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3 years ago

Find the interquartile range for 1117 1465 513 437 1038 954 650 893 456 1018

Mathematics

1 answer:

GalinKa [24]3 years ago

8 0

I think it would be 562!

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The value of (-8/13×26/-3)

ahrayia [7]

Answer:

Step-by-step explanation:

Given the expression -8/13×26/-3, getting the equivalent expression means multiplying the two fractions -8/13 and 26/-3 together as shown;

$\frac{-8}{13} * \frac{26}{-3}\\ \\= \frac{-8*26}{13*-3}\\ \\multiplying\ out \ the\ numerator\ and\ denominator;\\\\= \frac{-208}{-39} \\\\cancelling out the negative sign we will have;\\\\= \frac{208}{39}\\\\= 5.33\\The\ resulting\ value\ is\ 5.33$

3 0

3 years ago

Help please! ill mark brainliest!

Natasha_Volkova [10]

Answer:

0.225989

Step-by-step explanation:

3x+9+x-11=180

4x-3=180

4x=180-3

4x/4x=177/4x

0.225989

7 0

3 years ago

Read 2 more answers

If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____

Sauron [17]

<h2>Answer:</h2>

$2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}$

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

$\vec{u}=(5,-7) \ and \ \vec{v}=(-11,3)$

So we need to find two things:

$2\vec{v}-6\vec{u}$

and:

$||2\vec{v}-6\vec{u}||$

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

$2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}$

SECOND:

If we name:

$\vec{w}=2\vec{v}-6\vec{u}$

Then, $||2\vec{v}-6\vec{u}||$ is the magnitude of the vector $\vec{w}$ . Therefore:

$||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}$