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kramer
3 years ago
15

A recent physical inventory at a department store reports a merchandise value of $976,454. The merchandise book value was $1,150

,450. The merchandise sales were $2,950,675. What is the shrinkage as a percentage of sales?
a) 2.3%
b) 4.6%
c) 5.9%
d) 7.9%
Mathematics
2 answers:
Burka [1]3 years ago
5 0

Answer:

c) 5.9%

Step-by-step explanation:

iogann1982 [59]3 years ago
3 0
The answer for the percentage is C) 5.9%



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Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

7 0
3 years ago
I need to find a expression equivalent to this with the same numbers <br><br> 2•((4•x•3)•6)
Kryger [21]

Answer:

144 * x

Step-by-step explanation:

2*4*x*3*6

2*4*3*6*x

8*3*6*x

24*6*x

144*x

3 0
3 years ago
Help me pleasee<br> Questions r in picture
jok3333 [9.3K]
<h2>Answer:</h2>

\large \bf\implies\frac{12}{61}

<h2>Step-by-step explanation:</h2>

<h2>Given :</h2>

\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...

<h2>To Find :</h2>

  • nth term

<h2>Solution :</h2>

We have to add 1 in numerator and -10 in denominator because

\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]

\frac{8  \: + \:  1}{101 \:  - \:  10}  =  \frac{9}{91} \\\\  \frac{9 + 1}{91 - 10}  =  \frac{10}{81}  \\ \\ \frac{10 + 1}{81 - 10}  =  \frac{11}{71}  \\ \\ \frac{11 + 1}{71 - 10}  =  \frac{12}{61} ...

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.

4 0
2 years ago
How do you turn 245/360 into a decimal
erastovalidia [21]

Answer:

68.06%;

Step-by-step explanation:

6 0
3 years ago
In a certain sequence of numbers, each term after the first is found by doubling and then adding $3$ to the previous term. If th
eimsori [14]
Given a_7=125 you can find a_6:

a_6= \frac{125-3}{2} =61
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Answer: -1
6 0
3 years ago
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