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3 years ago

PLEASE HELP! 30 points

Mathematics

2 answers:

uranmaximum [27]3 years ago

7 0

Answer:

The median is 6

Step-by-step explanation:

Place all numbers in order from least to greatest:

2,3,5,5,6,6,6,6,7,7,7,9

Start by counting off the numbers evenly on each side:

5,5,6,6,6,6,7,7,

6,6,6,6,

6,6

If you are left with 2 numbers after evenly counting off numbers, then right what is in between the two numbers:

satela [25.4K]3 years ago

3 0

Two Numbers in the Middle. BUT, with an even amount of numbers things are slightly different. In that case we find the middle pair of numbers, and then find the value that is half way between them. This is easily done by adding them together and dividing by two.

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The length of a rectangle is 3 meters more then the width the perimeter of ractangle is 26 meters

Fudgin [204]

Answer:

Step-by-step explanation:

Let's say the width is x. If the length is 3 meters more than the width, the the length is x+3.

A rectangle has 4 sides. 2 widths and 2 lengths. If the perimeter is 26 meters, you can plug in your expressions for the width and the length to solve for x.

2(x+3) +2(x)=26

2x+6+2x=26

4x+6=26

4x=20

x=5.

If x=5, that means that your with is 5 meters, and your length is 8 meters (5+3)

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2 years ago

I will give brainleist

Kitty [74]

Answer:

= 28.27m 

Step-by-step explanation:

<h3> AREA OF CIRCLE = 2 π r</h3>

= 2 x 3.14 x 3m

= 28.27m

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2 years ago

If the ratio of a circle's sector to its total area is 1/3, what is the measure of

nalin [4]

Answer:

option A: 120°

Step-by-step explanation:

$Area \ of \ circle = \pi r^2\\\\Area \ of \ sector = \pi r^2 \frac{ \theta}{360}\\\\Ratio \ of \ sector \ to \ circle = \frac{1}{3}\\\\That is, \\\frac{1}{3} = \frac{\pi r^2 \frac{\theta}{360}} {\pi r^2}\\\\\\\frac{1}{3} = \frac{\theta}{360} \\\\\frac{360}{3} = \theta\\\\120 = \theta$

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3 years ago

Need help with a statistics question anything will help

stealth61 [152]

WHAT THE F-

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Afina-wow [57]

2 is one, i really don’t remember this. I think 6 and 4 3 i think.

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