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3 years ago

Unit Test, Part 2: Statistics

Mathematics

1 answer:

Lunna [17]3 years ago

8 0

Step-by-step explanation:

high: 91, 91, 91, 91, 90, 90, 92, 94

low:59, 60, 60, 60, 60, 60, 60, 61

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Name a pair of fractions that use the least common denominator and are equivalent to 9/10 and 5/6

VikaD [51]

Answer:

The Lowest Common Denominator is 30 9/10 = 27/30 5/6 = 25/30

6 0

3 years ago

There are 4 3/8 pounds of bricks in a bag. Each brick weighs 7/8 of a pound. How many bricks are in the bag?

goldfiish [28.3K]

Hello :) the answer to your question would be 5. There are 5 bricks in each bag. 35/8

7 0

1 year ago

Read 2 more answers

Jose bought a bag of 6 oranges for $2.82. He also bought 5 pineapples. He gave the cashier $20 and received $1.43 change. How mu

svp [43]

Your answer is: $3.15 each.

Here are the steps:

First, I added $2.82 and $1.43. Then, I subtracted the sum ($4.25) from $20. Lastly, I divided the difference($15.75) by five because there were five pineapples. Which gave me the answer: $3.15.

5 0

2 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

3 years ago

I NEED ANSWER ASAP !!!

aivan3 [116]

Answer:

Step-by-step explanation:

3+2x=5x

3=3x

x=1

PR=5

4 0

3 years ago