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3 years ago

Help plzzzz The minimum value os 58 The median value is 82 The third quartile is 58

Mathematics

2 answers:

iragen [17]3 years ago

4 0

Answer:

It is a box

Step-by-step explanation:

IT's a box ,because I had to put a stick in the box

andriy [413]3 years ago

3 0

This just labels the box plot hope this helps

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Compare 8/11 to 3/5 PLEASE HELP !!!

ad-work [718]

Rewrite both fractions with a common denominator:

8/11 x 5 = 40/55

3/5 x 11 = 33/55

40/55 is greater than 33/55 so 8/11 is greater than 3/5

5 0

3 years ago

What is the area of the rectangle? shoe your work! do not round anything until the very end and round your final answer to the t

weeeeeb [17]

Answer:

The area of the rectangle is 42 units^2

Step-by-step explanation:

we know that

The area of rectangle is equal to

$A=LW$

In this problem

AB=DC

BC=AD

see the attached figure with letters to better understand the problem

we have that

$L=BC\\W=AB$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

$A(2,-1),B(5,2),C(12,-5),D(9,-8)$

Find out the distance BC

we have

$B(5,2),C(12,-5)$

substitute in the formula

$d=\sqrt{(-5-2)^{2}+(12-5)^{2}}$

$d=\sqrt{(-7)^{2}+(7)^{2}}$

$d_B_C=\sqrt{98}\ units$

Find out the distance AB

we have

$A(2,-1),B(5,2)$

substitute in the formula

$d=\sqrt{(2+1)^{2}+(5-2)^{2}}$

$d=\sqrt{(3)^{2}+(3)^{2}}$

$d_A_B=\sqrt{18}\ units$

Find out the area

$A=(L)(W)$

we have

$L=d_B_C=\sqrt{98}\ units$

$W=d_A_B=\sqrt{18}\ units$

substitute

$A=(\sqrt{98})(\sqrt{18})=42.0\ units^2$

4 0

4 years ago

How do you use double number lines in equations?

VLD [36.1K]

Okay sorry but this question is actually a little simple if you think less

7 0

3 years ago

Find the value of x.

Lerok [7]

i dont knwo

really but u can try to solve it

Step-by-step explanation:

3 0

2 years ago

Please explain, thanks! :) Multiply DE

Anit [1.1K]

First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.

$\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]$

Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
$\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]$
We do the same for the bottom row of the first matrix.
First Column
2*2=4
0*1=0
4+0=4
Second Column
2*3=6
0*2=0
6+0=6
Third Column
2*4=8
0*3=0
8+0=8
Our final matrix is:
$\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]$

:)

8 0

3 years ago