Answer #1 please ASAP

Mathematics

2 answers:

Sladkaya [172]3 years ago

8 0

The correct answer is 4.5

Ad libitum [116K]3 years ago

6 0

Answer:

4.5

Step-by-step explanation:

To find the median, put the data in order from smallest to largest

1,2,2,3,4,4,5,5,8,10,10,11

The count up the number of data points

Divide by 2

12/2

Split the data into 2 groups, 6 in each, starting at the beginning

1,2,2,3,4,4, 5,5,8,10,10,11

Since there is an even number

Add the last number in first group and the first number in the last group and divide by 2

(4+5)/2 = 9/2 = 4.5

The median is 4.5

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Laura bought a square canvas to paint a picture of her cat one side measured 22 centimeters what is tha area of the canvas

Olin [163]

Hi there!

To start, squares are a shape in which all of its sides are congruent to one another. Knowing the length of one side would mean that you would know the length of all the other sides.

Because you know that one side measures 22 cm, you can conclude that all the other sides of the square are also at a length/width of 22 cm.

Next, you will find the area of the square.

Because the area formula is Area=Length * Width, you can easily find the area of the square by multiplying 22*22 since the length and width of any square is the same.

When you simplify 22*22, you should get 484.

Therefore, the area of the square canvas would be 484 cm squared.

Hope this helps and have a marvelous day! :)

8 0

3 years ago

25x2 + 5x

poizon [28]

Answer:

option C: 5x+1

Explaination:

take 5x common and simplify

3 0

2 years ago

Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.