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IRISSAK [1]
3 years ago
6

If a class of 27 students had a mean of 72 on a test, interpret the mean of 72 in the sense of a fair share measure of the cente

r of the test scores.
Mathematics
2 answers:
lions [1.4K]3 years ago
8 0

The fair share measure would mean that the score of 72 would be the score all 27 students got on the test if they all had the same score.

Ulleksa [173]3 years ago
5 0
Sooorrrry I am nice!!
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babymother [125]

You can probably look up a answer key for that worksheet. ;)

6 0
2 years ago
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Beinggreat78 answer fast please :')
Dovator [93]

Answer:

17662.5 sq. feet

Step-by-step explanation:

We know that the formula for Area of a Circle is A = \pi (r)^2.

r = radius; d = diameter, which is r*2


And so, in the question, it gives you that the diameter is 150, first we divide it by 2 to get the radius.

150/2 = 75

The question said to keep \pi =3.14, so then we plug in the equation-

A = 3.14 * 75^2


Then we use PEMDAS [Parenthesis; Exponents; Multiply & Divide; Add & Subtract] which would get us-

A = 3.14 * 5625 (first doing the Exponent)

And we continue-

A = 17662.5 (then multiple)

So our answer is-

17662.5 sq. feet

I hope this helps!
Please give brainliest!

Have a great day

3 0
2 years ago
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Please HELP!! It’s due today
Aleks04 [339]

Answer:

For C all she has to do is look at the chart-

0=0

1=4

2=8

3=12

4=16

4 0
3 years ago
Add (-12x + 6) + (4x - 12).
yulyashka [42]

Step-by-step explanation:

it's obviously the equation is equals to zero,

-12x + 16 = 0

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like terms cancel each other to remaining with x

thus,

x = 4/3

<h2>AND</h2>

4x - 12 = 0

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4. 4

like terms cancel each other to remaining (make x subject of the formula) with x

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2 years ago
Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(
BlackZzzverrR [31]
We can parameterize this part of a cone by

\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle

with 0\le u\le a and 0\le v\le2\pi. Then

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv

The area of this surface (call it \mathcal S) is then

\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi
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3 years ago
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