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Flauer [41]
3 years ago
10

What measures of spread is an average distance from the mean?

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

Step-by-step explanation:

The idea behind the standard deviation is to quantify the spread of a distribution by measuring how far the observations are from their mean. The standard deviation gives the average (or typical distance) between a data point and the mean.

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What is the next term in the sequence below?
andreyandreev [35.5K]

Answer: Option a.

Step-by-step explanation:

1. The sequence shown in the problem is a Geometric sequence, because it has a common ratio. This means that when you multiply one of those numbers by the common ratio, you obtain the next number.

2. Then, you can calculate the common ratio as following:

r=\frac{108}{-324}=-\frac{1}{3}

3. Therefor, the next term is:

(12)(-\frac{1}{3})=-4

7 0
3 years ago
Read 2 more answers
Write and equation with aa variable on both sides of the equal sign that has infinitely many solutions. solve the equation and e
LekaFEV [45]
X = x

to solve, subtract both sides by x:

0 = 0

x is always equals to x so infinite solutions
8 0
3 years ago
(3cosx-4sinx) + (3sinx+4cosx) = 5
Ksenya-84 [330]
(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0

Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º      (360º-53.13º=306.87)

sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º

The solutions when  0≤x≤360º are:  36.87º and 306.87º.


3 0
3 years ago
A pink crayon is made with 12 ML’s of red wax for every five ML of white wax
Papessa [141]

Answer:

kewl

Step-by-step explanation:

4 0
3 years ago
What is the value of the expression 1/4 ^-3?
zloy xaker [14]
(\frac{1}{4})^{-3}= (\frac{4}{1})^3\\\\=(4)^3\\\\=4\times 4\times 4\\\\=16\times 4\\\\=64

Your final answer is B. 64.
8 0
3 years ago
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