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Ierofanga [76]
3 years ago
11

Find the median, first quartile, third quartile, interquartile range, and any outliers for each set of data.

Mathematics
1 answer:
viktelen [127]3 years ago
6 0

Answer:The outlier is 68

The median is 102

First quartile: 87

Third Quartile: 115

The interquartile range is 87-115

Step-by-step explanation:

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A pair of equations is shown below:
horrorfan [7]

Answer:

A. x=2, y = 7.

B. (2, 7).

Step-by-step explanation:

A. You can eliminate y by subtracting the equations:

y =  8x - 9

y =  4x - 1       Subtract:

0 =  4x - 8

4x = 8

x = 8/4 = 2.

Now substitute for x in the first equation:

y = 8(2) - 9 = 7.

Check in the second equation:

y = 4x - 1

7 = 4(2) - 1 = 7.  Check is OK.

B. On the graph they will intersect at the point (2, 7).

They will intersect here because the values x=2 and y=7 satisfy both the 2 equations.

4 0
3 years ago
Part 2
Taya2010 [7]

9514 1404 393

Answer:

  a.  -4

Step-by-step explanation:

When using the "diamond method" for factoring quadratics, the bottom number is the coefficient of the linear term. In this quadratic, it is -4.

The bottom number is -4.

4 0
3 years ago
Read 2 more answers
Twice a number decreased by eight is sixteen.
Firlakuza [10]
The number is 12

16 plus 8 = 24/ 2= 12
7 0
2 years ago
Read 2 more answers
A mother explains to her child that the price on the sign is not the total price for an oil painting, because the price does not
dlinn [17]

Answer:

Or u cld simply just say 945 - 875 = 70

3 0
2 years ago
Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments
kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

7 0
1 year ago
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