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Gwar [14]
3 years ago
10

What is the absolute value of 0.6?

Mathematics
2 answers:
Nesterboy [21]3 years ago
7 0
The absolute value of 0.6 is 0.6 beacause its the same thing
Alex_Xolod [135]3 years ago
7 0

Answer:

0.6

Step-by-step explanation:

We are given that  0.6

We have to find the absolute value of 0.6

Absolute value of a number : it is defined as the magnitude of a number .

If a number x

Then, its absolute value is given by

\mid x\mid=x

It is always  positive because it is distance of a number  from zero .

Absolute value of 0.6 is given by

\mid 0.6\mid=0.6

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Solve for homework :)
katrin2010 [14]

Answer:

1) \frac{ {a}^{8} }{ {a}^{6} }  \\  = {a}^{8}  \times  {a}^{ - 6}  =  {a}^{2}  \\ 2) {m}^{2}  {m}^{ - 4}  =  {m}^{ - 2}  \\ thank \: you

6 0
2 years ago
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Please can anyone help me please​
Step2247 [10]
4.5, 4, 3.5, 3, 2.5, 2 and use the pencil and ruler, love a bit of MathsWatch
7 0
3 years ago
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A book claims that more hockey players are born in January through March than in October through December. The following data sh
denis-greek [22]

Complete question :

Birth Month Frequency

January-March 67

April-June 56

July-September 30

October-December 37

Answer:

Yes, There is significant evidence to conclude that hockey​ players' birthdates are not uniformly distributed throughout the​ year.

Step-by-step explanation:

Observed value, O

Mean value, E

The test statistic :

χ² = (O - E)² / E

E = Σx / n = (67+56+30+37)/4 = 47.5

χ² = ((67-47.5)^2 /47.5) + ((56-47.5)^2 /47.5) + ((30-47.5)^2/47.5) + ((37-47.5)^2/47.5) = 18.295

Degree of freedom = (Number of categories - 1) = 4 - 1 = 3

Using the Pvalue from Chisquare calculator :

χ² (18.295 ; df = 3) = 0.00038

Since the obtained Pvalue is so small ;

P < α ; We reject H0 and conclude that there is significant evidence to suggest that hockey​ players' birthdates are not uniformly distributed throughout the​ year.

5 0
3 years ago
Plzz help me simplify
Elza [17]

Step-by-step explanation:

\frac{8}{4} x {}^{8 - 4}

<h2><em>=</em><em>></em><em>2</em><em>x</em><em>⁴</em></h2>
4 0
3 years ago
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Can someone solve this?
ladessa [460]
  • First question:

Recall that \cos^2x+\sin^2x=1 and \sqrt{x^2}=|x| for all x. So

\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|

\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|

For 0, we expect both \cos x>0 and \sin x>0 (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, |x|=x if x>0.

So we have

\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}

making H the answer.

  • Second question:

C is always true, because the inequality reduces to x > y.

6 0
4 years ago
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