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kramer
3 years ago
13

Store A has a weekend sale. To the nearest hundredth of a percent, what discount must Store A offer for you to buy the system th

ere?

Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0
We are given the cost of the product along with the markup percentage.  The markup is how the store makes a profit.  To figure out the markup price, simply multiply the "cost to store" by 100% plus the markup percent.

So the markup prices per store is:

A = 162 * 1.4 = $226.80
B = 155 * 1.3 = $201.50
C = 160 * 1.25 = $200.00

Clearly, the best deal is store C.  So in order to choose store A, we want to pay less than the price of Store C, $200.  So if Store A cost $199.99, then we would purchase there instead of Store C.

Therefore, $226.80 - $199.99 = $26.81
If Store A offered a discount of $26.81, we would purchase from Store A.

$26.81/$226.80 = 0.1182

So we would need a discount of 11.82% from Store A.

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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
3 years ago
Let f(x) = 2x – 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following.
bonufazy [111]
4.
h(f(x)=h(2x-1)=
(2x-1)^2+1=
4x²-4x+1+1=
4x²-4x+2

5.
f(f(x))=2(2x-1)-1=4x-2-1=4x-3

6.
f o g (x)=f(g(x))
h o g (x)=h(g(x))=
(3x)^2+1=
9x²+1
4 0
3 years ago
Can you please help me
kirill [66]

Answer:

b

Step-by-step explanation:

3 0
3 years ago
What is the measure of x angles are not necessarily
lisov135 [29]

Answer:

X is 83

Step-by-step explanation:

Subtract 180 because thats what all of them eaqual and you subtract 97 from 180

7 0
2 years ago
Which expression is equivalent to ^5 square root 13^3
joja [24]

Answer:

\sqrt[5]{13^3} = 13^{\frac{3}{5}}

Step-by-step explanation:

5 0
2 years ago
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