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Archy [21]
3 years ago
5

{29, 29, 29, 28, 28, 27}

Mathematics
2 answers:
charle [14.2K]3 years ago
5 0

Answer:

Option: a is correct.

a .mean or median because there is no outlier

Step-by-step explanation:

Mean and median both try to measure the "central tendency" in a data set.

Clearly from the set of the values or data points:

{29 , 29, 29 , 28, 28, 27}

We could see that the data values are close too each other i.e. there is not outlier ( An outlier is a point that stand out of the rest points i.e. that value is either too high or low as compared to the other data points).

Here we have Mean=28.3333

and Median=28.5

Hence, the correct option is: option: a

a .mean or median because there is no outlier.

Tatiana [17]3 years ago
4 0
A is the answer hope this helps
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4 years ago
Coordinates will be fine PLs ASAP cause I need this rn<br> 13 POINTS!!!
grigory [225]

If the vertices of the triangle are J(-3,-5), K(-2,2) and L(1,-4) then its reflection in y axis are J(3,-5), K(2,2) and L(-1,-4)

Given,

The vertices of the triangle = J(-3,-5), K(-2,2) and L(1,-4)

We know,

In reflection in y axis (x,y)  will become (-x,y)

Then the vertices of the reflection in y axis = J(3,-5), K(2,2) and L(-1,-4)

Draw the triangle using the vertices.

Hence, if the vertices of the triangle are J(-3,-5), K(-2,2) and L(1,-4) then its reflection in y axis are J(3,-5), K(2,2) and L(-1,-4)

Learn more about reflection here

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1 year ago
Help!! Find missing value for each quadratic function
Svetllana [295]

I'll do the first one to get you started. The answer is -3

=================================================

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

------------

Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

------------

Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

------------

Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1

0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

------------

If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

------------

In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

------------

The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

y = -4 + 1

y = -3


3 0
4 years ago
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