Answer:
D
Step-by-step explanation:
Number of days in school = 180-7 days missed = 173
Percent = Part/Whole x 100%
Percent = 173/180 x 100%
Percent = 96%
It comes from integrating by parts twice. Let

Recall the IBP formula,

Let


Then

Apply IBP once more, with


Notice that the ∫ v du term contains the original integral, so that




Step-by-step explanation:

Given expression is

To, evaluate this limit, let we simplify numerator and denominator individually.
So, Consider Numerator

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.
So, using Sum of n terms of GP, we get


Now, Consider Denominator, we have

can be rewritten as

![\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B%7B%5Cdfrac%7Bn%20-%201%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7Bn%20-%202%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D)
![\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D)
Now, Consider

So, on substituting the values evaluated above, we get
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5Cdfrac%7B%20%7Bn%7D%5E%7Bn%7D%20%20-%201%7D%7B1%20-%20%20%5Cdfrac%7B1%7D%7Bn%7D%20%7D%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%20%7Bn%7D%5E%7Bn%7D%20%20-%201%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20-%20%5Cdfrac%7B1%7D%7B%20%7Bn%7D%5E%7Bn%7D%20%7D%20%5Cbigg%5D%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5Cbigg%5B1%20-%20%5Cdfrac%7B1%7D%7B%20%7Bn%7D%5E%7Bn%7D%20%7D%20%5Cbigg%5D%7D%7B%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B1%7D%7B%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
Now, we know that,
![\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}](https://tex.z-dn.net/?f=%5Cred%7B%5Crm%20%3A%5Clongmapsto%5C%3A%5Cboxed%7B%5Ctt%7B%20%5Cdisplaystyle%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cbigg%5B1%20%2B%20%5Cdfrac%7Bk%7D%7Bx%7D%20%5Cbigg%5D%5E%7Bx%7D%20%20%3D%20%20%7Be%7D%5E%7Bk%7D%7D%7D%7D%20)
So, using this, we get

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have





Hence,

The answer is a because i said so.
T = state income return (without sales tax)
106% of t = 4960.80
AKA
1.06t = 4960.80 --> Divide both sides by 1.06 to find the state income return without sales tax
t = 4680
Subtract t from the state income return with sales tax.
4960.80 - 4680 = 160.80
<u>She paid $160.80 in sales tax.</u>