how to create symbol parameters?
in mxnet symbolic graph considers every arguments as symbol, unlike in tensorflow where we declare our variables differently and input placeholder differently, so when you eval or run the graph you only have to pass your value of our input placeholder. while in mxnet you have to pass inputs as well …
yes, you can using a module.

you might have seen examples of module where a only a symbol is given, e.g. from the docs:

data = mx.sym.variable('data')

fc1 = mx.sym.fullyconnected(data, name='fc1', num\_hidden=128)

act1 = mx.sym.activation(fc1, name='relu1', act\_type="relu")

fc2 = mx.sym.fullyconn…
there are websites that give you technical analysis indicators (including golden crosses and death crosses) such as

https://www.stockmonitor.com/stock-screener/death-cross-50ma-cross-down-200ma/

or you can simply graph the moving averages yourself.

i am not a big fan of technical analysis indicators so i have never used them. hi

we can do that by defining a string variable in the very beginning of the process with assign activity or by defining that in variable panel in the studio at the bottom the window

but make sure that we are not assigning any value to that variable later across the workflow

if we assign later th… You can use SetCursorPos() to control the position of the cursor. It returns true is it succeeds and false if not. It takes 2 arguments int x and int y. You can use GetCursorPos() to retrieve the position of cursor. it returns the same as Set...() it takes one paramater a pointer to a structure of type POINT, which contains to values long x and long y. Also if you want to send Mouse clicks then you will need to look into the function SendInput() its well documented on msdn but feel free to ask for more information.



Chris the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points. Here is a sample of traversing a simple xml structure

 
```
    Dim myxml = <root>
                    <level id="1">
                        <prop n="PROPERTY" v="VALUE"/>
                    </level>
                    <level id="2">
                        <prop n="PROPERTY" v="VALUE"/>
                    </level>
                </root>

    For Each level In myxml.Elements
        Debug.WriteLine("id=" & level.@id.ToString)
        For Each prop In level.Elements
            Debug.WriteLine("n=" & prop.@n.ToString & "  v=" & prop.@v.ToString)
        Next
    Next

``` performance has always been an issue with dynamo, when working with large data sets, mainly due to the functional way in which a dynamo graph operates. you have to remember that data is never mutated. instead, it is always copied between nodes.

when you input something into a node, dynamo’s backend… thanks for your comment!

i definitely should’ve used mkshell, not sure why i wasn’t. it does seem to produce a different error:

make: entering directory '/build/source/doc/manual'

xsltproc --nonet --param section.autolabel 1 --param section.label.includes.component.label 1 --param html.stylesheet … histogram  divide data o equal width o no overlapping  bar chart o skew­ pulls direction of mean o symmetric numeric summary center, spread, and location  greek letters for population  roman letters for statistics 1. mean  µ= population mean  x bar= sanple mean o ∑ xi  all over n= sample mean  k=0 2. mean vs. median a. mean used the most b. n is small when outliers occur c. use median, mean is too affected spread 1. range (max­min) a. sensitive to outliers 2. iqr a. not sensitive to outliers variability­ xi­x bar def render("show.json", %{user: user} = assigns) do

%{

data: render\_one(user, admin.userview, "user.json", assigns)

}

end

there’s no “implicit” assigns map, which all functions would automatically inherit somehow. you need to explicitly pass the assigns map onwards into rend… so if you want to be able to run them as separate runs, you’ll still need some way of communicating between them. one mechanism we provide for that is by serializing daml values as json through the --input-file and --output-file options to daml script. so in your case, you should make sure that the …