n = \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}, \ b_n = \frac{1}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a_n%20%3D%20%5Cfrac%7B8n%5E3%20-%202n%5E2%20%2B%2019%7D%7B6%20%2B%203n%5E4%7D%2C%20%5C%20b_n%20%3D%20%5Cfrac%7B1%7D%7Bn%7D)
- Substitute in variables [LCT]:
![\displaystyle \lim_{n \to \infty} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \cdot n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B8n%5E3%20-%202n%5E2%20%2B%2019%7D%7B6%20%2B%203n%5E4%7D%20%5Ccdot%20n)
- Simplify:
![\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B8n%5E4%20-%202n%5E3%20%2B%2019n%7D%7B6%20%2B%203n%5E4%7D)
- [Limit] Evaluate [Coefficient Power Rule]:
![\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4} = \frac{8}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B8n%5E4%20-%202n%5E3%20%2B%2019n%7D%7B6%20%2B%203n%5E4%7D%20%3D%20%5Cfrac%7B8%7D%7B3%7D)
∴ Because
and the sum
diverges by DCT,
also diverges by LCT.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Convergence Tests (BC Only)
Book: College Calculus 10e