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Debora [2.8K]
3 years ago
13

Is it possible to obtain a negative value for the variance or the standard deviation?

Mathematics
1 answer:
Kaylis [27]3 years ago
4 0
Standard deviation is the square root of the variance, which is the average squared deviation from the mean and as an average of some squared numbers it can't be negative<span>. 

</span><span>The smallest value variance can reach is zero. This is when all the numbers in the data set are the same, therefore all the deviations from the mean are zero, all squared deviations are zero and their average is also zero.</span>
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12²+ 9 - 4<br> a) 119<br> b)147<br> c)137<br> d)221
ziro4ka [17]

Answer:

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Step-by-step explanation:

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
GuDViN [60]

Answer:

Euler's method is a numerical method used in calculus to approximate a particular solution of a differential equation. As a numerical method, we have to apply the same procedure many times, until get the desired result.

In first place, we need to know all the values the problem is giving:

  • The step size is 0.2; h = 0.2. This step size is a periodical increase of the x-variable, which will allow us to calculate each y-value to each x.
  • The problem is asking the solution y(1), which means that we have to find the y-value assigned for x = 1, through the numerical method.
  • The initial condition is y(0) = 9. In other words, x_{o} = 0\\y_{0}=9.

So, if the initial x-value is 0, and the step size is 0.2, the following x-value would be: x_{1}=0.2; then x_{2}=0.4; x_{3} =0.6; x_{4} =0.8;x_{5} =1; and so on.

Now, we have to apply the formula to find each y-value until get the match of x_{5}=1, because the problem asks the solution y(1).

According to the Euler's method:

y_{1} =y_{0} +hF(x_{0};y_{0})\\y_{2} =y_{1} +hF(x_{1};y_{1})\\y_{n} =y_{n-1} +hF(x_{n-1};y_{n-1})

Where F(x;y)=x^{2} y-12y^{2}, and x_{0} =0; y_{0} =9; h=0.2.

Replacing all values we calculate the y-value assigned to x_{1}:

y_{1} =9+0.2((0)^{2} 9-12(9)^{2})=-185.4.

Now, y_{1} =-185.4, x_{1} =0.2; h=0.2. We repeat the process with the new values:

y_{2} =y_{1} +hF(x_{1};y_{1})  \\y_{2} = -185.4+0.2((0.2)^{2} (-185.4)-12(-185.4)^{2} )\\y_{2}=-82682.47

Then, we repeat the same process until get the y-value for x_{5} =1, which is y_{5} = -1.0018, round to four decimal places.

Therefore, y(1)=-1.0018.

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