# how to create an array(1 by 5000) having random values in the range(9-11).

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sanky kumar on 10 Sep 2013
i have to put the array value as input for sine function, but i guess there are certain limitations for that. it would be great if you could help P.S: new to matlab excuse my immaturity.

Image Analyst on 10 Sep 2013
Edited: Image Analyst on 10 Sep 2013
Did you look in the help? It's example 1:
Example 1
Generate values from the uniform distribution on the interval [a, b]:
r = a + (b-a).*rand(100,1);
just change 100,1 to 1,5000, and change a to 9 and b to 11, like this:
r = 9 + 2 .* rand(1,5000);
By the way, what does "in sine" mean?
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Image Analyst on 10 Sep 2013
Edited: Image Analyst on 10 Sep 2013
I don't know if you mean that you want the random values to be the angle input to the sine,
out = sin(r);
or if you want to add those values to the sine wave:
x = linspace(0, 4*pi, length(r)); % Define angles.
noisy_y = amplitude * sin(x) + r;

### More Answers (1)

Azzi Abdelmalek on 10 Sep 2013
9+rand(1,5000)*2
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sanky kumar on 10 Sep 2013
thank you