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Artemon [7]
3 years ago
13

Why is the standard deviation usually preferred over the​ range?

Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0
Standard deviations, σ, is generally preferred over range. Reason being, it is calculated from all the data and when there is outliers it will not be impacted as much as the range. It is great to not also that, The IQR is preferred to the standard deviation when the distribution is very highly skewed or when there are severe outliers, because the IQR is less sensitive to this features than standard deviation.
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I will mark who ever gets it correct the brainiest.<br> Help Asap. See image below.
marysya [2.9K]
The transformation from the first equation to the second one can be found by finding
a
a
,
h
h
, and
k
k
for each equation.
y
=
a
|
x
−
h
|
+
k
y
=
a
|
x
-
h
|
+
k
Factor a
1
1
out of the absolute value to make the coefficient of
x
x
equal to
1
1
.
y
=
|
x
|
y
=
|
x
|
Factor a
1
1
out of the absolute value to make the coefficient of
x
x
equal to
1
1
.
y
=
|
x
|
−
4
y
=
|
x
|
-
4
Find
a
a
,
h
h
, and
k
k
for
y
=
|
x
|
−
4
y
=
|
x
|
-
4
.
a
=
1
a
=
1
h
=
0
h
=
0
k
=
−
4
k
=
-
4
The horizontal shift depends on the value of
h
h
. When
h
>
0
h
>
0
, the horizontal shift is described as:
g
(
x
)
=
f
(
x
+
h
)
g
(
x
)
=
f
(
x
+
h
)
- The graph is shifted to the left
h
h
units.
g
(
x
)
=
f
(
x
−
h
)
g
(
x
)
=
f
(
x
-
h
)
- The graph is shifted to the right
h
h
units.
Horizontal Shift: None
4 0
3 years ago
PLEASE HURRY I HAVE 10 MINUTES AND ILL GIVE 50 POINTS!
stealth61 [152]

Answer:

What Table!?!?

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Part A.
natali 33 [55]

Answer:

Part A

Martha is incorrect because she will pay 1/4 of the cost.

40%+0.35

0.40+0.35

0.75

0.25=25/100=1/4

Part B

Renee-$340

Susan-$297.50

Martha-$212.50

6 0
2 years ago
Make up a problem similar to an inverse variation problem.
ad-work [718]

Start with the formula:    h=k \t    

Substitute the values  :<span>   2=k\65         
</span><span>
then solve for</span> k:           k=130    

4 0
2 years ago
Let f(x) represent a function.
STatiana [176]

Answer:

For f(x+\frac{5}{4}), f(x) is translated \frac{5}{4} units left.

For f(x)-\frac{5}{4}, f(x) is translated \frac{5}{4} units down.

Step-by-step explanation:

The transformation of f(x)\rightarrow f(x+C) means that the graph shift to left by C units if C is positive number and shifts to right by C units if C is a negative number.

The transformation of f(x)\rightarrow f(x)+C means that the graph shifts up by C units if C is positive number and shifts down by C units if C is a negative number.

Here, in the transformation of f(x)\rightarrow f(x+\frac{5}{4}),C = \frac{5}{4}>0, so, the function translates left by \frac{5}{4} units.

Similarly, in the transformation of f(x)\rightarrow f(x)-\frac{5}{4},C = -\frac{5}{4}, so, the function translates down by \frac{5}{4} units.

3 0
3 years ago
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