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const2013 [10]
3 years ago
6

Pla help good amount of points will mark brainest

Mathematics
1 answer:
enot [183]3 years ago
3 0

Answer:

1) the general trend is skewed left because there is more data towards the left of the graph, there are more 1's 2's 3's and 4's than there are of any number on the right of the graph.

2) the median is 3 1122233344567 mark off numbers on either side one by one until you get to the center number which is the median

3) add up all the numbers divide by the number of numbers you have which is 13. you end up with 53/13 which is equal to 4.1

4) the mean because if you added 15 to the end of the data set the median would still be 3 but if you added 15 to the mean the mean would jump to 4.9

Step-by-step explanation:

its all explained for you, i hope this helps

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Solve for x. 7 + 11x + 60 = -13 + 3x
Snowcat [4.5K]

Answer:

x = -10

Step-by-step explanation:

Step 1: Write equation

7 + 11x + 60 = -13 + 3x

Step 2: Solve for <em>x</em>

  1. Combine like terms: 11x + 67 = -13 + 3x
  2. Subtract 3x on both sides: 8x + 67 = -13
  3. Subtract 67 on both sides: 8x = -80
  4. Divide both sides by 8: x = -10

Step 3: Check

<em>Plug in x to verify it's a solution.</em>

7 + 11(-10) + 60 = -13 + 3(-10)

67 - 110 = -13 - 30

-43 = -43

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3 years ago
hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a
nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

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By the divergence theorem,

\displaystyle\iint_{\partial\mathcal E}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_{\mathcal E}\nabla\cdot\mathbf f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

where \partial\mathcal E is the boundary of \mathcal E. We have

\nabla\cdot\mathbf f(x,y,z)=\dfrac{\partial(4x)}{\partial x}+\dfrac{\partial(xy)}{\partial y}+\dfrac{\partial(4xz)}{\partial z}=4+x+4x=5x+4

so the flux is

\displaystyle\int_{z=0}^{z=2}\int_{y=0}^{y=2}\int_{x=0}^{x=2}(5x+4)\,\mathrm dx\,\mathrm dy\,\mathrm dz=72
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