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3 years ago

Which of the following best describes the change in the mean when 1213 is

Mathematics

1 answer:

Rama09 [41]3 years ago

4 0

Answer:

Step-by-step explanation:

The mean is the sum of all the values divided by the number of values.

Mean = $\frac{34+75+26+81+65+38+49}{7}=52.57$

When 1213 is added to the data set, new mean is:

New Mean = $\frac{34+75+26+81+65+38+49+1213}{8}=197.625$

<u><em>Definitely</em></u>, the mean increases. Correct answer is A

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Please help with this ! will give brainless

irakobra [83]

Answer:

x^13

2n^2x-5n^7

-6x^10-x

60s^7

6v^7

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pls mark as brainliest if its right :)

7 0

2 years ago

Can someone please help me find the side length area, and please draw the shape on how it would look if it were finished

Mumz [18]

The side length is 5 units, and the area is 25 square units

<h3>What are areas?</h3>

The area of a shape is the amount of space it occupies

<h3>What is the area of a square</h3>

The area of a square is the product of its dimensions

From the graph, the coordinates of the square can be assumed to be:

$A = (0,0)$

$B = (4,3)$

The side length of the square is then calculated using the following distance formula

$d = \sqrt{(x_2- x_1)^2 + (y_2 -y_1)^2}$

So, we have:

$AB = \sqrt{(0 - 4)^2 + (0-3)^2}$

$AB = \sqrt{25}$

$AB = 5$

The area of the square is the calculated as:

$Area = AB^2$

This gives

$Area = 5^2$

$Area = 25$

Hence, the side length is 5 units, and the area is 25 square units

See attachment for the diagram of the shape

Read more about areas at:

brainly.com/question/24487155

6 0

2 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...