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Lana71 [14]
3 years ago
7

Just need the answer thanks :)

Mathematics
2 answers:
Oxana [17]3 years ago
8 0
For the first one  is the product is greater than 223.  The second one is the product is equal to 727. The third one is the product is less than 563. 
emmainna [20.7K]3 years ago
3 0
Answers
last one
second one
first one
Hope this helps!
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Graph the equation below by plotting the y-intercept and a second point on the line. <br> Y=3/2x-5
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Answer:

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Common denominator for 5/24 and 7/18
lisov135 [29]
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3 years ago
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Sam conducted a survey to determine the favorite flavor of ice cream in his hometown. Fifteen of the fifty people he
Rus_ich [418]

Answer:

1,050

Step-by-step explanation:

If 15 people out of 50 liked strawberry best.

The relative frequency of liking strawberry best in Sam's hometown will be:

Relative Frequency = \dfrac{15}{50}

Expected Outcome=Relative Frequency X Number of Trials

Therefore, out of the 3500 residents:

Number of People Expected to like strawberry best

= \dfrac{15}{50}X3500

=1,050

4 0
4 years ago
Please answer this correctly
Andrews [41]

Answer:

2 minutes 56 seconds

Step-by-step explanation:

3 0
4 years ago
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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