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BabaBlast [244]
3 years ago
11

What is the solution for -10 < x - 9? x < -1 x < -19 x > -1 x > -19

Mathematics
2 answers:
Annette [7]3 years ago
7 0

Answer:

x > -1

Step-by-step explanation:

-10 < x - 9

-10 + 9 < x - 9 + 9

- 1 < x

x > -1  is the correct answer.

Hope this helps.

d1i1m1o1n [39]3 years ago
7 0

Answer:

x > -1 Is your answer

Step-by-step explanation:

Hope this helps!

Have a nice day! ♥

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Determine whether or not the procedure described below results in a binomial distribution. If it is not​ binomial, identify at l
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a) it is not a binomial, it is hypergeometric distribution.

Step-by-step explanation:

a) Four hundred different voters in a region with two major political​ parties, A and​ B, are randomly selected from the population of 3000 registered voters.

it is not a binomial because the probability of succes changes between one trial and another one, so the trials are not independent.

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3 years ago
How many cubes with an edge length of 1/3 inches does it take to fill a box with width 2 and 2/3 inches, length 3 and 1/3 inches
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Step-by-step explanation:

6 0
3 years ago
The 3 by 3 grid below shows nine 1 cm × 1 cm squares and uses 24 cm of wire. What length of wire is required for a similar 20 by
Nady [450]

Answer:

840

Step-by-step explanation:

Hi there!

Usually, this kind of problem asks us the perimeter rather than the area. But when talking on this grid on how many cms are necessary to construct one made up of 9 squares 1cm x1 cm. We need to check one thing: Some line segments will be used twice. So this is not a classical perimeter question.

1) So we either manually and carefully count it avoiding not doing it twice the same common line segments.

Or.

2) By just applying an algorithm, Where n is the <em>nxn</em> grid

n(n+1)+n(n+1)=Length \:of \:the \:wire

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5 0
3 years ago
Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

Answer:1002

Step-by-step explanation:

\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )

and 1+\tan ^2\theta =\sec^2\theta

and \cos \theta =\frac{1}{\sec \theta }  

\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )

as \cos ^2\theta =\cos ^2(\pi -\theta )

Applying this we get

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]

every \thetathere exist \pi -\theta

such that \sin^2\theta +\cos^2\theta =1

therefore

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002                          

6 0
3 years ago
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