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zloy xaker [14]
3 years ago
12

At a gas station, all chips are marked down 10%. A customer brings a bag of chips with a regular price of $2.19 to the register.

After the 7% sales tax on the final price, how much does the customer pay for the bag of chips ?
Mathematics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer: $2.11

Step-by-step explanation:

A customer brings a bag of chips with a regular price of $2.19 to the register which is marked down 10% the price to pay will be:

= $2.19 - (10% × $2.19)

= $2.19 - $0.219

= $1.971

After a 7% sales tax on the final price, the amount paid by the customer will be:

= $1.971 + (7% × $1.971)

= $1.971 + $0.13797

= $2.11

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Distributive property of <br>4x 567​
Maksim231197 [3]

567 = 500+60+7

4*567 = 4*(500+60*7)

4*567 = 4*500 + 4*60 + 4*7 ... see note below

4*567 = 2000 + 240 + 28

4*567 = 2268

--------

note: multiply the outer term 4 by each term inside the parenthesis to use the distributive property. The general distributive property is a*(b+c) = a*b+a*c. This can be extended to a*(b+c+d) = a*b+a*c+a*d. You can have as many terms as you like inside the parenthesis.

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124 + 110 = 234
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Which is greater than or less than
Montano1993 [528]

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Step-by-step explanation:

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4 years ago
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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
4 years ago
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