Does anyone know the answer???

Allison paid $297.50 for a new laptop. Last week, the price of the laptop<br> was $350.

Thepotemich [5.8K]

15%

350x0.15=52.5

350-52.5=297.5

4 0

2 years ago

What’s the answer to this problem?

Alja [10]

Answer:

Figure: Amount of tiles:

1 5

2 8

3 11

4 14

5 17

6 20

7 23

8 26

9 29

10 32

11 35

12 38

13 41

The sequence/pattern is +3 tiles.

So Figure 13 will have 41 tiles.

I hope this helps!

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

3 0

2 years ago

I need help with this problem?

Cerrena [4.2K]

Answer:

Step-by-step explanation:

Volume of cylinder:

$\boxed{\text{Volume of cylinder=$\pi r^{2}h$}}$

<u>Bottom cylinder:</u>

1 inch = 2.54 cm

diameter = 14 in

r = 14/2 = 7 in

r = 7 *2.54 = 17.78 cm

h = 4 in = 4*2.54 = 10.16 cm

Volume = 3* 17.78 * 17.78 * 10.16

= 9635.59 in³

<u>Upper cylinder:</u>

diameter = 12 in

r = 12/2 = 6in

r = 6*2.54 = 15.24 cm

h = 12 in = 12*2.54 = 30.48 cm

Volume = 3 * 15.24 *15.24 * 30.48

= 21237.63 in³

Volume of the object = 9635.59 + 21237.63

= 30873 cm³

3 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

A local weather station collected the 12 p.m. temperature at 5 different locations in its town: Temperatures, °F: {63, 59, 60, 6

AveGali [126]

Ok so add all of them together and you get 305 then divide it by how many numbers there are for the mean and you get 61 then add all of the diffrences from each number to the mean and we get 6 then 61 divided by 6 gets us 10.1 but round it to 10. hope this helped.

7 0

3 years ago

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