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KengaRu [80]
3 years ago
7

*I WILL MARK U BRAINLIEST IF IT CORRECT* Using the following set of data, calculate the lower quartile, the upper quartile, and

the interquartile range.
20, 22, 25, 28, 29, 30, 32, 33, 34

Be sure to show your work for finding:

the lower quartile
the upper quartile
the interquartile range
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

1. 20

2. 34

3. 14

Step-by-step explanation:

20 is smallest number

34 is biggest number

34-20=14

range is 34-20

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Answer:

The answer is a

Step-by-step explanation:

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3 years ago
Which graph can be used to find the solution for the equation 4x + 2 = x + 3?
Marat540 [252]

Answer:

option C

Step-by-step explanation:

to find the solution for the equation 4x + 2 = x + 3, graph the equations

y= 4x+2  and y=x+3

Lets make a table for each equation

x          y= 4x+2

-1          4(-1) + 2= -4+2 = -2

0           4(0) + 2= 2

1             4(1) + 2= 6

plot all the points on graph

Lets make table for second equation

x          y= x+3

-1          -1+3 = 2

0           0+3 = 3

1             1+3 = 4

Plot the point on the graph and make a line

option C is correct

8 0
3 years ago
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Hmm, What is the answer?<br><br><br>PLEASE HELP ME!!!
Angelina_Jolie [31]

Answer:

y = kx.

Step-by-step explanation:

A proportional relationship between a quantity y and a quantity x that has a constant of proportionality k is represented by the equation y = kx.

4 0
2 years ago
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find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

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2 years ago
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Lina20 [59]

Answer:

3 and 4 cannot be evenly distributed

Step-by-step explanation:

No

3 0
3 years ago
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