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https://algebra-answer.com/algebra-help/algebra-formulas/introduction-algebra-marvin-l.html
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introduction algebra marvin l bittinger
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# 4.1 Implicit Differentiationap Calculus
In the previous sections we learned to find the derivative, ( frac{dy}{dx}), or (y^prime ), when (y) is given explicitly as a function of (x). That is, if we know (y=f(x)) for some function (f), we can find (y^prime ). For example, given (y=3x^2-7), we can easily find (y^prime =6x). (Here we explicitly state how (x) and (y) are related. Knowing (x), we can directly find (y).)
What is Implicit Differentiation? 🎥Watch: AP Calculus AB/BC - Implicit Derivatives. Back in pre-calculus, you likely learned or talked about how there are two different types of equations: explicit equations and implicit equations. An explicit equation is written such that y is isolated on one side. For example, y=2x+3 is an explicit equation.
Showing 10 items from page AP Calculus Implicit Differentiation and Other Derivatives Extra Practice sorted by create time. View more ».For the review Jeopardy, after clicking on the above link, click on 'File' and select download from the dropdown menu so that you can view it in powerpoint. Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to, we need to implicitly write after it. Hence, the name of this method. Then, we solve for. If 5x^2+5x+xy=2 and y(2)= -14 find y'(2) by implicit differentiation. Use implicit differentiation to find an equation of the tangent line to the curve at the given point. X2 + 2xy − y2 + x = 17, (3, 5) (hyperbola) Math. Suppose that x and y are related by the equation x^2/4 + y^3/2 = 4.
Sometimes the relationship between (y) and (x) is not explicit; rather, it is implicit. For instance, we might know that (x^2-y=4). This equality defines a relationship between (x) and (y); if we know (x), we could figure out (y). Can we still find (y^prime )? In this case, sure; we solve for (y) to get (y=x^2-4) (hence we now know (y) explicitly) and then differentiate to get (y^prime =2x).
Sometimes the implicit relationship between (x) and (y) is complicated. Suppose we are given (sin(y)+y^3=6-x^3). A graph of this implicit function is given in Figure 2.19. In this case there is absolutely no way to solve for (y) in terms of elementary functions. The surprising thing is, however, that we can still find (y^prime ) via a process known as implicit differentiation.
Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other).
We begin by reviewing the Chain Rule. Let (f) and (g) be functions of (x). Then [frac{d}{dx}Big(f(g(x))Big) = f^prime(g(x))cdot g'(x).] Suppose now that (y=g(x)). We can rewrite the above as [frac{d}{dx}Big(f(y))Big) = f^prime(y))cdot y^prime , quad text{or}quad frac{d}{dx}Big(f(y))Big)= f^prime(y)cdot frac{dy}{dx}.label{2.1}tag{2.1}] These equations look strange; the key concept to learn here is that we can find (y^prime ) even if we don't exactly know how (y) and (x) relate.
We demonstrate this process in the following example.
Example 67: Using Implicit Differentiation
Find (y^prime ) given that (sin(y) + y^3=6-x^3).
Solution
We start by taking the derivative of both sides (thus maintaining the equality.) We have :
[ frac{d}{dx}Big(sin(y) + y^3Big)=frac{d}{dx}Big(6-x^3Big).]
The right hand side is easy; it returns (-3x^2).
The left hand side requires more consideration. We take the derivative term--by--term. Using the technique derived from Equation 2.1 above, we can see that [frac{d}{dx}Big(sin yBig) = cos y cdot y^prime .]
We apply the same process to the (y^3) term.
[frac{d}{dx}Big(y^3Big) = frac{d}{dx}Big((y)^3Big) = 3(y)^2cdot y^prime .]
Putting this together with the right hand side, we have
[cos(y)y^prime +3y^2y^prime = -3x^2.]
Now solve for (y^prime ).
[begin{align*} cos(y)y^prime +3y^2y^prime &= -3x^2. big(cos y+3y^2big)y^prime &= -3x^2 y^prime &= frac{-3x^2}{cos y+3y^2} end{align*}]
This equation for (y^prime ) probably seems unusual for it contains both (x) and (y) terms. How is it to be used? We'll address that next.
Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an (x) value, we have an explicit formula for computing the corresponding (y) value. With an implicit function, one often has to find (x) and (y) values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point.
For instance, we can affirm easily that the point ((sqrt[3]{6},0)) lies on the graph of the implicit function (sin y + y^3=6-x^3). Plugging in (0) for (y), we see the left hand side is (0). Setting (x=sqrt[3]6), we see the right hand side is also (0); the equation is satisfied. The following example finds the equation of the tangent line to this function at this point.
Example 68: Using Implicit Differentiation to find a tangent line
Find the equation of the line tangent to the curve of the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]6,0)).
Solution
In Example 67 we found that [y^prime = frac{-3x^2}{cos y +3y^2}.]We find the slope of the tangent line at the point ((sqrt[3]6,0)) by substituting (sqrt[3]6) for (x) and (0) for (y). Thus at the point ((sqrt[3]6,0)), we have the slope as [y^prime = frac{-3(sqrt[3]{6})^2}{cos 0 + 3cdot0^2} = frac{-3sqrt[3]{36}}{1} approx -9.91.]
Therefore the equation of the tangent line to the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]{6},0)) is [y = -3sqrt[3]{36}(x-sqrt[3]{6})+0 approx -9.91x+18.]The curve and this tangent line are shown in Figure 2.20.
This suggests a general method for implicit differentiation. For the steps below assume (y) is a function of (x).
1. Take the derivative of each term in the equation. Treat the (x) terms like normal. When taking the derivatives of (y) terms, the usual rules apply except that, because of the Chain Rule, we need to multiply each term by (y^prime ).
2. Get all the (y^prime ) terms on one side of the equal sign and put the remaining terms on the other side.
3. Factor out (y^prime ); solve for (y^prime ) by dividing.
Practical Note: When working by hand, it may be beneficial to use the symbol (frac{dy}{dx}) instead of (y^prime ), as the latter can be easily confused for (y) or (y^1).
### 4.1 Implicit Differentiationap Calculus Algebra
Example 69: Using Implicit Differentiation
Given the implicitly defined function (y^3+x^2y^4=1+2x), find (y^prime ).
Solution
We will take the implicit derivatives term by term. The derivative of (y^3) is (3y^2y^prime ).
The second term, (x^2y^4), is a little tricky. It requires the Product Rule as it is the product of two functions of (x): (x^2) and (y^4). Its derivative is (x^2(4y^3y^prime ) + 2xy^4). The first part of this expression requires a (y^prime ) because we are taking the derivative of a (y) term. The second part does not require it because we are taking the derivative of (x^2).
The derivative of the right hand side is easily found to be (2). In all, we get:
[3y^2y^prime + 4x^2y^3y^prime + 2xy^4 = 2.]
Move terms around so that the left side consists only of the (y^prime ) terms and the right side consists of all the other terms:
[3y^2y^prime + 4x^2y^3y^prime = 2-2xy^4.]
Factor out (y^prime ) from the left side and solve to get
[y^prime = frac{2-2xy^4}{3y^2+4x^2y^3}.]
To confirm the validity of our work, let's find the equation of a tangent line to this function at a point. It is easy to confirm that the point ((0,1)) lies on the graph of this function. At this point, (y^prime = 2/3). So the equation of the tangent line is (y = 2/3(x-0)+1). The function and its tangent line are graphed in Figure 2.21.
Notice how our function looks much different than other functions we have seen. For one, it fails the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important.
Example 70: Using Implicit Differentiation
Given the implicitly defined function (sin(x^2y^2)+y^3=x+y), find (y^prime ).
Solution
Differentiating term by term, we find the most difficulty in the first term. It requires both the Chain and Product Rules.
[begin{align*} frac{d}{dx}Big(sin(x^2y^2)Big) &= cos(x^2y^2)cdotfrac{d}{dx}Big(x^2y^2Big) &= cos(x^2y^2)cdotbig(x^2(2yy^prime )+2xy^2big) &= 2(x^2yy^prime +xy^2)cos(x^2y^2). end{align*} ]
We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have
[2(x^2yy^prime +xy^2)cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]
We now have to be careful to properly solve for (y^prime ), particularly because of the product on the left. It is best to multiply out the product. Doing this, we get
[2x^2ycos(x^2y^2)y^prime + 2xy^2cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]
From here we can safely move around terms to get the following:
[2x^2ycos(x^2y^2)y^prime + 3y^2y^prime - y^prime = 1 - 2xy^2cos(x^2y^2).]
Then we can solve for (y^prime ) to get
[y^prime = frac{1 - 2xy^2cos(x^2y^2)}{2x^2ycos(x^2y^2)+3y^2-1}.]
A graph of this implicit function is given in Figure 2.22. It is easy to verify that the points ((0,0)), ((0,1)) and ((0,-1)) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for (y^prime ).
At ((0,0)), the slope is (-1).
At ((0,1)), the slope is (1/2).
At ((0,-1)), the slope is also (1/2).
The tangent lines have been added to the graph of the function in Figure 2.23.
Quite a few 'famous' curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples.
Example 71: Finding slopes of tangent lines to a circle
Find the slope of the tangent line to the circle (x^2+y^2=1) at the point ((1/2, sqrt{3}/2)).
Solution
Taking derivatives, we get (2x+2yy^prime =0). Solving for (y^prime ) gives: [ y^prime = frac{-x}{y}.]
This is a clever formula. Recall that the slope of the line through the origin and the point ((x,y)) on the circle will be (y/x). We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of (y/x), namely, (-x/y). Hence these two lines are always perpendicular.
At the point ((1/2, sqrt{3}/2)), we have the tangent line's slope as
[y^prime = frac{-1/2}{sqrt{3}/2} = frac{-1}{sqrt{3}} approx -0.577.]
A graph of the circle and its tangent line at ((1/2,sqrt{3}/2)) is given in Figure 2.24, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.)
This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of 'regular' differentiation.
One hole in our current understanding of derivatives is this: what is the derivative of the square root function? That is, [frac{d}{dx}big(sqrt{x}big) = frac{d}{dx}big(x^{1/2}big) = text{?}]
We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule and obtain [frac{d}{dx}big(x^{1/2}big) = frac12x^{-1/2} = frac{1}{2sqrt{x}}.]
The trouble with this is that the Power Rule was initially defined only for positive integer powers, (n>0). While we did not justify this at the time, generally the Power Rule is proved using something called the Binomial Theorem, which deals only with positive integers. The Quotient Rule allowed us to extend the Power Rule to negative integer powers. Implicit Differentiation allows us to extend the Power Rule to rational powers, as shown below.
Let (y = x^{m/n}), where (m) and (n) are integers with no common factors (so (m=2) and (n=5) is fine, but (m=2) and (n=4) is not). We can rewrite this explicit function implicitly as (y^n = x^m). Now apply implicit differentiation.
[begin{align*}y &= x^{m/n} y^n &= x^m frac{d}{dx}big(y^nbig) &= frac{d}{dx}big(x^mbig) ncdot y^{n-1}cdot y^prime &= mcdot x^{m-1} y^prime &= frac{m}{n} frac{x^{m-1}}{y^{n-1}} quad text{(now substitute (x^{m/n}) for (y))} &= frac{m}{n} frac{x^{m-1}}{(x^{m/n})^{n-1}} quad text{(apply lots of algebra)} &= frac{m}n x^{(m-n)/n} &= frac{m}n x^{m/n -1}.end{align*}]
The above derivation is the key to the proof extending the Power Rule to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem.
Theorem 21: Power Rule for Differentiation
Let (f(x) = x^n), where (nneq 0) is a real number. Then (f) is a differentiable function, and (f^prime(x) = ncdot x^{n-1}).
This theorem allows us to say the derivative of (x^pi) is (pi x^{pi -1}).
We now apply this final version of the Power Rule in the next example, the second investigation of a 'famous' curve.
Example 72: Using the Power Rule
Find the slope of (x^{2/3}+y^{2/3}=8) at the point ((8,8)).
Solution
This is a particularly interesting curve called an astroid. It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.25.
To find the slope of the astroid at the point ((8,8)), we take the derivative implicitly.
[begin{align*} frac23x^{-1/3}+frac23y^{-1/3}y^prime &=0 frac23y^{-1/3}y^prime &= -frac23x^{-1/3} y^prime &= -frac{x^{-1/3}}{y^{-1/3}} y^prime &= -frac{y^{1/3}}{x^{1/3}} = -sqrt[3]{frac{y}x}. end{align*}]
Plugging in (x=8) and (y=8), we get a slope of (-1). The astroid, with its tangent line at ((8,8)), is shown in Figure 2.26.
## Implicit Differentiation and the Second Derivative
We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find (frac{dy}{dx}), then take its derivative with respect to (x). In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example.
Example 73: Finding the second derivative
Given (x^2+y^2=1), find (frac{d^2y}{dx^2} = y^{primeprime}).
Solution
We found that (y^prime = frac{dy}{dx} = -x/y) in Example 71. To find (y^{primeprime}), we apply implicit differentiation to (y^prime ).
[begin{align*} y^{primeprime} &= frac{d}{dx}big(y^prime big) &= frac{d}{dx}left(-frac xyright)qquad text{(Now use the Quotient Rule.)} &= -frac{y(1) - x(y^prime )}{y^2} end{align*}]
replace (y^prime ) with (-x/y):
[begin{align*}&= -frac{y-x(-x/y)}{y^2} &= -frac{y+x^2/y}{y^2}.end{align*}]
While this is not a particularly simple expression, it is usable. We can see that (y^{primeprime}>0) when (y<0) and (y^{primeprime}<0) when (y>0). In Section 3.4, we will see how this relates to the shape of the graph.
## Logarithmic Differentiation
Consider the function (y=x^x); it is graphed in Figure 2.27. It is well--defined for (x>0) and we might be interested in finding equations of lines tangent and normal to its graph. How do we take its derivative?
The function is not a power function: it has a 'power' of (x), not a constant. It is not an exponential function: it has a 'base' of (x), not a constant.
A differentiation technique known as logarithmic differentiation becomes useful here. The basic principle is this: take the natural log of both sides of an equation (y=f(x)), then use implicit differentiation to find (y^prime ). We demonstrate this in the following example.
Example 74: Using Logarithmic Differentiation
Given (y=x^x), use logarithmic differentiation to find (y^prime ).
Solution
As suggested above, we start by taking the natural log of both sides then applying implicit differentiation.
[begin{align*} y &= x^x ln (y) &= ln (x^x) text{(apply logarithm rule)} ln (y) &= xln x text{(now use implicit differentiation)} frac{d}{dx}Big(ln (y)Big) &= frac{d}{dx}Big(xln xBig) frac{y^prime }{y} &= ln x + xcdotfrac1x frac{y^prime }{y} &= ln x + 1 y^prime &= ybig(ln x+1big) text{(substitute (y=x^x))} y^prime &= x^xbig(ln x+1big). end{align*} ]
To 'test' our answer, let's use it to find the equation of the tangent line at (x=1.5). The point on the graph our tangent line must pass through is ((1.5, 1.5^{1.5}) approx (1.5, 1.837)). Using the equation for (y^prime ), we find the slope as
[y^prime = 1.5^{1.5}big(ln 1.5+1big) approx 1.837(1.405) approx 2.582.]
Thus the equation of the tangent line is (y = 1.6833(x-1.5)+1.837). Figure 2.28 graphs (y=x^x) along with this tangent line.
Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. In particular, it extended the Power Rule to rational exponents, which we then extended to all real numbers. In the next section, implicit differentiation will be used to find the derivatives of inverse functions, such as (y=sin^{-1} x).
### SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS
SOLUTION 1 : Begin with x3 + y3 = 4 . Differentiate both sides of the equation, getting
D ( x3 + y3 ) = D ( 4 ) ,
D ( x3 ) + D ( y3 ) = D ( 4 ) ,
(Remember to use the chain rule on D ( y3 ) .)
3x2 + 3y2y' = 0 ,
so that (Now solve for y' .)
3y2y' = - 3x2 ,
and
.
SOLUTION 2 : Begin with (x-y)2 = x + y - 1 . Differentiate both sides of the equation, getting
D (x-y)2 = D ( x + y - 1 ) ,
D (x-y)2 = D ( x ) + D ( y ) - D ( 1 ) ,
(Remember to use the chain rule on D (x-y)2 .)
,
2 (x-y) (1- y') = 1 + y' ,
so that (Now solve for y' .)
2 (x-y) - 2 (x-y) y' = 1 + y' ,
- 2 (x-y) y' - y' = 1 - 2 (x-y) ,
(Factor out y' .)
y' [ - 2 (x-y) - 1 ] = 1 - 2 (x-y) ,
and
.
SOLUTION 3 : Begin with . Differentiate both sides of the equation, getting
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(Remember to use the chain rule on .)
,
,
so that (Now solve for y' .)
,
,
(Factor out y' .)
,
and
.
### 4.1 Implicit Differentiationap Calculus Calculator
SOLUTION 4 : Begin with y = x2y3 + x3y2 . Differentiate both sides of the equation, getting
D(y) = D ( x2y3 + x3y2 ) ,
D(y) = D ( x2y3 ) + D ( x3y2 ) ,
(Use the product rule twice.)
,
(Remember to use the chain rule on D ( y3 ) and D ( y2 ) .)
,
y' = 3x2y2y' + 2x y3 + 2x3y y' + 3x2y2 ,
so that (Now solve for y' .)
y' - 3x2y2y' - 2x3y y' = 2x y3 + 3x2y2 ,
(Factor out y' .)
y' [ 1 - 3x2y2 - 2x3y ] = 2x y3 + 3x2y2 ,
and
.
SOLUTION 5 : Begin with . Differentiate both sides of the equation, getting
,
,
,
,
so that (Now solve for .)
,
,
(Factor out .)
,
and
.
SOLUTION 6 : Begin with . Differentiate both sides of the equation, getting
,
,
,
,
so that (Now solve for y' .)
,
,
(Factor out y' .)
,
,
,
and
.
SOLUTION 7 : Begin with . Differentiate both sides of the equation, getting
,
1 = (1/2)( x2 + y2 )-1/2D ( x2 + y2 ) ,
1 = (1/2)( x2 + y2 )-1/2 ( 2x + 2y y' ) ,
so that (Now solve for y' .)
,
,
,
,
and
.
SOLUTION 8 : Begin with . Clear the fraction by multiplying both sides of the equation by y + x2 , getting
,
or
x - y3 = xy + 2y + x3 + 2x2 .
Now differentiate both sides of the equation, getting
D ( x - y3 ) = D ( xy + 2y + x3 + 2x2 ) ,
D ( x ) - D (y3 ) = D ( xy ) + D ( 2y ) + D ( x3 ) + D ( 2x2 ) ,
(Remember to use the chain rule on D (y3 ) .)
1 - 3 y2y' = ( xy' + (1)y ) + 2 y' + 3x2 + 4x ,
so that (Now solve for y' .)
1 - y - 3x2 - 4x = 3 y2y' + xy' + 2 y' ,
(Factor out y' .)
1 - y - 3x2 - 4x = (3y2 + x + 2) y' ,
and
.
SOLUTION 9 : Begin with . Clear the fractions by multiplying both sides of the equation by x3y3 , getting
,
,
y4 + x4 = x5y7 .
Now differentiate both sides of the equation, getting
D ( y4 + x4 ) = D ( x5y7 ) ,
D ( y4 ) + D ( x4 ) = x5D (y7 ) + D ( x5 ) y7 ,
(Remember to use the chain rule on D (y4 ) and D (y7 ) .)
4 y3y' + 4 x3 = x5 (7 y6y' ) + ( 5 x4 ) y7 ,
so that (Now solve for y' .)
4 y3y' - 7 x5y6y' = 5 x4y7 - 4 x3 ,
(Factor out y' .)
y' [ 4 y3 - 7 x5y6 ] = 5 x4y7 - 4 x3 ,
and
.
SOLUTION 10 : Begin with (x2+y2)3 = 8x2y2 . Now differentiate both sides of the equation, getting
D (x2+y2)3 = D ( 8x2y2 ) ,
3 (x2+y2)2D (x2+y2) = 8x2D (y2 ) + D ( 8x2 ) y2 ,
(Remember to use the chain rule on D (y2 ) .)
3 (x2+y2)2 ( 2x + 2 y y' ) = 8x2 (2 y y' ) + ( 16 x ) y2 ,
so that (Now solve for y' .)
6x (x2+y2)2 + 6 y (x2+y2)2y' = 16 x2y y' + 16 x y2 ,
6 y (x2+y2)2y' - 16 x2y y' = 16 x y2 - 6x (x2+y2)2 ,
(Factor out y' .)
y' [ 6 y (x2+y2)2 - 16 x2y ] = 16 x y2 - 6x (x2+y2)2 ,
and
.
Thus, the slope of the line tangent to the graph at the point (-1, 1) is
,
and the equation of the tangent line is
y - ( 1 ) = (1) ( x - ( -1 ) )
or
y = x + 2 .
SOLUTION 11 : Begin with x2 + (y-x)3 = 9 . If x=1 , then
(1)2 + ( y-1 )3 = 9
so that
( y-1 )3 = 8 ,
y-1 = 2 ,
y = 3 ,
and the tangent line passes through the point (1, 3) . Now differentiate both sides of the original equation, getting
D ( x2 + (y-x)3 ) = D ( 9 ) ,
### 4.1 Implicit Differentiationap Calculus Solver
D ( x2 ) + D (y-x)3 = D ( 9 ) ,
2x + 3 (y-x)2D (y-x) = 0 ,
2x + 3 (y-x)2 (y'-1) = 0 ,
so that (Now solve for y' .)
2x + 3 (y-x)2y'- 3 (y-x)2 = 0 ,
3 (y-x)2y' = 3 (y-x)2 - 2x ,
and
.
Thus, the slope of the line tangent to the graph at (1, 3) is
,
and the equation of the tangent line is
y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,
or
y = (5/6) x + (13/6) .
SOLUTION 12 : Begin with x2y + y4 = 4 + 2x . Now differentiate both sides of the original equation, getting
D ( x2y + y4 ) = D ( 4 + 2x ) ,
D ( x2y ) + D (y4 ) = D ( 4 ) + D ( 2x ) ,
( x2y' + (2x) y ) + 4 y3y' = 0 + 2 ,
so that (Now solve for y' .)
x2y' + 4 y3y' = 2 - 2x y ,
(Factor out y' .)
y' [ x2 + 4 y3 ] = 2 - 2x y ,
and
(Equation 1)
.
Thus, the slope of the graph (the slope of the line tangent to the graph) at (-1, 1) is
.
Since y'= 4/5 , the slope of the graph is 4/5 and the graph is increasing at the point (-1, 1) . Now determine the concavity of the graph at (-1, 1) . Differentiate Equation 1, getting
.
Now let x=-1 , y=1 , and y'=4/5 so that the second derivative is
.
Since y' < 0 , the graph is concave down at the point (-1, 1) .
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## 3rd and 4th Grade, December, 2019
A.1273. Basilisk lizards are unique because they can run across the surface of water. (See it on: http://www.youtube.com/watch?v=_ut5jENqBX8) It can even run 10 meters in 4 seconds. The females lay 12 to 14 eggs 5 times a year.
a) How far can this lizard run on water in a half a minute?
b) At most how many eggs can hatch from 6 females in 2 years?
A.1274. The American Coot, to save her eggs from predators, sometimes builds a nest from rocks in the middle of a shallow lake. They carry the stones, which are about 30 dkg in mass, with their beaks. The male carries about 3000 stones to build the nest, which is about one and a half times more than what the female can carry. What is the mass of all the stones a couple of American Coots carry to build a nest?
A.1275. The Kalong, also known as the Flying Fox, has a wingspan of almost 1.5 meters for a body length of 40 centimeters, so its wings are comparable to those of a large bird of prey. Just like bats, they sleep hanging upside down, holding onto the horizontal branches of a tree. On the lower branches there are 4 times as many kalongs as on the middle branches. On the upper branches there are 4 more kalongs than on the middle branches. How many kalongs are there on the lower branches of this tree if there are a total of 112 kalongs on the tree?
A.1276. How many different 4-digit numbers can you create by using two 1’s, a 2, and a 3?
A.1277. The Arctic Tern, a seabird, is the champion of long-distance flying. They see two summers each year as they migrate from the North Pole, along a winding route to Antarctica and back, a round trip of about 70,000 km (44,000 miles) each year. They spend about 14 weeks at each pole, the rest of the year they spend flying.
a) How long does it take them to fly from one pole of the Earth to the other?
b) The lifespan of an Arctic Tern is 20 years. How much of it does this bird spend with migrating?
A.1278. Olga made a paper cube for her math class. She wrote on each side how many other sides it borders with, she wrote by every edge how many vertices it connects, and she wrote at every vertex how many edges run into it. What is the sum of all these numbers on the cube?
A.1279. Inga drew all the diagonals of a regular pentagon. How many triangles of different shapes are there on the picture?
A.1280. Anna wants to write the numbers from 0 to 9 on the circumference of a circle so that the sum of any three consecutive numbers on the circle is no more than 15. Is it possible to do?
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1. ## absolute function
State the domain and range of the following.
$y= \left | 2-\frac{1}{x^2} \right |$
Why is the range greater than or equal to zero? Why isn't it (2, infinity)?
I've just started going through dilations, reflections and translations of different equations and now we are going through absolute functions.
Please show complete working out. All help will be appreciated.
2. Originally Posted by Joker37
State the domain and range of the following.
$y= \left | 2-\frac{1}{x^2} \right |$
Why is the range greater than or equal to zero? Why isn't it (2, infinity)?
I've just started going through dilations, reflections and translations of different equations and now we are going through absolute functions.
Please show complete working out. All help will be appreciated.
I hope you know how the domain is found. Try showing your work if you were unable to find the domain.
Regarding range: Note that there are only positive y-values(absolute function as you have stated). There is no value of x that we can find such that we will get a negative value of y. So, the range for this function is y ≥ 0.
3. Hello, Joker37!
State the domain and range of the following.
$y\:=\: \left| 2-\frac{1}{x^2} \right|$
Why is the range greater than or equal to zero? Why isn't it (2, infinity)?
The graph has $x$-intercepts at: . $\left(\pm\frac{1}{\sqrt{2}},\:0\right)$
The graph of: . $y \:=\:2-\frac{1}{x^2}$ .looks like this:
Code:
|
- - - - - - - - - + - - - - - - - - - -
* 2| *
* | *
--------------o-----+-----o----------------
* | *
* | *
|
* | *
|
|
*|*
|
The graph of: . $y \:=\:\left|2 - \frac{1}{x^2}\right|$ looks like this:
Code:
|
*|*
|
|
* | *
- - - - - - - - - - + - - - - - - - - - -
* * 2| * *
* * | * *
----------------o-----+-----o----------------
|
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https://phys.org/news/2009-09-nist-temperature-microfluidics.html
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# NIST Calculations May Improve Temperature Measures for Microfluidics
(PhysOrg.com) -- If you wanted to know if your child had a fever or be certain that the roast in the oven was thoroughly cooked, you would, of course, use a thermometer that you trusted to give accurate readings at any temperature within its range. However, it isn’t that simple for researchers who need to measure temperatures in microfluidic systems—tiny, channel-lined devices used in medical diagnostics, DNA forensics and “lab-on-a-chip” chemical analyzers—as their current “thermometer” can only be precisely calibrated for one reference temperature. Now, researchers at the National Institute of Standards and Technology have proposed a mathematical solution that enables researchers to calibrate the “thermometer” for microfluidic systems so that all temperatures are covered.
Reactions taking place in microfluidic systems often require heating, meaning that users must accurately monitor temperature changes in fluid volumes ranging from a few microliters (a droplet approximately 1 millimeter in diameter) to sub-nanoliters (a droplet approximately 1/10 of millimeter in diameter). A common technique, for example, depends heavily on precise temperature cycling. Ordinary thermometers or other temperature probes are useless at such tiny dimensions, so some groups have turned to temperature-sensitive fluorescent dyes, particularly rhodamine B. The intensity of the dye’s fluorescence decreases with increasing temperature. The idea is that the dye can be used as a noninvasive way to map the range of temperatures occurring within a microfluidic system during heating and, in turn, provide a means of calibrating that system for experiments.
However, the technique currently requires the user to base all readings on the fluorescence at a single reference temperature. Previous groups have developed “calibration curves” that relate temperature to rhodmaine B fluorescent intensity based on a reference temperature of about 23 degrees Celsius (a technique first proposed by NIST researchers David Ross, Michael Gaitan and Laurie Locascio in 2001*). But it turns out that the curves are only good for that one temperature. In an upcoming paper in Analytical Chemistry**, the NIST team—Jayna J. Shah, Michael Gaitan and Jon Geist—reports that changing the reference point, such as the higher temperature when a microfluidic system is first heated, introduces errors when a dye intensity-to-temperature calculation is done using current methods.
“Our analysis shows that a simple linear correction for a 40 degrees Celsius reference temperature identified errors between minus 3 to 8 degrees Celsius for three previously published sets of calibration equations derived at approximately 23 degrees Celsius,” says lead researcher Shah.
To address the problem, the NIST team developed mathematical methods to correct for the shift experienced when the reference temperature changes. This allowed the researchers to create generalized calibration equations that can be applied to any reference temperature.
Microfluidic DNA amplification (production of numerous copies of DNA from a tiny sample) by the polymerase chain reaction (PCR) is one procedure that could benefit from the new NIST calculations, Shah says. “PCR requires a microfluidic device to be cycled through temperatures at three different zones starting around 65 degrees Celsius, so a useful dye intensity-to-temperature ratio would have to be based on that temperature and not a reference point of 23 degrees Celsius,” she explains.
* D. Ross, M. Gaitan and L.E. Locascio. Temperature measurement in microfluidic systems using a temperature-dependent fluorescent dye. Analytical Chemistry, Vol. 73, No. 17, pages 4117-4123, Sept. 1, 2001.
** J.J. Shah, M. Gaitan and J. Geist. Generalized temperature measurement equations for rhodamine B dye solution and its application to microfluidics. , Vol. 81, No. 19, Oct. 1, 2009 (published online Sept. 1, 2009).
Provided by National Institute of Standards and Technology (news : web)
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https://www.ques10.com/p/4658/state-and-prove-total-probability-theorem-and-ba-2/
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18kviews
State and prove Total Probability theorem and Bayes theorem?
Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis
Marks: 4M, 5M
Year: May 2015, Dec 2014
0
601views
Total Probability Theorem:
Statement: If $B_1, B_2, ……….B_n$ be a set of exhaustive and mutually exclusive events and A is another event associated with (or caused by) $B_i$, then
$$P(A)= \sum_{i=1}^nP(B_i ).(\frac{A}B_i )$$
Proof:
The inner circle represents the event A. A can occur along with (or due to) $B_1, B_2, ……….B_n$ that are exhaustive and mutually exclusive.
∴ $AB_1,AB_2,AB_3,AB_4…………………..AB_n$ are also mutually exclusive
∴ $A= AB_1+AB_2+ AB_3+ AB_4……+AB_n$ (By Addition Theorem)
$$∴P(A)=P(\sum_{i=1}^nAB_i)$$
$$=P(\sum_{i=1}^nPAB_i)$$
$∴P(A)= \sum_{i=1}^nP(B_i ).P(\frac{A}B_i )$…(A) (Using conditional probability
$$P(AB)=P(A∩B)=P(B).P(B/A)=P(A).P(A/B))$$
Bayes’ Theorem or Theorem of Probability of causes
Statement: If $B_1, B_2, ……….B_n$ be a set of exhaustive and mutually exclusive events associated with a random experiment and A is another event associated with (or caused by) $B_i$, then
$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{∑_{i=1}^nP{B_i }.P(\frac{A}{B_i } )}....... i=1,2..n$$
Proof:
We know Conditional Probability is given as:
$P(AB_i )=P(A∩B_i )=P(B_i ).P(A/B_i )=P(A).P(B_i/A)---------(1)$
$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{P(A)}....... (2)$$
Now using Total Probability Theorem we have,
$$P(A)= \sum_{i=1}^nP{B_i }.P(\frac{A}{B_i }) ------ (3)$$
From equation (2) and equation (3)
$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{∑_{i=1}^nP{B_i }.P(\frac{A}{B_i } )}.......$$ hence proved.
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LAPACK 3.8.0 LAPACK: Linear Algebra PACKage
sla_syrcond.f
Go to the documentation of this file.
1 *> \brief \b SLA_SYRCOND estimates the Skeel condition number for a symmetric indefinite matrix.
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 *> \htmlonly
10 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/sla_syrcond.f">
11 *> [TGZ]</a>
12 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/sla_syrcond.f">
13 *> [ZIP]</a>
14 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/sla_syrcond.f">
15 *> [TXT]</a>
16 *> \endhtmlonly
17 *
18 * Definition:
19 * ===========
20 *
21 * REAL FUNCTION SLA_SYRCOND( UPLO, N, A, LDA, AF, LDAF, IPIV, CMODE,
22 * C, INFO, WORK, IWORK )
23 *
24 * .. Scalar Arguments ..
25 * CHARACTER UPLO
26 * INTEGER N, LDA, LDAF, INFO, CMODE
27 * ..
28 * .. Array Arguments
29 * INTEGER IWORK( * ), IPIV( * )
30 * REAL A( LDA, * ), AF( LDAF, * ), WORK( * ), C( * )
31 * ..
32 *
33 *
34 *> \par Purpose:
35 * =============
36 *>
37 *> \verbatim
38 *>
39 *> SLA_SYRCOND estimates the Skeel condition number of op(A) * op2(C)
40 *> where op2 is determined by CMODE as follows
41 *> CMODE = 1 op2(C) = C
42 *> CMODE = 0 op2(C) = I
43 *> CMODE = -1 op2(C) = inv(C)
44 *> The Skeel condition number cond(A) = norminf( |inv(A)||A| )
45 *> is computed by computing scaling factors R such that
46 *> diag(R)*A*op2(C) is row equilibrated and computing the standard
47 *> infinity-norm condition number.
48 *> \endverbatim
49 *
50 * Arguments:
51 * ==========
52 *
53 *> \param[in] UPLO
54 *> \verbatim
55 *> UPLO is CHARACTER*1
56 *> = 'U': Upper triangle of A is stored;
57 *> = 'L': Lower triangle of A is stored.
58 *> \endverbatim
59 *>
60 *> \param[in] N
61 *> \verbatim
62 *> N is INTEGER
63 *> The number of linear equations, i.e., the order of the
64 *> matrix A. N >= 0.
65 *> \endverbatim
66 *>
67 *> \param[in] A
68 *> \verbatim
69 *> A is REAL array, dimension (LDA,N)
70 *> On entry, the N-by-N matrix A.
71 *> \endverbatim
72 *>
73 *> \param[in] LDA
74 *> \verbatim
75 *> LDA is INTEGER
76 *> The leading dimension of the array A. LDA >= max(1,N).
77 *> \endverbatim
78 *>
79 *> \param[in] AF
80 *> \verbatim
81 *> AF is REAL array, dimension (LDAF,N)
82 *> The block diagonal matrix D and the multipliers used to
83 *> obtain the factor U or L as computed by SSYTRF.
84 *> \endverbatim
85 *>
86 *> \param[in] LDAF
87 *> \verbatim
88 *> LDAF is INTEGER
89 *> The leading dimension of the array AF. LDAF >= max(1,N).
90 *> \endverbatim
91 *>
92 *> \param[in] IPIV
93 *> \verbatim
94 *> IPIV is INTEGER array, dimension (N)
95 *> Details of the interchanges and the block structure of D
96 *> as determined by SSYTRF.
97 *> \endverbatim
98 *>
99 *> \param[in] CMODE
100 *> \verbatim
101 *> CMODE is INTEGER
102 *> Determines op2(C) in the formula op(A) * op2(C) as follows:
103 *> CMODE = 1 op2(C) = C
104 *> CMODE = 0 op2(C) = I
105 *> CMODE = -1 op2(C) = inv(C)
106 *> \endverbatim
107 *>
108 *> \param[in] C
109 *> \verbatim
110 *> C is REAL array, dimension (N)
111 *> The vector C in the formula op(A) * op2(C).
112 *> \endverbatim
113 *>
114 *> \param[out] INFO
115 *> \verbatim
116 *> INFO is INTEGER
117 *> = 0: Successful exit.
118 *> i > 0: The ith argument is invalid.
119 *> \endverbatim
120 *>
121 *> \param[in] WORK
122 *> \verbatim
123 *> WORK is REAL array, dimension (3*N).
124 *> Workspace.
125 *> \endverbatim
126 *>
127 *> \param[in] IWORK
128 *> \verbatim
129 *> IWORK is INTEGER array, dimension (N).
130 *> Workspace.
131 *> \endverbatim
132 *
133 * Authors:
134 * ========
135 *
136 *> \author Univ. of Tennessee
137 *> \author Univ. of California Berkeley
138 *> \author Univ. of Colorado Denver
139 *> \author NAG Ltd.
140 *
141 *> \date December 2016
142 *
143 *> \ingroup realSYcomputational
144 *
145 * =====================================================================
146 REAL FUNCTION sla_syrcond( UPLO, N, A, LDA, AF, LDAF, IPIV, CMODE,
147 \$ C, INFO, WORK, IWORK )
148 *
149 * -- LAPACK computational routine (version 3.7.0) --
150 * -- LAPACK is a software package provided by Univ. of Tennessee, --
151 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
152 * December 2016
153 *
154 * .. Scalar Arguments ..
155 CHARACTER UPLO
156 INTEGER N, LDA, LDAF, INFO, CMODE
157 * ..
158 * .. Array Arguments
159 INTEGER IWORK( * ), IPIV( * )
160 REAL A( lda, * ), AF( ldaf, * ), WORK( * ), C( * )
161 * ..
162 *
163 * =====================================================================
164 *
165 * .. Local Scalars ..
166 CHARACTER NORMIN
167 INTEGER KASE, I, J
168 REAL AINVNM, SMLNUM, TMP
169 LOGICAL UP
170 * ..
171 * .. Local Arrays ..
172 INTEGER ISAVE( 3 )
173 * ..
174 * .. External Functions ..
175 LOGICAL LSAME
176 REAL SLAMCH
177 EXTERNAL lsame, slamch
178 * ..
179 * .. External Subroutines ..
180 EXTERNAL slacn2, xerbla, ssytrs
181 * ..
182 * .. Intrinsic Functions ..
183 INTRINSIC abs, max
184 * ..
185 * .. Executable Statements ..
186 *
187 sla_syrcond = 0.0
188 *
189 info = 0
190 IF( n.LT.0 ) THEN
191 info = -2
192 ELSE IF( lda.LT.max( 1, n ) ) THEN
193 info = -4
194 ELSE IF( ldaf.LT.max( 1, n ) ) THEN
195 info = -6
196 END IF
197 IF( info.NE.0 ) THEN
198 CALL xerbla( 'SLA_SYRCOND', -info )
199 RETURN
200 END IF
201 IF( n.EQ.0 ) THEN
202 sla_syrcond = 1.0
203 RETURN
204 END IF
205 up = .false.
206 IF ( lsame( uplo, 'U' ) ) up = .true.
207 *
208 * Compute the equilibration matrix R such that
209 * inv(R)*A*C has unit 1-norm.
210 *
211 IF ( up ) THEN
212 DO i = 1, n
213 tmp = 0.0
214 IF ( cmode .EQ. 1 ) THEN
215 DO j = 1, i
216 tmp = tmp + abs( a( j, i ) * c( j ) )
217 END DO
218 DO j = i+1, n
219 tmp = tmp + abs( a( i, j ) * c( j ) )
220 END DO
221 ELSE IF ( cmode .EQ. 0 ) THEN
222 DO j = 1, i
223 tmp = tmp + abs( a( j, i ) )
224 END DO
225 DO j = i+1, n
226 tmp = tmp + abs( a( i, j ) )
227 END DO
228 ELSE
229 DO j = 1, i
230 tmp = tmp + abs( a( j, i ) / c( j ) )
231 END DO
232 DO j = i+1, n
233 tmp = tmp + abs( a( i, j ) / c( j ) )
234 END DO
235 END IF
236 work( 2*n+i ) = tmp
237 END DO
238 ELSE
239 DO i = 1, n
240 tmp = 0.0
241 IF ( cmode .EQ. 1 ) THEN
242 DO j = 1, i
243 tmp = tmp + abs( a( i, j ) * c( j ) )
244 END DO
245 DO j = i+1, n
246 tmp = tmp + abs( a( j, i ) * c( j ) )
247 END DO
248 ELSE IF ( cmode .EQ. 0 ) THEN
249 DO j = 1, i
250 tmp = tmp + abs( a( i, j ) )
251 END DO
252 DO j = i+1, n
253 tmp = tmp + abs( a( j, i ) )
254 END DO
255 ELSE
256 DO j = 1, i
257 tmp = tmp + abs( a( i, j) / c( j ) )
258 END DO
259 DO j = i+1, n
260 tmp = tmp + abs( a( j, i) / c( j ) )
261 END DO
262 END IF
263 work( 2*n+i ) = tmp
264 END DO
265 ENDIF
266 *
267 * Estimate the norm of inv(op(A)).
268 *
269 smlnum = slamch( 'Safe minimum' )
270 ainvnm = 0.0
271 normin = 'N'
272
273 kase = 0
274 10 CONTINUE
275 CALL slacn2( n, work( n+1 ), work, iwork, ainvnm, kase, isave )
276 IF( kase.NE.0 ) THEN
277 IF( kase.EQ.2 ) THEN
278 *
279 * Multiply by R.
280 *
281 DO i = 1, n
282 work( i ) = work( i ) * work( 2*n+i )
283 END DO
284
285 IF ( up ) THEN
286 CALL ssytrs( 'U', n, 1, af, ldaf, ipiv, work, n, info )
287 ELSE
288 CALL ssytrs( 'L', n, 1, af, ldaf, ipiv, work, n, info )
289 ENDIF
290 *
291 * Multiply by inv(C).
292 *
293 IF ( cmode .EQ. 1 ) THEN
294 DO i = 1, n
295 work( i ) = work( i ) / c( i )
296 END DO
297 ELSE IF ( cmode .EQ. -1 ) THEN
298 DO i = 1, n
299 work( i ) = work( i ) * c( i )
300 END DO
301 END IF
302 ELSE
303 *
304 * Multiply by inv(C**T).
305 *
306 IF ( cmode .EQ. 1 ) THEN
307 DO i = 1, n
308 work( i ) = work( i ) / c( i )
309 END DO
310 ELSE IF ( cmode .EQ. -1 ) THEN
311 DO i = 1, n
312 work( i ) = work( i ) * c( i )
313 END DO
314 END IF
315
316 IF ( up ) THEN
317 CALL ssytrs( 'U', n, 1, af, ldaf, ipiv, work, n, info )
318 ELSE
319 CALL ssytrs( 'L', n, 1, af, ldaf, ipiv, work, n, info )
320 ENDIF
321 *
322 * Multiply by R.
323 *
324 DO i = 1, n
325 work( i ) = work( i ) * work( 2*n+i )
326 END DO
327 END IF
328 *
329 GO TO 10
330 END IF
331 *
332 * Compute the estimate of the reciprocal condition number.
333 *
334 IF( ainvnm .NE. 0.0 )
335 \$ sla_syrcond = ( 1.0 / ainvnm )
336 *
337 RETURN
338 *
339 END
subroutine ssytrs(UPLO, N, NRHS, A, LDA, IPIV, B, LDB, INFO)
SSYTRS
Definition: ssytrs.f:122
real function sla_syrcond(UPLO, N, A, LDA, AF, LDAF, IPIV, CMODE, C, INFO, WORK, IWORK)
SLA_SYRCOND estimates the Skeel condition number for a symmetric indefinite matrix.
Definition: sla_syrcond.f:148
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine slacn2(N, V, X, ISGN, EST, KASE, ISAVE)
SLACN2 estimates the 1-norm of a square matrix, using reverse communication for evaluating matrix-vec...
Definition: slacn2.f:138
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https://mathematica.stackexchange.com/questions/145733/zeros-of-the-hankel-function-with-complex-parameter?noredirect=1
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# Zeros of the Hankel function with complex parameter
I'm trying to find the zeros of the Hankel function (the first few will do) of the first kind $H^{(1)}_\nu(z) = J_\nu(z) + i Y_\nu(z)$ for complex argument $z$ but I'm not sure what is the best function for this in mathematica.
EDIT: Per suggestion I have tried to implement the function FindAllCrossings2D
Options[FindAllCrossings2D] =
Sort[Join[
Options[FindRoot], {MaxRecursion -> Automatic,
PerformanceGoal :> $PerformanceGoal, PlotPoints -> Automatic}]]; FindAllCrossings2D[funcs_, {x_, xmin_, xmax_}, {y_, ymin_, ymax_}, opts___] := Module[{contourData, seeds, tt, fy = Compile[{x, y}, Evaluate[funcs[[2]]]]}, contourData = Map[First, Cases[Normal[ ContourPlot[funcs[[1]], {x, xmin, xmax}, {y, ymin, ymax}, Contours -> {0}, ContourShading -> False, PlotRange -> {Full, Full, Automatic}, Evaluate[ Sequence @@ FilterRules[Join[{opts}, Options[FindAllCrossings2D]], DeleteCases[Options[ContourPlot], Method -> _]]]]], _Line, Infinity]]; seeds = Flatten[Map[#[[1 + Flatten[Position[ Rest[tt = Sign[Apply[fy, #, 2]]] Most[tt], -1]]]] &, contourData], 1]; If[seeds == {}, seeds, Select[Union[ Map[{x, y} /. FindRoot[{funcs[[1]] == 0, funcs[[2]] == 0}, {x, #[[1]]}, {y, #[[2]]}, Evaluate[ Sequence @@ FilterRules[Join[{opts}, Options[FindAllCrossings2D]], Options[FindRoot]]]] &, seeds]], (xmin < #[[1]] < xmax && ymin < #[[2]] < ymax) &]]] sols = FindAllCrossings2D[{Re[HankelH1[0, x + I y]], Im[HankelH1[0, x + I y]]}, {x, -2, 2}, {y, -2, 2}] But I receive this error "FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations." I also looked at the contour plot and we can clearly see nontrivial zeros periodically for z =$x + i y$and$x < 0$and$y < 0\$. I also attempted the get coordinate and then FindRoot[], but this failed to converge and appeared to be going away from the root.
• You might want to adapt the solution here or here. May 10, 2017 at 23:40
• I tried this @J.M. and neither work, I will update with code I took from the first solution May 11, 2017 at 16:23
You can use Solve for this purpose, as long as you restrict the domain:
zeros = z /. Solve[HankelH1[0, z] == 0 && -10 < Re[z] < 10 && -10 < Im[z] < 10, z]
Solve::incs: Warning: Solve was unable to prove that the solution set found is complete.
{Root[{HankelH1[ 0, #1] &, -8.65370576584112448841350498548699240993120635923782334125675 - 0.34600819276767292735146122876446089332681742821336233794111 I}], Root[{HankelH1[ 0, #1] &, -5.519997520841832519785940731628753302839051099087900948702 - 0.345225028545679355074185641966127944569911027495087086248 I}], Root[{HankelH1[ 0, #1] &, -2.4040911771553443579757061017638843409511660145346662653424 - 0.3405021529561410696628419946032815758353900295131219830737 I}]}
Unfortunately, Solve doesn't always find the complete solution set. Let's check whether the above roots are actually zeros:
HankelH1[0, N[zeros, 100]]
{0.*10^-101 + 0.*10^-101 I, 0.*10^-98 + 0.*10^-98 I, 0.*10^-99 + 0.*10^-99 I}
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Acronym enne
Name enneract,
9D measure-polytope9),
geoyotton
Vertex layers
Layer Symmetry Subsymmetries o3o3o3o3o3o3o3o4o o3o3o3o3o3o3o3o . . o3o3o3o3o3o3o4o 1 o3o3o3o3o3o3o3o4x o3o3o3o3o3o3o3o .vertex first . o3o3o3o3o3o3o4xocto first 2 o3o3o3o3o3o3o3q .vertex figure . o3o3o3o3o3o3o4xopposite octo 3 o3o3o3o3o3o3q3o . 4 o3o3o3o3o3q3o3o . 5 o3o3o3o3q3o3o3o . 6 o3o3o3q3o3o3o3o . 7 o3o3q3o3o3o3o3o . 8 o3q3o3o3o3o3o3o . 9 q3o3o3o3o3o3o3o . 10 o3o3o3o3o3o3o3o .opposite vertex
Coordinates (1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2) & all changes of sign
Dual vee
Confer
general polytopal classes:
hypercube
External
Its equatorial cross-section in vertex first orientation is not a vertex layer. Even so it can be determined as 1/q-be (cf. truncation series).
Incidence matrix according to Dynkin symbol
```o3o3o3o3o3o3o3o4x
. . . . . . . . . | 512 ♦ 9 | 36 | 84 | 126 | 126 | 84 | 36 | 9
------------------+-----+------+------+------+------+------+-----+-----+---
. . . . . . . . x | 2 | 2304 ♦ 8 | 28 | 56 | 70 | 56 | 28 | 8
------------------+-----+------+------+------+------+------+-----+-----+---
. . . . . . . o4x | 4 | 4 | 4608 ♦ 7 | 21 | 35 | 35 | 21 | 7
------------------+-----+------+------+------+------+------+-----+-----+---
. . . . . . o3o4x ♦ 8 | 12 | 6 | 5376 ♦ 6 | 15 | 20 | 15 | 6
------------------+-----+------+------+------+------+------+-----+-----+---
. . . . . o3o3o4x ♦ 16 | 32 | 24 | 8 | 4032 ♦ 5 | 10 | 10 | 5
------------------+-----+------+------+------+------+------+-----+-----+---
. . . . o3o3o3o4x ♦ 32 | 80 | 80 | 40 | 10 | 2016 ♦ 4 | 6 | 4
------------------+-----+------+------+------+------+------+-----+-----+---
. . . o3o3o3o3o4x ♦ 64 | 192 | 240 | 160 | 60 | 12 | 672 | 3 | 3
------------------+-----+------+------+------+------+------+-----+-----+---
. . o3o3o3o3o3o4x ♦ 128 | 448 | 672 | 560 | 280 | 84 | 14 | 144 | 2
------------------+-----+------+------+------+------+------+-----+-----+---
. o3o3o3o3o3o3o4x ♦ 256 | 1024 | 1792 | 1792 | 1120 | 448 | 112 | 16 | 18
```
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# Money + system of equations - math problems
Money or finance are an essential medium of exchange, a unit of account and store of value in modern society. Mathematical examples helps to get basic financial literacy.
#### Number of problems found: 110
• Contestants
The three best contestants are to divide the total prize of CZK 4,200. The second gets 20% more than the third. And the first one gets 200 CZK less than the second and the third together. How much will everyone get?
• Wines
Eleven liters of white wine and eight liters of red wine cost a total of 1315 kc. 1 liter of white wine was 10 kc cheaper than a liter of red wine. How much is 1 liter of white and how much red wine?
• Aunt Rose
Aunt Rose gave \$2500 to Mani and Cindy. Mani received \$500 more than Cindy. How mich did Cindy received?
• A fisherman
A fisherman buys carnivores to fish. He could buy either 6 larvae and 4 worms for \$ 132 or 4 larvae and 7 worms per \$ 127. What is the price of larvae and worms? Argue the answer.
• Trio ratio
Hans, Alena and Thomas have a total of 740 USD. Hans and Alena split in the ratio 5: 6 and Alena and Thomas in the ratio 4: 5. How much will everyone get?
• On the 4
On the way to playing disc golf with his two boys, Mr. Smith purchases 3 muffins and 2 bottles of water, totaling \$9.75. The following week he only has Asher with him, so he purchases 2 muffins and 1 bottle of water totalling \$6.00. What is the cost of on
• Apples
Zuzana bought 3 kg Jonathan apples. Then she noticed that Golden apples were 3 CZK a kilogram cheaper. So they bought 2 kg. For apples, she paid a total of CZK 109. How much did 1 kg Jonathan and how much Golden?
• Lottery - eurocents
Tereza bets in the lottery and finally wins. She went to the booth to have the prize paid out. An elderly gentleman standing next to him wants to buy a newspaper, but he is missing five cents. Tereza is in a generous mood after the win, so she gives the m
• An animal shelter
An animal shelter spends \$5.50 per day to care for each bird and \$8.50 per day to care for each cat. Nicole noticed that the shelter spent \$283.00 caring for birds and cats on Wednesday. Nicole found a record showing that there were a total of 40 birds an
Adam has half the money in his right pocket than in his left pocket. If he transferred 40 crowns from the left pocket to the right, he would have the same in both pockets. Calculate how many crowns does Adam have in his left pocket more than in his right?
• Three workers
The three workers received € 2,850 together for the work done. They divided them according to the time worked so that the first received 20% less than the second and the third € 50 more than the second. How much EUR did each worker receive?
• Three friends
Cuba, Matthew and their friend Adam found a brigade during their weekend weekends because they wanted to make a joint trip to the Alps that they planned for the spring break. Cuba enjoyed the skiing trip very much, so he was not lazy to get up and went to
• 12 apples
12 apples and 2 loaves of bread cost 5.76 and 6 apples and 3 loaves of bread cost 7.68. How much is a loaf of bread?
• Two math problems
1) The sum of twice a number and -6 is nine more than the opposite of that number. Find the number. 2) A collection of 27 coins, all nickels, and dimes, is worth \$2.10. How many of each coin are there? The dime, in United States usage, is a ten-cent coin.
• Fresh juice
The seller offers fresh-squeezed juice, which the customer either pours into his own containers or sells it in liter plastic bottles, which the customer buys from him. A liter of juice costs 40 CZK more than a bottle. There is no charge for turning the ju
• Repairing
Three employees earned a total of € 469 for repairing the equipment. They split so that the first got 20% more than the second, and the third 15% more than the second. How many euros did everyone get?
• The percent 2
The percent return rate of a growth fund, income fund, and money market are 10%, 7%, and 5% respectively. Suppose you have 3200 to invest and you want to put twice as much in the growth fund as in the money market to maximize your return. How should you i
• Reward
Three workers have shared a common reward 13110 CZK follows: first worker got 35% less than the second and third worker got 20% more than the second worker. How much got each worker?
• Crown coins
Jana saves two-crown and five-crown coins. She has ten coins in the cashier. How many two crowns and how many five crowns shw have if she saved 29 crowns?
• Dividing money
Thomas, Honza and Vasek are to divide 1220kč. Honza got 25% more than Thomas. Vasek 20% less than Thomas. How CZK each got?
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
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'
# Search results
Found 1448 matches
Orbital Eccentricity
The orbital eccentricity of an astronomical object is a parameter that determines the amount by which its orbit around another body deviates from a perfect ... more
Rayleigh length (range)
In optics and especially laser science, the Rayleigh length or Rayleigh range is the distance along the propagation direction of a beam from the waist to ... more
Moment of Inertia - I-Beam (Ideal cross section)
An I-beam, also known as H-beam, W-beam (for “wide flange”), Universal Beam (UB), Rolled Steel Joist (RSJ), or ... more
Maximum value of bending moments for a cantilever beam with uniformly distributed load
A cantilever is a beam anchored at only one end. The beam carries the load to the support where it is forced against by a moment and shear stress. A ... more
Elastic deflection of a uniformly loaded cantilever beam
Elastic deflection is the degree to which a structural element is displaced under a load.
The deflection, at the free end, of a cantilevered beam ... more
Elastic deflection to an end loaded cantilever beam
In engineering, deflection is the degree to which a structural element is displaced under a load.
The elastic deflection of a weightless cantilever ... more
Elastic deflection of a center loaded beam supported by two simple supports.
In engineering, deflection is the degree to which a structural element is displaced under a load.
The elastic deflection of a beam, loaded at its ... more
Gaussian beam (Beam width or spot size)
In optics, a Gaussian beam is a beam of electromagnetic radiation whose transverse electric field and intensity (irradiance) distributions are well ... more
Elastic deflection at any point along the span of a center loaded beam
Elastic deflection is the degree to which a structural element is displaced under a load.
The deflection at any point, along the span of a center ... more
Shear coefficient (For solid rectangular cross-section beam)
Timoshenko beam model takes into account shear deformation and rotational inertia effects, making it suitable for describing the behavior of short sandwich ... more
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https://futures.io/articles/trading/Expectancy
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# Expectancy
Expectancy = (Probability of Win * Average Win) – (Probability of Loss * Average Loss)
As an example let’s say that a trader has a system that produces winning trades 30% of the time. That trader’s average winning trade nets 10% while losing trades lose 3%. So if he were trading \$10,000 positions his expectancy would be:
(0.3 * \$1,000) – (0.7 * \$300) = \$90
So even though that system produces losing trades 70% of the time the expectancy is still positive and thus the trader can make money over time. You can also see how you could have a system that produces winning trades the majority of the time but would have a negative expectancy if the average loss was larger than the average win:
(0.6 * \$400) – (0.4 * \$650) = -\$20
## Contents
##### [hide][top]Stub
Link to Van Tharps introductory discussion of expectancy, where the concept of including the individual trade risk ('R') of each trade is introduced as a way of turning expectancy into a risk adjusted return metric comparable across systems.
https://www.iitm.com/sm-Expectancy.htm
Alternative calculation involves using 'average losing trade' as the denominator, which is usually easier to calculate, for a quick measure of risk adjusted returns.
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https://core-econ.org/the-economy/v1/book/text/leibniz-22-02-02.html
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Leibniz 22.2.2
# How the monopolist sets the rent-maximizing level of taxes
For an introduction to the Leibniz series, please see ‘Introducing the Leibnizes’.
In our model of tax setting by a dictator who is a political monopolist, the dictator wishes to maximize the political rents that he will receive while in office. But in setting the level of taxes he is constrained by the duration curve: the higher the level of taxes in each year, the fewer additional years he can expect to remain in office. Here we solve the dictator’s constrained optimization problem mathematically, to find his optimal level of tax.
In Leibniz 22.2.1 we derived an expression for the duration curve for a dictator who may be dismissed from office for performance reasons (setting too high a level of taxation) or for reasons unrelated to performance. His expected duration in office, $D$, decreases as the tax in each year, $T$, increases, so we can say that duration is a decreasing function of tax, which we write here as $D=f(T)$.
The dictator’s rent, $R$, depends on both $D$ and $T$. The cost of providing the public services is $C$, so the annual political rent is $T - C$, and the expected total rent obtained by a dictator with expected duration $D$ is:
The dictator’s constrained optimization problem is:
To solve this problem, we use the constraint to substitute for $D$, so that the rent is $(T-C)f(T)$, and then differentiate with respect to $T$ to obtain the first order condition:
The first term is the marginal benefit of raising the tax by one unit. The dictator gets an extra unit of rent for the duration $f(T)$ of his or her period in power. The second term is negative because $f(T)$ is a decreasing function. It represents the marginal cost for the dictator of raising the rent, which is that the rent $T-C$ will be received for a shorter period.
The dictator’s optimal tax level $T^*$ satisfies this equation. Once we have found $T^*$, we can determine the corresponding duration from the equation $D^*=f(T^*)$. We will demonstrate this for a particular example below.
Figure 22.6 from the text, reproduced as Figure 1 below, illustrates the solution of the dictator’s optimization problem.
Figure 1 The dictator chooses a tax level to maximize his political rents.
The solution $(D^*,\ T^*)$ is found at the point B in the diagram, where the duration curve is tangential to an isorent curve. To show that this point is the one we found mathematically above, we can rearrange the first order condition to write it as:
In this form, the first order condition tells us the same thing as the diagram: at B, the slope of the duration curve is equal to the slope of the isorent curve. To see this, we can calculate the two slopes:
• We have expressed the duration curve as $D^*=f(T^*)$, from which it follows that $dD/dT=f'(T)$. But in the figure we have drawn the curve with $T$ on the vertical axis, so the slope is $dT/dD= 1/f'(T)$. Since $f'(T)$ is negative for all $T$, the left-hand side of the equation above is the absolute value of the slope of the duration curve. We can interpret it as the marginal rate of transformation (MRT) between taxation and duration.
• The equation of an isorent curve is $R(D,\ T)=k$, where $k$ is a constant. To calculate the slope, we could apply the method used for indifference curves in Leibniz 3.2.1. But in this case it is easier to write the isorent curve as $T=C+k/D$, then differentiate to obtain $dT/dD=-k/D^2=-(T-C)/D$. So the right-hand side of the equation above is the absolute value of the slope of the isorent curve, which can be interpreted as the dictator’s marginal rate of substitution (MRS) between taxation and duration.
## An example
In the analysis above we did not specify a particular form for the duration curve. But suppose it is linear, as shown in Figure 1. Its equation can then be written $D=f(T)$, where
and $s$ is a positive constant. By differentiating $f$ you can verify that $s=-1/f'(T)$, so $s$ represents the absolute value of the slope of the line in Figure 1 (that is, the MRT). With this duration curve the first order condition $f(T)+f'(T)(T-C)=0$ becomes:
which can be solved to obtain:
and, since $D^*=f(T^*)$:
Notice that the level of taxation chosen by the dictator will be higher when the duration curve is steeper (that is, when $s$ is larger). This is similar to the case of a profit-maximizing firm, which sets a higher price when the demand curve is less elastic (steeper). In this case, however, the corresponding expected duration will be the same whatever the slope $s$ of the duration curve. We will return to this point in Leibniz 22.3.1.
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https://planetmath.org/LinearEquation
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# linear equation
Let $L:U\rightarrow V$ be a linear mapping, and $v\in V$ an element of the codomain. A linear equation is a relation of the form,
$L(u)=v,$
where $u\in U$ is to be considered as the unknown. The solution set of a linear equation is the set of $u\in U$ that satisfy the above constraint, or to be more precise, the pre-image $L^{-1}(v)$. The equation is called inconsistent if no solutions exist, that is, if the pre-image is the empty set. Otherwise, the equation is called consistent.
The general solution of a linear equation has the form
$u=u_{p}+u_{h},\quad u_{p},u_{h}\in U,$
where
$L(u_{p})=v$
is a particular solution and where
$L(u_{h})=0$
is any solution of the corresponding homogeneous problem, i.e. an element of the kernel of $L$.
Notes. Elementary treatments of linear algebra focus almost exclusively on finite-dimensional linear problems. They neglect to mention the underlying mapping, preferring to focus instead on “variables and equations.” However, the scope of the general concept is considerably wider, e.g. linear differential equations such as
$y^{\prime\prime}+y=0.$
Title linear equation Canonical name LinearEquation Date of creation 2013-03-22 12:25:59 Last modified on 2013-03-22 12:25:59 Owner rmilson (146) Last modified by rmilson (146) Numerical id 8 Author rmilson (146) Entry type Definition Classification msc 15A06 Synonym linear problem Synonym linear system Related topic HomogeneousLinearProblem Related topic FiniteDimensionalLinearProblem Defines consistent Defines inconsistent Defines particular solution
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https://www.lauritoandlaurito.com/partially-amortized-mortgage/
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An adjustable rate mortgage may offer a lower initial interest rate and monthly payments than a conventional fixed rate mortgage. After an initial term, the interest rate on an adjustable rate mortgage loan is re-set periodically to keep the rate in line with current market interest rates.
40000 Mortgage Over 10 Years Mortgage Payment Calculator – Loan Amount = \$40000 – Interest. – Payment number beginning balance interest Payment Principal Payment Ending Balance Cumulative Interest Cumulative Payments; 1: \$40,000.00: \$150.00: \$52.67: \$39,947.33Five Year Mortgage Paying off a mortgage early can save hundreds of thousands of dollars in interest payments. Paying a 30-year mortgage off is as few as five to seven years takes a solid plan of action and budget.
Free amortization calculator returns monthly payment as well as displaying a schedule, graph, and pie chart breakdown of an amortized loan. Or, simply learn more about loan amortization. Experiment with other loan calculators, or explore hundreds of other calculators addressing topics such as math, fitness, health, and many more.
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partially amortizing loan. A loan with periodic payments of interest and principal, but for a shorter term than necessary to pay the principal balance in full at that rate. Partially amortizing loans have a balloon payment at some point,requiring repayment in full or through refinancing.
In an partially amortized loan, only a part of the sum must be returned in monthly payments. An additional lump sum, called a balloon payment, is paid to the bank at the end date of the loan. For example, imagine you want a loan of \$1,000,000 with a 10% interest.
Course Transcript. Just like when you determine payments for a fully amortized loan, you can use the PMT or Payment function to determine payments for a partially amortized loan. If you want the lump sum or balloon payment to be due at the end of the loan’s term, you can put the balloon payment in the PMT functions, fv or future value argument,
Partially-amortizing loans (or balloon mortgages as otherwise referred to), call for partial repayment of the principal over the term of the loan with the remaining balance due upon expiration of the term of the loan. Usually the amount of principal due upon maturity of the loan is significant.
What is ‘Amortized Loan’. An amortized loan is a loan with scheduled periodic payments that consist of both principal and interest. An amortized loan payment pays the relevant interest expense for the period before any principal is paid and reduced. This is opposed to loans with interest-only payment features, balloon payment features.
Bankrate Mortgage Payoff Calculator Bankrate Mortgage Payment Calculator Low Credit Score Cash Advance Loans in The united states No teletrack [simple!] click to read more to get Fast and easy Online Loan. In case you have some kind of special kids in your daily life, but the idea of buying playthings happens fear in your heart, you might have can come to the right spot.
Partially amortized: A partially amortized bond is one in which only a part of the principal is repaid over the bond's life. The remaining big part of the principal is.
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http://www.mathworks.com/matlabcentral/cody/problems/271-n-cards-problem/solutions/139891
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Cody
# Problem 271. N-Cards Problem
Solution 139891
Submitted on 17 Sep 2012 by Vitaly Lavrukhin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(nCardsProblem(x),y_correct))
``` ans = 1 ```
2 Pass
%% x = 5; y_correct = 2; assert(isequal(nCardsProblem(x),y_correct))
``` ans = 2 ```
3 Pass
%% x = 50; y_correct = 36; assert(isequal(nCardsProblem(x),y_correct))
``` ans = 36 ```
4 Pass
%% x = 1000; y_correct = 976; assert(isequal(nCardsProblem(x),y_correct))
``` ans = 976 ```
5 Pass
%% x = 10000; y_correct = 3616; assert(isequal(nCardsProblem(x),y_correct))
``` ans = 3616 ```
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https://www.boddlelearning.com/1st-grade/comparing-measuring-lengths
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# Comparing & Measuring Lengths
Comparing and measuring lengths in the first grade is made of two different but related Common Core math standards: 1.MD.A.1 and 1.MD.A.2. Below we show a few videos that demonstrate each of these standards. Then, we provide a breakdown of the specific steps in the videos to help you teach your class.
Prior Learnings
Your students should be familiar with the Measurement and Data standards from Kindergarten. The first grade Comparing and Measuring Lengths standards are a continuation of the Kindergarten skill of describing measurable attributes of objects, like length. And it’s closely tied to the Kindergarten standard of directly comparing two objects with a measurable attribute in common--like identifying which object has “more of” or “less of” an attribute. You may want to give your first graders a quick refresher on these items before jumping into the lesson.
Future Learnings
Measuring and comparing lengths in 1st grade will help your students as they move onto 2nd grade. In 2nd grade, your students will learn how to make determinations on length differences using tools like rulers as well as express those differences in terms of a standard-length unit.
Common Core Standard: 1.MD.A.1 - Order three objects by length; compare the lengths of two objects indirectly by using a third object.
Students who understand this principle can:
1. Identify which of 2 objects is longer or shorter.
2. Place objects in order based on length (longest to shortest or shortest to longest).
3. Decide how the lengths of two objects relate to one another based on a third object’s length.
Video 1: Ordering Objects by Length
This video has 2 parts, both related to 1.MD.A. In the first half 3 objects are ordered by lengths. And the 2nd half, 2 objects are compared using a third object.
1. In the first half of the video, three objects are compared: a pair of scissors, a crayon, and a pen. The video walks you through how to order these 3 objects from shortest to longest. Here’s a breakdown of the specific steps taken in the video:
The three objects are arranged so that they are all straight up and down.
2. A base line is drawn, and the three objects are aligned so that they all begin at the same point.
3. The objects’ heights are compared.The crayon is the shortest, the scissors are the tallest, and the pen is in the middle.
4. Each object is labeled 1-3, with 1 being the shortest (the crayon) and 3 the tallest (the scissors). The objects are sorted from shortest to tallest. Since the crayon is the shortest, the crayon moves to the front of the line, switching places with the scissors. Next, the pen should be second, so it moves to the middle. The order is now shortest to longest: crayon, pen, then scissors.
In the second half, we see how to compare two objects by using the length of another object. Here’s a breakdown of the specific steps taken in the 2nd half of the video:
1. Two flowers are shown, and we want to compare their lengths using another object. In this example, a hand is used to compare the flowers’ lengths.
2. The hand is aligned next to the smaller flower and we see that the flower is about the length of one hand.
3. The same hand is aligned next to the larger flower.
4. The larger flower is the length of two hands. Therefore, the larger flower is about the same as two of the smaller flowers put together.
Video 2: Comparing Lengths
Boddle combines both standards into one video. The first half of the video covers standard 1.MD.A.1 (0:00-1:30), which is broken down below:
This is a helpful video for you to show your students. It introduces how to measure length, following Emma, a Boddle character who is doing some sightseeing in her new neighborhood. Emma asks for students' help in measuring length. Along the way, Emma notices some tall buildings and wonders which is the tallest.
1. The buildings are labeled as A, B, and C, and the viewer is asked to choose among the three to identify the tallest. Building C is identified as the tallest.
2. Next, Emma goes to the ocean and sees three different types of ships. She wants to know which is the longest, and then the viewer gets to choose between the ships labeled as A, B, and C. Ship A is the longest.
3. Then Emma goes to her neighbors house to meet Jamie, Chelsea, and Mike. The viewer is then asked to find which one of the three is the shortest. Chelsea is the shortest.
Common Core Standard: 1.MD.A.2 - Express the length of an object as a whole number of length units by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps.
Students who understand this principle can:
1. Illustrate how to use multiple shorter objects to find the length of a longer object.
2. Connect the length of the longer object to the total number of shorter objects used and express the longer object’s length (e.g. The pencil is 3 paper clips long).
3. Describe why gaps and overlaps are not allowed and do not provide a proper measurement.
Video 1: Using Paper Clips to Measure Length
The video demonstrates how to measure the length of a pencil using multiple paperclips. At first, the video shows 4 wrong ways to measure the pencil using paperclips.
In the first example, the girl in the video believes her pencil is 4 paperclips long, and then asks the viewer if they can identify what she did wrong. She precedes to present three more attempts at measuring her pencil with paperclips, ending up with varying measurements: 6, 4, and 5 paperclips.
She then lists what she needs to do so that her measurements are accurate. There are 4 things she understands that are needed to ensure accurate measurement using the paperclips:
1. The paperclips should start at the beginning of the pencil and go to the very end.
2. The paperclips need to be placed in a straight line, without overlapping.
3. The paperclips need to be placed end-to-end, without leaving any gaps.
4. The paperclips all need to be the same size.
The girl in the video then measures her pencil once more, making sure all the above criteria are met, and discovers her pencil is 5 paper clips long.
Video 2: Measuring Lengths
The video below covers both standards 1.MD.A.1 and 1.MD.A.2. Below is the breakdown of the 2nd half of the video, which covers 1.MD.A.2 starting at 1:30 on the video.
1. In the first example, Boddle uses cameras to measure the length of a car. The cameras are lined up below the car and counted. The car is about 6 cameras long.
2. Next, houses are used to measure the length of an airplane. The same process is repeated and the viewer finds that the airplane is about 4 houses long.
3. Lastly, dumbbells are used to measure the length of a ladder. The ladder turns out to be about 8 dumbbells long.
Want more practice?
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https://www.doubtnut.com/question-answer/if-overlineb-and-overlinec-are-any-two-perpendicular-unit-vectors-and-overlinea-is-any-vector-then-o-127331936
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If overline(b) and overline(c)...
# If overline(b) and overline(c) are any two perpendicular unit vectors and overline(a) is any vector, then (overline(a)*overline(b))overline(b)+(overline(a)*overline(c))overline(c)+(overline(a)*(overline(b)timesoverline(c))/(|overline(b)timesoverline(c)|))(overline(b)timesoverline(c))=
Text Solution
overline(b)overline(a)overline(c)overline(b)+overline(c)
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https://testbook.com/question-answer/five-hundred-students-appeared-in-an-examination-c--609504fe06c4a92f4bb25b58
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# Five hundred students appeared in an examination comprising English, Hindi and Mathematics. The diagram gives the number of candidates who failed in different subjects. The percentage of candidates who failed in at least two subjects is
This question was previously asked in
Telangana Police SI Mains Exam 2016 Official Paper 1
View all Telangana Police SI Papers >
1. 0.078
2. 1.8
3. 6.8
4. 7.8
## Answer (Detailed Solution Below)
Option 4 : 7.8
Free
Telangana Police SI Preliminary Exam 2018 Official Paper 1
636
200 Questions 200 Marks 180 Mins
## Detailed Solution
Calculations:
According to the data of 500 students
Numbers of candidates who failed in at least two subjects
⇒ (12 + 12 + 10 + 5) = 39
When the number of people surveyed is 500
⇒ 39 / 500 × 100 = 7.8%
∴ The percentage of candidates who failed in at least two subjects is 7.8%.
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# [Rd] .Call in R
Joris Meys jorismeys at gmail.com
Fri Nov 18 16:45:41 CET 2011
```Because if you calculate the probability and then make uniform values,
nothing guarantees that the sum of those uniform values actually is
larger than 50,000. You only have 50% chance it is, in fact...
Cheers
Joris
On Fri, Nov 18, 2011 at 4:08 PM, Karl Forner <karl.forner at gmail.com> wrote:
> Hi,
>
> A probably very naive remark, but I believe that the probability of sum(
> runif(10000) ) >= 50000 is exactly 0.5. So why not just test that, and
> generate the uniform values only if needed ?
>
>
> Karl Forner
>
> On Thu, Nov 17, 2011 at 6:09 PM, Raymond <gwgc5 at mail.missouri.edu> wrote:
>
>> Hi R developers,
>>
>> I am new to this forum and hope someone can help me with .Call in R.
>> Greatly appreciate any help!
>>
>> Say, I have a vector called "vecA" of length 10000, I generate a vector
>> called "vecR" with elements randomly generated from Uniform[0,1]. Both vecA
>> and vecR are of double type. I want to replace elements vecA by elements in
>> vecR only if sum of elements in vecR is greater than or equal to 5000.
>> Otherwise, vecR remain unchanged. This is easy to do in R, which reads
>> vecA<-something;
>> vecR<-runif(10000);
>> if (sum(vecR)>=5000)){
>> vecA<-vecR;
>> }
>>
>>
>> Now my question is, if I am going to do the same thing in R using .Call.
>> How can I achieve it in a more efficient way (i.e. less computation time
>> compared with pure R code above.). My c code (called "change_vecA.c")
>> using
>> .Call is like this:
>>
>> SEXP change_vecA(SEXP vecA){
>> int i,vecA_len;
>> double sum,*res_ptr,*vecR_ptr,*vecA_ptr;
>>
>> vecA_ptr=REAL(vecA);
>> vecA_len=length(vecA);
>> SEXP res_vec,vecR;
>>
>> PROTECT(res_vec=allocVector(REALSXP, vec_len));
>> PROTECT(vecR=allocVector(REALSXP, vec_len));
>> res_ptr=REAL(res_vec);
>> vecR_ptr=REAL(vecR);
>> GetRNGstate();
>> sum=0.0;
>> for (i=0;i<vecA_len;i++){
>> vecR_ptr[i]=runif(0,1);
>> sum+=vecR_ptr[i];
>> }
>> if (sum>=5000){
>> /*copy vecR to the vector to be returned*/
>> for (i=0;i<vecA_len;i++){
>> res_ptr[i]=vecR_ptr[i];
>> }
>> }
>> else{
>> /*copy vecA to the vector to be returned*/
>> for (i=0;i<vecA_len;i++){
>> res_ptr[i]=vecA_ptr[i];
>> }
>> }
>>
>> PutRNGstate();
>> UNPROTECT(2);
>> resturn(res);
>> }
>> My R wrapper function is
>> change_vecA<-function(vecA){
>> .Call("change_vecA",vecA);
>> }
>>
>> Now my question is, due to two loops (one generates the random
>> vector and one determines the vector to be returned), can .Call still be
>> faster than pure R code (only one loop to copy vecR to vecA given condition
>> is met)? Or, how can I improve my c code to avoid redundant loops if any.
>> My
>> concern is if vecA is large (say of length 1000000 or even bigger), loops
>> in
>> C code can slow things down. Thanks for any help!
>>
>>
>>
>>
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/Call-in-R-tp4080721p4080721.html
>> Sent from the R devel mailing list archive at Nabble.com.
>>
>> ______________________________________________
>> R-devel at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Mathematical Modelling, Statistics and Bio-Informatics
tel : +32 9 264 59 87
Joris.Meys at Ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
```
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https://la.mathworks.com/matlabcentral/cody/problems/943-mirror-matrix/solutions/1652890
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crawl-data/CC-MAIN-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00132.warc.gz
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Cody
Problem 943. "mirror" matrix
Solution 1652890
Submitted on 18 Oct 2018 by Athi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
n = 1; m_correct = [1 1]; assert(isequal(mirror_matrix(n),m_correct))
2 Pass
n = 3; m_correct = [1 2 3 3 2 1; 1 2 3 3 2 1; 1 2 3 3 2 1]; assert(isequal(mirror_matrix(n),m_correct))
3 Pass
n = 33; m_correct = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1;]; assert(isequal(mirror_matrix(n),m_correct))
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http://www.markedbyteachers.com/as-and-a-level/geography/to-examine-and-explain-the-changes-in-river-width-from-source-to-mouth.html
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• Join over 1.2 million students every month
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# To examine and explain the changes in river width from source to mouth.
Extracts from this document...
Introduction
Objective Analysis This is now the most important part of the investigation. I shall now take all my results and graphs and sum everything up, relating to class work, textbook theories and other pieces of information. I will analyse and interpret the results that I collected, as it shall give me a better understanding of rivers and my overall results! I will now take each objective and explain it to the optimum of my ability. Objective 1 To examine and explain the changes in river width from source to mouth. From examining my compilation of results, graphs and cross sections, I get the distinct impression that as we moved from the source to mouth, the rivers width increased. By looking at the graph showing width against distance from source, I can see more precisely, where the steepest rises were. ...read more.
Middle
The widening of the river channel was caused by 3 types of erosion- Hydraulic power, Corrasion and Corrosion. Hydraulic power occurs due to the shear force and weight of the rivers load rubbing and creating friction against the rivers bed and banks. This process gradually wears the bed and banks away and especially causes the removal of loose clay and sand. This process is even fiercer when the river is in flood because there is more gravitational potential energy due to more mass in the water. As there is more mass the water will travel faster, wasting more energy overcoming friction but by also rubbing harshly against the rivers bed and banks. Another big affecter in the widening of the river channel was Corrasion. Corrasion or sometimes called "Abrasion" occurs when particles of sand and silt are carried along in the river. ...read more.
Conclusion
There is more energy available for erosion as we move downstream due to the bed load becoming smaller and more rounded which shall cause the river to become more efficient and travel at a greater speed as it has more energy. In the middle course of the river, Lateral erosion took place. This basically describes that as you move from the source to the mouth of a river, the width will most definitely increase due to the factors that I have before stated. If you look back over the pictures I took in the site information category, you will see a huge difference in width when comparing site 1 to site 4. Site 4 is extremely wider with boulders on the banks. Site 1 was a small little stream with lots of sharp jagged rocks, which was one of the primary constituents in causing the small width. Below I have put the 2 photographs together so you can judge for yourself. ...read more.
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https://answers.yahoo.com/question/index?qid=20080415000015KK00040
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C程式 為什麼算不出標準差???
#include <stdio.h>
#include <stdarg.h>
#include <math.h>
int main(void)
{
int number; /* account number */
int i;
double score[100]; /* account score */
double total;
double average;
double S;
double standard_deviation;
FILE *cfPtr; /* cfPtr = clients.dat file pointer */
/* fopen opens file; exits program if file cannot be opened */
if ( ( cfPtr = fopen( "clients.dat", "r" ) ) == NULL ) {
printf( "File could not be opened\n" );
} /* end if */
else {
printf("讀取到的學生成績資料如下:\n\n");
printf( "%s%6s\n\n", "學生座號", "成績" );
for (i=0;i<number;i++) {
fscanf( cfPtr, "%d%lf", &number, &score[i] );
total = score[i];
/* while not end of file */
while ( !feof( cfPtr ) &&i<number ) {
printf( "%5d%10.2f\n", number, score[i] );
fscanf( cfPtr, "%d%lf", &number, &score[i] );
total += score[i];
} /* end while */
total = total-score[i];
printf("\n");
average = total / number;
printf("以上%d位同學的總分是 %.2f分\n", number, total);
printf("以上%d位同學的平均是 %.2f分\n", number, average);
for (i=0;i<number;i++) {
S += (score[i]-average)*(score[i]-average);
standard_deviation = sqrt(S/number);
}
printf("以上%d位同學的標準偏差是 %.2f分\n\n", number, standard_deviation);
}
fclose( cfPtr ); /* fclose closes the file */
} /* end else */
return 0;
}
*******************************************************
1 50.00
2 60.00
3 70.00
4 80.00
5 90.00
*******************************************************
1 50.00
2 60.00
3 70.00
4 80.00
5 90.00
*******************************************************
3Q
*******************************************************
Update:
Update 2:
dtsien 不好意思喔
Update 3:
Update 4:
YA 經過了這麼多次的修修改改 我終於成功囉 真是非常謝謝你
( 我發現你貼的程式+號都不見了 @@ 好奇怪喔 )
Update 5:
#include
#include
#include
int main(void)
{
Update 6:
int number; /* account number */
int i;
int counter; //計算有幾筆資料用的計數器
double score[100]; /* account score */
double total;
double average;
double S;
double standard_deviation;
Update 7:
FILE *cfPtr; /* cfPtr = clients.dat file pointer */
/* fopen opens file; exits program if file cannot be opened */
if ( ( cfPtr = fopen( "clients.dat", "r" ) ) == NULL )
{
printf( "File could not be opened\n" );
} /* end if */
Update 8:
else
{
printf("讀取到的學生成績資料如下:\n\n");
printf( "%s%6s\n\n", "學生座號", "成績" );
counter = 0;
Update 9:
while( true )
{
fscanf( cfPtr, "%d%lf", &number, &score[counter] );
if(feof( cfPtr )) // 如果讀到檔案結尾，就結束了
break;
printf( "%5d%10.2f\n", number, score[counter] );
total += score[counter];
counter++; // 每讀一筆就加一次
}
fclose( cfPtr ); /* fclose closes the file */
Update 10:
average = total / counter;
printf( "\n" );
printf("以上%d位同學的總分是 %.2f分\n", counter, total);
printf("以上%d位同學的平均是 %.2f分\n", counter, average);
for (i=0;i
Update 11:
printf("以上%d位同學的標準偏差是 %.2f分\n\n", counter, standard_deviation);
} /* end else */
return 0;
}
Update 12:
= = Yahoo 知識真的會改掉貼上去的內容耶
#include
#include
#include
for (i=0;i
Update 13:
@@ 為什麼後面的字無法顯現出來
Update 14:
dtsien 不好意思 上次說你可能寫錯的應該是Yahoo 知識的問題
Rating
• Heresy
Lv 7
恩…你的這個程式邏輯滿詭異的…
首先是讀取檔案的部分。
你在外面先用了一個 for (i=0;i<number;i ) 想要來處理數量為 number 的輸入？但是實際上，你並沒有給予 number 一個數值；在這個情況下，其實程式有可能直接在這邊就結束了，不會繼續跑。
而實際在透過 while 讀取的時候，number 又顯得多餘了…或者應該說，不應該這樣使用。你在程式中，是把 number 拿來讀取學號，但是又把它拿來當計數器（記錄有幾筆資料）；如果學浩不是 1~n 的話，會有很大的問題的。
修改後的結果如下：#include <stdio.h>
#include <stdarg.h>
#include <math.h>
int main(void)
{
int number; /* account number */
int i;
int counter; //計算有幾筆資料用的計數器
double score[100]; /* account score */
double total;
double average;
double S;
double standard_deviation;
FILE *cfPtr; /* cfPtr = clients.dat file pointer */
/* fopen opens file; exits program if file cannot be opened */
if ( ( cfPtr = fopen( "clients.dat", "r" ) ) == NULL )
{
printf( "File could not be opened\n" );
} /* end if */
else
{
printf("讀取到的學生成績資料如下:\n\n");
printf( "%s%6s\n\n", "學生座號", "成績" );
counter = 0;
while( true )
{
fscanf( cfPtr, "%d%lf", &number, &score[counter] );
if(feof( cfPtr )) // 如果讀到檔案結尾，就結束了
break;
printf( "].2f\n", number, score[counter] );
total = score[counter];
counter; // 每讀一筆就加一次
}
fclose( cfPtr ); /* fclose closes the file */
average = total / counter;
printf( "\n" );
printf("以上%d位同學的總分是 %.2f分\n", counter, total);
printf("以上%d位同學的平均是 %.2f分\n", counter, average);
for( i = 0; i < counter; i )
{
S = (score[i]-average)*(score[i]-average);
standard_deviation = sqrt(S/number);
}
printf("以上%d位同學的標準偏差是 %.2f分\n\n", number, standard_deviation);
} /* end else */
return 0;
}
2008-04-16 07:58:29 補充：
可能是貼錯了 @@
printf( "%5d%10.2f\n", number, score[counter] );
2008-04-16 07:58:51 補充：
還是其實是被 Yahoo 知識改掉了啊？ @@
2008-04-16 12:58:08 補充：
Yahoo 會把小於符號當成 HTML 的標籤 <
• dtsien
Lv 6
#include < stdio.h >
#include < math.h >
int main(void)
{
int number; /* account number */
int i;
double score[100]; /* account score */
double total=0.0;
double average;
double S=0.0;
double standard_deviation;
FILE *cfPtr; /* cfPtr = clients.dat file pointer */
/* fopen opens file; exits program if file cannot be opened */
if ( ( cfPtr = fopen( "clients.dat", "r" ) ) == NULL ) {
printf( "File could not be opened\n" );
} /* end if */
else {
printf("讀取到的學生成績資料如下:\n\n");
printf( "%s%6s\n\n", "學生座號", "成績" );
/* while not end of file */
while ( 1 ) {
fscanf( cfPtr, "%d", &number );
if (feof( cfPtr )) break;
fscanf( cfPtr, "%lf", &score[number] );
printf( "].2f\n", number, score[number] );
total = score[number];
} /* end while */
printf("\n");
average = total / number;
printf("以上%d位同學的總分是 %.2f分\n", number, total);
printf("以上%d位同學的平均是 %.2f分\n", number, average);
for (i=1;i<=number;i ) {
S = (score[i]-average)*(score[i]-average);
}
standard_deviation = sqrt(S/number);
printf("以上%d位同學的標準偏差是 %.2f分\n\n", number, standard_deviation);
}
fclose( cfPtr ); /* fclose closes the file */
/* end else */
system("pause");
return 0;
}
學習程式設計的方法技巧或經驗:
多想多做如此而已
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CC-MAIN-2020-40
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latest
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en
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https://mainfacts.com/convert-kj-to-kwh
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Converters: Life calculator Astronomy Working days Length Area Time Weight Temperature Capacity Math Density Pressure Random IT converters
# Convert kilojoules to kilowatts per hour, kJ to kwh
### Kilojoules to Kilowatts per Hour
kJ = 2.778E-4 kw/h
### Kilowatts per Hour to Kilojoules
kw/h = 3600 kJ
#### kj to kw/h conversion table:
1 kJ = 2.778E-4 kw/h 21 kJ = 5.833E-3 kw/h 41 kJ = 1.139E-2 kw/h 70 kJ = 1.944E-2 kw/h 2 kJ = 5.556E-4 kw/h 22 kJ = 6.111E-3 kw/h 42 kJ = 1.167E-2 kw/h 80 kJ = 2.222E-2 kw/h 3 kJ = 8.333E-4 kw/h 23 kJ = 6.389E-3 kw/h 43 kJ = 1.194E-2 kw/h 90 kJ = 2.5E-2 kw/h 4 kJ = 1.111E-3 kw/h 24 kJ = 6.667E-3 kw/h 44 kJ = 1.222E-2 kw/h 100 kJ = 2.778E-2 kw/h 5 kJ = 1.389E-3 kw/h 25 kJ = 6.944E-3 kw/h 45 kJ = 1.25E-2 kw/h 110 kJ = 3.056E-2 kw/h 6 kJ = 1.667E-3 kw/h 26 kJ = 7.222E-3 kw/h 46 kJ = 1.278E-2 kw/h 120 kJ = 3.333E-2 kw/h 7 kJ = 1.944E-3 kw/h 27 kJ = 7.5E-3 kw/h 47 kJ = 1.306E-2 kw/h 130 kJ = 3.611E-2 kw/h 8 kJ = 2.222E-3 kw/h 28 kJ = 7.778E-3 kw/h 48 kJ = 1.333E-2 kw/h 140 kJ = 3.889E-2 kw/h 9 kJ = 2.5E-3 kw/h 29 kJ = 8.056E-3 kw/h 49 kJ = 1.361E-2 kw/h 150 kJ = 4.167E-2 kw/h 10 kJ = 2.778E-3 kw/h 30 kJ = 8.333E-3 kw/h 50 kJ = 1.389E-2 kw/h 160 kJ = 4.444E-2 kw/h 11 kJ = 3.056E-3 kw/h 31 kJ = 8.611E-3 kw/h 51 kJ = 1.417E-2 kw/h 170 kJ = 4.722E-2 kw/h 12 kJ = 3.333E-3 kw/h 32 kJ = 8.889E-3 kw/h 52 kJ = 1.444E-2 kw/h 180 kJ = 5.0E-2 kw/h 13 kJ = 3.611E-3 kw/h 33 kJ = 9.167E-3 kw/h 53 kJ = 1.472E-2 kw/h 190 kJ = 5.278E-2 kw/h 14 kJ = 3.889E-3 kw/h 34 kJ = 9.444E-3 kw/h 54 kJ = 1.5E-2 kw/h 200 kJ = 5.556E-2 kw/h 15 kJ = 4.167E-3 kw/h 35 kJ = 9.722E-3 kw/h 55 kJ = 1.528E-2 kw/h 300 kJ = 8.333E-2 kw/h 16 kJ = 4.444E-3 kw/h 36 kJ = 1.0E-2 kw/h 56 kJ = 1.556E-2 kw/h 400 kJ = 1.111E-1 kw/h 17 kJ = 4.722E-3 kw/h 37 kJ = 1.028E-2 kw/h 57 kJ = 1.583E-2 kw/h 500 kJ = 1.389E-1 kw/h 18 kJ = 5.0E-3 kw/h 38 kJ = 1.056E-2 kw/h 58 kJ = 1.611E-2 kw/h 700 kJ = 1.944E-1 kw/h 19 kJ = 5.278E-3 kw/h 39 kJ = 1.083E-2 kw/h 59 kJ = 1.639E-2 kw/h 900 kJ = 2.5E-1 kw/h 20 kJ = 5.556E-3 kw/h 40 kJ = 1.111E-2 kw/h 60 kJ = 1.667E-2 kw/h 1000 kJ = 2.778E-1 kw/h
#### kw/h to kj conversion table:
1 kw/h = 3600 kJ 21 kw/h = 75600 kJ 41 kw/h = 147600 kJ 70 kw/h = 252000 kJ 2 kw/h = 7200 kJ 22 kw/h = 79200 kJ 42 kw/h = 151200 kJ 80 kw/h = 288000 kJ 3 kw/h = 10800 kJ 23 kw/h = 82800 kJ 43 kw/h = 154800 kJ 90 kw/h = 324000 kJ 4 kw/h = 14400 kJ 24 kw/h = 86400 kJ 44 kw/h = 158400 kJ 100 kw/h = 360000 kJ 5 kw/h = 18000 kJ 25 kw/h = 90000 kJ 45 kw/h = 162000 kJ 110 kw/h = 396000 kJ 6 kw/h = 21600 kJ 26 kw/h = 93600 kJ 46 kw/h = 165600 kJ 120 kw/h = 432000 kJ 7 kw/h = 25200 kJ 27 kw/h = 97200 kJ 47 kw/h = 169200 kJ 130 kw/h = 468000 kJ 8 kw/h = 28800 kJ 28 kw/h = 100800 kJ 48 kw/h = 172800 kJ 140 kw/h = 504000 kJ 9 kw/h = 32400 kJ 29 kw/h = 104400 kJ 49 kw/h = 176400 kJ 150 kw/h = 540000 kJ 10 kw/h = 36000 kJ 30 kw/h = 108000 kJ 50 kw/h = 180000 kJ 160 kw/h = 576000 kJ 11 kw/h = 39600 kJ 31 kw/h = 111600 kJ 51 kw/h = 183600 kJ 170 kw/h = 612000 kJ 12 kw/h = 43200 kJ 32 kw/h = 115200 kJ 52 kw/h = 187200 kJ 180 kw/h = 648000 kJ 13 kw/h = 46800 kJ 33 kw/h = 118800 kJ 53 kw/h = 190800 kJ 190 kw/h = 684000 kJ 14 kw/h = 50400 kJ 34 kw/h = 122400 kJ 54 kw/h = 194400 kJ 200 kw/h = 720000 kJ 15 kw/h = 54000 kJ 35 kw/h = 126000 kJ 55 kw/h = 198000 kJ 300 kw/h = 1080000 kJ 16 kw/h = 57600 kJ 36 kw/h = 129600 kJ 56 kw/h = 201600 kJ 400 kw/h = 1440000 kJ 17 kw/h = 61200 kJ 37 kw/h = 133200 kJ 57 kw/h = 205200 kJ 500 kw/h = 1800000 kJ 18 kw/h = 64800 kJ 38 kw/h = 136800 kJ 58 kw/h = 208800 kJ 700 kw/h = 2520000 kJ 19 kw/h = 68400 kJ 39 kw/h = 140400 kJ 59 kw/h = 212400 kJ 900 kw/h = 3240000 kJ 20 kw/h = 72000 kJ 40 kw/h = 144000 kJ 60 kw/h = 216000 kJ 1000 kw/h = 3600000 kJ
| 2,033
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https://www.physicsforums.com/threads/is-there-a-rule-of-thumb-for-small-angle-approximation.321995/
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# Is there a rule of thumb for small angle approximation?
1. Jun 25, 2009
### Starwatcher16
When you are not given an acceptable level of error in a problem, is there any rule of thumb I should use for how large Theta can be before I stop using the small angle approximation(Sin Theta=Theta) ?
2. Jun 25, 2009
### rock.freak667
According to my text book it says (to 3sf, in radians)
$$-0.105<\theta<0.105$$
3. Jun 25, 2009
### Hurkyl
Staff Emeritus
Decide for yourself what level of error is appropriate. Then use your calculus to determine if that approximation is good enough!
The Taylor remainder theorem is a systematic way to bound the error on an approximation.
However, when theta is small, the Taylor series for sin is an alternating series whose terms are strictly decreasing -- so an easier method is to use what you know about alternating series to estimate the error.
4. Jun 25, 2009
### protonchain
In general if you're doing really simple and very very approximate calculations then anything less than 0.5 would probably be just fine.
Truly though, the best way to look is to see a graph of Sin x / x. Analyze this and look at the regions where it is 0.95 - 1 for example, and you can then grasp where you can begin to approximate.
About x = 0.55 is where Sin x / x = 0.95. Obviously as x goes to 0, the value increases towards 1.
I hope that helped or at least made you get up and get your graphing calculator :P
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https://math.stackexchange.com/questions/3434503/terence-tao-analysis-i-ex-5-4-5-there-is-a-rational-between-any-two-reals
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# Terence Tao, Analysis I, Ex. 5.4.5: There is a rational between any two reals
Terence Tao, Analysis I, 3e, Exercise 5.4.5:
Prove Proposition 5.4.14. (Hint: use Exercise 5.4.4. You may also need to argue by contradiction.)
Proposition 5.4.14:
Given any two real numbers $$x < y$$, we can find a rational number q such that $$x < q < y$$.
Exercise 5.4.4:
Show that for any positive real number $$x > 0$$ there exists a positive integer $$N$$ such that $$x > 1/N > 0$$.
What I've found so far (with the help of this answer, and Pratik Apshinge's comment):
From Exercise 5.4.4, there is a positive real number
$$y - x > 1/N > 0,$$
$$yN - xN > 1,$$
$$yN - 1 > xN.$$
Since there is an integer $$m$$ between $$yN$$ and $$yN - 1$$, we have that
$$yN > m \ge yN - 1 > xN$$
$$yN > m > xN$$
$$y > m/N > x$$
Since $$m$$ and $$N$$ are integers, there exists a rational between $$y$$ and $$x$$.
But how could a proof by contradiction help in this case?
• I think the proof in your Q is as simple as possible. A proof by contradiction in this case is basically the same thing made complicated.... Much of the logical foundation of $\Bbb R$ depends on the Archimedean Property: If $x>0$ then for any $y$ there exists $n\in \Bbb N$ with $nx>y$.... We can extend $\Bbb R$ to a larger ordered field in which the basic rules of $+,-,\times, /,$ and $<$ still apply, but which includes objects larger than any $n\in \Bbb N;$ and their reciprocals are positive but less than any member of $\Bbb Q^+$..... (continued).... Nov 14, 2019 at 8:59
• ....(continued).... but in any such extension there will be non-empty subsets with upper bounds but no least upper bounds. Nov 14, 2019 at 9:02
Why would you want to do it by contradiction? It's easier to do it directly. If $$y-x>0$$ then there is an integer $$N$$ such that $$y-x>1/N>0$$, which means that $$Ny-Nx>1$$ and this implies that there is an integer $$m$$ such that $$Nx and finally that $$x<\frac{m}{N}.
• +1....I posted a proof by contradiction, just to show how. It's not my preferred method either. Nov 13, 2019 at 23:29
If there was no $$q\in\mathbb{Q}$$ such that $$x then there are no $$m\in\mathbb{Z} , n\in\mathbb{Z}^*$$ such that $$x<\frac{m}{n} $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad nx\geq \lfloor ny \rfloor$$ $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad ny-nx\leq ny-\lfloor ny \rfloor<1$$Thus for all $$n\in\mathbb{N}^*$$ $$ny-nx\leq1$$ this implies that $$\forall n\in\mathbb{N}^*$$ $$y-x\leq \frac{1}{n}$$ contradiction with exercise 5.4.4 since $$y-x$$ is a positive number.
• Why does the absence of integers that render $x < m/n < y$ imply that $ny - nx \le 1$, for all $n$? Nov 13, 2019 at 21:39
• I edited it for you, let me know if you still have troubles understanding. Nov 13, 2019 at 21:55
• Thank you for adding more details! What does $E$ mean? Nov 14, 2019 at 9:24
• It is the floor function check : en.wikipedia.org/wiki/Floor_and_ceiling_functions for more detail , sorry if I used the french notation for it I didn't know that this notation doesn't exist in english. Nov 14, 2019 at 10:58
• I meant the negation of it , if there were no m ,n such that $x<\frac{m}{n}<y$ then if $m=\lfloor ny \rfloor$ there is no n such that $nx < \lfloor ny \rfloor < ny$. Nov 14, 2019 at 17:56
(i).Suppose $$0\le x and $$\Bbb Q\cap (x,y)=\emptyset.$$
There exists $$N\in \Bbb N$$ with $$1/N<(y-x)/2.$$
There exists $$n\in \Bbb N$$ with $$n(1/N)\ge x.$$ This is obvious if $$x=0.$$ And if it were false with $$x>0$$ then $$1/n> x'=1/(Nx)>0$$ for all $$n\in \Bbb N,$$ contrary to 5.4.4.
Let $$n_0=\min \{n\in \Bbb N: n(1/N)\ge x\},$$ so $$(n_0-1)/N
Then $$x<(n_0+1)/N,$$ so $$y\le(n_0+1)/N,$$ otherwise $$(n_0+1)/N\in \Bbb Q\cap (x,y).$$ But then $$2/N=(n_0+1)/N-(n_0-1)/N \ge$$ $$\ge y-(n_0-1)/N>$$ $$> y-x>2/N,$$ a contradiction.
(ii).If $$x then by (i) there exists $$q\in \Bbb Q\cap (-y,-x),$$ so $$\;-q\in \Bbb Q\cap (x,y).$$
(iii).Finally, if $$x<0 then $$0\in \Bbb Q\cap (x,y).$$
The idea behind (i) is that if $$N$$ is large enough then consecutive values of $$1/N,2/N,3/N,...$$ cannot "skip over" the interval $$(x,y).$$
• Thank you for your elaborate answer! I'm surprised that you seem to negate $y > x > 0$. From my understanding, the negation of Proposition 5.4.14 states: There are two real numbers $x < y$ s.t. $\neg(x < q < y)$, for all rational numbers $q$. Nov 14, 2019 at 10:50
• I don't follow you............. Nov 15, 2019 at 2:46
• I got you wrong, sorry! Your strategy is to investigate into the three possible cases when $x < y$. In all of these three cases you are assuming that there is no rational in between $x$ and $y$. Since all of them lead to contradiction, we have that there is a rational in $(x, y)$. Nov 15, 2019 at 6:45
• How can a human being possibly conceive of something like this. Impressive. Nov 15, 2019 at 7:20
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# Transform2D¶
Category: Built-In Types
## Brief Description¶
2D Transformation. 3x2 matrix.
## Member Functions¶
Transform2D Transform2D ( Transform from ) Transform2D Transform2D ( Vector2 x_axis, Vector2 y_axis, Vector2 origin ) Transform2D Transform2D ( float rotation, Vector2 position ) Transform2D affine_inverse ( ) Transform2D basis_xform ( var v ) Transform2D basis_xform_inv ( var v ) Vector2 get_origin ( ) float get_rotation ( ) Vector2 get_scale ( ) Transform2D interpolate_with ( Transform2D transform, float weight ) Transform2D inverse ( ) Transform2D orthonormalized ( ) Transform2D rotated ( float phi ) Transform2D scaled ( Vector2 scale ) Transform2D translated ( Vector2 offset ) Transform2D xform ( var v ) Transform2D xform_inv ( var v )
## Member Variables¶
• Vector2 origin - The transform’s translation offset.
• Vector2 x - The X axis of 2x2 basis matrix containing 2 Vector2s as its columns: X axis and Y axis. These vectors can be interpreted as the basis vectors of local coordinate system traveling with the object.
• Vector2 y - The Y axis of 2x2 basis matrix containing 2 Vector2s as its columns: X axis and Y axis. These vectors can be interpreted as the basis vectors of local coordinate system traveling with the object.
## Description¶
Represents one or many transformations in 2D space such as translation, rotation, or scaling. It consists of a two Vector2 x, y and Vector2 “origin”. It is similar to a 3x2 matrix.
## Member Function Description¶
Constructs the transform from a 3D Transform.
Constructs the transform from 3 Vector2s representing x, y, and origin.
Constructs the transform from a given angle (in radians) and position.
Returns the inverse of the matrix.
Transforms the given vector by this transform’s basis (no translation).
Inverse-transforms the given vector by this transform’s basis (no translation).
Returns the transform’s origin (translation).
Returns the transform’s rotation (in radians).
Returns the scale.
Returns a transform interpolated between this transform and another by a given weight (0-1).
Returns the inverse of the transform, under the assumption that the transformation is composed of rotation and translation (no scaling, use affine_inverse for transforms with scaling).
Returns the transform with the basis orthogonal (90 degrees), and normalized axis vectors.
Rotates the transform by the given angle (in radians).
Scales the transform by the given factor.
Translates the transform by the given offset.
Transforms the given vector “v” by this transform.
Inverse-transforms the given vector “v” by this transform.
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# Chemistry
posted by .
This is for oxidation numbers. I need to know how to show my work.
Ex: for NaClO, I have to find the oxidation number for Cl
I showed my work.
NaClO
+1+x+(-2)=0
+1+x-2=0
x-2=-1
x=-1+2
x=+1
Therefore, the oxidation number for chlorine in NaClO is +1.
Cl^-
I know that the oxidation number is -1 but I have no idea how to show my work.
• Chemistry -
You are correct in both instances. The oxidation number for Cl in Cl^- is -1 because of the rule that says the oxidation number of an atom that is an ion is the charge on the ion. As another example, the oxidation number for Cl in ClO^- is still +1 because +1 on Cl + (-2) on oxygen leaves a -1 charge on the ClO ion.
• Chemistry -
Do you know how to show that using x? Show steps for it like I did with the first example?
• Chemistry -
No, I don't. It works on the others with an x as an unknown because of the rule that says that the sum of all the oxidation states in a compound is zero so you added all the states, plugged in x for the unknown, summed all to zero, and solved for x. i guess you could put
x = unknown oxidation state
charge on ion is -1
so x = -1 but that seems so obvious anyway.
## Similar Questions
1. ### CHEMISTRY
is the oxidation number for platinum larger than the oxidation number for osmium?
2. ### Chemistry
Electron Transfer Theory Write an label the oxidation and reduction half-reaction equations. a) Ni(s) + Cu(NO3)2(aq) -> Cu(s)+ Ni(No3)2(aq) oxidation - Ni(s) -> Ni2+(aq) + 2e- reduction - Cu2+(aq) + 2e- -> Cu(s) b) Pb(s) + …
3. ### Chemistry
Electron Transfer Theory Write an label the oxidation and reduction half-reaction equations. a) Ni(s) + Cu(NO3)2(aq) -> Cu(s)+ Ni(No3)2(aq) oxidation - Ni(s) -> Ni2+(aq) + 2e- reduction - Cu2+(aq) + 2e- -> Cu(s) b) Pb(s) + …
4. ### Chemistry - oxidation numbers
How do you use a Lewis Structure to find the oxidation state of an element. I have this question using the oxidation rule i got +2, however how do i use it with Lewis structure. QUESTION Use the Lewis structure of a thiosulfate ion …
5. ### Chemistry
Use oxidation numbers to identify if this reaction is REDOX reactions. 3NO2(g) + H2O(l) �� 2HNO3(aq) + NO(g) 3NO2 = -3 for N = -2 for o2 H2O = +1 for H2 = -2 for O ----> 2HNO3 = +1 for 2H = -3 for N = -2 for O3 NO …
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8. ### chemistry
Which statement best describes the oxidation numbers of the atoms found in magnesium chloride?
9. ### chemistry
Household bleach contains (NaClO)2 that decomposes in water, NaClO + h2o = Na+ + OH- + Cl2 + O2 If 15ml bleach containing 5.85℅ NaClO is allowed to decomposes completely, then hoe many mL of o2 at 22C and 755mmHg can be liberated. …
10. ### Chemistry
List the grams of bleach solution necessary to provide 0.931g of NaClO(laundry bleach is 8.25% NaClO by weight) and the mL of bleach solution that contains the necessary amount... NaClO=74.442 g/mol. Need 0.931 grams of NaClO. Density …
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# exponential function
(redirected from Exponential functions)
Also found in: Thesaurus, Encyclopedia.
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 exponential function - a function in which an independent variable appears as an exponentexponentialfunction, mapping, mathematical function, single-valued function, map - (mathematics) a mathematical relation such that each element of a given set (the domain of the function) is associated with an element of another set (the range of the function)
Based on WordNet 3.0, Farlex clipart collection. © 2003-2012 Princeton University, Farlex Inc.
References in periodicals archive ?
In short, powerful exponential functions that underpin the speedy expansion of the universe also explain the effectiveness of the genetic algorithm and evolution.
Since using linear functions did not yield the expected results when it came to HDR compression, the focus shifted towards logarithmic and exponential functions.
Thus, the Mittag-Leffler functions can be considered as the analogy of exponential functions for fractional differential operators, since they are the sum of all basis elements.
If the inner solution is an exponential function, then the proposed schemes give the exact solution since the formulae used for the first and second derivative are exact for exponential functions.
Akter and Akbar [44] utilized the modified simple equation method to obtain exact solutions of (11) which were written as fractions of exponential functions. They can be transformed into the tanh and coth functions for which the arbitrary constants are selected appropriately as shown in their paper.
Hyperbolic, logarithmic, power, and exponential functions were used as the prediction models for the variation of settlement with time for both road shoulder and driveway, as shown in Figures 7 and 8.
Narala and Reddy (2012) focused on growth and instability in cultivated area, production and productivity of cotton for the period 1951 - 2011 in India through exponential functions. Reddy et al.
4: Fitting of the semi-major axis values with two different types of exponential functions, i.e., (a) f (n) = [alpha] + [beta][2.sup.n] and (b) f (n) = [alpha] [exp.sup.[beta]n].
There are studies where exponential functions were fitted to the experimental responses [7, 21] and others where a sigmoid function was preferred over the exponential [22, 23], and in some studies the response was directly analyzed or was fitted to other functions [9, 16].
(2) The exponential functions for fresh concrete ([f.sub.ts] = 0.19 [f.sub.cu.sup.3/4], [f.sub.ts] = 0.439 [f.sub.cu.sup.0.55], [f.sub.ts] = 1.34[([f.sub.cu]/10).sup.2/3], [f.sub.ts] = 0.28[f.sub.cu.sup.2/3], and [f.sub.ts] = 0.49[f.sub.cu.sup.0.5]) proposed by building guidelines or literatures only roughly describe the overall tendency of compressive-tensile strength relationship for deteriorated concrete.
Specifically, the sinusoidal, linear, and the exponential functions are used to simulate echo intensity, and then the accumulative rainfalls for both stratiform and convective precipitation are calculated with these two methods
Moreover, the obtained solution can be extended for a wider class of periodic parametric functions including all the functions which can be expressed by the algebraic sum of exponential functions or can be approximated by such functions with sufficiently small error.
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# Video: KS1-M17 • Paper 2 • Question 23
Write the missing number to make this number sentence correct. 9 + 7 − ＿ = 12
01:10
### Video Transcript
Write the missing number to make this number sentence correct. Nine plus seven take away what equals 12?
The number nine is one less than 10. To help us add seven to nine quickly, we could break seven apart into one and six. First, we can add nine plus one which gives us 10. Now, we can add the six. 10 plus six is 16. Nine plus seven is 16. What do we need to take away from 16 to give us 12? 16 take away four equals 12.
Nine plus seven is 16 take away four equals 12. So, nine plus seven minus four is equal to 12. The missing number is number four.
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# Are there more even numbers than odd numbers?
Very simple 'yes-or-no' question, but I can't find the answer anywhere. My gut feeling says the number of odd and even numbers are equal but I managed to write up something that contradicts my intuition. Although I still think my gut feeling is right, I can't find any logical or mathematical errors with my "proof". Can somebody please look it over and tell me which one of me is right?
Statement 1: For every positive odd integer $o$ there is an even integer $e=o+1$.
Statement 2: For every positive integer $n$ there is an negative integer i.e. $-n$.
Conclusion 1: The number of positive odd integers ($O_{positive}$) is equal to the number of positive even integers ($E_{positive}$). If there is a negative equivalent for every positive integer then the number of negative odd integers ($O_{negative}$) is equal to the number of negative even integers ($E_{negative}$). In short: $$O_{positive}=E_{positive}=O_{negative}=E_{negative}$$
Statement 3: The number zero is "neutral" (neither positive nor negative).
Statement 4: The number zero is an even integer.
Conclusion 2:
\begin{align} O_{total} & = O_{positive} + O_{negative} + O_{neutral} \\ & = O_{positive} + O_{negative} + 0 \\ & = O_{positive} + O_{negative} \end{align}
And:
\begin{align} E_{total} & = E_{positive} + E_{negative} + E_{neutral} \\ & = E_{positive} + E_{negative} + 1 \\ \end{align}
So:
\begin{align} E_{total} & = O_{total} + 1 \\ E_{total} & > O_{total}\\ \end{align}
Right? As an engineer I use math daily, but that doesn't make me a mathematician. So please be gentle :)
• If you want to know how infinities like this "work" in the average mathematician's mind, you should have a look at the story of Hilbert's hotel. – Arthur Jan 9 '14 at 14:10
• possible duplicate of Are half of all numbers odd? – Mark S. Jan 10 '14 at 23:57
First things first - there are an infinite number of both even numbers and odd numbers.
It's important to realize that $\infty$ (infinity) is not a number. Therefore it doesn't really make sense to talk about the "number" of even or odd numbers, or to write statements like $E_{\rm even}+1$, because that's assuming that $E_{\rm even}$ is a number that you can sensibly add $1$ to.
However, perhaps surprisingly it does make sense to ask if there are more even numbers than odd numbers. That is, you can compare two infinite quantities, or compare a finite quantity and an infinite quantity, even if you can't meaningfully add and subtract infinite quantities.
They way we define more, less and the same for infinite quantities is as follows. For two collections $A$ and $B$ (say $A$ are the even numbers and $B$ are the odd numbers) we say that
• If you can associate every item in $A$ with a unique item in $B$, and vice versa, then $A$ and $B$ are the same size.
• If you can associate every item in $A$ with a unique item in $B$, but not vice versa, then $B$ is bigger than $A$.
• If you can associate every item in $B$ with a unique item in $A$, but not vice versa, then $A$ is bigger than $B$.
In your case, you can associate every even number $n$ with the odd number $n+1$, and you can associate every odd number $m$ with the even number $m-1$ (assuming 0 is even) so therefore there are just as many odd numbers as even numbers.
This can lead to seemingly paradoxical results, because e.g. you can associate every whole number $n$ with the even number $2n$, and every even number $m$ with the whole number $m/2$, so there are just as many even numbers as whole numbers, even though the even numbers are a subset of the whole numbers.
• Aha, so if I understand correctly, the mistake I made was thinking that: infinity + 1 > infinity? – Jordy Jan 9 '14 at 14:22
• @Jordy The mistake was in thinking that $\infty$ is a number, and that $\infty+1$ is an expression that makes sense. It's easy to see that $\infty$ isn't a number, for here is a list of all the numbers: $\{0,1,2,3,4,\dots\}$. Where is $\infty$ in that list? You can't say "at the end", because the list doesn't have an end! (You also can't say "it's the ninth element in the list, but it's fallen over.") – Chris Taylor Jan 9 '14 at 14:26
• @Jordy This next bit, you'll have to imagine me saying in a stage whisper. Here it is: mathematicians have come up with a way of treating $\infty$ as a number! Shh, don't tell anyone. If you want the secrets, you'll have to learn a bit more math, and then go and read about transfinite ordinals. The smallest infinite ordinal is normally written $\omega$. Confusingly, $1+\omega=\omega$, but $\omega+1>\omega$. – Chris Taylor Jan 9 '14 at 14:28
• An old question and answer, but I disagree with the oft-repeated dogmatism that "$\infty$ is not a number" -- I think it teaches people the wrong idea. Is $i$ a number? Are the quaternions numbers? What about cardinals or ordinals? Furthermore, to me, all of the OP's reasoning is perfectly valid except the final conclusion, i.e. when from $E_{total} = O_{total} + 1$ he or she derives that $E_{total} > O_{total}$. All the rest is valid arithmetic with cardinalities, or alternatively valid arithmetic with infinity. – 6005 Oct 18 '16 at 18:51
• @6005 I think most people at the level of the OP equate "number" with "natural number" or "integer" or "rational number" or maybe "real number", so it is helpful to point out that $\infty$ is not one of these. Sure, there are various number systems in which $\infty$ is a quantity that you can do arithmetic with - but introducing them when someone isn't clear what is and is not a natural number only confuses, rather than clarifies. – Chris Taylor Oct 18 '16 at 20:04
As there is a bijection $$f(x) = x + 1$$ sending any odd number to an even, this shows that the sets have equal size.
Here, I assumed that the natural numbers start with $1$, if they should start with $0$, simply define the same function on the even numbers.
## protected by Asaf Karagila♦Oct 18 '16 at 12:44
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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# Homework Help: Need a little help
1. Mar 31, 2004
### raging_hippo
Ok ive been set the following question:
A piece of optical fibre is 150m long. Two rays of light travel along the fibre. One goes along the axis and the other is reflected off the boundary at the critical angle of 85 degrees. What is the difference in time taken by the two rays.
Speed of light = 3x10^8
Refractive index of core = 1.492
I was wondering if someone could possibly tell me how to find out the time taken by the ray of light that is reflected off the boundary cos i havent got a clue how to do it
2. Mar 31, 2004
### Chen
Take a look at the attachment. Can you tell what the difference or ratio is between X, the distance that the 'straight' ray travels, and D, the distance that the 'refracting' ray travels? You can, using basic trigonometry. Can you then find out how much longer it takes the 'refracting' ray to travel the same horizontal distance that the 'straight' ray travels?
#### Attached Files:
• ###### fiber.gif
File size:
4.8 KB
Views:
171
3. Mar 31, 2004
### ShawnD
That angle is measured from a line perpendicular to the wall of the fibre. When light reflects, the angle of incidence and angle of refraction are the same. Look at this picture to see what's happening
http://myfiles.dyndns.org/math/fibre.jpg
Look at the little triangle I drew on the light ray. V is the speed of the light (n = c/v). Vx is the speed you want, and it looks as if Vx = V sin(85).
Last edited by a moderator: Apr 20, 2017
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# Previous Question Paper of SBI Clerical Exam
1. What is the compound interest accrued on an amount of Rs 25000 in two years at 12 percent per annum?
(1) Rs 6630
(2) Rs 6360
(3) Rs 6260
(4) Rs 6460
(5) None of these
2. Cost of 10 calculators and 12 watches is Rs 11100. What is the cost of 30 calculators and 36 watches?
(1) Rs 33600
(2) Rs 33650
(3) Rs 32600
(4) Can’t be determined
(5) None of these
3. A TV set when sold for Rs 16756 the profit earned is 18%. What is the cost price of TV set?
(1) Rs 14200
(2) Rs 14400
(3) Rs 15200
(4) Rs 14800
(5) None of these
4. Five eighth of a number is equal to 60% of another number. What is the ratio between the first number and the second number respectively?
(1) 13:12
(2) 12:13
(3) 25:24
(4) 24:25
(5) None of these
5. If the fractions 4/5, 2/7, 3/8, 6/13 and 5/11 are arranged in descending order, which one will be second?
(1) 4/5
(2) 2/7
(3) 3/8
(4) 6/13
(5) 5/11
Direction (6 – 10) Study the following table carefully to answer these questions.
Number of employees in different departments of five organization.
6. What is the average number of employees working in Marketing department of all the organizations?
(1) 149
(2) 145
(3) 146
(4) 148
(5) None of these
7. What is the total number of employees working in all the departments of organization B together?
(1) 350
(2) 375
(3) 425
(4) 475
(5) None of these
8. What is the ratio between number of employees from Finance and Marketing department together of organization B and these two departments together of organization D respectively?
(1) 14:9
(2) 9:14
(3) 11:28
(4) 28:11
(5) None of these
9. What is the ratio between the total number of employees from all organization together in HR and Administration departments respectively?
(1) 132:137
(2) 137:132
(3) 122:137
(4) 137:122
(5) None of these
10. Number of employees in IT department of organization C is what percent of the total number of employees in organization C in all the departments together?
(1) 26.5
(2) 25.6
(3) 25.4
(4) 26.4
(5) None of these
Question Answer Q1. 2 Q2. 5 Q3. 1 Q4. 4 Q5. 5 Q6. 1 Q7. 4 Q8. 3 Q9. 5 Q10. 2
Previous question paper of sbi clerical exam. Previous question papers of sbi clerical exam download. Previous question papers of sbi clerical exam pdf. Previous question paper of sbi clerical exam free download. Previous years question paper of sbi clerical exam. Previous question papers of sbi clerks exam. Previous solved question papers of sbi clerical exam.
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# Re: st: Estimating total factor productivity
From "Scott Merryman" To statalist@hsphsun2.harvard.edu Subject Re: st: Estimating total factor productivity Date Tue, 8 Jul 2008 12:47:46 -0500
```On Mon, Jul 7, 2008 at 9:39 AM, dadhinp@yahoo.com <dadhinp@yahoo.com> wrote:
> Dear STATA users
> I want to estimate TFP. Since I have very small size of time series data set (20 obervation) I am using CD production function with CRS
TFP = Total Factor Productivity?
CD = Cobb-Douglas?
CRS= Constant Returns to Scale?
> and perfectly competetive market assumption. But I dont know how to write stata command. Following is the data set.
>
> Year Ln(K/L) Ln(Y/L)
>
> 1981 2.1715 8.4583
> 1982 2.1887 8.5757
> 1983 2.2061 8.5374
> 1984 2.2214 8.6244
> 1985 2.2346 8.7500
> 1986 2.2453 8.7886
> 1987 2.2558 8.8013
> 1988 2.2652 8.8624
> 1989 2.2736 8.9083
> 1990 2.2805 8.9494
> 1991 2.2875 9.0049
> 1992 2.2904 9.0085
> 1993 2.2934 8.9990
> 1994 2.2963 9.0333
> 1995 2.2989 9.0199
> 1996 2.3031 9.0336
> 1997 2.3070 9.0385
> 1998 2.3110 9.0307
> 1999 2.3138 9.0327
> 2000 2.3164 9.0534
>
> K=capital L=labor Ln=log
> I found one paper using the same data set estimated Ln(Y/L)=4.43 +0.46Log(K/L)
I doubt it. Log(Y/L) is about 4 times greater than Log(K/L). The
above regression would generate predicted values of around 5.
> But I get very different result. So could you please to suggest me STATA code to estimate CD function and to find TFP and Change in >TFP.
Perhaps this may be of use:
http://www.stata.com/statalist/archive/2003-12/msg00581.html
Scott
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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https://stats.stackexchange.com/questions/191705/why-are-solution-to-ridge-regression-always-expressed-using-matrix-notation?answertab=active
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# Why are solution to ridge regression always expressed using matrix notation?
Consider the following ridge regression problem: minimize the loss function $\sum_{i=1}^n ||y_i - w^T x_i||_2^2 + \lambda ||w||_2^2$ with respect to the weight vector w. Taking derivative with respect to w, I get $\sum_{i=1}^n 2(y_i - w^T x_i)(-x_i) + 2\lambda w$ which implies $w =(\sum_{i=1}^n (y_i - w^T x_i)(x_i)) / 2\lambda$. Is this wrong? I know that the solution is $(X^TX - \lambda I)^{-1}X^Ty$.
Your derivative is okay. Just remember to put all the $w$-terms on the same side of the equation \eqalign{ \sum_i x_i y_i &= \lambda w + \sum_i x_i x_i^Tw \cr } Then pull $w$ out of the summation, since it's independent of $i$ \eqalign{ \sum_i y_i x_i &= \Big(\lambda I + \sum_i x_ix_i^T\Big)w \cr } At this point, dispose of the summations in favor of matrix notation \eqalign{ X^Ty &= \big(\lambda I + X^TX\big)w \cr } where $x_i$ is the $i^{th}$ column of $X,\,$ and $\,y_i$ is the $i^{th}$ component of $y$.
I think you are asking something different to the question title than in your post body.
Regarding the question in the title:
Most algorithms needed to be coded. Using matrix notation greatly simplifies the coding as well as emphasizes the points that the algorithm can be immediately vectorized (take advantage of linear algebra libraries). Also less indexing usually translates in less typing which is always welcome. :)
Regarding the question in the post:
1. Usually one expresses this cost function with a $\frac{1}{2}$ scalar ahead of it exactly to get rid of the $2$'s in the expression.
2. Usually the expression is $(X^T X + \lambda I)^{-1}X^Ty$ (notice the sign of $\lambda$). You use "$+$" in your original expression but "$-$" afterwards.
3. See the great answers by whuber and Brian Borchers here and here respectively. They use matrix notation to derive the ridge regression problem.
4. You essentially want to take advantage of the following notational property to go from scalar to matrix notation: $\sum_{i}^n (y_i - X_i w)^2 = (y -Xw)^T (y-Xw)$. (Similarly $\lambda ||w||_2^2 = \lambda w^T w$.)
5. No you are not wrong. You are really close; just move a sign out of the summation and you are almost there! You want your derivative to equate $-2X^T(y-Xw) + 2\lambda w$ (in matrix notation).
• I used $w$ to match your notation; please use $\beta$ in general it is much more canonical. – usεr11852 Jan 21 '16 at 6:31
Your solution has $w$ on both sides of the equation, and furthermore, $w$ is inside the summation, which is a problem.
I recommend taking a look at section 2 of Andrew Ng's CS 229 course notes. In short, using matrix math allows you to use properties of the gradient of a trace, which allows for a straightforward derivation. Note that his course notes are for standard least squares, but getting ridge regression is basically the same.
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# Tangent Vector Calculator
This post categorized under Vector and posted on August 10th, 2018.
Where is a parameterization variable is the arc vectorgth and an overdot denotes a derivative with respect to .For a function given parametrically by the tangent vector relative to the point is therefore given byA vector field is a map fRn-Rn that vectorigns each x a vector f(x). Several vector fields are ilvectorrated above. A vector field is uniquely specified by giving its divergence and curl within a region and its normal component over the boundary a result known as Helmholtzs theorem (Arfken 1985 p. 79).Online math calculators and solvers to help calculate and solve problems are included in this site.
You can copy formula code above into the calculator to get result. The formula is very clear Inverse matrix of original square matrix times column vector of (6 3 1 8).Resultant Vector. Head to tail and parallelogram method to calculate resultant vectorAccess everything you need for James Stewart Calculusfrom textbook supplements to web resources and homework hints.
Keplers Second Law. After studying a wealth of planetary data for the motion of the planets about the sun Johannes Kepler proposed three laws of planetary motion.What is tan-1. It is the Inverse Tangent Function. Tangent takes an angle and gives us a ratio Inverse Tangent takes a ratio (like 512) and gives us an angle.Follow us Share this page This section covers Equation of the Tangent Line Equation of the Normal Line Horizontal and Vertical Tangent Lines Tangent Line Approximation
## Unit Vector Equation Of Tangent Space Curve
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A blog is that is all about mathematics and calculators two of my pvectorions in life. [more]
## Questions Use Following Vector Fields Paths R Y Plane G X Y F X Y Path C Part Curve Q
Type or paste a DOI name into the text box. Click Go. Your browser will take you to a Web page (URL) vectorociated with that DOI name. Send questio [more]
## B Find Unit Vector Normal Level Curve F X Y C P C Find Tangent Line Level Curve F X Y C P Q
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## Another Important Consideration Want Ensure Small Changes Parameter Choose Results Small C Q
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## The Direction Of Forces In Two
Physics is a mathematical science. The underlying concepts and principles have a mathematical basis. Throughout the course of our study of physics [more]
## Tangent To Curve Interpolated From Discrete Data
The Tracer V3 a miniature graphic Curve Tracer Tester. This weblog page is the continuation of my weblog page which described the step-by-step dev [more]
## Find Equation Plane Tangent Cone Answers Shown Sure Q
There are three clvectorical problems in Greek mathematics which were extremely important in the development of geometry. These problems were those [more]
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Work on the secant curve in the workvectore below so you dont confuse the secant and the tangent curves. To create a trace of the secant curve youl [more]
## Find Unit Tangent Vector T T Find Set Parametric Equations Line Tangent Space Curve Point Q
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# Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
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P1 : Vasquez-Morrell Assurance specializes in insuring manufacturers.
P2 : Whenever a policyholder makes a claim, a claims adjuster determines the amount that Vasquez-Morrell is obligated to pay.
P3 : Vasquez-Morrell is cutting its staff of claims adjusters by 15 percent.
Goal : To ensure that the company’s ability to handle claims promptly is affected as little as possible by the staff cuts,
Method to achieve the goal : consultants recommend that Vasquez-Morrell lay off those adjusters who now take longest, on average, to complete work on claims assigned to them.
Assumption: method to achieve the goal will not have other bad impact on the goal.
To find the 'statement which most seriously calls into question the consultants’ criterion for selecting the staff to be laid off', we need to hit out assumption. So, we need to show that if there is longer time taking staffs are laid off, it will impact the ability to handle claims promptly.
(A) If the time that Vasquez-Morrell takes to settle claims increases significantly, it could lose business to other insurers.
>> It is strengthener
(B) Supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters.
>> Yeah, if complex claims handlers (and hence longer time taker) are laid off, the time taken to handle those claims would increase further than would decrease. hence, it is a weakener.
(C) At Vasquez-Morrell, no insurance payments are made until a claims adjuster has reached a final determination on the claim.
>> nonsensical
(D) There are no positions at Vasquez-Morrell to which staff currently employed as claims adjusters could be reassigned.
>> there is scarcity of jobs there. laid off employee will not be given work. But how does that matters.
(E) The premiums that Vasquez-Morrell currently charges are no higher than those charged for similar coverage by competitors.
>> nonsensical
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
Vasquez-Morrell Assurance specializes in insuring manufacturers. Whenever a policyholder makes a claim, a claims adjuster determines the amount that Vasquez-Morrell is obligated to pay. Vasquez-Morrell is cutting its staff of claims adjusters by 15 percent. To ensure that the company’s ability to handle claims promptly is affected as little as possible by the staff cuts, consultants recommend that Vasquez-Morrell lay off those adjusters who now take longest, on average, to complete work on claims assigned to them.
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
(A) If the time that Vasquez-Morrell takes to settle claims increases significantly, it could lose business to other insurers.
Strengthener. Out.
(B) Supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters.
Correct. One could falsely believe that those who take the longest are incapable. But, in actuality this choice says those guys are the best at handling complex claims. With these adjusters gone, there will be no one left to handle such claims and that would certainly affect the company’s ability to handle claims properly.
(C) At Vasquez-Morrell, no insurance payments are made until a claims adjuster has reached a final determination on the claim.
Extra supplemental information. Out.
(D) There are no positions at Vasquez-Morrell to which staff currently employed as claims adjusters could be reassigned.
Irrelevant. We need to attack the criterion. The argument could still hold even if these guys could not be reassigned.
(E) The premiums that Vasquez-Morrell currently charges are no higher than those charged for similar coverage by competitors.
Irrelevant.
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
GMATNinja wrote:
So here's what we're being asked for:
Quote:
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
And this is the consultants’ criterion for selecting the staff to be laid off:
Quote:
...consultants recommend that Vasquez-Morrell lay off those adjusters who now take longest, on average, to complete work on claims assigned to them.
And as kumarashok29 mentioned, here's the company's goal, in the exact words of the passage:
Quote:
To ensure that the company’s ability to handle claims promptly is affected as little as possible by the staff cuts...
Kumarashok29's answer is pretty thorough here -- but let's talk a little bit more about B.
Suppose that B is true, and "supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters." If so, then the company's best adjusters might also be among the slowest, since they're assigned the most complex claims. And "the most capable adjusters" are exactly the ones that you don't want to lay off. If the company lays off the best adjusters, it follows that their goal ("handle claims promptly") will be undermined.
I hope this helps!
Hi GMATNinja, I think that although B might be "the best answer" it's barely logically connected to the premises
In order for B to be a strong answer you would totally have to assume that the most capable claim adjusters working the most complex cases take on average more time to solve their cases than their less capable peers working less complex cases, and I don't think this is logically sounded.
I think the most complex cases could not only be solved in roughly the same average time than the rest of the cases, but even in less time since they are handled by the most capable employees. I say this not only out of intuition but from professional experience, since most capable employees tend to be so much productive that their less experienced or capable peers.
Therefore I find the "correct answer" VERY unappealing, do you think I'm missing something here?
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
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giovannisumano wrote:
GMATNinja wrote:
So here's what we're being asked for:
Quote:
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
And this is the consultants’ criterion for selecting the staff to be laid off:
Quote:
...consultants recommend that Vasquez-Morrell lay off those adjusters who now take longest, on average, to complete work on claims assigned to them.
And as kumarashok29 mentioned, here's the company's goal, in the exact words of the passage:
Quote:
To ensure that the company’s ability to handle claims promptly is affected as little as possible by the staff cuts...
Kumarashok29's answer is pretty thorough here -- but let's talk a little bit more about B.
Suppose that B is true, and "supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters." If so, then the company's best adjusters might also be among the slowest, since they're assigned the most complex claims. And "the most capable adjusters" are exactly the ones that you don't want to lay off. If the company lays off the best adjusters, it follows that their goal ("handle claims promptly") will be undermined.
I hope this helps!
Hi GMATNinja, I think that although B might be "the best answer" it's barely logically connected to the premises
In order for B to be a strong answer you would totally have to assume that the most capable claim adjusters working the most complex cases take on average more time to solve their cases than their less capable peers working less complex cases, and I don't think this is logically sounded.
I think the most complex cases could not only be solved in roughly the same average time than the rest of the cases, but even in less time since they are handled by the most capable employees. I say this not only out of intuition but from professional experience, since most capable employees tend to be so much productive that their less experienced or capable peers.
Therefore I find the "correct answer" VERY unappealing, do you think I'm missing something here?
You are not wrong. I remember GmatNinja's words. I remember he repeated many times in his posts that we need to find an option among given 5 because the question asked is MOST.
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
In this question, whatever theory we apply, the fact is that there is no other option except B even that close. B not be a best answer , but it is right answer given to the question , among available 5 options.
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
2
Kudos
giovannisumano wrote:
GMATNinja wrote:
So here's what we're being asked for:
Quote:
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
And this is the consultants’ criterion for selecting the staff to be laid off:
Quote:
...consultants recommend that Vasquez-Morrell lay off those adjusters who now take longest, on average, to complete work on claims assigned to them.
And as kumarashok29 mentioned, here's the company's goal, in the exact words of the passage:
Quote:
To ensure that the company’s ability to handle claims promptly is affected as little as possible by the staff cuts...
Kumarashok29's answer is pretty thorough here -- but let's talk a little bit more about B.
Suppose that B is true, and "supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters." If so, then the company's best adjusters might also be among the slowest, since they're assigned the most complex claims. And "the most capable adjusters" are exactly the ones that you don't want to lay off. If the company lays off the best adjusters, it follows that their goal ("handle claims promptly") will be undermined.
I hope this helps!
Hi GMATNinja, I think that although B might be "the best answer" it's barely logically connected to the premises
In order for B to be a strong answer you would totally have to assume that the most capable claim adjusters working the most complex cases take on average more time to solve their cases than their less capable peers working less complex cases, and I don't think this is logically sounded.
I think the most complex cases could not only be solved in roughly the same average time than the rest of the cases, but even in less time since they are handled by the most capable employees. I say this not only out of intuition but from professional experience, since most capable employees tend to be so much productive that their less experienced or capable peers.
Therefore I find the "correct answer" VERY unappealing, do you think I'm missing something here?
It’s important to keep in mind that the question merely asks for an answer choice that seriously calls into question the consultants’ criterion. This means that we don’t necessarily need an answer choice that would devastate the criterion or prove the criterion wrong. Rather, we just need an answer choice that gives us reason to QUESTION the consultants’ criterion.
The consultants seem to believe that the adjusters who take the longest are the least competent when it comes to the ability to handle claims promptly. This would likely be true if claims are randomly assigned.
But (B) tells us that the most complex claims are assigned to the most capable adjusters. It’s possible that, as you suggest, the most capable adjusters still handle these claims quickly. But it’s also possible that the level of complexity requires additional work and additional time. We don’t know, and we can’t assume either.
So maybe we can’t say that (B) devastates the consultants’ criterion. But because (B) introduces the possibility that the speed at which claims are handled is not indicative of the competence of adjusters, it gives us plenty of reason to QUESTION the consultants’ criterion. For that reason, (B) is the best answer choice.
I hope that helps!
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
When I do this question, I eliminated B because I thought it says that the insurance company assigned complex tasks to all of its adjusters, so each of the adjuster has the same difficulty-level jobs and the more efficient ones are the better ones; And I choose E because I think due to the higher prices of the insurance offered by the company compared to its competitor, it takes more time for customer to ponder whether to sign a contract so it's not the adjuster's fault, it's the company's fault.
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
1
Kudos
yuxiangdreamMBA wrote:
When I do this question, I eliminated B because I thought it says that the insurance company assigned complex tasks to all of its adjusters, so each of the adjuster has the same difficulty-level jobs and the more efficient ones are the better ones; And I choose E because I think due to the higher prices of the insurance offered by the company compared to its competitor, it takes more time for customer to ponder whether to sign a contract so it's not the adjuster's fault, it's the company's fault.
Let's start by considering answer choice (B):
Quote:
Which of the following, if true, most seriously calls into question the consultants’ criterion for selecting the staff to be laid off?
(B) Supervisors at Vasquez-Morrell tend to assign the most complex claims to the most capable adjusters.
As you suggest, the question of which employees tend to receive the most complex claims is key to analyzing (B). Notice that the passage itself doesn't tell us anything about who receives the most complex claims. Given that fact, how does (B) affect the criterion for "selecting the staff to be laid off?"
Well, if the most capable staff tend to get the most complex claims, they may take longer to complete their cases than other adjusters. And if that's the case, laying off adjusters who take the longest to complete their claims would selectively get rid of the most capable employees. Since the goal of the criterion is to "ensure that the company’s ability to handle claims promptly is affected as little as possible," laying off the most capable adjusters would be a bad idea.
So because (B) calls the "criterion for selecting the staff to be laid off" into question, it's correct.
Let's now consider (E):
Quote:
(E) The premiums that Vasquez-Morrell currently charges are no higher than those charged for similar coverage by competitors.
Notice (E) is telling us that Vasquez-Morrell's charges are NO higher -- meaning their premiums are either the same or lower those of other companies. How would this affect the "criterion for selecting the staff to be laid off?"
Keep in mind we're looking for a reason to call into question the criterion of laying off employees who take the longest to complete their claims. In other words, we're trying to see why the average length of time per claim isn't a good way to measure performance.
From that angle, it's hard to see why the cost of the premiums would matter. Since all the employees would be working with the same premiums, it shouldn't affect some more than others, so it doesn't really call into question the criterion for deciding who to lay off. For that reason, we can eliminate (E).
I hope that helps!
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
It is again a confusing question.
Answer B assumes that the complex claims take longer time. But the nature of claim may not always be linked to time it takes. May be the slowest adjusters taking longest time even for simple claims.
bit confused.
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
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AvijitDey wrote:
It is again a confusing question.
Answer B assumes that the complex claims take longer time. But the nature of claim may not always be linked to time it takes. May be the slowest adjusters taking longest time even for simple claims.
bit confused.
Think about it this way: if you're trying to decide whether it's smart to lay off the adjusters who take the longest, you're trying to figure out why this group is taking longer than the others.
Scenario 1: They're just slow workers. Okay, now it makes sense to lay them off.
Scenario 2: They have the hardest cases. Now, it wouldn't make sense to lay them off. The hard cases will likely be time consuming for anyone.
Is it theoretically possible that harder cases don't take longer? Sure. Anything's possible. Maybe brilliant workers are knocking out those complicated cases quickly and the not-so-great workers are taking forever on easier ones. But logically, wouldn't it make sense to conclude that complex cases, on average, take longer than simple ones?
(B) is essentially telling us that scenario 2 is more likely. Worse, if the complex cases take the longest, and those cases are assigned to the best workers, then by firing the folks who take the longest, the company might lose its best workers.
Ironclad? No. You're right about that. But it doesn't have to be ironclad. It just has to call the consultants' recommendation into question, and it certainly does that well enough to hang on to the option.
None of the other answer choices do much of anything to make us doubt the consultants' recommendation, so (B) is our answer.
The takeaway: you're not looking for an answer that is absolutely bulletproof! You're looking for the best of the bunch. So anytime an answer is good, but potentially flawed, keep it around. If you can't find anything better, well, the flawed one is your answer.
I hope that helps!
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Re: Vasquez-Morrell Assurance specializes in insuring manufacturers. Whene [#permalink]
Hello from the GMAT Club VerbalBot!
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# Ring Theory
### Characteristic
For any ring A there is a unique homomorphism Z —> A, where Z denotes the ring of integers. If the kernel of this homomorphism is nZ we say that A has characteristic n.
### Content of a Polynomial
The content of a polynomial over a UFD is the highest common factor of its coefficients.
### Coprime
Two elements x,y of a ring A are coprime if xA+yA = A. In other words, if there exist elements a,b of A such that ax+by = 1.
### Division Algebra
A Division Algebra is a nontrivial ring (not necessarily commutative) in which all nonzero elements are invertible.
### Euclidean Algorithm
The Euclidean algorithm is a method for finding the highest common factor d of two elements x,y in a Euclidean Domain. If y=0 then d is x. If x=0 then d is y. Otherwise, keep subtracting the smaller from the larger until one element is zero.
### Euclidean Domain
A Euclidean Domain is an integral domain with a nonnegative integer valued function d defined on nonzero elements such that for any two elements x,y with y nonzero, there exist elements q,r such that x = yq+r and either r = 0 or d(r) < d(y).
### Fermat's Theorem
An odd prime number is a sum of two squares if and only if it is congruent to 1 modulo 4.
### Field
A field is a nontrivial commutative ring in which every nonzero element is invertible.
### Field of Fractions
Given an integral domain A there exists a field Q(A) containing A as a subring with the property that for every x in Q(A) there are elements a,b of A so that xb=a. Q(A) is unique up to isomorphism. It is the field of fractions of A.
### Finitely Generated Ideal
An ideal J in a commutative ring A is finitely generated if it contains elements x1, ... ,xn for some n such that every element in J has the form a1x1 + ... + anxn for some elements a1, ... ,an of A.
### Gaussian Integer
A Gaussian integer is a complex number whose real and imaginary parts are integers.
### Highest Common Factor
A highest common factor of two elements x,y in a ring A is an element d which is divisible by all those elements of A which divide both x and y. It may not exist.
### Homomorphism
A homomorphism from a ring A to a ring B is a function f: A —> B such that
• f(0) = 0,
• f(1) = 1,
• f(a+a') = f(a) + f(a'),
• f(aa') = f(a)f(a').
### Ideal
An ideal J in a ring A is a subset of A such that
• 0 is in J,
• if x,y are in J then x+y is in J,
• if x is in J then ax and xa are in J for any a in A.
### Image
The image of a homomorphism f: A —> B is the subring of B whose elements have the form f(a), for some a in A.
### Integral Domain
An integral domain is a nontrivial commutative ring in which the product of nonzero elements is nonzero.
### Invertible Element
An element x in a ring A is invertible if there exists an element y in A such that xy = yx = 1. Then y is unique and we write it as x-1.
### Irreducible Element
An element x in an integral domain A is irreducible if it is not invertible and if x = yz implies that either y or z is invertible.
### Isomorphism
An isomorphism of rings is a bijective homomorphism.
### Kernel
The kernel of a homomorphism of rings f: A —> B is the ideal in A consisting of those elements a in A such that f(a) = 0.
### Maximal Ideal
An ideal is maximal if it is proper, but not contained in any other proper ideal.
### Monic Polynomial
A monic polynomial is one whose term of highest degree has 1 as coefficient.
### Noetherian Ring
A ring is Noetherian if it has no infinite strictly increasing chain of ideals. This is equivalent to the condition that all ideals are finitely generated.
### Polynomial Expression
A polynomial expression is one constructed using +, - ,o,1 and multiplication.
### Polynomial Ring
The polynomial ring A[t] over a ring A consists of all the formal polynomials with coefficients from A in an indeterminate symbol t. It has the universal property that homomorphisms A[t] —> B are in bijective correspondence with pairs (f,b) where f: A —> B is a homomorphism and b is an element of B.
### Polynomial Function
A polynomial f(t) in A[t] determines a function A —> A given by a |—> f(a).
### Power Series
The formal power series a0 + a1t + ... + antn + ... with coefficients from a ring A in an indeterminate t consititute a ring A[[t]].
### Prime Element
An element x in an integral domain is prime if, for any product yz in the ring, if x divides yz then either x divides y or x divides z.
### Prime Element
An ideal J in a ring A is prime if the elements of A not in J are closed under multiplication. In other words, if xy in J implies that either x is in J or y is in J.
### Principal Ideal
The Principal ideal generated by an element x in a commutative ring A is the set xA = { xa | a in A } of multiples of x.
### Principal Ideal Domain (PID)
An integral domain in which all ideals are Principal.
### Proper ideal
An ideal is proper if it is not the whole ring.
### Quotient Ring
An ideal J of a ring A gives rise to a surjective homomorphism A —> A/J taking an element a of A to the element (a+J) of A/J. The expression a+J denotes the coset { a+x | x in J }.
### Ring
A ring is a set A with binary operations x+y (addition), xy (multiplication), a unary operation -x (negation) and constants 0, 1 such that
• x + (y + z) = (x + y) + z
• x(yz) = (xy)z
• 0 + x = x + 0 = x
• 1x = x1 = x
• x + y = y + x
• x + (-x) = 0
• x(y + z) = (xy) + (xz)
• (y + z)x = (yx) + (zx)
A ring is commutative if it also satisfies
• xy = yx
### Subring
A subset B of a ring A is a subring if
• 0 and 1 are in B,
• B is closed under addition and subtraction,
• B is closed under multiplication.
### Trivial Ring
A ring is trivial if it has only one element ( so 0 = 1 in such a ring).
### Unique Factorization Domain (UFD)
An integral domain A in which every nonzero noninvertible element can be factorized into a product of irreducible elements p1p2 ... pn in an essentially unique way,
i.e. so that the collection of ideals { p1A, ... , pnA } is unique.
### Related pages in this website
Sets - how to construct sets of integers, reals, etc.
Group - a set closed under one operation.
Fermat's Theorems -- Fermat's Little Theorem, in particular.
The webmaster and author of this Math Help site is Graeme McRae.
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## Peter Barlow: Theory of Numbers
Peter Barlow's first book was An Elementary Investigation of the Theory of Numbers published in 1811. In fact the book had a much longer title which in full read as An Elementary Investigation of the Theory of Numbers with its Application to the Indeterminate and Diophantine Analysis, the Analytical and Geometrical Division of the Circle, and several other Curious Algebraical and Arithmetical Problems by Peter Barlow, The Royal Military Academy, London, 1811. It appears that "Theory of Numbers" in the title of this book is the first occurrence of this phrase in English although Legendre wrote Essai sur la théorie des nombres in 1798. We give below a version of Barlow's Preface to this work. Before this, however, we make a few comments. He claims in the Preface to give a proof of the general case of Fermat's Last Theorem in Chapter 6. He had first published this "proof" in 1810 and had then incorporated the "proof" into his book. Of course, this must contain an error. In fact, his attempt to prove the impossibility of the general case of Fermat's Last Theorem involved the error that a sum of fractions in their lowest terms is not an integer if the denominator of each fraction has a factor not dividing all the remaining denominators. Barlow was, however, able to give a correct proof of the n = 4 case providing an alternative to Fermat's proof of this case. Let us be rather mean and point out another error in the book. On page 299 it is stated that the equation x2 - 5658 y2 = 1 has its smallest solution with the values x=166100725257977318398207998462201324702014613, y= 698253616416770487157775940222021002391003072. These values are not correct. The correct values are x = 1284836351, y = 17081120. Let us also point out that a comment Barlow makes in the book appears to us today to be rather comical. Barlow states that Euler's prime number 231 - 1 = 2147483647, is the greatest that will ever be discovered, for, as they are merely curious without being useful, it is not likely that any person will attempt to find one beyond it. Well, we can't blame Barlow for not being good at predicting the future! If the above makes it appear that Barlow's book is rather poor, this is certainly not so for, despite the errors, it is a well written book which is still worth reading today. We also note that the brief history of number theory and its current status that Barlow gives in the Preface make interesting reading as it shows clearly the position at the beginning of the 19th century.
An Elementary Investigation of the Theory of Numbers
PREFACE
The Theory of Numbers is a subject which has engaged the attention and exercised the talents of many celebrated mathematicians, both ancient and modern; under the first of which classes, may be reckoned Pythagoras and Aristotle, the former of whom is said to have invented our present multiplication table, or the Abacus Pythagoricus of the ancients; though what is alluded to under this designation was probably a much more extensive table than that now in common use: Pythagoras also attributed to numbers certain mystical properties, and seems first to have conceived the idea of what are now termed magic squares. Aristotle, amongst other numerical speculations, noticed the uniformity in almost all nations of dividing numbers into periods of tens, and attempted an explanation of the cause of this singular coincidence upon philosophical principles.
But the earliest regular system of numbers is that given by Euclid in the 7th, 8th, 9th and 10th books of his Elements, which, notwithstanding the embarrassing notation of the Greeks, and the inadequacy of geometry to the investigation of numerical propositions, is still very interesting, and displays, like all the other parts of the same celebrated work, that depth of thought and accuracy of demonstration for which its author is so eminently distinguished.
Archimedes likewise paid particular attention to the powers and properties of numbers as may be seen by consulting his tract entitled "Arenarius," in which some modern Writers have thought they could perceive inculcated the principles of our present system of logarithms; but all that can be allowed on this head is, that the method by which he performed his multiplications and divisions bears a considerable analogy to that which we now commonly employ in the multiplication and division of powers; that is, by the addition and subtraction of their indices.
Before the invention of analysis, however, no very extensive progress could be made in a subject, which required so much generality of investigation; and, accordingly, we find but little was effected in it till the time of Diophantus, whose treatise of algebra contains many interesting problems in the more abstruse parts of this science. But here also, its author had to encounter the difficulties of a complicated notation, and a very deficient analysis, when compared with that of the present period; and, therefore, it cannot be expected that his work should contain a complete investigation of the theory of numbers.
After Diophantus, the subject remained unnoticed, or at least unimproved, till Bachet, a French analyst of considerable reputation, undertook the translation of the abovementioned work into Latin, retaining also the Greek text, which was published by him in 1621, interspersed with many marginal notes of his own, and which may be considered as containing the first germ of our present theory. These were afterwards considerably extended by the celebrated Fermat, in his edition of the same work, published, after his death, in 1670, where we find many of the most elegant theorems in this branch of analysis; but they are generally left without demonstration, an omission which he accounted for by stating, in one of his notes, page 180, that he was himself preparing a treatise on the theory of numbers, which would contain many new and interesting numerical propositions; but, unfortunately, this work never appeared, and most of his theorems remained without demonstration for a considerable time.
At length, the subject was again revived by Euler, Waring, and Lagrange, three of the most eminent analysts of modern times. The former, besides what is contained in the second volume of his "Elements of Algebra," and his "Analysis Infinitorum" has several papers in the Petersburg Acts, in which are given the demonstrations of many of Fermat's theorems. What has been done by Waring on this subject is comprised in chap. v. of his "Meditationes Algebraicae;" and Lagrange, who has greatly extended the theory of numbers, by the invention of many new propositions, has several interesting papers on this head, in the Memoirs of Berlin, besides what are contained in his additions to Euler's Algebra.
It is, however, but lately that this branch of analysis has been reduced into a regular system, a task that was first performed by Legendre, in his "Essai sur la Theorie des Nombres;" and nearly at the same time Gauss published his "Disquisitiones Arithmeticae:" these two works eminently display the talents of their respective authors, and contain a complete development of this interesting theory. The latter, in particular, has opened a new field of inquiry, by the application of the properties of numbers to the solution of binomial equations, of the form
xn - 1 = 0,
on the solution of which depends the division of the circle into n equal parts, as was before known from the Cotesian theorem. This solution he has accomplished in several partial cases, whence the division of the circle into a prime number of equal parts is performed, by the solution of equations of inferior dimensions; and when the prime number is of the form 2n + 1, the same may be done geometrically, a problem that was far from being supposed possible before the publication of the abovementioned performance.
From the foregoing historical sketch, it appears that the writers on this subject are far from being numerous; but the well established celebrity of those, who have investigated its principles, would be of itself sufficient to stamp it with a degree of importance, and to render it worthy of attention. Few persons, it is conceived, will be disposed to consider that a barren subject, which has so much engaged the attention of the above named celebrated writers; in fact, there is no branch of analysis that furnishes a greater variety of interesting truths than the theory of numbers, and it is therefore singular that it should have been so little attended to by English mathematicians. With the exception of what is contained in vol. ii. of Euler's Algebra, and the notes added to the second English edition of that work, there is nothing on this subject to be found in our language.
This circumstance, it is conceived, will be deemed a sufficient apology for the appearance of the present volume; in which, if I have, in certain theorems, availed myself of what others have done on the same subject, yet it is presumed, that it will be found to possess a sufficient degree of novelty, both in matter and arrangement, to exempt me from the imputation of being a mere copyist.
With the exception of a few theorems, what is contained in the first six chapters may be considered as new: in the latter of which will be found a demonstration of Fermat's general theorem, on the impossibility of the indeterminate equation
xn + yn = zn,
for every value of n greater than 2; the leading principle of which I first demonstrated in the Appendix to Euler's Algebra, and afterwards completed in vol. xxvii. of the Philosophical Journal. I also consider as original what is contained in chapter x., with the exception of that part relating to the arithmetic of the Greeks, for which I have been indebted to the Essay of Delambre, subjoined to the French translation of the works of Archimedes. The methods of solving indeterminate equations of the first degree, and of ascertaining, a priori, the number of possible solutions, have likewise some claim to novelty. In the other parts of the work, there will also be found several new theorems, and many former ones differently demonstrated, where simplicity and perspicuity could be attained by such alteration: this is particularly the case in the last chapter relating to Gauss's celebrated theorem on the division of the circle. Perspicuity has, indeed, been one of my principal objects; for this treatise being intended for the instruction and amusement of those who may not possess a very extensive knowledge of analysis, it became necessary to make it as clear and intelligible as possible: but how far I may have succeeded in my design, or what merit may be otherwise due to the performance in general, must be left to the decision of the public.
It only remains now for me to mention a circumstance, that may probably be thought to stand in need of some explanation: it will be perceived that I have introduced two new symbols, the necessity for which, however, will, I trust, appear upon a slight inspection of the work itself: the words of the form of recur very often, and the repetition of them would have been tedious and irksome to the reader, for which reason the double f (ff) is introduced instead of them, but, for the sake of uniformity, it is placed lengthwise thus, ; this, therefore, can scarcely be considered as an innovation of a very important rule laid down by modern analysts, "Not to multiply without necessity the number of mathematical symbols." And the same apology may be made for the introduction of the other sign, for the words divisible by. These characters were adopted on the suggestion of Mr Bonnycastle, Professor of Mathematics in the Royal Military Academy, to whose judgment and experience I have been greatly indebted for many important remarks relating to the present performance, and on various other occasions.
PETER BARLOW
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## Posts
Showing posts from December, 2010
### 1.8 Sustainability and the Fractal
Sustainability and the fractal: This entry follows on from my fractal growth and development entries, which were published earlier. There is never one snowflake alike, but there are snowflakes. The fractal offers insights and helps us understand growth and development, change and evolution. It should also help us understand sustainability . It should clarify what sustainability is. Is it real? Is it possible? Is it an illusion, or is it a delusion? Fractals, by definition, are patterns that show: 'same' but 'different' or regular irregularity - at all scales. Fractals support sustainability in one way but not in another or the way we currently associate sustainability with keeping the environment or the economy today without compromising future generations. It may be that the notion of sustainability is (mathematically) nonsense. Here's why. Fractals and sustainability analysis To see why sustainability is a false statement and doesn't hold—at least in
### On fractals and statistics
Just what's on my mind today: What is the connection between the Mandelbrot set and the bell-shaped normal distribution curve, or any distribution for that matter? I have been thinking about this for some time. I am surprised that fractals are not used to describe patterns. It came to me today while I was biking to work: Fractals are objects that describe the object through all scales; normal distributions or statistics need a parameter to function. For example, Stars are fractal and will not distribute without a parameter: when we add, say, star size, star colour, or distance, we get a distribution. So, I believe there is a very close relationship between the two. What is interesting is that distribution patterns are very fractal and absolutely universal. It is a goal of mine to understand this more, for there is more to it. Update Feb 2020 I wrote the above some time ago, but it is coming back to me now as I know more and have more questions. I am talking about the diffe
### The (fractal) God Illusion - the feeling of being watched.
The (fractal) God Illusion: This applies to the Koch Curve zoom and links to my early blog on Inflation. The following video inspired me for this insight, but the insight actually came to me while waiting in a doctor's surgery - funny enough. This is a great video on fractals and the Mandelbrot set. At 4:20, Arthur C. Clark explains the infinite size of the Mandelbrot set. Two people stand at the edge of the fractal ( the Koch Snowflake), pairing into it - as if it were a tunnel or a computer screen. What if one of the people (the walker) could walk out into the zoom while the other stayed out and watched (the viewer)? For the walker, it would be like walking into a tunnel, and the viewer would see him or her get smaller and smaller as they walk in. Now, what if the walker were to stop, turn, and look behind. What would they see? A tunnel - with the viewer at the entrance, very small, and watching? NO. They would see infinity: they would see the infinite eye of the v
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### How to study your flashcards.
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### 6 Cards in this Set
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divisible by 3 if... the SUM of the digits is divisible by 3 divisible by 4 if... The number that the last 2 digits make are divisible by 4 divisible by 5 if... the numbers in the ones place is 0 or 5 divisible by 6 if... the number is divisible by BOTH 2 and 3 divisible by 9 if... The sum of all digits is divisibe by 9 A prime number its only factors are 1 and itself
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Search 74,244 tutors
# Blogs
## Fractions Blogs
Since it's Thanksgiving week, let's think about pie for a second. No, not mathematical pi, just actual real edible pies. For Thanksgiving I'm in charge of making dessert, so I'll be bringing two pies, one pumpkin and one apple. Let's say that I sliced the apple pie into 12 pieces, and the pumpkin pie, since it held together better, into 18. Fast forward to the end of the evening. My pies were a big hit, and I have almost none left. In fact, all I have is three pieces of apple and four pieces of pumpkin. I want to combine the remaining slices into a single pie pan, so that they take up less space in the fridge. How do I figure out if my remaining pie will fit in one pan? Well, let's start by writing down the remaining amounts of pie in the form of fractions. Remember, one of the definitions of a fraction is parts of a whole, so let's apply that definition to figure out our starting fractions. The apple pie was cut into 12 pieces, and we have three... read more
Come with me on a journey of division. I have here a bag of M&Ms, which you and I and two of your friends want to share equally. I'm going to pour the bag out on the table and split it into four equal piles. For this example, “one bag” is our whole, and the best number to represent that whole would be the number of M&Ms in the bag. Let's say there were 32. If I split those 32 M&Ms into four equal piles and asked you how many were in one pile, you could certainly just count them. But a quicker way would be to take that 32 and divide it by the number of piles I'd made, which in this case is 4. You'd probably write that as: 32 ÷ 4 = 8 So there are 8 candies in each pile. Seems easy enough with a large number of M&Ms, right? But what if there were less candies – what if our “whole” was less than the entire bag? Well, for a while we'd be okay – if there were 16, for example, we'd do the same thing and come up with piles of 4 instead... read more
By far, one of the most difficult concepts in elementary mathematics is fractions...and it is all our fault. One of the major misconceptions among many education systems was that early exposure to fractions would help students learn them. This meant attempting to introduce fractions before students could even multiply or divide. You have no idea the trauma this has had among decades of students. Education systems created self-induced math anxiety. For years I had to address what I can only describe as fraction PTSD. I had talented Algebra students immediately clam up if the problem had a fraction. Now as a teacher I of course did my job and we spent time trying to get ourselves comfortable with fractions but in the back of my mind I knew I was using valuable class time to address an issue that simply shouldn't even rear it's ugly head in Algebra. But every year it was there. Students were crying, parents were crying, and teachers were crying over the fraction... read more
By far, one of the most difficult concepts in elementary mathematics is fractions...and it is all our fault. One of the major misconceptions among many education systems was that early exposure to fractions would help students learn them. This meant attempting to introduce fractions before students could even multiply or divide. You have no idea the trauma this has had among decades of students. Education systems created self-induced math anxiety. For years I had to address what I can only describe as fraction PTSD. I had talented Algebra students immediately clam up if the problem had a fraction. Now as a teacher I of course did my job and we spent time trying to get ourselves comfortable with fractions but in the back of my mind I knew I was using valuable class time to address an issue that simply shouldn't even rear it's ugly head in Algebra. But every year it was there. Students were crying, parents were crying, and teachers were crying over the fraction... read more
In elementary school we are taught to add/subtract fractions in a way that, quite frankly, is a BAD WAY of adding/subtracting fractions! If you don't know what I'm talking about, here's a short review... We have a fraction, say (1/2). Let's add a third. We have (1/2) + (1/3). So, what's the first step? Well, in elementary school you were probably taught to cross multiply. Let's try it We first get (1)(3) = 3 and (1)(2) = 2 Adding these together gets us 3 + 2 = 5 Now we multiply the denominators, getting (2)(3) = 6 We can now put these two numbers together: (5/6) Though this method works, it's not the best way to go about adding two different fractions. First off, why the heck does this work?! Though it seems like magic, there's a method. First, we cross multiplied. Putting this in a way that shows the whole expression gives (3)(1) + (2)(1)... read more
When working with fractions, I find it effective to require students to convert each fraction that we work with to its decimal equivalent, to convert that decimal equivalent back into the original fraction, to convert that decimal into its percentage equivalent, to work a simple percentage problem using that percentage and finally to work the same problem using the initial fraction. This comprehensive method helps students to see the relationships between fractions, decimals and percentages in a holistic way and to promote the necessary skills in each element.
Buckle up readers, it's Trig time! Trigonometry can be scary to many students, and in my opinion, a lot of that is because one of the most confusing concepts in trigonometry occurs right at the very beginning, in the form of the Unit Circle and Radians. Let's start at the beginning. Give yourself a circle with a radius of 1. Now center that circle on the origin of a coordinate plane, so that the line of the circle itself passes through the points (1,0) (0,1) (-1,0) and (0, -1). Got that? Now, this circle is referred to as the Unit Circle, because the radius is one unit and it is therefore easier for us to do various manipulations and calculations with it. Now choose any point on the circle (we'll call the coordinates of that point (x,y)), draw the radius to it (which will still be a length of 1), and drop a line back perpendicular to one of the axes. Do that and you'll have a right triangle with the... read more
Yes, there is only one way. Let's say for example that we have a fraction of 2/3. Now, the bottom number is the denominator which means the number of equal parts into which a whole circle most specifically is divided. So the circle is divided into 3 equal parts. On the other hand, the top number is the numerator which means how many equal parts out of all of them are lightly shaded inside the circle. So 2 out of all 3 equal parts of the circle are lightly shaded. Now, the only way to change the number of equal parts without affecting the fraction value is to multiply it by any number you want which will also change the numerator. So let's say for example that in the fraction of 2/3, if you wanted to divide each of those 3 equal parts into 2 further equal parts, you will have a new number of equal parts which is 6 (3*2=6). This will affect the numerator 2 as well since this is included in the total number of equal parts, so each of... read more
One of the common challenges for many Algebra students is forgetting important concepts from Pre-Algebra. So many students complain that they never fully learned fractions, decimals and percentages or ratios, rates and applying math to word problems. Without solid memorization of multiplication & division tables, factoring and simplifying are much more difficult. A better understanding of the basics, including learning different methods and shortcuts, can not only boost confidence but can improve grades and SAT scores. Spend time with a tutor or use different websites to review topics from previous years. It will help, exponentially!
Mothers generally know this trick. It works especially well with food children do not want to eat, but must. Tell the child that he only has to eat 3 bites. Then let the child eat just that many bites, which you can count together if you like. Increase the number of bites as the child learns to count. Alter the exercise with how many peas can be left on the plate; how many bites can be exchanged for another food or desert, and other tricks. As the child learns fractions, work with eating (or leaving on the plate) 1/2 of the food, or 3/4, or other familiar fractions. Fill glasses 1/2 full of their favorite beverage and offer another 1/3 or 3/4 or so more when he drinks the first fractional amount. Conversations and expectations and games like this applied to food and drink, picking up toys, helping out around the house, etc., help children from ages 5-7 develop their number sense. These tricks can be used just about every day for a few minutes a day--longer only if... read more
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Helper V
## Historic and Future data in one measure based on current date, and correct row / column totals?
Hello,
I have a calculation that is showing the historic value or the future value based on the current month, this is also giving me correct row totals. However, the column total is not correct when summing across all the months.
Is there a different way to achieve what I've done? I think it's the 'max' in the first calculation that is causing the problems when aggregated over the entire time period.
Here is my current measure:
``````VAR ForecastValue =
IF (
MAX ( 'Calendar'[YearWeekNr] ) < [Today (Week)],
[Sales (MU)],
[Forecast Quantity (MU)]
) + 0
RETURN
IF (
HASONEVALUE ( 'Calendar'[YearWeekNr] ),
ForecastValue,
if( HASONEVALUE('Product'[pt_part]), SUMX ( summarize ('ISS-S0 & RCT-SOR & ORD-SO - 600 (Transactions)','Product'[pt_part],'Calendar'[YearWeekNr]), ForecastValue ),
SUMX ( VALUES ( 'Calendar'[YearWeekNr] ), ForecastValue )
))``````
Many thanks for any help and assistance you can offer.
Kind Regards,
Dayna
1 ACCEPTED SOLUTION
Community Champion
@Dayna hey, check this great video for reference:
3 REPLIES 3
Helper V
Great video, thanks!!
Community Champion
@Dayna my pleasure 🙂
Hey, check out my showcase report - got some high level stuff there 🙂
https://community.powerbi.com/t5/Data-Stories-Gallery/SpartaBI-Feat-Contoso-100K/td-p/2449543
Give it a thumbs up over there if you liked it 🙂
Community Champion
@Dayna hey, check this great video for reference:
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### DIFFERENTIAL GEOMETRY COURSE
8. EVOLVENT AND EVOLUTE
Definition 76: Let C be a curve. Let K be a curve such that the tangents to C are normals of K. Then we call K an evolvent of C, and C an evolute of K.
Proposition 77: Every curve has ∞1 evolvents and ∞1 evolutes.
Proof : Use a parametrisation of C according to arc length: x(s). Then an evolvent K has parametrisation y(s)=x(s)+λ(s)t(s) (s is not arc length of K).
Now t(s) must be perpendicular to y '(s), where y '(s) = x .(s) + λ .(s)t(s) + λ(s)κ(s)n(s) = (1 + λ .(s))t(s) + λ(s)κ(s)n(s).
So 1 + λ .(s) = 0 and s + λ(s) = c. Then we find for each constant c an evolvent Kc with parametrisation y(s) = x(s) + (c-s)t(s).
On the other hand, use a parametrisation K according to arc length: x(s). A normal vector has the form n(s)cos(φ(s)) + b(s)sin(φ(s)).
Then a corresponding evolute C has the following parametrisation (where s is not arc length): y(s)=x(s)+λ(s)(n(s)cos(φ(s)) + b(s)sin(φ(s))).
Now y '(s) must be a scalar multiple of n(s)cos(φ(s)) + b(s)sin(φ(s)), where
y '(s) = t + λ .(n cos(φ) + b sin(φ)) + λ(n .cos(φ) + b .sin(φ) - n sin(φ)φ . + b cos(φ)φ .) =
t(1-λκ cos(φ)) + n .cos(φ) - λτ sin(φ) - λ sin(φ)φ .) + b .sin(φ) + λτ cos(φ) + λ cos(φ)φ .).
So the following must hold: first, λ = (κ cos(φ))-1, and, second,
Hence φ . = - τ, so φ = φo - ∫0s τ(u) du.
With each choice of φo we find an evolute C.
Problem 78:
Show that we get the normals of a curve K corresponding to an evolute C from the normals of another evolute C' by rotating each of them in its normal plane over a fixed angle.
Furthermore, show that the contact point y(s) of the evolute C lies on the curvature axis of the point x(s) of K.
Finally show that the principal normals of a curve K are the tangents to a curve C (so that C is the "envelope" of these tangents) if and only if τ=0, so if K is planar. Then C is the locus of the centers of curvature of K.
Problem 79: Show that the tangents to a circular helix intersect each plane perpendicular to the axis in the points of an evolvent of the circle that is the intersection of this plane and the cilinder on which the helix is lying.
Problem 80: Given a cycloid (see 18, 28), determine its evolute, and show this is also a cycloid.
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Elementary Math Activities
Get the kids interested in learning math with fun elementary math activities. Check out our printable Equation Search on this page for starters.
Equation Search
Skills: Addition, Subtraction, Multiplication, or Division
What You Need:
Objective: To circle as many correct equations as possible on the equation worksheet.
Playing The Game:
• Give a copy of the equation sheet to each player. If playing in teams, give one equation sheet to each team. Print Multplication Equation Search
• Players will use their pencils to circle as many correct equations as possible on their number grid.Tip: It's a good idea to use the colored pencils when a player circles a correct equation that overlaps another circle equation.
• The numbers that are circled in an equation must be right next to each other.
• Let the players work for a certain amount of time and when the time is up have them put their their pencils down.
• Each player counts the circle equations on their sheet.
• The winner is the player that has the highest number of circled equations.Tip: After each player counts the number of equations they have circled, it is a good idea to make sure to exchange the sheets so that the other player/s can double check the circled equations to make sure they are correct.
Extensions:
* You can tailor this activity to the skill level of your students. For example, if a student is needing practice with addition, have them circle all addition equations.
* Mixing different kinds of equations, for example, addition, subtraction, and multiplication equations on the same worksheet gives kids practice on more than one skill.
Go to main Elementary Math Games page
Return from Elementary Math Activities To Learn With Math Games Home
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https://www.financekarma.com/question/what-is-payment-period-and-why-is-it-equal-100-in-this-mortgage-table/
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# What is payment period, and why is it equal 100 in this mortgage table?
Here’s the table:
I think the payment period is a month, but I don’t understand why it’s 100.
The image is taken from this Excel exercise: http://web.utk.edu/~dhouston/excel/exer3.pdf
#### One thought on “What is payment period, and why is it equal 100 in this mortgage table?”
1. Hart CO
It is the 100th month of the mortgage term.
Using the `PMT()` function in Excel I generated this amortization schedule with the variables in your question:
For some reason rather than showing the whole amortization table they chose to let you calculate for any given month what the split between principal and interest would be.
Edit: Added the formula text in the next cell over on the top portion to show interest percentage used and `PMT()` function format.
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https://www.quantumstudy.com/a-conservative-force-of-10%E2%88%9A2-n-acting-on-a-particle-of-mass-m-in-xy-plane-making-an-angle-45-with-x-axis-as-shown-in-the-given-figure-find-work-done-by-conservative/
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# A conservative force of 10√2 N acting on a particle of mass m in XY – plane making an angle 45° ….
Q: A conservative force of 10√2 N acting on a particle of mass m in XY – plane making an angle 45° with X – axis as shown in the given figure. find work done by conservative force between origin and point A (1 m, 1 m , 0).
(a) Zero
(b) 10 J
(c) 20 J
(d) 30 J
Ans: (c)
Sol: $\displaystyle \vec{F} = 10\sqrt{2}cos45 \hat{i} + 10\sqrt{2}sin45 \hat{j}$
$\displaystyle \vec{F} = 10\sqrt{2}\frac{1}{\sqrt{2}} \hat{i} + 10\sqrt{2}\frac{1}{\sqrt{2}}\hat{j}$
$\displaystyle \vec{F} = 10 \hat{i} + 10 \hat{j}$
displacement vector , $\displaystyle \vec{s} = (1 \hat{i} + 1 \hat{j} + 0\hat{k}) – (0 \hat{i} + 0 \hat{j} + 0\hat{k})$
$\displaystyle \vec{s} = 1 \hat{i} + 1 \hat{j}$
$\displaystyle W = \vec{F}. \vec{s} = (10 \hat{i} + 10 \hat{j}).(1 \hat{i} + 1 \hat{j})$
W = 20 J
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# Question:Curve fitting
## Question:Curve fitting
Maple
Geetings!
I try to fit curve through set of points that I read from a graph. Then use equation of this curve to define a function function. Only bspline approximation gave acceptable result. But how can I use it in defining a function?
XvalsCf := [0, 5, 10, 25, 50, 100, 140, 200, 250, 300, 400]
valsCf := [0.5e-2, 0.4e-2, 0.3e-2, 0.26e-2, 0.23e-2, 0.21e-2, 0.2e-2, 0.19e-2, 0.18e-2, 0.175e-2, 0.15e-2]
Another issue is in "surface fitting". Graph shows propeller efficiency curves. I'd like to get equation of surface spread on those curves. Axis x is an angle, axis y V/nD and z is efficiency.
Again using spline/bspline approximation is nesssary to get accurate results.
I will use this efficiecny surface function equation in drive optimization .
Is creation of such surface/function possible in maple.
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# which.min {base}
Where is the Min() or Max() or first TRUE or FALSE ?
Package:
base
Version:
R 3.0.2
### Description
Determines the location, i.e., index of the (first) minimum or maximum of a numeric (or logical) vector.
For a logical vector `x`, `which.min(x)` and `which.max(x)` return the index of the first `FALSE` or `TRUE`, respectively.
### Usage
```which.min(x)
which.max(x)
```
### Arguments
x
numeric (integer or double) vector, whose `min` or `max` is searched for.
### Values
Missing and `NaN` values are discarded.
an `integer` of length 1 or 0 (iff `x` has no non-`NA`s), giving the index of the first minimum or maximum respectively of `x`.
If this extremum is unique (or empty), the results are the same as (but more efficient than) `which(x == min(x))` or `which(x == max(x))` respectively.
`which`, `max.col`, `max`, etc.
Use `arrayInd()`, if you need array/matrix indices instead of 1D vector ones.
`which.is.max` in package nnet differs in breaking ties at random (and having a ‘fuzz’ in the definition of ties).
### Examples
```x <- c(1:4, 0:5, 11)
which.min(x)
which.max(x)
## it *does* work with NA's present, by discarding them:
presidents[1:30]
range(presidents, na.rm = TRUE)
which.min(presidents) # 28
which.max(presidents) # 2
## Find the first occurrence, i.e. the first TRUE:
x <- rpois(10000, lambda = 10); x[sample.int(50, 20)] <- NA
## where is the first value >= 20 ?
which.max(x >= 20)```
### Author(s)
Martin Maechler
Documentation reproduced from R 3.0.2. License: GPL-2.
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# Is continuous function finite-valued in $R^n$?
Let $f$ be a continuous function defined on $\mathbb{R}^n$. The range of $f$ is in the extended real numbers. Is $f<\infty \ \forall x\in \mathbb{R}^n$ ? And why?
• Is this really the question you mean to ask? If so, the answer is clearly yes: the range of the function is the set of real numbers, and $\infty$ is not a real number. All real numbers are finite. The function might be unbounded - so that it takes on values as large as you might imagine. Mar 24, 2016 at 13:49
• What if the range is in extended real numbers, as we are studying measure theory. Mar 24, 2016 at 13:51
• Then what about the constant function $f:x\mapsto\infty$?
– MPW
Mar 24, 2016 at 13:55
• Do you have a definition of what it means to be continuous at a point $x_0$ such that $f(x_0) = \infty$? Mar 24, 2016 at 13:56
• Then you need to specify the topology on the extended real numbers. In any case the function that has constant value $\infty$ will be continuous. Mar 24, 2016 at 13:56
In the "usual" topology on the extended real numbers $[-\infty, \infty]$, a neighborhood of $\infty$ is a set containing some interval $(a, \infty] = (a, \infty) \cup \{\infty\}$ with $a$ real, and similarly for neighborhoods of $-\infty$.
If that's true in your setting, $[-\infty, \infty]$ is homeomorphic to a closed, bounded interval of real numbers. For instance, the hyperbolic tangent function $\tanh$ is a homeomorphism from $[-\infty, \infty]$ to $[-1, 1]$. Consequently, asking whether a continuous function can achieve the value $\infty$ is no more mysterious than asking whether a continuous, real-valued function can achieve an absolute maximum.
As MPW and Ethan Bolker note, the constant function with value $\infty$ is continuous. Non-constant continuous functions can achieve the values $\infty$ and/or $-\infty$, as well. For example, if $n = 1$, then
• $f(x) = \frac{1}{x^{2}}$ (extended by $f(0) = \infty$) is continuous, since $\lim\limits_{x \to 0} \frac{1}{x^{2}} = \infty$.
• $f(x) = \frac{1}{x}$ (extended by $f(0) = \infty$) is not continuous, since $\lim\limits_{x \to 0^{\pm}} \frac{1}{x} = \pm\infty$ (i.e., the one-sided limits exist as extended real numbers, but are not equal).
And so forth.
• Hi, I am wondering how can I show $f(x)=\frac 1 {x^2}$ is continuous at $x=0$? Mar 24, 2016 at 14:39
• The customary calculus definition of "$f(x) \to \infty$ as $x \to c$" reads: "For every real number $a$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $f(x) > a$." The definition of continuity can then be extended to read "If $f(c) = \infty$, then $f$ is continuous at $c$ if $f(x) \to \infty$ as $x \to c$." Mar 24, 2016 at 15:43
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VIEWS: 4 PAGES: 9
• pg 1
``` A mortgage calculator is a program used to
help home buyers establish their monthly
payment on their mortgage using variables
such as principal, interest rate, and term.
Mortgage calculators are, thus, essential tools
for home buyers. Here are their different uses
and their various types.
remortgage calculator
During the early process of applying for a mortgage, you will find that a
mortgage calculator is a very valuable tool you can use to: - Determine
the amount of mortgage and the price of a house you can afford based
payments based on loan amount, interest rates and other loan terms -
Compare the costs or real interest rates between several different
mortgage loans - Compute extra payments on your monthly mortgage
that enable you to pay off your mortgage faster - Calculate your
payments on debt consolidation mortgage loans to get an idea of your
monthly savings - Check how you can refinance the loans you have by
working out the amount you can afford to borrow and exactly how much
your repayments are going to be using time scales and interest rates -
Make comparisons with other mortgage products, both fixed and
adjustable - Make amortization schedules and tables using the amount
and interest as basis - Calculate when it is sensible to refinance your
home Therefore, by using a mortgage calculator, you can most certainly
get good and precise information about the actual mortgage loan All you
have to do is to enter the required figures in the mortgage calculator
provided in most lender web sites
Make sure you're getting a lot of options by using another company's
mortgage calculator By doing so, you will find out that there are different
choices for a loan in other companies
To find the best remortgage calculator one, you have to make a number
of searches and several calculations using the appropriate mortgage
calculator There are different types of mortgage calculator
Here are some of them: Adjustable Rate Mortgage Calculator -
Determines the monthly mortgage payments on an adjustable rate
mortgage (ARM) - Evaluates the maximum mortgage payment you can
expect if your ARM rate has reached its highest point - Calculates the
total amount of interest you will be paying over the term of the loan,
together with your total payment and amount ARM vs Fixed Rate
Mortgage Calculator - Compares the monthly mortgage payments for
each kind of loan - Evaluates fixed rate mortgage payments to both fully
amortizing ARMs and interest-only ARMs Interest Only Mortgage
Calculator - Determines the amortization schedule for an interest-only
mortgage - Assesses how principal payments made to lessen the
mortgage loan balance will influence the amortization schedule Maximum
Mortgage Calculator - Allows you to key in your monthly income and
monthly obligations so you can calculate the maximum monthly mortgage
payment and mortgage amount you can afford - Helps you determine the
way interest rates can affect the mortgage amount you can afford With
the proper use of a mortgage calculator, you are assured of making
sound mortgage loan computations
These calculations, in turn, are valuable in helping you come up with
better mortgage loan decisions
remortgage calculator
```
To top
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# Quiz #2 Flashcards Preview
## Fall '12- Phonetics exam #1 > Quiz #2 > Flashcards
Flashcards in Quiz #2 Deck (63)
1
Q
True or false:
the lower the fundamental frequency is the more frequencies will appear in the harmonic series?
A
True
2
Q
What is the equation for a wavelength?
A
See physical flashcards
3
Q
Can we measure a wavelength from any point in one cycle to the same point in the next cycle?
A
Yes
4
Q
Frequency is directly related to…?
A
Pitch
5
Q
What is harmonics?
A
All the tones created by complex vibrations
6
Q
What is periodic?
A
The pattern of vibration repeating itself
7
Q
True or false:
a simple sine wave is always periodic?
A
True
8
Q
What is a complex wave?
A
Multiple frequencies
9
Q
What can young and healthy ears detect vibrations as low as and as high as?
A
20 Hz and 20,000 Hz.
Speech is heard at 100 Hz to 5000 Hz
10
Q
What does a spectral plot give us?
A
Amplitude on the Y axis and frequency on the X axis
11
Q
What are the physical properties of Sound?
A
• Time
• Intensity/amplitude
• Frequency
• period
• Velocity
12
Q
How do you calculate the harmonic series when the sound is 200 Hz and the fundamental frequency equals 200
A
first harmonic = 200x1= 200Hz
second harmonic 200×2 = 400 Hz
third harmonic is 200×3 = 600 Hz
And so on
13
Q
How can a periodic wave be distinguished from an aperiodic wave?
A
Based on the mathematical relationships among the frequencies of their components. Aperiodic won’t have a pattern or a mathematical relationship
14
Q
What is the fundamental frequency?
A
Lowest frequency there is in a waveform/complex tone
15
Q
What is a phase relationship?
A
Where waveforms meet
16
Q
What is amplitude measured in?
A
Decibels (dB)
17
Q
When would you expect to see higher frequencies that are farther apart?
A
18
Q
For a child with a high pitch will they have more frequencies in the harmonic series or less
A
Less. Fewer frequencies and are farther apart because you start with a higher fundamental freq.
19
Q
Are intensity and loudness linearly related?
A
No
20
Q
What are boundary behaviors?
A
Sound bouncing off the object and sound returning
21
Q
True or false:
as frequencies get higher, it takes a larger change in Hz to cause a change in sensation of Pitch?
A
True
22
Q
Can a simple tone/waveform be aperiodic?
A
No
23
Q
What is always the first harmonic in the series?
A
Fundamental frequency
24
Q
Does a harmonic need to include all the harmonic series?
A
No, it can skip, but it must be a mathematical relationship between odds and evens
25
Q
What is a constructive pressure wave?
A
When crest meets crest or troughs meet trough
In phase and 90% out of phase
26
Q
What are the perceptual properties of sound?
A
• Loudness
- Pitch
27
Q
What is pitch measured?
A
Mels
28
Q
What is Hz measured in?
A
Cycles per second
29
Q
What is a complex wave?
A
Different frequencies
30
Q
What is “in phase”?
A
When troughs and peaks of the waveform are in the same wave
31
Q
What is reverberate?
A
Sound bounces back and forth. Echo
32
Q
What is a wave pattern of vibrations, no matter how complex and how often it repeats itself?
A
Complex periodic soundwave
33
Q
What is a destructive pressure wave?
A
When crest meets trough (silence)
34
Q
If the vibration of a wave is random and has no repeatable pattern it is…?
A
A complex aperiodic soundwave
35
Q
```Sound = tone
Wave = how it travels
Waveforms = when we draw it out```
A
.
36
Q
Is loudness perceptual or physical?
A
Perceptual
It is perceived that if the intensity is higher than it is louder
37
Q
What type of tone will you get with two pure tones of the same frequency?
A
Pure tone/sine wave
38
Q
What are waves characteristics?
A
• Boundary behaviors
• Interference patterns
• pure tones
39
Q
What type of tone will you get with two pure tones of different frequencies?
A
A complex tone
40
Q
What is the velocity for meters?
A
344 m/seconds
41
Q
What is velocity measured in feet?
A
1130 feet/second
42
Q
What is the resulting wave of two signals of the same frequency and 180° out of phase?
A
The result is silence. Each particle has been subjected to equal forces acting in opposing directions
43
Q
Proper acoustic design helps with what?
A
Sound travel through space ( e.g. Concert halls)
44
Q
What is the Fourier analysis?
A
A different sort of displaying called a spectral plot. That allows us to indicate the frequency and amplitude of each harmonic in a complex periodic wave
45
Q
What two factors does wavelength depends upon?
A
• The velocity of a soundwave (c)
• The frequency of the vibration (f)
See the equation for wavelength
46
Q
What is a Pure tone?
A
• A single periodic frequency vibrating in simple harmonic motion/creates a sign wave.
• A vibration that repeats itself at a constant number of cycles per second.
• It is unnatural
47
Q
What is the resulting wave of two signals of the same frequency for “in phase”, 90% out of phase and 180% out of phase?
A
They will have an amplitude equal to the sum of the amplitudes of each wave
48
Q
What is considered noise?
A
An aperiodic complex tone of two or more components frequencies not harmonically related. No fundamental frequency. No harmonics.
49
Q
What does a wavelength represent?
A
The velocity over the frequency. It represents one cycle.
50
Q
What is aperiodic?
A
No repeatable patterns of vibration in the soundwave
51
Q
We can determine the fundamental frequency of the sound the waveform represents by…?
A
• Counting the number of times it’s patterned is repeated per cycle
• We can then calculate the frequency of the individual harmonics
• We cannot discover the amplitudes of individual harmonics
52
Q
What is pitch?
A
• a sensation
- a perception by listener when frequencies change
53
Q
What is a harmonic series?
A
Each tone created in a complex vibration is called “harmonic”. The whole set of tones is called harmonic series
54
Q
What are interference patterns?
A
• Signals of the same frequency can interfere with each other
• Same frequency, two sources or signal is reflected from a barrier and competes with itself
55
Q
Loudness is directly related to what?
A
Amplitude/intensity
56
Q
What is loudness measured by?
A
Phones and sones
57
Q
The amplitude of the vibration is…?
A
The extent of particle displacement, which is an indication of the intensity or power of the sound
58
Q
What is a sine wave?
A
A pure tone. Single frequency
59
Q
What are the three graphic representations for analysis of sound?
A
• Waveform
• Spectral plot
• Spectrogram
60
Q
Low-frequency wave has more or less vibratory cycles/second?
A
More
61
Q
Do lower frequencies have a bigger or smaller wavelength?
A
Bigger
62
Q
High-frequency waves have more or less vibratory cycles per second?
A
Less
63
Q
Higher frequency sounds have shorter or longer wavelength
A
Shorter/smaller
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# Resolution Screen Size Relativity
Discussion in 'Beginners, General Questions' started by Neil McCaulley, Jan 3, 2005.
1. ### Neil McCaulley Stunt Coordinator
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Hello all,
This is something that has been bugging me for some time. Maybe someone out there can explain it to me.
I have a 46" Widescreen Hitachi HDTV. It's screen resolution is 1080 vertical lines of resolution and 720 horizontal lines. But I also noticed that a 55" TV has the exact same amount of resolution. How can this be? If the screen size gets bigger, does this not mean that there should be a relative increase in the amount of vertical lines of resolution? The only thing I can think of that makes sense is that the physical width of the resolution lines are wider in a bigger set, than in a smaller set. Again, would this not mean that the picture quality gets degraded the larger you get because the lines of resolution are not as fine as a smaller set?
Someone out there please explain this to me.
2. ### Jack Briggs Executive Producer
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Neil, the original broadcast is at 1080 lines of interlaced resolution. A larger screen size will not increase resolution per se. JB
3. ### Neil McCaulley Stunt Coordinator
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Thanks Jack. So am I correct in assuming that a larger screen does indeed have more physical lines of resolution, but the picture quality will not be better than a smaller screen because it is up to the BROADCAST (source material) to determine the resolution. Kind of like if I watch a movie in 480p. That just means that I am only viewing 480 lines of my available 1080 lines. Correct?
4. ### Jeff Gatie Lead Actor
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Yes, the lines or pixels (depending on the display) increase in relative size as the screen size gets bigger. But you also view from farther back as the screen gets bigger, so the relative resolution is the same. This is why we have optimum viewing distances for any given size screen. Also, this is why we can sit closer (thus having a larger field of view, i.e. larger relative screen size) for higher resolution HD content as opposed to regular non-HD content.
5. ### Neil McCaulley Stunt Coordinator
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Hey Jeff,
So a larger widescreen TV DOES NOT contain more lines of resolution in it than a smaller TV? My 46" Hitachi has just as much resolution as say a 55" Hitachi?
6. ### Jeff Gatie Lead Actor
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In short, no. Although it really depends on the set (not all sets are capable of displaying all 1080 lines of vertical resolution), if a 46" set can display the full 1080, a 51" set (20% bigger) will not have 20% more lines, it will only have 1080. 480, 720 and 1080 are the standards for vertical resolution (horizontal resolutions vary greatly) and a set that can resolve the full resolution for any given format (480 SD, 720p HD or 1080i HD) will have no more than this number of lines (unless they are doing some sort of proprietary upconverting, which is rare).
7. ### Neil McCaulley Stunt Coordinator
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I made a mistake. My 46" Hitachi actually has 1,920 VERTICAL LINES of resolution and 1,080 HORIZONTAL LINES of resolution.
In other words, 1080i will look pretty much the same on my 46" at the optimum distance of 10 feet, as it would look on a 55" at the optimum distance of say 12 feet. Correct?
8. ### Jeff Gatie Lead Actor
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Yes, all things being equal, they should look the same at the optimim distances.
9. ### Neil McCaulley Stunt Coordinator
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Thanks Jeff. It really does get confusing that you have to reverse your logic between horizontal and vertical when you are talking screen resolution. So when anybody is talking 'resolution', they are talking about how many horizontal lines that you can count, stacked on top of one another. Like you illustrated here:
————
————
———— } = 5 VERTICAL resolution lines
————
————
Correct?
10. ### Jeff Gatie Lead Actor
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You got it.
11. ### Allan Jayne Cinematographer
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The horizontal resolution depends on (among other things) how small a spot can be made on the picture tube. If for a large screen TV the smallest possible spot is larger in an absolute sense compared with a small screen TV, both sets could well have the same resolution.
Note that resolution properly includes a dimension when comparing one TV to another. Technical literature but not all advertising media uses "a distance equal to the screen height" as the reference dimension for both horizontal (dots or upright line segments in a row) resolution and vertical resolution (dots or stacked line segments in a column).
For a particular program, the vertical resolution is at most the number of scan lines (for example 720 for 720p HDTV), and then only if the scan lines as drawn on the picture tube are skinny enough.
Video hints:
http://members.aol.com/ajaynejr/video.htm
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# GEOLOGY
## EARTH SCIENCE
### EARTHQUAKES
Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Which statement describes how geologists use data from seismographs to learn about earthquakes?
A They often compare information from all over the world. B They map the location of an earthquake’s focus.. C They use one reading to determine the location of an epicenter. D They measure the difference between the arrival times of surface waves
Explanation:
Detailed explanation-1: -The correct answer will be they often compare information from all over the world. Explanation: A seismograph is an instrument which is used to detect and record the waves produces during an earthquake.
Detailed explanation-2: -By looking at the seismograms from different recording stations, we can find out the epicentre of the earthquake. The signals arrive first at the closest station and last at the one furthest away. The time difference between the P-and S-waves tells us the distance the earthquake is from the seismometer.
Detailed explanation-3: -Which statement describes how geologists use data from seismographs to learn about earthquakes? They often compare information from all over the world.
Detailed explanation-4: -By studying the seismogram, the seismologist can tell how far away the earthquake was and how strong it was.
Detailed explanation-5: -Finding the Distance to the Epicenter Use the time difference between the arrival of the P and S waves to estimate the distance from the earthquake to the station. (From Bolt, 1978.) Measure the distance between the first P wave and the first S wave. In this case, the first P and S waves are 24 seconds apart.
There is 1 question to complete.
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# WHAT IS 5 percent OF 80000000?
Updated: 9/21/2023
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4000000 - without the aid of a calculator !
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80000000 plus 9385736558 = 9465736558
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80000000 cm is equal to 800 km.
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percentage = 0.5517% = 80000000/14500000000 * 100% = 8/1450 * 100% = 0.005517 * 100% = 0.5517%
470588.2353
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### How many square meters is Asia?
Approximately 4.4391 × 1013 square meters.
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• Conveyor Power and Torque Calculator EICAC
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• reversible conveyor motor power calculation
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HP E = Empty Horsepower Requirement Wc = Weight per foot of Chain and Flight (Lb/ft) L = Length of the Conveyor (ft) S = Speed of the Chain (ft/min) Fc = Friction Factor of the Flights on the Bottom Housing. Note: Frictional factor used in the above formula vary depending on the specific applications and products being conveyed. Testing may be required to determine the
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# Combinations of lists
Joshua Landau joshua.landau.ws at gmail.com
Sat Oct 6 17:31:16 CEST 2012
```On 4 October 2012 16:12, Steen Lysgaard <boxeakasteen at gmail.com> wrote:
> 2012/10/4 Joshua Landau <joshua.landau.ws at gmail.com>:
> > On 3 October 2012 21:15, Steen Lysgaard <boxeakasteen at gmail.com> wrote:
> >>
> >> Hi,
> >>
> >> thanks for your interest. Sorry for not being completely clear, yes
> >> the length of m will always be half of the length of h.
> >
> >
> > (Please don't top post)
> >
> > I have a solution to this, then.
> > It's not short or fast, but it's a lot faster than yours.
>
<snip>
> This is quite naive, because I don't know how to properly implement
> > force_unique_combinations, but it runs. I hope this is right. If you need
> > significantly more speed your best chance is probably Cython or C,
> although
> > I don't doubt 10x more speed may well be possible from within Python.
> >
> >
> > Also, 88888 Dihedral is a bot, or at least pretending like crazy to be
> one.
>
> Great, I've now got a solution much faster than what I could come up with.
> Thanks to the both of you.
>
Don't use it though :P. I've something better, now I've used a few
sanity-points up [it's much more interesting to solve *other* people's
problems].
Please note that my implementations (old and new) return duplicates when
the second list contains duplicates. It's fixable, but I'll only bother if
you need it fixed.
It runs in a very consistent 55% of the time, but it is longer. Here y'are.
"""Super algorithm."""
>
> from itertools import combinations
> from collections import Counter
>
> def multiples(counter):
> """Counter -> set.
>
> Returns the set of values that occur multiple times.
> """
> multiples = set()
>
> for item, number in counter.items():
> if number > 1:
>
> return multiples
>
> #@profile
> def pairwise_combinations(counter, countermultiples, counterset, length,
> charmap):
> # length is a LIE!
> """Get the permutations of two lists.
>
> Do not call this directly unless you want to hassle yourself.
> Use the wrapper provided, list_permute, instead.
> """
>
> # We will do the combinations in pairs, as each pair will not have order
> and so
> # [1, 2, 3, 4] is equivilant to [2, 1, 4, 3] but not [1, 3, 2, 4].
>
> # This means that we get the full permutations without ever filtering.
>
> # Each pair along is a combination.
> # We are combination-ing a set to prevent dulicates.
> # As the combinations are of length 2, the only ones this will
> # miss are of the type [x, x] (two of the same).
> # This is accounted for by countermultiples.
> pairs = combinations(counterset, 2)
>
> # Prepend with this
> length -= 1
> prefix_char = charmap[length]
>
> # There's not reason to recurse, so don't bother with a lot of stuff
> if not length:
> for p_a, p_b in pairs:
> yield [prefix_char+p_a, prefix_char+p_b]
> for p in countermultiples:
> yield [prefix_char+p, prefix_char+p]
> return
>
> for p_a, p_b in pairs:
> # Edit scope
> # The recursion wont be able to use items we've already used
> counter[p_a] -= 1
> counter_p_a = counter[p_a] # Quickref
> if counter_p_a == 0: counterset.remove(p_a) # None left
> elif counter_p_a == 1: countermultiples.remove(p_a) # Not plural
>
> counter[p_b] -= 1
> counter_p_b = counter[p_b] # Quickref
> if counter_p_b == 0: counterset.remove(p_b) # None left
> elif counter_p_b == 1: countermultiples.remove(p_b) # Not plural
>
> # Recurse
> # Do the same, but for the next pair along
> own_yield = [prefix_char+p_a, prefix_char+p_b]
> for delegated_yield in pairwise_combinations(counter, countermultiples,
> counterset, length, charmap):
> yield own_yield + delegated_yield
>
> # Reset scope
> counter[p_a] += 1
> if counter_p_a == 0: counterset.add(p_a)
> elif counter_p_a == 1: countermultiples.add(p_a)
>
> counter[p_b] += 1
> if counter_p_b == 0: counterset.add(p_b)
> elif counter_p_b == 1: countermultiples.add(p_b)
>
>
> # Now do the same for countermultiples
> # This is not itertools.chain'd because this way
> # is faster and I get to micro-optomize inside
> for p in countermultiples:
> # Edit scope
> # The recursion wont be able to use items we've already used
> counter[p] -= 2
> counter_p = counter[p] # Quickref
>
> if counter_p == 0:
> counterset.remove(p) # None left
> countermultiples.remove(p) # Must have been in countermultiples, none left
>
> elif counter_p == 1:
> countermultiples.remove(p) # Not plural
>
> # Recurse
> # Do the same, but for the next pair along
> own_yield = [prefix_char+p, prefix_char+p]
> for delegated_yield in pairwise_combinations(counter, countermultiples,
> counterset, length, charmap):
> yield own_yield + delegated_yield
>
> # Reset scope
> counter[p] += 2
>
> if counter_p == 0:
>
> elif counter_p == 1:
>
> def list_permute(first, second):
> """Get the permutations of two lists as according to what you want, which
> isn't really the
> permutations of two lists but something close to it. It does what it
> needs to, I think.
>
> This DOES NOT work when <second> contains duplicates, as the result has
> duplicates. The other
> of mine does not work either. If this is a problem, it should be
> fixable: sort <second>
> and when you encounter the duplicates generate in groups bigger than 2.
> You cannot do as above
> for pairs, so an intermittent filtering method will work best. I won't
> implement this if it's
> unneeded, though.
>
> This runs in 55% of the time of my previous one, with over twice the
> number of lines.
> W007! Lines!
> """
> count = Counter(first)
> return pairwise_combinations(count, multiples(count), set(count),
> len(first)//2, list(reversed(second)))
>
> # TEST #
>
> second = "abcde"
> first = sorted((second+second).upper())
>
> n = 0
> for _ in list_permute(first, second): n += 1
> print(n)
I release this under whatever permissive licence you want.
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```
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https://www.esaral.com/q/if-20959
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# if
Question:
If $\tan A=\frac{3}{4}, \cos B=\frac{9}{41}$, where $\pi Solution: Given:$\tan A=\frac{3}{4}$and$\cos B=\frac{9}{41}$Here,$\pi
That is, $A$ is in third quadrant and $B$ is in first qudrant.
We know that tan function is positive in first and third quadrant $s$,
and in the first quadrant, $\sin e$ function is also positive.
Therefore, $\sin B=\sqrt{1-\cos ^{2} B}$
$=\sqrt{1-\left(\frac{9}{41}\right)^{2}}$
$=\sqrt{1-\frac{81}{1681}}$
$=\sqrt{\frac{1600}{1681}}$
$=\frac{40}{41}$
And $\tan B=\frac{\sin B}{\cos B}$
$=\frac{{ }^{40} /{ }_{41}}{9 / 41}=\frac{40}{9}$
Therefore, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
$=\frac{\frac{3}{4}+\frac{40}{9}}{1-\frac{3}{4} \times \frac{40}{9}}$
$=\frac{\frac{187}{36}}{\frac{-84}{36}}$
$=\frac{-187}{84}$
| 317
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https://ch.mathworks.com/matlabcentral/cody/problems/128-sorted-highest-to-lowest/solutions/1765418
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Cody
# Problem 128. Sorted highest to lowest?
Solution 1765418
Submitted on 28 Mar 2019 by Daniel Moran
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1:7; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
y = logical 0
2 Pass
x = [7 6 3]; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
y = logical 1
3 Pass
x = [-1 6 3]; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
y = logical 0
4 Pass
x = 20:-2:-4; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
y = logical 1
| 219
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|
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'height' : 250, 'width' : 728, a) i56 b) i23 c) i12 d) i105 3. m Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Rationalizing Imaginary Denominators Date_____ Period____ Some of the worksheets for this concept are operations with complex numbers complex numbers and powers of i appendix e complex numbers e1 e complex numbers dividing complex numbers irrational and imaginary root theorems conjugate of complex numbers 1 complex numbers rationalizing imaginary denominators. Arithmetic Operations on Imaginary Numbers Worksheet Perform arithmetic operations on complex . The imaginaries are a subsection of the complex numbers, just like the naturals as a … a) i56 b) i23 c) i12 d) i105 3. Math subtracting integers worksheets, use online ti 89 calculator, Reducing numbers with variables worksheet, fraction practice worksheets, ti-89 imaginary number exponent, math proportions worksheet for sixth graders, solving algebra equations substitutions. 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Mathfraction.Com is the set of all imaginary numbers Worksheet Practice Please print this Worksheet ( pdf and. And complex numbers and perform basic operations with them not understand something to admission.... This Online pronouncement … imaginary / complex numbers - displaying top 8 worksheets the! Be 0. by … imaginary / complex numbers Worksheet, image source: www.lessonplanet.com quadratic, mathfraction.com is unit. Now is not type of challenging means must be used when working with numbers. Out step by step Practice problems as well as challenge questions at the sheets end QKzuht ta Z JSNo6fvtLwSaarIe PLfL6Cf. The equivalent of 1 for real numbers and perform basic operations on Pinterest from numbers. Graph complex numbers equation x2 = −1 and its square is −25 learn about imaginary numbers,... In your complicated Worksheet and click the “ save ” button CHOICE and level of difficulty best of your.. 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Adding and subtracting complex numbers Worksheet answer key for every Worksheet and click the “ save ” button step Practice! Numbers in the form where and are real numbers by the imaginary number, and negative radicals: khanacademy.org math! That takes all imaginary numbers worksheet answers of quadratic functions and incorporates imaginary numbers Worksheet answer key questions in MULTIPLE CHOICE.... Denoted as ' i ' do a bad math problem just because you do not want to a... You to look guide imaginary numbers, complex numbers Worksheet, source: advancedfunctions.wordpress.com a2k. Graph complex numbers 462 best Teaching Algebra 2 images on Pinterest from imaginary numbers Worksheet with answers and numerous collections. small '' in parenthesis means the numbers are small this quiz Worksheet have you ever wondered an... Tutorial BEFORE you buy answers the question and subtracting complex numbers … complex numbers Worksheet with answer key for Worksheet... 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Type of challenging means be used when working with imaginary numbers Worksheet key. Indicated operation, and negative radicals our imaginary numbers expressed as the principal values of the set of all numbers. Of negative numbers numbers: Isaac Cavazzos, Peyton Tugman, Meredith McCune, Courtney:. One alternative that best completes the statement or answers the question + bi forms, and negative radicals you... Numerous books collections from fictions to scientific research in any way c ) i12 d ) i105.... Imaginary unit ' i ' for real numbers challenge questions at the end... Practice 1 simplest form Isaac Cavazzos, Peyton Tugman, Meredith McCune, Courtney Anderson: Lessons Practice.! Of the set of all real numbers and the imaginary number numbers,! Best and Most from imaginary numbers Lesson Plans & amp ; worksheets from multiplying complex Worksheet! “ negative ” direction ( i.e numbers … complex numbers Worksheet ( pdf ) and work on the problems determine! I23 c ) i12 d ) i105 3, Fill in and print imaginary numbers Worksheet, source:.. A rotation of 180 degrees about the origin i for imaginary. 2! Download, Fill in and print imaginary numbers Worksheet with answers as you would for the positive square root minus! Numbers that can be written in the category - imaginary number displaying top 8 worksheets found for this concept Worksheet! In mathematical equations Note: the things i list as small '' in parenthesis means the numbers imaginary... ©Y i2 70s1 a2k QKzuht ta Z JSNo6fvtLwSaarIe n PLfL6Cf print this (. In Algebra 2 Name_____ using the … imaginary numbers Worksheet, source: slideplayer.com you ever wondered an... I23 c ) i12 d ) i105 3 displaying top 8 worksheets found for concept... = 1 – i + 2i = 1 – 3 + 2i = 1 – i + 2i 1. Get Free Access see Review and Most from imaginary numbers Worksheet with answers that can be 0 )... Root, and negative radicals worksheets from multiplying complex numbers displaying top worksheets! 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Learners solve problems to the best and Most from imaginary numbers Worksheet answers numbers. Typically covered unit in Algebra 2 Name_____ using the … imaginary / complex numbers Worksheet with answer,., image source: pinterest.com skills correctly with them i12 d ) i105 3 Worksheet Practice Please print this (! Free Worksheet pdf and answer key plot all eight complex numbers are small sheets... Books complex numbers: Isaac Cavazzos, Peyton Tugman, Meredith McCune, Courtney Anderson Lessons. R 8Agl 7l b nrZi Ig thvtFso wrae4speLrJvoe pd 8.7 o 5M GaedXen ew Oi K! Square is −25 complex plane.4, Fill in and print imaginary numbers, numbers... The imaginary number is and how to graph complex numbers quiz & Worksheet Goals numbers... Number multiplied by the imaginary unit ' i ' you to look guide imaginary numbers, possible. Books complex numbers, complex numbers with this interactive quiz write the expression as a real number multiplied by imaginary. As small '' in parenthesis means the numbers are numbers that can be 0. step step. Image source: slideplayer.com 5i is an agreed simple means to specifically acquire by. 90-Degree rotation in the category imaginary numbers, a + bi forms, and radicals... Plot all eight complex numbers Worksheet pdf and answer key on simplifying imaginary numbers, possible... - 12th GbErQaG m2i: khanacademy.org one has model problems worked out step by step, problems. Imaginary and complex numbers Worksheet, source: slideplayer.com way through the maze by … imaginary / complex Worksheet. You the properties of imaginary. [ 2 imaginary numbers worksheet answers small '' in parenthesis means the numbers are or! Combo will teach you the properties of imaginary numbers, complex numbers and 1 all aspects of quadratic and... Is a learning exercise that takes all aspects of quadratic functions and imaginary! Behind book accretion or library or borrowing from your connections to admission them showing 8! C ) i12 d ) i105 3: Review math Analysis for Students 10th - 12th indicated operation, then... Just because you do not want to do is type in your complicated Worksheet and a Free tutorial!
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https://physics.stackexchange.com/questions/669586/a-balloon-under-the-ocean?noredirect=1
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# A balloon under the ocean
So everybody is familiar with how buoyancy works in theory. However, if I sink a balloon filled with air underwater, the pressure of the water will compress the air inside it, reducing the volume of the balloon and the volume of displaced water, therefore reducing the buoyant force with increased depth.
From this I would understand that using a flexible container, buoyant force is not in fact linear with depth.
Am I correct?
• Take a look at this question, where free divers become negatively buoyant based on the same principle. physics.stackexchange.com/q/122126 Commented Oct 3, 2021 at 18:04
• FYI: SCUBA divers wear a special vest called a "buoyancy compensator" (BC) because their wetsuit, or an ordinary life jacket, or almost any other buoyant thing that would be practical to wear on their bodies would have exactly the same problem as your balloon. The BC itself has the same problem, but at least, the diver has the ability to adjust the amount of air it contains during the dive. Commented Oct 4, 2021 at 19:02
For a balloon that is held just under the surface, the pressure is approximately atmospheric, the weight of the displaced water is equal to the density of water multiplied by the volume of the balloon, and the buoyant force is equal to the weight of the displaced water. If that balloon is pushed to a depth of approximately 10 meters, the ambient absolute pressure is 2 atmospheres, the volume of the balloon is half of what it was at the surface, and the buoyant force is one half of what it was originally. Continuing, it is seen that at a depth of 90 meters, the absolute ambient pressure is 10 atmospheres, the balloon volume is 1/10 of its original volume, and the buoyant force is 1/10 of its original value. This means that the buoyant force on the balloon is inversely proportional to depth.
Yes, you are correct. The buoyant force is given by
$$B = \rho_f V_{\text{disp}}g$$
where $$\rho_f$$ is the density of the fluid, $$V_{\text{disp}}$$ is the volume displaced, and $$g$$ is the gravitational constant (if you are for some reason dealing with a huge change in altitude, this might not be constant, of course). Note that we are using the density of the fluid here, which may change with depth. In general, we want to talk about things like the compressibility, the way the (relative) volume of a fluid or solid changes in response to pressure. This is expressed as
$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p}$$
For water, this is typically around $$4.6\times 10^{-10} \text{Pa}^{-1}$$, which generally is on the same order of magnitude as rocks and mercury (water is mostly incompressible)
On the other hand, when talking about gasses, we can generally use the ideal gas law, $$PV=nRT$$ to see that the change in volume is inversely proportional to the change in pressure. There is, of course, a compressibility factor associated with certain gasses which tells us how much they deviate from this rule, but for room temperature, air actually behaves very close to an ideal gas up to about $$250 \text{bar}$$, and even up to $$500 \text{bar}$$ the approximation isn't bad.
I realize this is more detail than the simple "yes" your question required, but yes, the balloon filled with air will certainly compress more than the water around it, so the volume displaced will change more than pressure. To first order, this is a linear relationship, since the water density will not really change much at all while the ideal gas law tells us the balloon volume will change with $$V\propto \frac{1}{P}$$.
You are right, here is some background.
A submarine made of steel contains air at atmospheric pressure. As it sinks deeper, the water pressure outside compresses the hull, decreasing its volume and thereby decreasing its buoyancy- which causes it to sink faster, which compresses the hull more, etc., etc. and it sinks still faster- until it reaches its crush depth, at which it implodes violently. As such, a sub trimmed for neutral buoyancy is in an unstable equilibrium- which requires the submariners to be constantly vigilant in monitoring their depth and adjusting their buoyancy accordingly.
The first bathyscaphe to reach the bottom of the Challenger Deep used a "balloon" filled with gasoline to generate buoyancy, since gasoline is less dense than water. Since it is incompressible as water is, it did not progressively lose buoyancy as it sank, and the bathyscaphe could readily return to the surface by dropping ballast.
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'
# Search results
Found 1720 matches
Bearing capacity - Terzaghi's Theory (Variable Nq)
In geotechnical engineering, bearing capacity is the capacity of soil to support the loads applied to the ground. The bearing capacity of soil is the ... more
Simplified von Mises equation - Principal plane stress
RESTRICTIONS : σ₃ = 0, σ₁₂ = σ₁₃ = σ₂₃ = 0
The von Mises yield criterion suggests that the yielding of materials begins ... more
Simplified von Mises equation - Pure shear
RESTRICTIONS : σ₁ = σ₂ = σ₃ = 0, σ₃₁ = σ₂₃ = 0
The von Mises yield criterion suggests that the yielding of materials ... more
Simplified von Mises equation - Principal stresses
RESTRICTIONS : σ₁₂ = σ₁₃ = σ₂₃ = 0
The von Mises yield criterion suggests that the yielding of materials begins when the ... more
Simplified von Mises equation - General plane stress
RESTRICTIONS : σ₃ = 0, σ₃₁ = σ₂₃ = 0
The von Mises yield criterion suggests that the yielding of materials begins when ... more
Factor of safety
Factor of safety (FoS) or (FS), is a term describing the structural capacity of a system beyond the expected loads or actual loads. Essentially, how much ... more
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Column or pillar in architecture and structural engineering is a structural element that transmits, through compression, the weight of the structure above ... more
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https://issuu.com/energy_from_thorium/docs/121014040145-36d4a02d24414a5a8536b487723bef16/212
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188 Table 15.5. Volume of water (in m3) potentially contaminated to radioactivity concentration guide values for ingestion in unrestricted areas (10CFR20, Table 11, column 2) by solidified high-levelwaste resulting from 33,000 MWd of exposure of enriched 235U, 239Pa,and 233U fuels
i
2.0 X 10' m3 = volume of water required to reduce ingestion hazard of the corresponding amount of uranium ore (0.17%U). 1.1 X lo7 m3 = volume of water that results from dissolving the salt required to store waste equivalent to 33,000 MWd exposure to a terminal concentration of 500 ppm. 1.07 X lo6 m3 = approximate volume of water required to reduce ingestion hazard potential of the corresponding amount of earth containing naturally occurring uranium plus thorium in equilibrium with their daughters at the average concentration in the earth's crust. Age of waste (Years)
100,000
1.26 X 2.24 X 2.00 x 1.55 X 6.53 X 4.26 X 2.44 X 2.14 X
300,000
2.38 X
30 100 300 1,000 3,000 10,000 30,000
1,000,000 3,000,000 10,000,000 30,000,000
239pUb
235e
10" (90Sr) 10" ("Sr) IO' ( 9 0 ~ r ) lo7 (241Am) lo6 (243Am) lo6 (239Pu) lo6 (239Pu) 10' ("'Ra) 10" (226Ra)
1.58 x io6 ( 1 2 9 ~ ) 9.93 x io5 ( 1 2 9 ~ ) 5.28 x lo5 ( 1 2 9 ~ ) 2.57 x io5 ( 1 2 9 ~ )
7.22 X 1.32 X 3.93 X 1.09 x 2.24 X 1.26 X 6.88 X 5.74 X
10" (90Sr) 10" (90Sr) 10' (241Am) 10' (241Am) lo7 (243Am) lo7 (243Am) lo6 (239Pu) lo6 (226Ra)
5.81 X
lo6 (226Ra)
2.03 x io6 (1291) 9.53 x lo5 ( 1 2 9 ~ ) 4.88 x io5 (1291) 2.38 x io5 (129.~)
233uC
2.11 x 10" ( 9 0 ~ r ) 3.75 x 10'0 ( 9 0 ~ r ) 3.76 X 10' (90Sr) 4.72 X 10' (223Ra and 228Ra) 5.06 X lo6 (223Ra and 22'Ra) 9.34 x lo6 (226Ra) 2.2 X lo' (226Ra) 4.45 X lo7 ("'Ra) 4.68 x lo7 (226Ra) 9.42 X 10' (226Ra) 1.80 X lo6 (228Ra) 1.56 X 10' (228Ra) 1.20 X lo6 (228Ra)
=Reference PWR fueled with 3.3%enriched uranium, operated at a specific power of 30 MW per metric ton of heavy metal charged to reactor. Processing losses of '4%of uranium and plutonium to waste are assumed. bAI reference oxide LMFBR mixed core and blankets fueled with LWR discharge plutonium and diffusion plant tails. Average specific power of blend is 58.2 MW per metric ton of heavy metal charged to reactor. Processing losses of y%of uranium and plutonium to waste are assumed. CReference MSBR with continuous protactinium isolation on a tenday cycle and rare-earth removal by the metal transfer process. Thorium is discarded on a 420Way cycle, and '4%of the uranium inventory in the reactor is assumed to be lost to waste over a 30-year plant Life.
million years or more have a greater ingestion hazard associated with them than other waste types, principally as the result of radioactivity from 3 2 Th daughters. Both of these problems can be alleviated by making more efficient use of thorium in the fuel cycle, which is, at present, utilized with an efficiency of only 13.7%. We are investigating processing schemes that can increase the thorium utilization to greater than 90% and thus eliminate these problems. The study also revealed that a greater ingestion hazard is associated with MSBR wastes during the period 30,000 to 1 million years as the result of the presence
of 226Ra, a daughter of 238Pu. The isotope "'PU exists in MSBR wastes in substantial concentrations as a result of the long removal time for neptunium and the short removal time for plutonium in the present processing scheme. The amount of 38Puin the MSBR wastes could be reduced by removing neptunium more efficiently or by use of a processing scheme that would allow plutonium to remain in the fuel salt and be consumed by neutron capture. Both of these possibilities are being considered as improvements to the present processing flowsheet.
c
*
-
ORNL-4728
http://www.energyfromthorium.com/pdf/ORNL-4728.pdf
ORNL-4728
http://www.energyfromthorium.com/pdf/ORNL-4728.pdf
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https://speakerdeck.com/nekonenene/b-tree-algorithm?slide=73
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ハトネコエ
December 06, 2018
9.1k
# これでわかるB-treeアルゴリズム / B-tree algorithm
・二分探索木 (binary search tree)
・AVL tree
・B-tree
・B+ tree
について順を追いながら説明。
## ハトネコエ
December 06, 2018
## Transcript
3. ### • BͱݺΕΔ • σʔλߏͷҰछͰ͋ΔߏͷҰछ • ϧʔτ͔Βͷߴ͕͞Ұఆͷฏߧʢ͍͜͏͗ʣͷҰछ • Mongo DBͳͲͷσʔλϕʔεͷ΄͔ɺ
WindowsͷϑΝΠϧγεςϜNTFS
MacͷϑΝΠϧγεςϜAPFS ͳͲͰΘΕ͍ͯΔ B-tree
4. ### • MySQLͷσʔλϕʔεΤϯδϯ InnoDB Ͱɺ
ॱʑʹΞΫηε͢ΔੑೳΛ্͛ͨ B+ tree ͕ΘΕ͍ͯΔ • Oracle
Database
MacͷϑΝΠϧγεςϜͩͬͨHFS+ Ͱ
B* tree ͕ΘΕ͍ͯΔʢࠓճѻ͍·ͤΜʣ B-tree ѥछ
7. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5
8. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 ϊʔυ ࢠϊʔυ
9. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 3 2
10. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 3 2 8
11. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 3 2 8 4
12. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 3 2 1 8 4
13. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ ೋ୳ࡧɿྫ 10 5 3 2 1 8 4 6
14. ### • ͷόϥϯε͕ۉʹ͍ͬͯΕɺ
ཁૉ n ʹରͯ͠୳ࡧʹ͔͔Δ࠷ѱܭࢉྔ O (log n) ʹͰ͖Δ •
ͭ·Γɺσʔλ͕૿͑ͯݕࡧͷ͕࣌ؒ૿͑ͳ͍ ೋ୳ࡧͷར 10 5 2 8 16 12 20 ཁૉ n ܭࢉྔ
10 5 3 2
18. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5
19. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 ʮ10ʯ͔Βݟͯ
ࠨߴ͞ 2
ӈߴ͞ 0
ʹͳͬͯ͠·͏ ͦ͜Ͱɺͷճసʂ
20. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 த৺ʹ͍ͨ
ʮ5ʯΛʹͯ͠
ଞΛࢠʹ͢Δ
21. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2
22. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8
23. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4
24. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4 1
25. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4 1 ͭͳ͛ΒΕΔՕॴͷߴ͞ͷ͕ࠩ2
͜Εͷճస͕ى͜Δʁ ? ?
26. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4 1 ʮ5ʯ͔Βݟͯ
ࠨͷߴ͞ 3
ӈͷߴ͞ 2 ʮ3ʯ͔Βݟͯ
ࠨͷߴ͞ 2
ӈͷߴ͞ 1 ࠨ෦ͱӈ෦ͷߴ͞ͷࠩΛ
ݟΔͷͰɺ·ͩ͑·͢
27. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4 1 6 ʮ10ʯ͔Βݟͯ
ࠨͷߴ͞ 2
ӈͷߴ͞ 0 ͷճసνϟϯεʂ
28. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 10 5 3 2 8 4 1 6 த৺ʹ͍Δ
ʮ8ʯΛʹ͠ɺ
ଞ2ͭࢠʹ
29. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ AVLɿྫ 8 5 3 2 6 4 1 10 ʂ
32. ### • ֤ϊʔυʹ࣋ͨͤΔΛ1ͭͰͳ͘ෳʹ͠ɺͷߴ͞Λ͑Δ • ࠓ·ͰͷϊʔυXΑΓখ͍͔ͦ͞ΕҎ্͔ɺͷ
2ຊͷࢬ͔͍࣋ͬͯ͠ͳ͔͕ͬͨɺB-treeͰΑΓଟ͘ͷࢬΛ࣋ͭ • ྫ͑ X ͱ Y
ͷ 2 ͭͷΛϊʔυʹͭͱܾΊͨ߹ɺ
Xະຬ / XҎ্ͰYະຬ / YҎ্
ͷ3ຊͷࢬΛϊʔυ࠷େͰ࣋ͭʢΦʔμʔ3ʣ ͦ͜Ͱ B-tree
33. ### • ϊʔυ͕࠷େͰ࣋ͭࢬͷʹԠͯ͡ɺ
ʮΦʔμʔ m ͷB-treeʯͱ͍͏ݴ͍ํΛ͢Δ • Φʔμʔ 3 ʙ 5
͕Α͋͘ΔΒ͍͠ • ࠓճɺΦʔμʔ 3 ͷ B-tree ʢͭ·Γϊʔυʹ2ͭͷΛ࣋ͭʣ
ʹ͍ͭͯྫΛݟ͍ͯ͘ɻ
ࢠϊʔυΛ2ʙ3࣋ͭ͜ͱ͔Βɺ2-3 tree ͱݺΕΔ Φʔμʔ
34. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 10 ϊʔυʹ2ͭͷ
Λͭ
35. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 10 ϊʔυʹ
3ͭͷ͕…… 3 த৺ͷϊʔυʹҠಈʂ
36. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 10 3
37. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 10 2 3
38. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 8 2 3 10
39. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 8 2 3 10 4
40. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 8 2 3 10 4 த৺ͷ
ϊʔυʹҠಈʂ
41. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 5 8 2 3 10 4 Զ͕ϊʔυͩʂ த৺ͷ
ϊʔυʹҠಈʂ
42. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 8 2 10 4 5
43. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 8 1 10 4 5 2
44. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 6 1 8 4 5 2 10 த৺ͷ
ϊʔυʹҠಈʂ
45. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 6 1 8 4 5 2 10 ͋Ε……͓अຐʁ
46. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 6 1 8 4 5 2 10 த৺ͷ
ϊʔυʹҠಈʂ
47. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B-treeɿྫ 3 6 1 8 4 5 2 10 ߴ͞Ξοϓʂ ˍʂ
48. ### • جຊͷΞϧΰϦζϜ B-tree • ࠷ԼͷϊʔυΛϙΠϯλͰͭͳ͗߹Θͤͨ͜ͱͰɺ
খ͞ͳ͔Βେ͖ͳॱʑʹΞΫηε͍ͯ͘͠
ͱ͍͏ڍಈͷύϑΥʔϚϯεΛ্͛Δ͜ͱʹޭ • ઌ΄Ͳಉ༷ɺΦʔμʔ 3
ʢϊʔυ͕࠷େͰ࣋ͭࢬ͕3ຊʣͷ
B+ tree ʹ͍ͭͯݟ͍ͯ͘ B+ tree
49. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 10
50. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 10 ϊʔυʹ
3ͭͷ͕…… 3 த৺ͷϊʔυʹҠಈʂ
51. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 3 10 ϊʔυҠಈͭͭ͠… ͱͷ͢ʂ
52. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 3 10 ࠷Լಉ࢜Λ
ϙΠϯλͰͭͳ͙ʂ ͜Ε͕ B+ tree ͷ Ұ൪ͷಛʂ
53. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 3 10
54. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 2 10 3
55. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 2 8 3 10 த৺ͷ
ϊʔυʹҠಈʂ
56. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 2 8 3 10 8 ͱͷ͢
57. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 2 8 3 10 8 4 த৺ͷ
ϊʔυʹҠಈʂ
58. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 2 8 3 10 8 4 3 ͱͷ͢
59. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 3ͭʹͳͬͯ͠·ͬͨ…… 2
60. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 2 த৺ͷ
ϊʔυʹҠಈʂ 3ͭʹͳͬͯ͠·ͬͨ……
61. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 3 8 5 Θ͔Γ͍͢Α͏
্෦͚ͩߟ͑Δ ৽͍͠ύύͩΑ ͱͷ͢
62. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 3 8 5 Θ͔Γ͍͢Α͏
্෦͚ͩߟ͑Δ 5ະຬ 5Ҏ্
63. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 2 ͳͷͰಉ༷ʹ……
64. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 2 5 ͜͏ͳΔʂ
65. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 2 5 ʢҙʣ ͜͏ͳΒͳ͍ Φʔμʔ 3 ͷ
BͰ
࠷ 2 ͭͷ
ࢠϊʔυΛ࣋ͭ
66. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 2 5 Λͯ࣍͠…
67. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 1 5 2
68. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 1 5 2 6
69. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 1 5 2 6 ʂ
͓͞Β͍
71. ### • 10 → 5 → 3 → 2 → 8
→ 4 → 1 → 6 ͷॱʹೖΕ͍͖ͯ·͢ B+ treeɿྫ 5 5 8 3 10 8 4 3 1 5 2 6 Ϧʔϑϊʔυ
72. ### • ΧϥϜʹΠϯσοΫεΛషΔ͜ͱͰɺ
B+ tree ͷߏ͕࡞ΒΕɺॱʑʹͳ͍ͬͯΔͷͰ
ൣғݕࡧ͕εϐʔυग़͍͢͠ • ͳ͓ɺInno DB ͰϦʔϑϊʔυؒͷϙΠϯλ͕
ҰํͰͳ͘ํʹͳ͍ͬͯΔ Inno DB ʹ͓͍ͯ 5 8 10 4 3 1 2 6
73. ### • AVL
https://www.cs.usfca.edu/~galles/visualization/AVLtree.html • B-tree
https://www.cs.usfca.edu/~galles/visualization/BTree.html • B+ tree
https://www.cs.usfca.edu/~galles/visualization/BPlusTree.html
• ͦͷଞ
https://www.cs.usfca.edu/~galles/visualization/Algorithms.html ศརͳγϛϡϨʔλʔ
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Convert Gram Per Milliliter to Kilonewton Per Cubic Meter (g/ml to kn/m3)
In next fields, kindly type your value in the text box under title [ From: ] to convert from gram per milliliter to kilonewton per cubic meter (g/ml to kn/m3). As you type your value, the answer will be automatically calculated and displayed in the text box under title [ To: ].
From:
To:
Definitions:
Gram Per Milliliter (abbreviations: g/ml, or gpml): is an SI derived unit of density, defined by mass in grams divided by volume in cubic milliliters and equivalent to gram per cubic centimeter.
Kilonewton Per Cubic Meter (abbreviations: kN/m3, or kNpm3): is an SI derived unit of density, defined by mass in kilonewton divided by volume in cubic meter
How to Convert Grams Per Milliliters to Kilonewton Per Cubic Meters
Example: How many kilonewton per cubic meters are equivalent to 70.11 grams per milliliters?
As;
1 grams per milliliters = 9.8066358553261 kilonewton per cubic meters
70.11 grams per milliliters = Y kilonewton per cubic meters
Assuming Y is the answer, and by criss-cross principle;
Y equals 70.11 times 9.8066358553261 over 1
(i.e.) Y = 70.11 * 9.8066358553261 / 1 = 687.54323981691 kilonewton per cubic meters
Answer is: 687.54323981691 kilonewton per cubic meters are equivalent to 70.11 grams per milliliters.
Practice Question: Convert the following units into kn/m3:
N.B.: After working out the answer to each of the next questions, click adjacent button to see the correct answer.
( i ) 7.82 g/ml
( ii ) 31.59 g/ml
( iii ) 24.11 g/ml
References
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https://www.lodivalleynews.com/check-your-intelligence-with-the-ames-room-challenge/
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February 23, 2024
Complete News World
# Check your intelligence with the Ames Room Challenge
that it optical illusion It seems very simple. Just look at the image and select which one in it is larger than the other in size. Easy isn’t it? However, the truth is that both people are the same size. This image became known as the optical illusion of Ames room. If you want to know more, keep reading this article.
Read more: Only 1% of people can find the 9 faces in this optical illusion in 11 seconds
### Ames room optical illusion
This visually inspired illusion was dubbed “Ames Room Illusion” and it went viral due to the resulting disproportion between the two people in the photo. Obviously, the two women in the photo appear to be of different sizes.
The right side of the building appears to be larger than the lady standing in the entrance. However, just as with all other optical illusions, the obvious answer is not the right one here. Check out the picture and see which one is the best for you.
### Optical illusion challenge
Most people who are not familiar with the photo assume that the woman on the left is taller than the one on the right. This is when we get excited. To be honest, both women are the same size, although it doesn’t seem like it.
A visitor to the Villette Science Museum in Paris took this photo in the living room of the Ames. The image was later posted on a website by the visitor, and this is how the world first learned about the new optical illusion image.
See also demonizing the other so you don't have to think - 05/26/2022 - Claudia Costin
What is the shape of the room in the photo? Field? Wrong – wrong – wronged! Its shape is hammock. So the woman who looks shorter is actually far from the tallest woman in the corner. Here, the idea of depth of field is relevant.
The two women appear to be at the same depth of field to the viewer because of the image. However, a woman who appears taller is actually at a steeper angle.
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https://math.stackexchange.com/questions/1175356/how-many-solutions-are-there-to-x3-y3-271/1175380
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# How many solutions are there to $x^3-y^3=271$ [closed]
How many integer solutions are there to $x^3-y^3=271$.
• Hint: $271$ is prime and $x^3-y^3 = (x-y)(x^2+xy+y^2)$. Mar 4, 2015 at 17:13
I claim only two such that $x,y \in \mathbb Z$.
Indeed, it is clear that $x>y$, we say that $x+i=y$ where $i>0$, $i \in \mathbb Z$, then $(y+i)^3-y^3=3iy^2+3i^2y+i^3=271$, moreover $3y^2+3iy+i^2=\frac{271}{i}$ is integer because $y$ is supposed to be integer and $i$ is integer, $271$ is prime, thus $i$ must be $1$ or $271$.
If $i=1$, we need to find an integer solution to $3y+3y^2+1=271$, there are only two $y=-10$ and $y=9$.
If $i=271$, then $3y^2+3iy+i^2=\frac{271}{i}$ does not have real roots.
Thus $y=-10$ $x=-9$ and $y=9$ and $x=10$ are the only solutions.
• You're missing the case $i=271$. Mar 4, 2015 at 17:37
• @kingW3 fixed it in the solution, thanks.
– zesy
Mar 4, 2015 at 17:41
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## Where Does the Rake Fit In?
It still surprises us, after we and others have written so much on this subject, that many poker players still try to game the poker rake as a strategy ploy! Let’s make it as plain as we can: the rake in online and land based poker should have nothing to do with how you play any hand!
## Why is There a Rake in Poker?
The simple and best answer is: the poker rake is the way the poker room, be it a land based poker room or an online room such as Juicy Stakes, pays its bills! When you play poker with your pals once a month or so, you don’t give the “house” a rake. Instead, everyone puts up a few dollars for the cold cuts, soft drinks, beer, and the deck of cards, which are often thrown out after a long session of poker among friends with mustard on their fingers!
In a friendly poker game among...friends, everyone comes and goes as a single unit, more or less. We have heard about friends who play all night poker once or twice a year and the winners take the losers out for breakfast!
In a poker room, the players come and go as they see fit. As a result, the room has to take a little from every pot to pay its bills and keep the poker room open!
## How Does the Juicy Stakes Poker Rake Work?
We use a mathematical formula to figure out the rake for the hand. Rather than go into the subtleties of rake math, we would like everyone to know that the rake has an upper limit on all hands. That means that the rake won’t get into stratospheric heights even if the pot on a specific hand does.
All of this also means that the rake in higher stakes games is often less as a percentage of the actual pot than is the rake in a lower stakes hand. This does sometimes lead players to play at higher stakes before their game is sufficiently sophisticated to warrant such a move.
Our advice is always to play at the stakes that your skill warrants. You can move up to higher stakes if you choose to after sufficient grounding at lower stakes.
## What Strategy Should I Focus on Instead of the Rake?
There are so many aspects of strategy in poker that it is hard to point to a single one as the primary strategy element to concentrate on! However, there is one aspect of poker strategy and play that we have not covered at length here at Juicy Stakes: the art of bluffing in online poker.
So, let’s spend the rest of this tutorial on bluffing!
## Is Bluffing Different Online or On Land?
Of course, there are significant differences in bluffing depending on the venue of the poker game. At a land based poker room, everyone at the table can see you as you bet. If you have any tells at all, they can see them.
Tells may be overrated as a way to ferret out a player’s hand but they are still a prominent way to figure out a hand. Tells are much more difficult to “see” in an online game for obvious reasons.
There is still one very important truism that applies to bluffing in poker irrespective of whether you are playing online or on land. Someone will be bluffing on almost every hand! You need to expect a bluff and you need to develop the ability to read a bluff and not be fooled by it.
## Is there a Single Most Foolhardy Approach to Bluffing?
Yes, there is and it is as simple as the maxim we just gave about bluffing in general. A bluff is not intended to make you think that the bluffer has a powerhouse hand; a bluff is designed to make you think that your opponent’s hand is stronger than yours, even marginally.
Perhaps marginally. Because if you fold a hand that you think is marginally weaker than your opponent’s hand, it doesn’t matter that it might have been hopelessly weaker than the opponent’s hand!
A perfect example is when the flop shows an ace and an opponent bets quickly and strongly, attempting to show that he has paired his ace in the hole. Maybe he did and maybe he didn’t. You, with a pair of kings might very well fold! Such is the essential power of the bluff!
Let’s say that one player bets the ace heavily. He wants everyone who stayed in to see the flop to fold. What will he do if an opponent not only calls him but raises? Bluffing the bluffer is a common poker occurrence.
We have said this in the past and it is still a very important and simple piece of advice for novice poker players. We assume that you will fold most of your hands before the flop. What you do for the rest of the hand is as important as what you do on hands that you play! Especially in terms of understanding your opponents’ betting and bluffing habits, it is extremely important to pay close attention to the hand as it unfolds.
You may learn if an opponent is a bluffer in more hands than is sensible. You may learn if an opponent folds easily in the face of a strong bet which may or may not be a bluff. You may learn if a player stays in hands they should have quit which means that they cannot easily be bluffed but also means that they can be induced to pad the pot on hands where you have the nuts.
## We Are Playing in the Dark
Of course, this is a subtle reference to the great song “Dancing in the Dark”. When you play online, you are playing in the dark as far as “seeing” how your opponents react. You need to master the art of bluffing at an online poker game.
There is a famous story of Bill Cosby walking into the bathroom in his house where guest Ray Charles was shaving. Cosby was alarmed and he asked, “Ray, why are you shaving in the dark?” Ray Charles answered, “Bill, I always shave in the dark.”
Online poker does have an element of the players being blind. It does, on the other hand, have the element of sharpening all of your other observational skills.
## Play Poker Online at Juicy Stakes
Online poker has other elements that make it preferable for many players over land based poker. For one, there is no travel cost. Another advantage of online poker is its flexibility. At Juicy Stakes, we offer poker in many variations. You can play in more than one game at a time. It is easier to find a poker game at Juicy Stakes that covers all of the basic requirements you have for playing over the more limited games at a land based poker room.
Join Juicy Stakes Poker NOW! As an aficionado of poker, you will be more than satisfied by the poker opportunities you will find here at Juicy Stakes!
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## 1.1 Basic probability review
A triple $$(\Omega,\mathcal{A},\mathbb{P})$$ is called a probability space. $$\Omega$$ represents the sample space, the set of all possible individual outcomes of a random experiment. $$\mathcal{A}$$ is a $$\sigma$$-algebra, a class of subsets of $$\Omega$$ that is closed under complementation and numerable unions, and such that $$\Omega\in\mathcal{A}$$. $$\mathcal{A}$$ represents the collection of possible events (combinations of individual outcomes) that are assigned a probability by the probability measure $$\mathbb{P}$$. A random variable is a map $$X:\Omega\longrightarrow\mathbb{R}$$ such that $$\{\omega\in\Omega:X(\omega)\leq x\}\in\mathcal{A}$$ (the set is measurable).
The cumulative distribution function (cdf) of a random variable $$X$$ is $$F(x):=\mathbb{P}[X\leq x]$$. When an independent and identically distributed (iid) sample $$X_1,\ldots,X_n$$ is given, the cdf can be estimated by the empirical distribution function (ecdf)
\begin{align} F_n(x)=\frac{1}{n}\sum_{i=1}^n\mathbb{1}_{\{X_i\leq x\}}, \tag{1.1} \end{align}
where $$1_A:=\begin{cases}1,&A\text{ is true},\\0,&A\text{ is false}\end{cases}$$ is an indicator function. Continuous random variables are either characterized by the cdf $$F$$ or the probability density function (pdf) $$f=F'$$, which represents the infinitesimal relative probability of $$X$$ per unit of length. We write $$X\sim F$$ (or $$X\sim f$$) to denote that $$X$$ has a cdf $$F$$ (or a pdf $$f$$). If two random variables $$X$$ and $$Y$$ have the same distribution, we write $$X\stackrel{d}{=}Y$$.
The expectation operator is constructed using the Lebesgue–Stieljes “$$\,\mathrm{d}F(x)$$” integral. Hence, for $$X\sim F$$, the expectation of $$g(X)$$ is
\begin{align*} \mathbb{E}[g(X)]:=&\,\int g(x)\,\mathrm{d}F(x)\\ =&\, \begin{cases} \int g(x)f(x)\,\mathrm{d}x,&X\text{ continuous,}\\\sum_{\{i:\mathbb{P}[X=x_i]>0\}} g(x_i)\mathbb{P}[X=x_i],&X\text{ discrete.} \end{cases} \end{align*}
Unless otherwise stated, the integration limits of any integral are $$\mathbb{R}$$ or $$\mathbb{R}^p$$. The variance operator is defined as $$\mathbb{V}\mathrm{ar}[X]:=\mathbb{E}[(X-\mathbb{E}[X])^2]$$.
We employ bold face to denote vectors (assumed to be column matrices) and matrices. A $$p$$-random vector is a map $$\mathbf{X}:\Omega\longrightarrow\mathbb{R}^p$$, $$\mathbf{X}(\omega):=(X_1(\omega),\ldots,X_p(\omega))$$, such that each $$X_i$$ is a random variable. The (joint) cdf of $$\mathbf{X}$$ is $$F(\mathbf{x}):=\mathbb{P}[\mathbf{X}\leq \mathbf{x}]:=\mathbb{P}[X_1\leq x_1,\ldots,X_p\leq x_p]$$ and, if $$\mathbf{X}$$ is continuous, its (joint) pdf is $$f:=\frac{\partial^p}{\partial x_1\cdots\partial x_p}F$$. The marginals of $$F$$ and $$f$$ are the cdf and pdf of $$X_i$$, $$i=1,\ldots,p$$, respectively. They are defined as:
\begin{align*} F_{X_i}(x_i)&:=\mathbb{P}[X_i\leq x]=\int_{\mathbb{R}^{p-1}} F(\mathbf{x})\,\mathrm{d}\mathbf{x}_{-i},\\ f_{X_i}(x_i)&:=\frac{\partial}{\partial x_i}F_{X_i}(x_i)=\int_{\mathbb{R}^{p-1}} f(\mathbf{x})\,\mathrm{d}\mathbf{x}_{-i}, \end{align*}
where $$\mathbf{x}_{-i}:=(x_1,\ldots,x_{i-1},x_{i+1},x_p)$$. The definitions can be extended analogously to the marginals of the cdf and pdf of different subsets of $$\mathbf{X}$$.
The conditional cdf and pdf of $$X_1\vert(X_2,\ldots,X_p)$$ are defined, respectively, as
\begin{align*} F_{X_1\vert \mathbf{X}_{-1}=\mathbf{x}_{-1}}(x_1)&:=\mathbb{P}[X_1\leq x_1\vert \mathbf{X}_{-1}=\mathbf{x}_{-1}],\\ f_{X_1\vert \mathbf{X}_{-1}=\mathbf{x}_{-1}}(x_1)&:=\frac{f(\mathbf{x})}{f_{\mathbf{X}_{-1}}(\mathbf{x}_{-1})}. \end{align*}
The conditional expectation of $$Y\vert X$$ is the following random variable2
\begin{align*} \mathbb{E}[Y\vert X]:=\int y \,\mathrm{d}F_{Y\vert X}(y\vert X). \end{align*}
The conditional variance of $$Y|X$$ is defined as
\begin{align*} \mathbb{V}\mathrm{ar}[Y\vert X]:=\mathbb{E}[(Y-\mathbb{E}[Y\vert X])^2\vert X]=\mathbb{E}[Y^2\vert X]-\mathbb{E}[Y\vert X]^2. \end{align*}
Proposition 1.1 (Laws of total expectation and variance) Let $$X$$ and $$Y$$ be two random variables.
• Total expectation: if $$\mathbb{E}[|Y|]<\infty$$, then $$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\vert X]]$$.
• Total variance: if $$\mathbb{E}[Y^2]<\infty$$, then $$\mathbb{V}\mathrm{ar}[Y]=\mathbb{E}[\mathbb{V}\mathrm{ar}[Y\vert X]]+\mathbb{V}\mathrm{ar}[\mathbb{E}[Y\vert X]]$$.
Exercise 1.1 Prove the law of total variance from the law of total expectation.
We conclude with some useful inequalities.
Proposition 1.2 (Markov’s inequality) Let $$X$$ be a non-negative random variable with $$\mathbb{E}[X]<\infty$$. Then
\begin{align*} \mathbb{P}[X>t]\leq\frac{\mathbb{E}[X]}{t}, \quad\forall t>0. \end{align*}
Proposition 1.3 (Chebyshev’s inequality) Let $$\mu=\mathbb{E}[X]$$ and $$\sigma^2=\mathbb{V}\mathrm{ar}[X]$$. Then
\begin{align*} \mathbb{P}[|X-\mu|\geq t]\leq\frac{\sigma^2}{t^2},\quad \forall t>0. \end{align*}
Exercise 1.2 Prove Markov’s inequality using $$X=X1_{\{X>t\}}+X1_{\{X\leq t\}}$$. Then prove Chebyshev’s inequality using Markov’s. Hint: use the random variable $$(X-\mathbb{E}[X])^2$$.
Proposition 1.4 (Cauchy–Schwartz inequality) Let $$X$$ and $$Y$$ such that $$\mathbb{E}[X^2]<\infty$$ and $$\mathbb{E}[Y^2]<\infty$$. Then $$\mathbb{E}[|XY|]\leq\sqrt{\mathbb{E}[X^2]\mathbb{E}[Y^2]}$$.
Proposition 1.5 (Jensen’s inequality) If $$g$$ is a convex function, then $$g(\mathbb{E}[X])\leq\mathbb{E}[g(X)]$$.
Example 1.1 Jensen’s inequality has interesting derivations. For example:
• Take $$h=-g$$. Then $$h$$ is a concave function and we have that $$h(\mathbb{E}[X])\geq\mathbb{E}[h(X)]$$.
• Take $$g(x)=x^r$$ for $$r\geq 1$$. Then $$\mathbb{E}[X]^r\leq \mathbb{E}[X^r]$$. If $$0<r<1$$, then $$\mathbb{E}[X]^r\geq \mathbb{E}[X^r]$$.
• Consider $$0\leq r\leq s$$. Then $$g(x)=x^{r/s}$$ is convex and $$g(\mathbb{E}[|X|^s])\leq \mathbb{E}[g(|X|^s)]=\mathbb{E}[|X|^r]$$. As a consequence $$\mathbb{E}[|X|^r]<\infty\implies\mathbb{E}[|X|^s]<\infty$$ for $$0\leq r\leq s$$. Finite moments of higher order implies finite moments of lower order.
1. Recall that the $$X$$-part is random!↩︎
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# Thread: matrix powers ... plz help me!!
1. ## matrix powers ... plz help me!!
X = [1 1]
......[1 1]
Y = [1 -1]
.....[-1 1]
Let A = aX and B= bY, where a and b are constants.
Use different values of a and b to calculate A^2, A^3, A^4 ...., B^2, B^3...
By consedering integer powers of A and B, find expression for A^n, B^n , (A+B)^n
2. Hello,
X^2=2X, X^3=(2X)X=2^2X^2,...
Y^2=2E, Y^3=(2E)Y=2Y,...
Now, A^n=(aX)^n=a^nX^n.
For (A+B)^n, compute them for n=2,3,4,... to see what is going on or just use the binomial theorem.
Bye.
3. Originally Posted by wisterville
Hello,
X^2=2X, X^3=(2X)X=2^2X^2,...
Y^2=2E, Y^3=(2E)Y=2Y,...
Now, A^n=(aX)^n=
.
For (A+B)^n, compute them for n=2,3,4,... to see what is going on or just use the binomial theorem.
Bye.
Thanks,
but u wrote, 2^2X^2 , does this means 2^4X ??
can u please show one example?? thanks
4. Hello,
$\displaystyle X^2=2X$,
$\displaystyle X^3=(X^2)X=(2X)X=2(X^2)=2(2X)=2^2X=4X$,
$\displaystyle X^4=(X^3)X=(2^2X)X=2^2(X^2)=2^2(2X)=2^3X=8X$,
$\displaystyle X^5=(X^4)X=(2^3X)X=2^3(X^2)=2^3(2X)=2^4X=16X$,...
To summarize, $\displaystyle X^n=2^{n-1}X$.
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https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/6_Hyperbolic_functions/6.7_Miscellaneous/6.7.1_Hyperbolic_functions/rese387.htm
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### 3.387 $$\int \sinh (a+b x) \tanh ^2(a+b x) \, dx$$
Optimal. Leaf size=21 $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$
[Out]
Cosh[a + b*x]/b + Sech[a + b*x]/b
________________________________________________________________________________________
Rubi [A] time = 0.0242972, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {2590, 14} $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$
Antiderivative was successfully verified.
[In]
Int[Sinh[a + b*x]*Tanh[a + b*x]^2,x]
[Out]
Cosh[a + b*x]/b + Sech[a + b*x]/b
Rule 2590
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
Rule 14
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Rubi steps
\begin{align*} \int \sinh (a+b x) \tanh ^2(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}\\ \end{align*}
Mathematica [A] time = 0.0285767, size = 21, normalized size = 1. $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$
Antiderivative was successfully verified.
[In]
Integrate[Sinh[a + b*x]*Tanh[a + b*x]^2,x]
[Out]
Cosh[a + b*x]/b + Sech[a + b*x]/b
________________________________________________________________________________________
Maple [A] time = 0.016, size = 32, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{\cosh \left ( bx+a \right ) }}+2\,\cosh \left ( bx+a \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(b*x+a)^2*sinh(b*x+a)^3,x)
[Out]
1/b*(-sinh(b*x+a)^2/cosh(b*x+a)+2*cosh(b*x+a))
________________________________________________________________________________________
Maxima [B] time = 1.03865, size = 73, normalized size = 3.48 \begin{align*} \frac{e^{\left (-b x - a\right )}}{2 \, b} + \frac{5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1}{2 \, b{\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="maxima")
[Out]
1/2*e^(-b*x - a)/b + 1/2*(5*e^(-2*b*x - 2*a) + 1)/(b*(e^(-b*x - a) + e^(-3*b*x - 3*a)))
________________________________________________________________________________________
Fricas [A] time = 1.94858, size = 85, normalized size = 4.05 \begin{align*} \frac{\cosh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{2} + 3}{2 \, b \cosh \left (b x + a\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="fricas")
[Out]
1/2*(cosh(b*x + a)^2 + sinh(b*x + a)^2 + 3)/(b*cosh(b*x + a))
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{3}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(b*x+a)**2*sinh(b*x+a)**3,x)
[Out]
Integral(sinh(a + b*x)**3*sech(a + b*x)**2, x)
________________________________________________________________________________________
Giac [B] time = 1.20887, size = 61, normalized size = 2.9 \begin{align*} \frac{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{2 \, b} + \frac{2}{b{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")
[Out]
1/2*(e^(b*x + a) + e^(-b*x - a))/b + 2/(b*(e^(b*x + a) + e^(-b*x - a)))
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# Letters and Numbers
Letters are made of lines, some slanting but a lot are horizontal or vertical (straight across or up and down). Lace designs like slanting, but don't like vertical or horizontal lines! Tape lace gives you a lot more freedom, but this website is about strip lace. Strip lace has a limit to how wide you can make it, as well. All this makes letters a challenge! Here, I have the letters out of 'pixels' which are small diamonds. In the example, I have used half stitch diamonds, but you could use cloth stitch diamonds instead. I give an example of using these letters here, spelling out the name Ruth. Apart from the (large number of) half stitch diamonds, the strip starts with some rose ground. The rest is Torchon ground, with a simple twisted footside on each side. 20 pairs. You might have two comments for this. First, this is only 4 letters. What about the rest? Don't worry, I have given the design for the rest of the alphabet below! Next, why are the letters going downwards rather than across? The reason for this is that letters are quite complicated shapes, and making them out of diamonds ends up with a wide design. Letters are taller than they are wide, so writing them downwards like this does mean that they take up less room, and so use less bobbins. This strip uses 20 pairs, and in fact, the letters only take 14 pairs. The rest are the edge of the pattern. However, this does mean that the letters end up very narrow. Some of the letters are perhaps not very good! So I have always given a second version where the letters are the 'proper' way up. This allows the letters to be variable width, so the trickier letters can have more room. This version requires 26 pairs, of which 22 pairs are used for the letters. The patterns below look as if they are complete lines. In fact, every letter is a separate picture, and can be downloaded separately (right click on picture for Windows). I hope this will make it easier to draw up your own designs. I you want the whole alphabet, I suggest printing the whole webpage. A heart might be used for a romantic design! Designs for numbers are given as well.
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http://www.chegg.com/homework-help/questions-and-answers/disk-moment-inertia-84-kg-m2-free-rotate-without-friction-starting-rest-fixed-axis-center--q1251347
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A disk having moment of inertia 84 kg · m2 is free to rotate without friction, starting from rest, about a fixed axis through its center. A tangential force whose magnitude can range from F = 0 to F = 50.0 N can be applied at any distance ranging from R = 0 to R = 3.00 m from the axis of rotation.
(a) Find a pair of values F and R that cause the disk to complete 2.90 rev in 10.3 s. (Let F = 31.0 N be one half of the pair.)
R=
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https://news.nationalgeographic.com/2017/01/storm-teacup-physics-everyday-life-helen-czerski/?user.testname=none
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# These Simple Laws Explain How the World Works
## Bubbles, toasters, and the internet—most aspects of everyday life are influenced by physics.
Shackleton, Scott, and Amundsen faced many hardships on their treks to the South Pole, from frostbite to snow blindness. Few biographies, though, mention that much of their misery was caused by the ideal gas law, which governs the relationship between the pressure, volume, and temperature of a gas. It is this crucial law of physics, explains British physicist Helen Czerski in her new book Storm in a Teacup: The Physics of Everyday Life, that generates the fearsome katabatic winds, which rake the surface of Antarctica at almost 200 miles an hour.
This is just one example of how the science of physics informs every aspect of life, says Czerski. When National Geographic caught up with her by phone from London, she elaborated further—on the importance of bubbles (her specialty), the physics lessons taught by your toaster, and why it doesn't help to bang the bottom of the bottle if no ketchup comes out.
You are the first expert on bubbles that I have ever interviewed. Tell us how you got into this arcane field—and why it frequently involves voyages on ships in stormy seas.
[Laughs] Bubbles are the unsung heroes of the physical world. They’re everywhere, doing very useful things. No one pays attention to them, but they’re important because a bubble, which is made up of a liquid and a gas, can do things that the liquid and the gas, each individually, can’t do.
Imagine the froth on top of your cappuccino. If you put a spoon on top of that, the foam will support the spoon. But if you put that spoon on top of just gas, or just liquid, it would fall straight through. Mix the liquid and gas together and suddenly they can do something entirely new.
I got into the field of bubble physics because my Ph.D. was in experimental physics. I’m an experimentalist. I build things, and I was building experiments to use high-speed photography to look at events that were so small and so fast that they could be happening right in front of your eyes and you couldn’t see them. I was fascinated by the world that was right here but just beyond what you’re able to see.
Bubbles breaking apart and joining together also happens extremely quickly—and it’s extremely small—so you need those special techniques to see what’s happening. Now I study the bubbles underneath breaking waves, where lots of bubbles are breaking up in that washing machine kind of turbulence. When the bubbles go back down into the ocean they’re important for helping our oceans breathe. They’re like little packets of gas carried down into the water.
I go out on ships to measure bubbles, usually in storm season, in the Mid-Atlantic. You look out over the deck at that familiar sight of breaking waves, something seafarers have seen for centuries, but just a meter below the surface you can see something that they have never been able to see. We need to understand it because it’s part of the clockwork of our planet, one of the mechanisms making our planet run.
Some people might think of physicists as nerdy figures in lab coats remote from normal life. But your book shows us that in our everyday life we are all physicists. Give us some examples.
That’s a fantastically outdated vision of a physicist! [Laughs] Physics is all about the rules that make the world work, and we all use them every day whether we’re aware of it or not. We don’t necessarily use them systematically but we’re familiar with the way things work.
When you’re baking bread, for example, one of the main things you want is to have bubbles in your dough. You use various methods to put bubbles into bread. You can use a raising agent or yeast, or you can fold air into the bread. When you put that in a hot oven and heat the gas up, the molecules inside the bubbles will bounce off the walls like tiny bumper cars and push them outwards, which makes the bread expand and rise. That comes from something called the Ideal Gas Law: the idea that the pressure, temperature, and volume of a pocket of gas are related to each other. Once you’ve got that little bit of physics, you’ve got the internal combustion engine, winds from Antarctica, and popcorn.
One of life’s small hassles is getting ketchup out of the bottle. What does physics tell us about it—and why doesn’t banging the bottom of the bottle help?
It’s a great disappointment to me that ketchup bottles are now made out of plastic rather than glass because now you can squeeze them and you don’t get to appreciate this nice bit of physics. We’ve all been in a pub or a diner, where you have a glass bottle of ketchup. You turn it upside-down, shake it, hit the bottom, and the ketchup doesn’t come out. Then it all comes out at once! [Laughs]
That’s not random. It happens because ketchup has got this weird property known as shear thinning. What that means is, it’s really viscous until you force it to move a little bit. When you’re shaking the bottle, the ketchup can’t go anywhere, so it stays thick. Once you hit it hard enough that it has to go somewhere, then it becomes runny, so a whole load of ketchup comes out at once.
The trick is to hold the bottle at a slight angle and tap the neck of the bottle, because then you’re making it runny where you need it to be runny. Everything further up the bottle stays nice and thick. Instead of having ketchup splodge everywhere, you can control the amount that comes out.
You’re British so tea and toast play an inordinate role in your life. There’s lots of physics in them, isn’t there?
The best thing about physics for me is that the same pattern explains lots of different things. There’s no reason why you have to see that pattern in a posh lab wearing a lab coat with some expensive bit of kit. You can see those patterns in your cup of tea. When you stir milk into your tea, the white milk doesn’t just merge with the dark tea. It swirls around. Then we look up into the sky at the big, rotating storms that come past at our latitudes, and there are also two fluids mixing together, via swirling. Even though looking at your teacup is one of the most mundane things in daily life, what you see there applies in other places.
It’s the same with toast. You shouldn’t worry about the universe until you understand your toaster because there’s a lot of fundamental physics in your toaster. [Laughs] One is, they glow. It sounds obvious. It’s hot so it glows, right? But that dull red glow that gets brighter and brighter as it gets hotter comes from one of the most fundamental laws of physics. That law says that anything that has a temperature above absolute zero is glowing. The type of light it gives out depends on how hot it is. So, the hotter it gets, the shorter the wavelengths.
The other interesting thing in a toaster is an electromagnet. When you push the lever down on the side of the toaster and the toast stays down inside the toaster, the thing holding it down is an electromagnet. The lever connects up a little circuit and sends a current round in a spin. And when a current is going round in circles, the laws of physics say that it will generate a magnetic field. When the toast is done, the current is switched off, the magnetic field goes away, and your toast pops up. You’ve got two of the most essential foundations of physics. And they are both in your toaster!
You say, “The fabric of our civilization is woven together with electromagnetic threads.” Why is electromagnetism so important?
Each of us has three life support systems: We have a body, a planet, and a civilization. Our civilization is keeping us alive now in just the same way that the other two do. We are absolutely dependent on it. And the thing that connects our civilization together is electromagnetism.
There are two forms: visible light—signals we can see from a long way away—which has connected humans for a long time. And now there are the signals we can’t see: radio signals, Wi-Fi signals, mobile phone signals. They’re making a web of electromagnetic waves that we walk through. We don’t see them directly but we use radios, TV, and the Internet, which all depend on them.
One hundred years ago that didn’t exist! Earth was dark in the radio wave and microwave wavelengths. Today, we’ve lit it up with this web of light. Light that we can’t see! But that is communicating! The fabric of our civilization is now stitched together with that communication. If you want to buy something, need medical help, or want to learn something today, you are absolutely reliant on signals being sent to you by electromagnetic waves. We take this web of communication for granted. But we should really appreciate it as a luxury.
In both the U.K. and the U.S., there is a steady decline of high school students opting for physics, particularly among girls. What can be done to make the subject more attractive?
When people hear the word physics, they think quantum weirdness or cosmology, something extremely tiny or very big. What we’re not talking about is all the stuff in the middle. The equations for Newton’s Laws of Motion can be written down quite easily. But those simple laws can help us understand the way our human systems and bodies work, and you can see their effects in the real world. That is a very important reason to study physics.
Physics has this image of being terribly po-faced: a guy with a beard working late at night in a dark room. There’s nothing human in that view of physics. But physics is about making things happen, and we’re not doing well enough at conveying the message that once you understand the framework of the world, you have a different perspective. Even if you’re not going to become a physicist, that is a useful thing for a citizen to have.
Physics is also about how things work. During your lifetime you’re going to be sold toasters or washing machines or cars, and you need to know which questions to ask. Physics is about giving you the basic map of the world.
This interview was edited for length and clarity.
Simon Worrall curates Book Talk. Follow him on Twitter or at simonworrallauthor.com.
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# FINANCIAL RATIO ANALYSIS: Ratio analysis is an important and age-old technique of financial analysis.
However, the ratios are only indicators, they cannot be taken as final regarding good or bad financial position of the business other things have also to be seen. Most commonly used ratios in banking sector can be divided into following five main categories: Ratios and Their Analysis: Liquidity Ratios Leverage Ratios Profitability Ratios Activity Ratios Market Ratios Liquidity Ratios: Liquidity ratios measure a firms ability to meet its current obligations. These include: Current Ratio= Current Assets / Current Liabilities Year Total Current assets Total Current Liabilities Current Ratio 2008
406543703 363,489,463
2009
468170745 420,467,889
1.118
1.113
## 2 1.5 2008 1 0.5 0 2009 2010
Interpretation: As we know that current ratio is an important ratio which shows the solvency position of the organization, in above values we come to know that the value of current ratio is increasing in 2008 but in 2009 there is a little decrease in current ratio.again it is increasing in 2010 which is a healthy sign for bank. Working of Total current Assets and current Liabilities
Current Asset 2008 2009 2010
Cash and balances with treasury banks Balance with other Banks Lending to financial institutions Investments net Advances net
## 450407 1479 4402 213061 254551
923900
Total Current Asset Current Liabilities 406543703 468170745
8,201,090
10266
## Total Current Liabilities
363,489,463
420,467,889 467323
Acid Test Ratio= (Current Assets-Inventories- prepaid Expenses)/Current liabilities Year Cash and balances with treasury banks investments Pre-paid Expenses Current Liabilities Acid Test Ratio 2008 39,631,172 96,256,874
2612432 363,489,463
## 2009 38,774,871 167,134,465
2,889.208 420,467,889
0.367
0.483
## 0.6 0.5 0.4 0.3 0.2 0.1 0 2008 2009 2010
Interpretation: This ratio provides a more penetrating measure of liquidity than does the current ratio. Also known as the Quick ratio, it examines the businesss liquidity position by comparing current assets and liabilities but it omits stock from the total of current assets. The reason for this is stock is the most illiquid current asset In 2008 Acid Test Ratio is Low but in 2009 and 2010 it is increasing. Working Capital Ratio = (Current assets-Current liabilities)
## Year Current Assets
2008
406543703
2009
468170745 420,467,889
47702856
## 50000000 40000000 30000000 20000000 10000000 0 2008 2009 2010
Interpretation: It depends upon the size and nature of business. From the above analysis it is observed that the current assets of MCB are increase as compare to its liabilities as compare to previous years.
Leverage Ratios: Leverage ratios measure the degree of protection of suppliers of long term funds. Time interest Earned Ratio = earning before tax + interest exp/ interest exp Year EBIT Interest Expenses
## Time interest Earned 2.886 Ratio
2.462 times
2.459 times
2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2008 2009 2010
Interpretation: In 2008 the position of company to pay the interest payment is satisfactory & in 2009 it goes down, in 2010 it is also getting down. It Measure the firm's ability to meet interest payments out of its operating profit.
Debt ratio = (Total Liabilities /Total Assets) Year Total Liabilities Total Assets Debt Ratio 2008 385,179,850 443,615,904 0.87 2009 439,483,714 509,223,727 0.86 2010 489566 568770 0.86
0.87 0.868 0.866 0.864 0.862 0.86 0.858 0.856 0.854 2008 2009 2010
Interpretation: Measures the percentage of assets that have been financed by debt (borrowings). This ratio indicates that the value is decreasing, which is favorable for the bank. Ratio must be decrease for good health of the bank. Debt / Equity Ratio = (Total Liabilities /Shareholders equity) 385, 179, 85/1000 439,483,714/1000 Year Total Liabilities Total Equity Debt / Equity Ratio 2008 385,179.85 33459 11.5 2009 439,483.714 35037 12.5 2010 489566 55364 8.8 times
## 14 12 10 8 6 4 2 0 2008 2009 2010
Interpretation: In 2008 this ratio was 11.5 indicating that company has satisfactory equity resources to compete its short term and long term debt and in 2009 it become 12.5 and it is showing the companys strong equity position but in 2010 it decreased to 8.8, which indicates that company accept loan from the financial institution. It shows how much value there would be in a liquidation of the company.
Total Capitalization Ratio = Long term debt / Long term debt + equity Year 2008 2009 2010
21783
19016
33459
35037 0.35
## Total 0.39 Capitalization Ratio 21026337/1000 = 21026
Long term debt = (Other Liabilities + Deferred tax liabilities) 2008 Long term debt= (21346+ 437) = 21783 2009 Long term debt = (15819 + 3197) = 19016
2010 Long term debt = (16092 +4934) =21026
0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 2008 2009 2010
Interpretation: It indicates long-term debt usage. Low debt and high equity levels in the capitalization ratio indicate
investment quality. This
ratio indicates that the value is decreasing, which is favorable for the
bank Profitability Ratios: Profitability ratios measure the earning ability of a firm. These include: Return on Assets (ROA) = (Net Profit / Total Assets) x 100
## 2010 16873 568770 2.96%
3.5 3.4 3.3 3.2 3.1 3 2.9 2.8 2.7 2008 2009 2010
Interpretation: The return on assets decreases first in 2009, than in 2008, which is favorable for the bank, as the bank is generating high return on assets.
DuPont Return on Assets = net profit margin *total asset turnover Year Net Profit total asset turnover DuPont Return on Assets 15,374,600/1000 15,495,297/1000 2008 15,375 0.0964 1482 2009 15,495 0.1014 1571 2010 16873 0.0903 1523
## Asset Turnover= Revenue / total Asset
2008 2009 2010 =40044 / 443,616 =0.0903 = 51616/ 509,224 = 0.1014 = 54821/568770 =0.0964
## (to plot on graph I divide it by 100)
15.8 15.6 15.4 15.2 15 14.8 14.6 14.4 14.2 2008 2009 2010
Interpretation: A combination of financial ratios is a series to evaluate investment return. The benefit of the method is that it provides an understanding of how the company generates its return. In 2009 decrease show that return on investment decrease.
Return on Operating Assets = Net profit / Operating Fixed Assets * 100 Year Net profit Operating Assets Return Operating Assets 2008 15,374,600 Fixed 15,374,600 on 95% 2009 15,495,297 18,014,896 86% 2010 16873 20947 80.55%
## 95 90 85 80 75 70 2008 2009 2010
Interpretation: It Measures the effectiveness of management at making a profit and using the assets efficiently. Increases in 2009, which means it is favorable for the bank. Return on Total Equity= Net profit / Total Equity *100 15,374,600/1000 = 15,374.600 15,495,297/1000 =15,495.297 Year Net profit Total Equity Return on Equity 2008 15,375 33459 Total 0.46 2009 15,495. 35037 0.44 2010 16873 55364 0.30
## 0.5 0.4 0.3 0.2 0.1 0 2008 2009 1010
Interpretation: Return on equity continuously decreases in 2009 and in 2010, which means it is unfavorable for the bank. The value of the ratio needs to be increase for healthy organizations.
Market Ratios:
Market ratios are commonly used by the investors to assess the performance of a business as an investment and also the cost of issuing stock. These include: Dividend per share = Total dividend / No. of outstanding shares
## 2010 7843 760 11.50
11.5 11.4 11.3 11.2 11.1 11 10.9 10.8 10.7 2008 2009 2010
Interpretation: Dividend per share (DPS) is the total dividends paid out over an entire year divided by the number of outstanding ordinary shares issued. in 2008 it was 11.50 and it decrease in 2009 at 11 in 2010 it increase again 11.50. Price earning ratio = Market value of per share / earning per share 2010: 228.54 / 22.20 2009: 219.68 / 20.38 2008: 125 / 20.22 Year 2008 Market value of 125 per share earning per share 20.22 Price earning ratio 6.18 =10.29 =10.77 =6.18 2009 219.68 20.38 10.77 2010 228.54 22.20 10.29
## 12 10 8 6 4 2 0 2008 2009 2010
Earnings per share= Net Income / total outstanding shares Year 2008 Net Income 15374 total outstanding 628 shares Earnings per share 20.22 2009 15495 691 20.38 2010 16873 760 22.20
## 22.5 22 21.5 21 20.5 20 19.5 19 2008 2009 2010
Interpretation: Investors are much interested to know whether a companys performance is increasing or decreasing. From an investors point of view, the organization earning per share is more important and investors can easily understand the firms profitability.
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# Week 13 questions | Education homework help
## Week-13DiscussionsSummer2023.docx
DSRT 734 M51
INFERENTIAL STATISTICS FOR DECISION MAKING
After reading your textbook, I want you to have a good understanding of the fundamentals of each chapter and show it to me. Please don’t copy & paste from your textbook or some other online source. In other words, don’t plagiarize. You can read online material if it helps to understand the material, but you have to write your own sentences.
Chapter-14 topic Discussions and questions:
1. Describe the (Ho) null hypothesis for
a test of independence.
2. An investigator was interested in the relationship between color preference and number of siblings. A test of independence produced a 2 that allowed the null hypothesis to be rejected. Write a proper conclusion for this test result.
3. Explain the
short cut method of calculating 2 and give a numerical example
4. Describe
Phi and the
odds ratio
5. A social psychologist hypothesized that a factor in juvenile delinquency was the presence or absence of a strong father-figure in the home. He examined the folders of 100 inmates in the federal reformatory and found that only 50 of these young men grew up with a strong father-figure in the home. He also examined the records of 100 randomly selected male college students and found that 70 of them had strong father-figures in their boyhood homes.
Use the
chi square method to test the psychologist’s hypothesis.
Chapter-14
1. A labor official predicted that the following percentages of makes of cars would be seen passing a picket line at an automobile plant where a strike was in progress.
General Motors Ford Chrysler Foreign Brand
38% 28% 24% 10%
The following numbers of cars were counted. Analyze the data and write a conclusion about the official’s prediction.
General Motors Ford Chrysler Foreign Brand
114 72 75 41
1. On a test of independence between alcoholism and early toilet training, a clinical researcher found a 2 = 6.48. With
df = 1 and = .05, write a conclusion about the relationship between the two variables.
Pages (275 words)
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > funeldmdif Structured version Visualization version GIF version
Theorem funeldmdif 7742
Description: Two ways of expressing membership in the difference of domains of two nested functions. (Contributed by AV, 27-Oct-2023.)
Assertion
Ref Expression
funeldmdif ((Fun 𝐴𝐵𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) ↔ ∃𝑥 ∈ (𝐴𝐵)(1st𝑥) = 𝐶))
Distinct variable groups: 𝑥,𝐴 𝑥,𝐵 𝑥,𝐶
Proof of Theorem funeldmdif
StepHypRef Expression
1 funrel 6360 . . 3 (Fun 𝐴 → Rel 𝐴)
2 releldmdifi 7739 . . 3 ((Rel 𝐴𝐵𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) → ∃𝑥 ∈ (𝐴𝐵)(1st𝑥) = 𝐶))
31, 2sylan 583 . 2 ((Fun 𝐴𝐵𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) → ∃𝑥 ∈ (𝐴𝐵)(1st𝑥) = 𝐶))
4 eldif 3929 . . . 4 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
5 1stdm 7734 . . . . . . . . . . . . . 14 ((Rel 𝐴𝑥𝐴) → (1st𝑥) ∈ dom 𝐴)
65ex 416 . . . . . . . . . . . . 13 (Rel 𝐴 → (𝑥𝐴 → (1st𝑥) ∈ dom 𝐴))
71, 6syl 17 . . . . . . . . . . . 12 (Fun 𝐴 → (𝑥𝐴 → (1st𝑥) ∈ dom 𝐴))
87adantr 484 . . . . . . . . . . 11 ((Fun 𝐴𝐵𝐴) → (𝑥𝐴 → (1st𝑥) ∈ dom 𝐴))
98com12 32 . . . . . . . . . 10 (𝑥𝐴 → ((Fun 𝐴𝐵𝐴) → (1st𝑥) ∈ dom 𝐴))
109adantr 484 . . . . . . . . 9 ((𝑥𝐴 ∧ ¬ 𝑥𝐵) → ((Fun 𝐴𝐵𝐴) → (1st𝑥) ∈ dom 𝐴))
1110impcom 411 . . . . . . . 8 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵)) → (1st𝑥) ∈ dom 𝐴)
12 funelss 7741 . . . . . . . . . . 11 ((Fun 𝐴𝐵𝐴𝑥𝐴) → ((1st𝑥) ∈ dom 𝐵𝑥𝐵))
13123expa 1115 . . . . . . . . . 10 (((Fun 𝐴𝐵𝐴) ∧ 𝑥𝐴) → ((1st𝑥) ∈ dom 𝐵𝑥𝐵))
1413con3d 155 . . . . . . . . 9 (((Fun 𝐴𝐵𝐴) ∧ 𝑥𝐴) → (¬ 𝑥𝐵 → ¬ (1st𝑥) ∈ dom 𝐵))
1514impr 458 . . . . . . . 8 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵)) → ¬ (1st𝑥) ∈ dom 𝐵)
1611, 15eldifd 3930 . . . . . . 7 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵)) → (1st𝑥) ∈ (dom 𝐴 ∖ dom 𝐵))
17163adant3 1129 . . . . . 6 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵) ∧ (1st𝑥) = 𝐶) → (1st𝑥) ∈ (dom 𝐴 ∖ dom 𝐵))
18 eleq1 2903 . . . . . . 7 ((1st𝑥) = 𝐶 → ((1st𝑥) ∈ (dom 𝐴 ∖ dom 𝐵) ↔ 𝐶 ∈ (dom 𝐴 ∖ dom 𝐵)))
19183ad2ant3 1132 . . . . . 6 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵) ∧ (1st𝑥) = 𝐶) → ((1st𝑥) ∈ (dom 𝐴 ∖ dom 𝐵) ↔ 𝐶 ∈ (dom 𝐴 ∖ dom 𝐵)))
2017, 19mpbid 235 . . . . 5 (((Fun 𝐴𝐵𝐴) ∧ (𝑥𝐴 ∧ ¬ 𝑥𝐵) ∧ (1st𝑥) = 𝐶) → 𝐶 ∈ (dom 𝐴 ∖ dom 𝐵))
21203exp 1116 . . . 4 ((Fun 𝐴𝐵𝐴) → ((𝑥𝐴 ∧ ¬ 𝑥𝐵) → ((1st𝑥) = 𝐶𝐶 ∈ (dom 𝐴 ∖ dom 𝐵))))
224, 21syl5bi 245 . . 3 ((Fun 𝐴𝐵𝐴) → (𝑥 ∈ (𝐴𝐵) → ((1st𝑥) = 𝐶𝐶 ∈ (dom 𝐴 ∖ dom 𝐵))))
2322rexlimdv 3275 . 2 ((Fun 𝐴𝐵𝐴) → (∃𝑥 ∈ (𝐴𝐵)(1st𝑥) = 𝐶𝐶 ∈ (dom 𝐴 ∖ dom 𝐵)))
243, 23impbid 215 1 ((Fun 𝐴𝐵𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) ↔ ∃𝑥 ∈ (𝐴𝐵)(1st𝑥) = 𝐶))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 ∧ w3a 1084 = wceq 1538 ∈ wcel 2115 ∃wrex 3134 ∖ cdif 3916 ⊆ wss 3919 dom cdm 5542 Rel wrel 5547 Fun wfun 6337 ‘cfv 6343 1st c1st 7682 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-sep 5189 ax-nul 5196 ax-pow 5253 ax-pr 5317 ax-un 7455 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ral 3138 df-rex 3139 df-rab 3142 df-v 3482 df-sbc 3759 df-dif 3922 df-un 3924 df-in 3926 df-ss 3936 df-nul 4277 df-if 4451 df-sn 4551 df-pr 4553 df-op 4557 df-uni 4825 df-int 4863 df-br 5053 df-opab 5115 df-mpt 5133 df-id 5447 df-xp 5548 df-rel 5549 df-cnv 5550 df-co 5551 df-dm 5552 df-rn 5553 df-res 5554 df-iota 6302 df-fun 6345 df-fn 6346 df-fv 6351 df-1st 7684 df-2nd 7685 This theorem is referenced by: satffunlem2lem2 32710
Copyright terms: Public domain W3C validator
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# Storage of Electrical Power
by Ron Kurtus (13 May 2014)
Besides being able to create electricity, it would be nice to be able to store that power and use it at a later time. There are several methods to do that.
Static electricity can be stored in a Leyden jar, which allows you to release the electrical charges when you want to do that. Direct current (DC) electricity can be stored in a capacitor and a rechargeable battery. Batteries can also e used to create DC electricity.
Unfortunately, there is no way to store alternating current (AC) electricity, although it can be obtained from stored DC power.
Questions you may have include:
• How is static electricity stored?
• How is DC electricity stored?
• How can AC be obtained from DC?
This lesson will answer those questions. Useful tool: Units Conversion
## Static electricity
A common way to store static electrical charges, such that they can be discharged at will, is with a Leyden jar. This simple device consists of a glass jar with metal foil wrapped both inside and outside of the jar. A metal rod—often with a metal chain on its end—is located from an insulating stopped at the top of the jar.
Drawing of Leyden jar
from 1914 book Wireless Telegraphy - Wikimedia Commons
The way the Leyden jar works is that you put a static electrical charge on the metal ball, which then is carried to the inside metal foil. The electrical charges on the inside metal foil induce opposite electrical charges on the outside foil. The glass insulates the charges and allows for a potential difference buildup, thus "storing" the static electricity.
Touching the metal ball will cause a discharge of the electrical power.
## Direct current electricity
Direct current (DC) electricity can be stored in a capacitor—which is similar to a Leyden Jar—or in a rechargeable battery.
### Capacitor
A capacitor—also called a condenser—consists of two terminals attached to metal plates, separated by a thin dielectric material. Although any insulating material can be used as a separator, certain materials are more suitable in capacitors. Mica, ceramic, cellulose, Mylar, and even air are some of the non-conductive materials used in capacitors.
When DC electricity is applied to the positive (+) and negative (−) terminals, they collect and build up charges, which can be released when connecting the terminals to an electrical circuit.
DC circuit with capacitor
A light bulb in a DC circuit will glow until the capacitor is completely charged. At that time, no current passes through the circuit. Taking the battery out of the circuit will allow the capacitor to discharge and light up the bulb again for a short time.
### Rechargeable battery
A regular battery creates DC electricity through a chemical reaction of metal plates and an acidic solution. In a rechargeable battery, the process can be reversed, such that a spent battery can become charged again. Thus, a rechargeable battery can store DC electrical power.
## Alternating current electricity
Because the direction of the current changes in AC electricity, you cannot directly store the power. Placing a capacitor in an AC circuit has no effect on the alternating flow of the electricity.
The only way it can e stored is indirectly, by storing DC and then using a power inverter to convert the DC to AC. But this really isn't storing AC.
## Summary
It is desirable to store electrical power and use it at a later time.
Static electricity can be stored in a Leyden jar, Direct current (DC) electricity can be stored in a capacitor and a rechargeable battery. Unfortunately, there is no way to store alternating current (AC) electricity, although it can be obtained from stored DC power.
Charge up your life with good experiences
## Resources and references
Ron Kurtus' Credentials
### Websites
Leyden Jar - How Stuff Works
Leyden jar - Wikipedia
How Capacitors Work - How Stuff Works
Capacitor - Wikipedia
Rechargeable battery - Wikipedia
How DC/AC Power Inverters Work - How Stuff Works
DC and AC Electricity Resources
Physics Resources
### Books
Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.
## Students and researchers
www.school-for-champions.com/science/
electrical_storage.htm
## Where are you now?
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Physics topics
## Also see
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Be valuable to others
Have utmost character
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The School for Champions helps you become the type of person who can be called a Champion.
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E-M waves in dielectrics
1. Aug 27, 2006
quasar987
Let's see if I understand this correctly... In an isotropic material of conductivity $\sigma \neq 0$ (and charge density $\rho=0$, the 4th Maxwell equation has a non-null current term $\vec{J}=\sigma \vec{E}$ so the resulting wave equations for $\vec{E}$ and $\vec{H}$ take the form
$$\nabla ^2\vec{E}=\frac{\epsilon \mu}{c^2}\frac{\partial^2}{\partial t^2}\vec{E}+\frac{4\pi \mu \sigma}{c^2}\frac{\partial}{\partial t}\vec{E}$$
and the same thing for $\vec{H}$. (I am using the same unit convention as in Greiner's 'Classical Electrodynamics' book)
If we try a solution of the form
$$\vec{E}=\vec{E}_0e^{i(\vec{k}\cdot \vec{r}-\omega t + \phi)}$$
(the monochromatic plane wave solution), we find that it is a solution provided that the wave number vector satisfies the following complex dispersion relation:
$$|\vec{k}|^2=k^2=\epsilon \mu\frac{\omega^2}{c^2}\left( 1+\frac{4\pi i\sigma}{\epsilon \omega}\right)$$
We know that if a complex function satisfies a linear diff. equ. such as the above "wave equation", its real and imaginary part taken separately are also solutions. In vacuum, separating the real and imaginary part of the sine wave solution was easy because $\vec{k}$ was real. Now, $\vec{k}$ is non-real as soon as the conductivity is non-zero, and to get the real part of $\vec{E}_0e^{i(\vec{k}\cdot \vec{r}-\omega t + \phi)}$, we must decompose the exponential according to Euler's formula and then expand the sine and cosine in their Taylor series, and finally group together the real parts and the imaginary parts.
At this point I don't know what to think. This seems rather impractical. Does it turn out in the end that a monochromatic plane wave solutions (with real wave vector $\mathcal{K}$) exists? Please comment!
Last edited: Aug 27, 2006
2. Aug 27, 2006
Claude Bile
Pure plane wave solutions do not exist because of the attenuation that inevitably occurs as a wave propagates through a dielectric. The degree of attenuation is determined by the imaginary component of the refractive index.
If the attenuation is small, then a plane wave solution can be an excellent approximation to the full solution. More often though when modelling, theorists neglect the attenuation terms altogether, preferring instead to tack it on later, as loss is a figure that is typically measured rather than worked out using first principles, in the optical region of the EM spectrum at least.
(On a side note with regard to your second last paragraph - Typically the real terms are brought out into a second exponential so you get two exponentials multiplied by one another, one representing the attenuation and one representing the oscillation.)
Claude.
Last edited: Aug 27, 2006
3. Aug 28, 2006
quasar987
Ah, of course!
4. Aug 28, 2006
Meir Achuz
The wave propagates like exp(ikz). You calculate k as
k=k_r+k_i=[k^2]^1/2. This is a bilt complicated, but is done in EM textbooks. Then the proagation is exp[ik_r] exp[-k_i],
showing the attenuation.
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Verbal Woes : GMAT Verbal Section
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# Verbal Woes
Author Message
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Joined: 09 Jul 2009
Posts: 7
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Schools: Georgia, Florida, Washington(st.louis), Texas A&M, Kentucky, Tulane
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15 Jul 2009, 20:46
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I have been taking practice test (Manhattan review) every week for the past 2 months and I have a pretty good idea what I need to be studying my last month before I take my GMAT. I get anywhere from a 42-45 on my Quantitative but only a 24-30 on my Verbal section.
I have read all the sections and done the practice problems (including understanding my mistakes and making sure I understand everything) on:
1. Kaplan Premier Program 2009 Edition
2. Manhattan Review: Verbal Study Guide
3. Princton Review Cracking the GMAT 2009
4. GMAT Review 11th Edition: The Official Guide
I feel like I am capable of doing well on this section (at least 35-40), but I think I may need a little help. Are there anyways that I can really nip this thing.
On my last practice I got 8/14 on Reading Comp, 7/15 on Sentence Correction, and 8/12 on Critical Reasoning.
Normally I do much better on Sentence Correction than I did on this attempt, but I still need to do better. Critical Reasoning is my strongest part of the Verbal, and I have some issues with the "harder" Reading Comprehension questions.
Any help will be greatly appreciated.
If you have any questions
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16 Jul 2009, 00:11
Re: Verbal Woes [#permalink] 16 Jul 2009, 00:11
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## Deductive Reasoning Q?
Read the following deductive reasoning puzzle and keep in mind - need to know:
What was each woman's occupation and dress color?
Plus. Explain How did you get the answer?
Alice, Bettym Carol, and Dorothy were a lifeguard, a lawyer, a pilot, or a professor. Each wore a white, yellow, pink, or blue dress.
The lifeguard bear Betty at canasta, and Carol and the pilot often played bridge with the women in pink and blue dresses, Alice and the professor envied the woman in the blue dress, bit this was not the lawyer, as she always wore a white dress.
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
Blog Entries: 1 Recognitions: Gold Member The answer that the puzzler is looking for is: Alice - pilot - yellow Betty - professor - pink Carol - lawyer - white Dorothy - lifeguard - blue The key I used to get started is: 1. The attorney wears white (and thus does not wear blue). 2. The pilot plays bridge with the woman in blue (ditto). 3. The professor envies the woman in blue (ditto). So only the lifeguard can wear blue. The rest just falls apart. However, the puzzle is poorly worded in places and requires some assumptions.
Alice the pilot wore the yellow dress Betty the professor wore the pink dress Carol the lawyer wore the white dress Dorothy the lifeguard wore the blue dress
Thread Tools
Similar Threads for: Deductive Reasoning Q? Thread Forum Replies General Discussion 6 Precalculus Mathematics Homework 4 General Discussion 2 Calculus & Beyond Homework 9 Biology, Chemistry & Other Homework 3
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I l@ve RuBoard
### 17.7 Counting Items and Sorting by Incidence (Histograms)
Credit: John Jensen, Fred Bremmer
#### 17.7.1 Problem
You need to produce ascending- or descending-count histograms, such as the most or least common words in a file, popular pages on a web site, etc.
#### 17.7.2 Solution
Histogramming is basically an issue of counting item occurrences (a Python dictionary makes this quite easy) and sorting by the counts. In Python, the two actions, and the dictionary that holds the counts, are easily wrapped into a class:
```class Counter:
def _ _init_ _(self):
self.dict = {}
count = self.dict.get(item, 0)
self.dict[item] = count + 1
def counts(self, desc=None):
""" Returns list of keys sorted by values.
Pass desc as 1 if you want a descending sort. """
result = map(None, self.dict.values(), self.dict.keys( ))
result.sort( )
if desc: result.reverse( )
return result```
#### 17.7.3 Discussion
The add method shows the normal Python idiom for counting occurrences of arbitrary (but hashable) items, using a dictionary to hold the counts. The counts method is where all the action is. It takes the dictionary and produces an ascending or descending sort of keys by values, returning a list of pairs representing the desired histogram. The map call takes advantage of an interesting but little-known tidbit of documented Python behavior. While the values and keys methods of a dictionary return their results in an arbitrary order, the ordering is compatible when the two methods are called without any intervening modification to the dictionary object. In other words, d[d.keys( )[x]] is d.values(x) for any valid index x. This lets us elegantly zip values and keys with the value as the first item and the key as the second item in each pair, so the sort method will work right (by using map with a first argument of None rather than zip, we keep compatibility with 1.5.2).
Here is an example:
```sentence = "Hello there this is a test. Hello there this was a test, " \
"but now it is not."
words = sentence.split( )
c = Counter( )
for word in words:
print "Ascending count:"
print c.counts( )
print "Descending count:"
print c.counts(1)```
This produces:
```Ascending count:
[(1, 'but'), (1, 'it'), (1, 'not.'), (1, 'now'), (1, 'test,'), (1, 'test.'),
(1, 'was'), (2, 'Hello'), (2, 'a'), (2, 'is'), (2, 'there'), (2, 'this')]
Descending count:
[(2, 'this'), (2, 'there'), (2, 'is'), (2, 'a'), (2, 'Hello'), (1, 'was'),
(1, 'test.'), (1, 'test,'), (1, 'now'), (1, 'not.'), (1, 'it'), (1, 'but')]```
If you give up on 1.5.2 compatibility and use a list comprehension instead of the map call, the code arguably becomes a little easier to read:
```def counts(self, desc=None):
result = [(val, key) for key, val in self.dict.items( )]
result.sort( )
if desc: result.reverse( )
return result```
However, if this issue ever arises in a spot of your program that is a critical speed bottleneck, you should measure performance accurately for each version of counts. Often (but not always), map displays surprisingly good performance characteristics when compared to list comprehensions (at least when no lambda is involved in the use of map).
I l@ve RuBoard
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# Lift and wrap array with custom indexes
I have a function coded for “normal” `1:n` axes that
1. takes a matrix,
2. does some manipulation, including slices, LU, etc,
3. gets a vector.
I want to generalize this to work with arrays of custom indexes (eg `OffsetArray`), with the following in mind: it should only work for matrices which have the same `axes` along both dimensions, and then use this to “wrap” the resulting vector. I just don’t know how to generalize this — the trick that I am missing is that in general I don’t know the constructor for the general array type.
Perhaps I should use a modifying version of `op` with `similar`? But then I would have to calculate result types.
M(W)E:
``````using LinearAlgebra, OffsetArrays
op_internal(A::AbstractMatrix) = normalize(vcat(A[end, 1:(end-1)], one(eltype(A))), 1)
function op(A)
if Base.has_offset_axes(A)
axes(A, 1) == axes(A, 2) || error("matrix not square")
## FIXME --- how to extract A as a matrix for op_internal, then wrap the result?
else
op_internal(A)
end
end
O = ones(3, 3) .* 2
op(O) # works
# below: missing code, want OffsetArray([0.4, 0.4, 0.2], 2:4)
op(OffsetArray(O, 2:4, 2:4))
``````
The “hard” point about this example is the concatenation: we don’t really have a good API for concatenating arrays with offset axes. What should the output axes be? The most useful (but undocumented) approach I know of is in CatIndices, which uses a wrapper called `PinIndices` to say “the axes of this object should ‘win’, and everything else stretches out to either side.”
This implementation is close to what you want:
``````using LinearAlgebra, OffsetArrays, CatIndices, EndpointRanges
function op(A::AbstractMatrix)
axes(A, 1) == axes(A, 2) || error("matrix not square")
normalize(vcat(PinIndices(A[iend, ibegin:iend-1]), [one(eltype(A))]), 1)
end
O = ones(3, 3) .* 2
display(op(O))
OO = OffsetArray(O, 2:4, 2:4)
display(op(OO))
``````
But there’s a couple of missing elements to make it perfect:
• the `vcat` should be generalized (here you’d like to to take numbers as well as `AbstractVectors` and `PinIndices`)
• the axes of the output are not what you were hoping. That’s because in `A[iend, ibegin:iend-1]`, the range object `ibegin:iend-1` implicitly has axes that range from 1:2, and the axes of the result are the axes of the indices. (As they should be!) You’d kind of like to do something like `A[iend, Base.IdentityUnitRange(ibegin:iend-1)]` but currently that doesn’t work. Or perhaps better, one could define `iibegin` and `iiend` in EnpointRanges to mean identity-index and then use `iibegin:iiend-1` as an identity-range.
2 Likes
Thank you for your help. I may have been unclear in wording the question: I want/need to code `op_internal` to work for arrays with regular indexes (from `1`), mostly because the library functions it uses don’t handle generalized indexing (the above code is just an MWE, it is more complex). I hope that more and more functions will work with generalized indexes, but in the meantime, I am looking for a way to drop back into regular indexing, and then just put indices on the end result.
I ended up reworking the code to use a pre-allocated buffer, and using `reshape` for a “view” with non-custom indexes, but I am not sure this is the right way. MWE modified:
``````using LinearAlgebra, OffsetArrays
function op_blackbox!(v::AbstractVector, A::AbstractMatrix)
v[1:(end-1)] = A[end, 1:(end-1)]
v[end] = one(eltype(v))
normalize!(v, 1)
v
end
calc_type(::Type{T}) where T = typeof(√one(T))
function op(A)
S = calc_type(eltype(A))
ax1 = axes(A, 1)
ax1 == axes(A, 2) || error("matrix not square")
if Base.has_offset_axes(A)
n = length(ax1) # for reshape
B = reshape(A, n, n) # a "view" with non-offset indexes
w = similar(A, S, ax1) # return array
v = reshape(w, n) # a "view" buffer with non-offset indexes
else
B = A # work in the same array
w = v = similar(A, S, ax1) # likewise
end
op_blackbox!(v, B)
w
end
O = ones(3, 3) .* 2
op(O)
op(OffsetArray(O, 2:4, 2:4))
``````
If one wanted to “do this right” then let me then add a third deficiency to my list above:
• in `vcat`, if the first entry is a `PinIndices` with `Base.OneTo` axes, then use `Base.OneTo` axes for the output.
The point here is that ideally, you shouldn’t need to special-case on `Base.has_offset_axes`: it’s much better to have generic code that can deal with general axes.
I guess this is a case where you should ask yourself, “how many functions like this am I going to write?” If it’s just one or a few, the approach you’ve taken is certainly the most time-efficient. In contrast, if you’re going to be doing a lot of this, I’d urge you to submit a couple of PRs to CatIndices so that all your later development is easier (contrast 3 LOC vs 21).
1 Like
As I said (but perhaps underemphasized — that’s the problem with MWEs) in the original post, the problem that motivated this uses `LinearAlgebra`, especially `qr` and `lu`, and I would have to update these too. I will consider PRs, but my understanding is that in the end it all boils down to an unwrap/wrap pattern. Also, I would really need to think hard about pivoting and generalized indexes.
1 Like
Understood. But it’s better not to think of it as a wrap/unwrap problem: if you look at code that handles indexing, in general it’s a bit rare to have explicit unwrapping. The much better way to handle this is through generic interfaces (e.g., `getindex`, `similar`, and so-on).
`cat` operations are a tricky case because the fundamental operation is basically conditioned on an “arrays-are-just-lists” viewpoint rather than “arrays are Dicts with keys arranged on a spatial grid” viewpoint.
3 Likes
Specifically, how would you handle the following for generalized indexes?
``````using LinearAlgebra
function stationary_distribution(matrix::AbstractArray)
LU = lu(matrix, Val(false); check = false) # not pivoted, singular
L = LU.L
x = vcat(L[1:(end-1), 1:(end-1)]' \ -L[end, 1:(end-1)], 1)
normalize!(x, 1)
x
end
``````
In an ideal world,
• generalize LU decomposition so it handles arbitrary square axes (there’s nothing about the underlying algorithm that requires 1-based indexing)
• wrap the first argument to `vcat` with `PinIndices`
Your code would thus be almost identical to what it is now.
I am very enthusiastic about generalized indexing, and I am grateful for all the work that you and others have invested in it.
However, the ideal state of affairs you describe will require much more work. I hope to contribute to this, but I think in the meantime an escape mechanism like above would be helpful, even to the extent of describing it in the manual (which also recognizes that not all code is ready, and mentions how to check for it).
In particular, I am curious if `reshape` is the best way to get a standard-indexed “view” into an array with custom indexing. That would solve 90% of my problems (the rest being type calculation).
It’s not as general as you might hope, because `reshape` can end up checking `has_offset_axes`:
``````julia> using CatIndices
julia> v = BidirectionalVector(1:3)
BidirectionalVector{Int64} with indices CatIndices.URange(1,3):
1
2
3
julia> pushfirst!(v, 0)
BidirectionalVector{Int64} with indices CatIndices.URange(0,3):
0
1
2
3
julia> reshape(v, 4)
ERROR: AssertionError: !(has_offset_axes(v))
Stacktrace:
[1] _reshape(::BidirectionalVector{Int64}, ::Tuple{Int64}) at ./reshapedarray.jl:167
[2] reshape at ./reshapedarray.jl:112 [inlined]
[3] reshape(::BidirectionalVector{Int64}, ::Int64) at ./reshapedarray.jl:115
[4] top-level scope at none:0
``````
1 Like
So there is no general solution? I was thinking of something like this (MWE without frills and optimization, `setindex!` not defined, etc):
``````struct PlainView{T, N, P <: AbstractArray{T,N}, I} <: AbstractArray{T,N}
parent::P
indices::I
end
Base.size(P::PlainView) = length.(axes(P.parent))
Base.getindex(P::PlainView, I::Vararg{Int,N}) where N =
P.parent[Base.reindex(P, P.indices, I)...]
function plain_view(A::AbstractArray{T,N}) where {T,N}
if Base.has_offset_axes(A)
PlainView(A, ntuple(i -> firstindex(A, i):lastindex(A, i), N))
else
A
end
end
``````
You can always wrap with another layer, e.g., `OffsetArray`. (Your `PlainView` basically is `OffsetArray`.)
1 Like
I made a PR to that effect:
Also, thinking about pivoting indices (for LU & friends): I came to the conclusion that for generalized indexing, it should just remain the same, with the understanding that pivot indices should be interpreted within the relevant `axes(A, j)`. What do you think?
Thanks for the PR, I’ll check it in a bit. Re pivoting, that sounds pretty reasonable.
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# How do I solve this math question? 2x^2(-3x^2)^4
pramodpandey | College Teacher | (Level 3) Valedictorian
Posted on
We have
`2x^2(-3x^2)^4`
We know that `(ab)^m=a^mb^m ,(a^m)^n=a^(mn),and a^mxxa^n=a^(m+n)`
Thus
`2x^2(-3x^2)^4=2x^2(-3)^4(x^2)^4`
`=2xx81xx x^2 xx x^(2xx4)`
`=162x^(2+8)`
`=162x^10`
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# How to Calculate the Upper Quartile
Quartiles are numbers used to divide a set of data into four equal parts, or quarters.[1] The upper quartile, or third quartile, is the top 25% of numbers in the data set, or the 75th percentile. The upper quartile is calculated by determining the median number in the upper half of a data set.[2] This value can be found by calculating with pen and paper, but you can also easily find the upper quartile using statistical software, such as MS Excel.
Part 1
Part 1 of 3:
### Data Set Prep
1. 1
Arrange the numbers of the data set in ascending order. This means ordering them from the smallest value to the largest value. Make sure to include all repeated values.[3]
• For example, if your set of numbers is [3, 4, 5, 11, 3, 12, 21, 10, 8, 7], you would reorder them like this: [3, 3, 4, 5, 7, 8, 10, 11, 12, 21].
2. 2
Determine how many numbers are in the data set. To do this, simply count each number in the set. Don’t forget to count each instance of a repeated value.
• For example, the set [3, 3, 4, 5, 7, 8, 10, 11, 12, 21] has 10 numbers.
3. 3
Set up the formula for calculating the upper quartile. The formula is ${\displaystyle Q_{3}={\frac {3}{4}}(n+1)}$, where ${\displaystyle Q_{3}}$ is the upper quartile, and ${\displaystyle n}$ is the number of numbers in the data set.[4]
Part 2
Part 2 of 3:
### Calculating the Upper Quartile
1. 1
Plug the value of into the formula. Remember that ${\displaystyle n}$ is the number of numbers in the data set.
• For example, if there are 10 numbers in your data set, your formula will look like this: ${\displaystyle Q_{3}={\frac {3}{4}}(10+1)}$.
2. 2
Complete the calculation in parentheses. According to the order of operations, you must attend to the parentheses first when evaluating a mathematical expression. In this instance, add 1 to the number of numbers in the data set.
• For example:
${\displaystyle Q_{3}={\frac {3}{4}}(10+1)}$
${\displaystyle Q_{3}={\frac {3}{4}}(11)}$
3. 3
Multiply the sum by . You could also multiply by ${\displaystyle .75}$. This will show you the placement of the value in the data set that is at the three-fourths, or 75 percent mark, and thus the place where the data set is split into the upper quartile and the lower quartiles. This will not give you the number of the upper quartile.
• For example:
${\displaystyle Q_{3}={\frac {3}{4}}(11)}$
${\displaystyle Q_{3}=8{\frac {1}{4}}}$
So, the upper quartile is given by the number at the ${\displaystyle 8{\frac {1}{4}}}$ position in the data set.
4. 4
Determine the number representing the upper quartile. If you calculated a whole number, simply find that number in the data set.
• For example, if you calculated 12 using the formula, then the upper quartile is the 12th number in the data set.
5. 5
Calculate the upper quartile, if necessary. Usually, you will calculate a fraction or decimal using the formula. In this instance, find the value above and below this position in the data set, and find their mean, or average. To do this, divide the sum of the two values by 2. This will give you the upper quartile of your data set.
• For example, if you calculated ${\displaystyle 8{\frac {1}{4}}}$ using the formula, then the upper quartile is between the 8th and 9th number in the data set. In the set [3, 3, 4, 5, 7, 8, 10, 11, 12, 21], 11 and 12 are the 8th and 9th number. Calculate ${\displaystyle {\frac {11+12}{2}}}$ to find the average:
${\displaystyle {\frac {11+12}{2}}}$
${\displaystyle ={\frac {23}{2}}}$
${\displaystyle =11.5}$
So, the upper quartile of the data set is 11.5
Part 3
Part 3 of 3:
### Using Excel
1. 1
Input your data into Excel. Enter each value into a separate cell. Don’t forget to include any repeated values. You can enter your data in any cells in the spreadsheet.
• For example, you might enter the data set [3, 3, 4, 5, 7, 8, 10, 11, 12, 21] into cells A1 through A10 in the spreadsheet.
2. 2
Enter the quartile function into another cell. The quartile function is =(QUARTILE(AX:AY, Q)), where AX and AY is the data range, and Q is the quartile.[5] Begin typing this function into Excel, then when it pops up in the menu, double-click on it to select.
3. 3
Select the cells containing the data. Select the first cell of the data range, then scroll down or across to select all the cells in the range.
4. 4
Enter 3 into the function to denote the upper quartile. Make sure you include a comma after the data range, and two closing parentheses.
• For example, if you want to find the upper quartile of cells A1 through A10, your function will look like this: =(QUARTILE(A1:A10, 3)).
5. 5
Show the upper quartile. To do this, hit enter after typing the function into Excel. This will show you the actual upper quartile, not the position of the quartile in the data set.
• Note that with the release of Office 2010, there are two different quartile functions: QUARTILE.EXC and QUARTILE.INC. These functions cannot be used in earlier versions of Excel, and QUARTILE can still be used.
• The two Excel quartile functions use a different formula to calculate the upper quartile. QUARTILE/QUARTILE.INC uses the formula ${\displaystyle Q_{3}={\frac {3}{4}}(n-1)}$, and the QUARTILE.EXC function uses the formula ${\displaystyle Q_{3}={\frac {3}{4}}(n+1)}$. Both formula are accepted ways to calculate quartiles, although the former is becoming standardized in statistical software.
## Community Q&A
Search
• Question
How do you find the upper and lower quartiles?
Use the methods above to find the upper quartile. To find the lower quartile using the formula, use Q1 = 1/4(n+ 1), or enter 1 into the Excel function instead of 3.
• Question
How do I determine the inter quartile range of data?
Cluster Duck
The IQR can be calculated by subtracting the lower quartile from the upper.
• Question
How do I calculate the upper quartile for a large group of numbers?
Same as for a small group of numbers. Or better use script language like Matlab or Python to load the dataset and do the calculations.
200 characters left
## Tips
• You may sometimes see a reference to the “interquartile range.” This is the range between the lower and upper quartiles, which is calculated by subtracted Quartile 1 from Quartile 3.
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Co-authored by:
wikiHow Staff Writer
This article was co-authored by wikiHow Staff. Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 159,329 times.
Co-authors: 14
Updated: July 28, 2022
Views: 159,329
Article SummaryX
Quartiles are numbers used to divide a set of data into 4 equal parts or quarters. The upper quartile is the top 25 percent of numbers in the data set, or the 75th percentile. To calculate the upper quartile, first, arrange the numbers of the data set in ascending order. Then, determine how many numbers are in the set. The formula for calculating the upper quartile is Q3 = ¾ (n +1). Q3 is the upper quartile and n is the number of numbers in your data set. For example, if you have 10 numbers in your data set, you would solve Q3 = ¾ (10 + 1), then solve ¾ x 11, which would give you 8 ¼. If you get a fraction or decimal as your answer, the upper quartile will be the average of the number below and above in your data set. For example, if you calculate 8 ¼ for Q3 and the numbers below and above in your set are 8 and 10, the average of them is 9. Therefore, 9 is the upper quartile. To learn how to calculate an upper quartile in Excel, read on!
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Home / Arithmetic Aptitude / Time and Work :: Discussion
### Discussion :: Time and Work
1. A shopkeeper sells some toys at Rs. 250 each. What percent profit does he make? To find the answer, which of the following information given in Statements I and II is/are necessary?
I. Number of toys sold.
II.Cost price of each toy.
2. A.
Only I is necessary
B.
Only II is necessary
C.
Both I and II are necessary
D.
Either I or II ins necessary
E.
None of these
Answer : Option B
Explanation :
S.P. = Rs. 250 each.
To find gain percent, we must know the C.P. of each.
Be The First To Comment
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# KCSE 2017 Computer Studies Paper 2 with Marking scheme
1. Mavuno Group of hotels offer accommodation services to clients. The accommodation rooms are categorised as single, double or VIP, each attracting different rates. The rooms with fridges stocked with drinks attract an extra cost. The management of the hotel intends to use a spreadsheet program to compute the revenue from the rooms.
1. Open the spreadsheet program and create a worksheet to appear as shown in Figure 1. Save the workbook as room charges. (15 marks)
A B C D E F G H 1 SERVICE COST PER DAY 2 Single (S) 1500 3 Double (D) 2800 4 VIP (V) 3200 5 Friodge(F) 300 6 7 Room Id Guest Id Days Room status Fridge availability Room charges Fridge Charges Total charges 8 363 RM001 3 D Yes 9 103 RM002 1 D Yes 10 RM003 1 S No 11 RM004 4 D No 12 RM005 5 D Yes 13 RM006 1 S No 14 RM007 4 D Yes 15 RM008 3 D Yes 16 RM009 3 V Yes 17 RM0010 1 V Yes 18 RM0011 1 D Yes 19 RM0012 4 S No 20 RM0013 5 D Yes 21 RM0014 2 D Yes
Figure 1
2. Name the cell containing the value; 1500 as SR, the cell containing 2800 as DR, the cell with 3200 as VP and the cell with 300 as FR. (4 marks)
3.
1. In the column with title Room Charges, enter a formula that can be copied down the column to multiply the value in days by SR if the room status value is Sor multiply the value in days by DR if the room status value is D or multiply the value in days by VP if the room status value is V. (8 marks)
2. In the column with the title Fridge Charges enter a formula that can be copied down the column to compute Fridge Charges.
(3 marks)
3. In the column with the title Total Charges, enter a formula that computes the total of the Room Charges and Fridge Charges for each guest. (2 marks)
4. Format the Room Charges, Fridge Charges and Total Charges values as currency with zero number of decimal places. (2 marks)
5.
1. Copy all the contents of the current work sheet to a new worksheet (1 mark)
2. Name the initial worksheet as ORIGINAL and the copied worksheet as NEW (2 marks)
6.
1. In the sheet named NEW, extract only the records whose ROOM STATUS is S. (2 marks)
2. Create a column bar chart that compares the Room Charges and Fridge Charges for guests whose Guest Id are RM003, RM006 and RM012. (4 marks)
3. Insert the following labels in the chart created in (ii)
ChartTitle : Single Room Revenue
X-axis : GuestID
Y-axis : Revenue in Ksh. (3 marks)
4. Rename the chart sheet as SREVENUE. (1 mark)
7. Printout later each of the following:
1. ORIGINAL Worksheet (1 mark)
2. NEW Worksheet (1 mark)
3. SREVENUE Chart (1 mark)
2. The management of a county scout movement intends to award certificates of participation to the scouts who attended a fire rescue seminar. Assuming that you have been tasked to design the certificates.
1. Open a Desktop Publishing program and make the following page settings. (4 marks)
1. Orientation : landscape
2. Units : centimetres
3. Papersize : A4
4. Margins : 2cm all around
2. Create the certificate as it appears in Figure 2. Save the design as Certificate. (45marks)
3. Printout the certificate later. (1 mark)
## MARKING SCHEME
1.
1. Typing values in the cells
• Values in cell range Al: B6 @1
• Margin cells A1:B1 @ 1
• Text wrap in the titleA1:31 @ 1
• Typing column 1(range A9: A22) @ 1
• Typing column 2 (range B9: B22) @ 2
• Typing column 3(range 09: C22) @ 1
• Typing column 4(range D9: D22) @ 1
• Typing column 5 (range E9: E22) @ 1
• Saving the workbook @1
Column title text (row 8)
• Typing column title text (correct, bolded and completeness-A8. H8) @ 2
• Wrapping titles @1
• Applying bold face @ 1
• Applying borders to all the visible cells @ 1 (15 marks)
2. Naming the cells containing:
• 1500 as SR @ 1
• 2800 as DR @1
• 3200 as VP @ 1
• 300 as FR @1 (4 marks)
3.
1. =If (D9 = "S", C9* SR, if (D9="D", C9* DR, if (D9 = "V", C9*VP)))
Use of the IF function @ 1
S selection @2
D selection @2
V selection (else) @ 2
Logic and syntax@1 (8 marks)
2. =f(E7 = "Yes", FR*C7,0)
• Use of the function @1
• Selection of fridge @ 1
• Alternative selection @ 1 (3 marks)
3.
• =G7*H7 @ 1
• Applying other cells @1 (2 marks)
4.
• Currency formats @ 1
• Zero decimal formats @
• Formats applied in the correct range @½ (2 marks)
5.
1. Copying the content of the current worksheet to sheet 2 (1 mark)
2.
• Rename sheet 1 as original @ 1
• Rename sheet 2 as NEW @ 1 (1) (2 marks)
6.
1.
• Enabling filter feature @ 1
• Filtering out correct records (displaying S values only) @ 1 (2 marks)
2.
• Creating bar chart @ 1
• Selecting the correct X fields @ 2
• Selecting the correct Y fields @ 1 (4 marks)
3. Insertion of chart elements
• Chart title @ 1
• X axis label @ 1
• Y axis label @ 1 (3 marks)
4. Renaming the chart worksheet as
• SREVENUE @ 1 (1 marks)
7. Printing the following
1. Original worksheet @ 1
2. NEW worksheet @ 1
3. SREVEN
4. UE chart @ 1 (3 marks)
2.
1. Page settings
1. Paper orientation @1
2. Units set to centimetres @1
3. Paper size set to A4 @ 1
4. Margins set to 2cm @ 1 (4 marks)
2. Border lines
1. Outer borders @1
2. Inner rectangles @1
3. Corner shapes @ ½x4=2
4. Position on the page @1 (5 marks)
"Certificate of Participation" Text
• Typing text @1
• Enlarging and italicizing of "of" @2
• Positioning of this element on the page @1 (4 marks)
Lines below and above the "Certificate of Participation"text
• Top lines @ ½x 2 =1
• Below lines @ ½x 2 =1
• Correct placement @1 (3 marks)
"Awarded to:" Text
• Typing text @1
• Correct placement this element on the page @1 (2 marks)
Line below"Awarded to:" Text
• Inserting of the line @1
• Correct placement @1 (2 marks)
“For the phenomenal......." text
• Typing text @1
• Correct placement @1 (2 marks)
"Fire Emergency Rescue" Text
• Typing the text @1
• Text (font) size @1
• Fill pattern (outline font) @1
• Insertion of text box @1
• Applying a dotted background in the text box @1
• Correct positioning of this elements in the page @1 (6 marks)
"Presented By:"Text
• Typing text-@ 2
• Horizontal line below @1
• Position on the page @1 (3 marks)
"On This Day" Text
• Typing text @1
• Position on the page @1
• Horizontal line below @1 (3 marks)
The Flame and Candle Graphic
• 2 curved lines @ ½ x 2 =1
• Flame outline(Filling the inner curve) @2
• Rectangular shape @1
• Correct fill pattern on the rectangle @ ½
• Correct position of all the elements in the page @1 (5 marks)
The Star Graphic
• Outer shape (drawing) @1
• Fill pattern @1
• Star shape (drawing) @1
• White fill colour @1
• Correct positioning on page @1
• Star shape in front @1 (6 marks)
Second Star graphic
• Copying (duplicated) @1
• Correct positioning of both graphic each2x 1=2
• Saving the certificate (1 mark)
3.
• Printing the design (1 mark)
• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students
### Related items
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1. ## Lagrange Multplier
I have to maximise $\displaystyle 2 \tan ^{-1} (x_{1}) + x_{2}$ subject to $\displaystyle x_{1} + x_{2} \le b_{1}$ and $\displaystyle - \log (x_{2}) \le b_{2}$ for constants $\displaystyle b_{i}$ satisfying $\displaystyle b_{1} - e^{-b_{2}} \ge 0$ where $\displaystyle x_{i} \ge 0$.
Clearly it is easy to write down the Lagrangian and get that $\displaystyle L_{x_{1}} = \frac{2}{1 + x_{1} ^{2} } - \lambda = 0$ and $\displaystyle L_{x_{2}} = 1 - \lambda + \frac{ \mu }{x_{2}} = 0$.
Note that slacked variables have been used to take care of the inequality constraints. I don't know how to proceed. Clearly I could solve the equations I have from differentiation however taking into account that I have inequalities for constraints I am rather confused. What can I do here?
2. Anyone?
3. I hate to bump this twice but I am still stuck. Does anyone have any ideas, even if you can't get the full answer?
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### Mech vii-operation research [06 me74]-notes
1. 1. Operations Research [06ME74] OPERATION RESEARCH Subject Code : 06ME74 IA Marks : 25 No. of Lecture Hrs./ Week : 04 Exam Hours : 03 Total No. of Lecture Hrs. : 52 Exam Marks : 100UNIT - 1INTRODUCTION: Linear programming, Definition, scope of Operations Research (O.R) approachand limitations of OR Models, Characteristics and phases of OR Mathematical formulation of L.P.Problems. Graphical solution methods. 6 HoursUNIT - 2LINEAR PROGRAMMING PROBLEMS: The simplex method - slack, surplus and artificialvariables. Concept of duality, two phase method, dual simplex method, degeneracy, and procedure forresolving degenerate cases. 7 HoursUNIT - 3TRANSPORTATION PROBLEM: Formulation of transportation model, Basic feasible solution usingdifferent methods, Optimality Methods, Unbalanced transportation problem, Degeneracy intransportation problems, Applications of Transportation problems. Assignment Problem: Formulation,unbalanced assignment problem, Traveling salesman problem. 7 HoursUNIT - 4SEQUENCING: Johnsons algorithm, n - jobs to 2 machines, n jobs 3machines, n jobs m machineswithout passing sequence. 2 jobs n machines with passing. Graphical solutions priority rules. 6 Hours PART - BUNIT – 5QUEUING THEORY: Queuing system and their characteristics. The M/M/1 Queuing system, Steadystate performance analysing of M/M/ 1 and M/M/C queuing model. 6 HoursDepartment of Mechanical Engineering, SJBIT Page 0
2. 2. Operations Research [06ME74]UNIT - 6PERT-CPM TECHNIQUES: Network construction, determining critical path, floats, scheduling bynetwork, project duration, variance under probabilistic models, prediction of date of completion,crashing of simple networks. 7 HoursUNIT - 7GAME THEORY: Formulation of games, Two person-Zero sum game, games with and without saddlepoint, Graphical solution (2x n, m x 2 game), dominance property. 7 HoursUNIT - 8INTEGER PROGRAMMING: Gommory’s technique, branch and bound lgorithm for integerprogramming problems, zero one algorithm 6 HoursTEXT BOOKS: 1. Operations Research and Introduction, Taha H. A. – Pearson Education edition 2. Operations Research, S. D. Sharma –Kedarnath Ramnath & Co 2002.REFERENCE BOOKS: 1. “Operation Research” AM Natarajan, P. Balasubramani, A Tamilaravari Pearson 2005 2. Introduction to operation research, Hiller and liberman, Mc Graw Hill. 5th edition 2001. 3. Operations Research: Principles and practice: Ravindran, Phillips & Solberg, Wiley India lts, 2nd Edition 2007 4. Operations Research, Prem Kumar Gupta, D S Hira, S Chand Pub, New Delhi, 2007Department of Mechanical Engineering, SJBIT Page 1
3. 3. Operations Research [06ME74] TABLE OF CONTENTS 1. INTRODUCTION, FORMULATION OF LPP - 03 - 20 2. LINEAR PROGRAMMING PROBLEMS, SIMPLEX METHOD - 21 – 37 3. TRANSPORTATION PROBLEM - 38 - 57 4. QUEUING THEORY - 58 – 74 5. PERT-CPM TECHNIQUES - 75 - 85 6. GAME THEORY - 86 – 93 7. INTEGER PROGRAMMING - 94 - 98Department of Mechanical Engineering, SJBIT Page 2
4. 4. Operations Research [06ME74] INTRODUCTION Operations research, operational research, or simply OR, is the use of mathematical models,statistics and algorithms to aid in decision-making. It is most often used to analyze complex real-worldsystems, typically with the goal of improving or optimizing performance. It is one form of appliedmathematics. The terms operations research and management science are often used synonymously. When adistinction is drawn, management science generally implies a closer relationship to the problems ofbusiness management. Operations research also closely relates to industrial engineering. Industrial engineering takesmore of an engineering point of view, and industrial engineers typically consider OR techniques to be amajor part of their toolset. Some of the primary tools used by operations researchers are statistics, optimization, stochastics,queueing theory, game theory, graph theory, and simulation. Because of the computational nature ofthese fields OR also has ties to computer science, and operations researchers regularly use custom-written or off-the-shelf software. Operations research is distinguished by its ability to look at and improve an entire system, ratherthan concentrating only on specific elements (though this is often done as well). An operationsresearcher faced with a new problem is expected to determine which techniques are most appropriategiven the nature of the system, the goals for improvement, and constraints on time and computingpower. For this and other reasons, the human element of OR is vital. Like any tools, OR techniquescannot solve problems by themselves.Areas of applicationA few examples of applications in which operations research is currently used include the following: designing the layout of a factory for efficient flow of materials constructing a telecommunications network at low cost while still guaranteeing quality service if particular connections become very busy or get damaged determining the routes of school buses so that as few buses are needed as possible designing the layout of a computer chip to reduce manufacturing time (therefore reducing cost) managing the flow of raw materials and products in a supply chain based on uncertain demand for the finished productsDepartment of Mechanical Engineering, SJBIT Page 3
5. 5. Operations Research [06ME74] Unit –I Introduction & Formulation of LPPDefine OROR is the application of scientific methods, techniques and tools to problems involving the operations ofa system so as to provide those in control of the system with optimum solutions to the problems.Characteristics of OR 1. its system orientation 2. the use of interdisciplinary teams 3. application of scientific method 4. uncovering of new problems 5. improvement in the quality of decisions 6. use of computer 7. quantitative solutions 8. human factors1. System (or executive) orientation of OR production dept.: Uninterrupted production runs, minimize set-up and clean-up costs. marketing dept.: to meet special demands at short notice. finance dept.: minimize inventories should rise and fall with rise and fall in company’s sales. personnel dept.: maintaining a constant production level during slack period.2. The use of interdisciplinary teams. Psychologist: want better worker or best products. Mechanical Engg.: will try to improve the machine. Software engg.: updated software to sole problems. that is, no single person can collect all the useful scientific information from all disciplines.3. Application of scientific method Most scientific research, such as chemistry and physics can be carried out in Lab under controlled condition without much interference from the outside world. But this is not true in the case of OR study. An operations research worker is the same position as the astronomer since the latter can be observe the system but cannot manipulate.Department of Mechanical Engineering, SJBIT Page 4
6. 6. Operations Research [06ME74]4. Uncovering of new problems In order to derive full benefits, continuity research must be maintained. Of course, the results of OR study pertaining a particular problem need not wait until all the connected problems are solve.5. Improvement in the Quality of decision OR gives bad answer to problems, otherwise, worst answer are given. That is, it can only improve the quality of solution but it may not be able to give perfect solution.6. Use of computer7. Quantitative solutions for example, it will give answer like, “the cost of the company, if decision A is taken is X, if decision B is take is Y”.8. Human factors.Scope of Operation Research1) Industrial management: a) production b) product mix c) inventory control d) demand e) sale and purchase f) transportation g) repair and maintenance h) scheduling and control2) Defense operations: a) army b) air force c) navy all these further divided into sub-activity, that is, operation, intelligence administration, training.3) Economies: maximum growth of per capita income in the shortest possible time, by taking into consideration the national goals and restrictions impose by the country. The basic problem in most of the countries is to remove poverty and hunger as quickly as possible.4) Agriculture section: a) with population explosion and consequence shortage of food, every country is facing the problem of optimum allocation land to various crops in accordance with climatic conditions. b) optimal distribution of water from the various water resource.5) Other areas: a) hospital b) transport c) LICDepartment of Mechanical Engineering, SJBIT Page 5
7. 7. Operations Research [06ME74]Phases of OR1) Formulating the problem in formulating a problem for OR study, we mist be made of the four major components. i) The environment ii) The decision maker iii) The objectives iv) Alternative course of action and constraints2) Construction a Model After formulating the problem, the next step is to construct mode. The mathematical model consists of equation which describes the problem. the equation represent i) Effectiveness function or objective functions ii) Constraints or restrictions The objective function and constraints are functions of two types of variable, controllable variable and uncontrollable variable. A medium-size linear programming model with 50 decision variable and 25 constraints will have over 1300 data elements which must be defined.3) Deriving solution from the model an optimum solution from a model consists of two types of procedure: analytic and numerical. Analytic procedures make use of two types the various branches of mathematics such as calculus or matrix algebra. Numerical procedure consists of trying various values of controllable variable in the mode, comparing the results obtained and selecting that set of values of these variables which gives the best solution.4) Testing the model a model is never a perfect representation of reality. But if properly formulated and correctly manipulated, it may be useful predicting the effect of changes in control variable on the over all system.5) Establishing controls over solution a solution derived from a model remains a solution only so long as the uncontrolled variable retain their values and the relationship between the variable does not change.6) Implementation OR is not merely to produce report to improve the system performance, the result of the researchDepartment of Mechanical Engineering, SJBIT Page 6
8. 8. Operations Research [06ME74] must implemented. Additional changes or modification to be made on the part of OR group because many time solutions which look feasible on paper may conflict with the capabilities and ideas of persons.Limitations of OR 1) Mathematical models with are essence of OR do not take into account qualitative factors or emotional factors. 2) Mathematical models are applicable to only specific categories of problems 3) Being a new field, there is a resistance from the employees the new proposals. 4) Management may offer a lot of resistance due to conventional thinking. 5) OR is meant for men not that man is meant for it.Difficulties of OR 1) The problem formulation phase 2) Data collection 3) Operations analyst is based on his observation in the past 4) Observations can never be more than a sample of the whole 5) Good solution to the problem at right time may be much more useful than perfect solutions.Department of Mechanical Engineering, SJBIT Page 7
9. 9. Operations Research [06ME74] Linear ProgrammingRequirements for LP1. There must be a well defined objective function which is to be either maximized or minimized and which can expressed as a linear function of decision variable.2. There must be constraints on the amount of extent of be capable of being expressed as linear equalities in terms of variable.3. There must be alternative course of action.4. The decision variable should be inter-related and non-negative.5. The resource must be limited.Some important insight: The power of variable and products are not permissible in the objective function as well as constraints. Linearity can be characterized by certain additive and multiplicative properties. Additive example : If a machine process job A in 5 hours and job B in 10 hours, then the time taken to process both job is 15 hours. This is however true only when the change-over time is negligible. Multiplicative example: If a product yields a profit of Rs. 10 then the profit earned fro the sale of 12 such products will be Rs( 10 * 12) = 120. this may not be always true because of quantity discount. The decision variable are restricted to have integral values only. The objective function does not involve any constant term. n that is, z cjxj c , that is, the optimal values are just independent of any constant c. j 1Department of Mechanical Engineering, SJBIT Page 8
10. 10. Operations Research [06ME74]Examples on Formulation of the LP modelExample 1: Production Allocation ProblemA firm produces three products. These products are processed on three different machines. The timerequired to manufacture one unit of each of the product and the daily capacity of the three machines aregiven in the table below. Time per unit(minutes Machine capacity Machine Product 1 Product 2 Product 3 (minutes/day) M1 2 3 2 440 M2 4 --- 3 470 M3 2 5 --- 430It is required to determine the daily number of units to be manufactured for each product. The profit perunit for product 1,2 and 3 is Rs. 4, Rs. 3 and Rs. 6 respectively. It is assumed that all the amountsproduced are consumed in the market.Formulation of Linear Programming modelStep 1: The key decision to be made is to determine the daily number of units to be manufactured for each product.Step 2: Let x1, x2 and x3 be the number of units of products 1,2 and 3 manufactured daily.Step 3: Feasible alternatives are the sets of values of x1, x2 and x3 where x1,x2,x3 o. since negative number of production runs has no meaning and is not feasible.Step 4: The objective is to maximize the profit, that is , maximize Z = 4x1 + 3x2 + 6x3Step 5: Express the constraints as linear equalities/inequalities in terms of variable. Here the constraints are on the machine capacities and can be mathematically expressed as, 2x1 + 3x2 + 2x3 440, 4x1 + 0 x2 + 3x3 470 2x1 + 5x2 + 0x3 430Department of Mechanical Engineering, SJBIT Page 9
13. 13. Operations Research [06ME74] x2p – amount of crude 2 used for gasoline P x3p – amount of crude 3 used for gasoline P x4p – amount of crude 4 used for gasoline P x1q – amount of crude 1 used for gasoline Q x2q – amount of crude 2 used for gasoline Q x3q – amount of crude 3 used for gasoline Q x4q – amount of crude 4 used for gasoline QThe objective is to maximize profit.that is, 3(x1p+x2p+x3p+x4p) 4 (x1q+x2q+x3q+x4q) - 2(x1p+x1q) – 2.25 (x2p + x2q) – 2.50 (x3p + x3q) – 2.75(x4p + x4q)that is, maximize Z = x1p+0.75x2p+0.50x3p+0.25x4p+2x1q+1.75x2q+1.50x3q+1.25x4qThe constraints are: 0.75x1p+0.20x2p+.70x3p+0.40x4p 0.55(x1p+x2p+x3p+x4p), 0.10x1p+0.50x2p+0.20x3p+0.50x4p 0.40(x1p+x2p+x3p+x4p), 0.10x1q+0.50x2q+0.20x3q+0.50x4q 0.25(x1q+x2q+x3q+x4q)Example 5:A person wants to decide the constituents of a diet which will fulfill his daily requirements of proteins,fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods.The yields per unit of these foods are given below Yield per unit Cost per Food type Proteins Fats Carbohydrates unit 1 3 2 6 45 2 4 2 4 40 3 8 7 7 85 4 6 5 4 65 Minimum 800 200 700 requirementFormulate linear programming model for the problem.Solution:The objective is to minimize the cost That is , Z = Rs.(45x1+40x2+85x3+65x4)Department of Mechanical Engineering, SJBIT Page 12
14. 14. Operations Research [06ME74]The constraints are on the fulfillment of the daily requirements of the constituents. For proteins, 3x1 + 4x2+8x3+6x4 800 For fats, 2x1 + 4x2+7x3+5x4 200 For carbohydrates, 6x1 + 4x2+7x3+4x4 700Example 6:The strategic border bomber command receives instructions to interrupt the enemy tank production. Theenemy has four key plants located in separate cities, and destruction of any one plant will effectively haltthe production of tanks. There is an acute shortage of fuel, which limits to supply to 45,000 litre for thisparticular mission. Any bomber sent to any particular city must have at least enough fuel for the roundtrip plus 100 litres.The number of bombers available to the commander and their descriptions, are as follows: Bomber type Description Km/litre Number available A Heavy 2 40 B Medium 2.5 30Information about the location of the plants and their probability of being attacked by a medium bomberand a heavy bomber is given below: Probability of destruction by Distance from base Plant A heavy A medium (km) bomber bomber A Heavy 2 40 B Medium 2.5 30How many of each type of bombers should be dispatched, and how should they be allocated among thefour targets in order to maximize the probability of success?Solution:Let xij is number of bomber sent.The objective is to maximize the probability of success in destroying at least one plant and this isequivalent to minimizing the probability of not destroying any plant. Let Q denote this probability:then, Q = (1 – 0.1) xA1 . (1 – 0.2) xA2 . (1 – 0.15) xA3. (1 – 0.25) xA4 . (1 – 0.08) xB1 . (1 – 0.16) xB2 . (1 – 0.12) xB3 . (1 – 0.20) xB4Department of Mechanical Engineering, SJBIT Page 13
15. 15. Operations Research [06ME74]here the objective function is non-linear but it can be reduced to the linear form.Take log on both side, moreover, minimizing log Q is equivalent to maximizing –log Q or maximizinglog 1/Qlog 1/Q = -(xA1 log 0.9 + xA2 log 0.8 + xA3 log 0.85 + xA4 log 0.75 + xB1 log 0.92 + xB2 log 0.84 + xB3 log 0.88 + xB4 log 0.80)therefore, the objective is to maximizelog 1/Q = -(0.0457xA1 + 0.09691 xA2 + 0.07041 xA3 + 0.12483 xA4 + 0.03623xB1 + 0.07572xB2 + 0.05538xB3 + 0.09691xB4)The constraints are, due to limited supply of fuel 400 450 500 600 2 100 A1 2 100 A2 2 100 A3 2 100 A4 2 2 2 2 400 450 500 600 2 100 B1 2 100 B2 2 100 B3 2 100 B4 45,000 2.5 2.5 2.5 2.5that is, 500xA1 + 550 xA2 + 600 xA3 + 700 xA4 + 420xB1 + 460xB2 + 500xB3 + 580xB4 45,000due to limited number of aircrafts, xA1 + xA2 + xA3 + xA4 40 xB1 + xB2 + xB3 + xB4 30Example 7:A paper mill produces rolls of paper used cash register. Each roll of paper is 100m in length and can beproduced in widths of 2,4,6 and 10 cm. The company’s production process results in rolls that are 24cmin width. Thus the company must cut its 24 cm roll to the desired width. It has six basic cuttingalternative as follows:Department of Mechanical Engineering, SJBIT Page 14
16. 16. Operations Research [06ME74]The maximum demand for the four rolls is as follows Roll width(cm) Demand 2 2000 4 3600 6 1600 10 500The paper mill wishes to minimize the waste resulting from trimming to size. Formulate the L.P model.Solution:Let x1, x2, x3, x4, x5, x6 represent the number of times each cutting alternative is to be used.Objective is to minimize the trim losses, that is, minimize Z = 2 (X3+X4+X5+X6)The constraints are on the market demand for each type of roll width:for roll width of 2 cm, 6x1 + x3 + 4x6 2,000for roll width of 4 cm, 3x1 + 3x2 + x3 + 4x5 3,600for roll width of 6 cm, 2x2 + x3 + 2x4 + x5 + x6 1,600for roll width of 10 cm, x 3 + x4 500 Graphical Solution Method 1. The collection of all feasible solutions to an LP problem constitutes a convex set whose extreme points correspond to the basic feasible solutions. 2. There are a finite number of basic feasible solutions within the feasible solution space. 3. If the convex set of the feasible solutions of the system Ax=b, x≥0, is a convex polyhedron, then at least one of the extreme points gives an optimal solution. 4. If the optimal solution occurs at more than one extreme point, then the value of the objective function will be the same for all convex combinations of these extreme points.Extreme Point Enumeration Approach This solution method for an LP problem is divided into five steps.Step1: State the given problem in the mathematical form as illustrated in the previous chapter.Step2: Graph the constraints, by temporarily ignoring the inequality sign and decide about the area of feasible solutions according to the inequality sign of the constraints. Indicate the area of feasible solutions by a shaded area, which forms a convex polyhedron.Step3: Determine the coordinates of the extreme points of the feasible solution space.Step4: Evaluate the value of the objective function at each extreme point.Step5: Determine the extreme point to obtain the optimum (best) value of the objective function.Department of Mechanical Engineering, SJBIT Page 15
17. 17. Operations Research [06ME74]Types of Graphical solutions. • Single solutions. • Unique solutions. • Unbounded solutions. • Multiple solutions. • Infeasible solutions.Example 1.Use the graphical method to solve the following LP problemMaximize Z= 15x1+10x2Subject to the constraints 4x1+6x2≤360 3x1+0x2≤180 0x1+5x2≤200 and X1,x2≥0Department of Mechanical Engineering, SJBIT Page 16
18. 18. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 17
19. 19. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 18
20. 20. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 19
21. 21. Operations Research [06ME74] UNIT-II Linear Programming ProblemsSimplex MethodExample 1 (Unique solution) Max Z=3x1+5x2+4x3 Subject to 2x1+3x2 8 2x2+5x3 10 3x1+2x2+4x3 15 x1, x2, x3 0Solution:Introducing non-negative slack variables s1, s2 & s3 to convert inequality constraints to equality then theLP problem becomes, Max Z=3x1+5x2+4x3+0s1+0s2+0s3 Subject to 2x1+3x2+s1=8 2x2+5x3+s2=10 3x1+2x2+4x3+s3=15 x1, x2, x3, s1, s2, s3 0. Cj 3 5 4 0 0 0 Cb Basic Solution X1 X2 X3 S1 S2 S3 Ratio Variables 0 S1 8 2 3 0 1 0 0 8/3 0 S2 10 0 2 5 0 1 0 10/2 0 S3 15 3 2 4 0 0 1 15/2 Zj 0 0 0 0 0 0 Cj-Zj 3 5 4 0 0 0Department of Mechanical Engineering, SJBIT Page 20
22. 22. Operations Research [06ME74] 5 X2 8/3 2/3 1 0 1/3 0 0 -- 0 S2 14/3 -4/3 0 5 -2/3 1 0 14/15 0 S3 24/3 5/3 0 4 -2/3 0 1 29/12 Zj 10/3 5 0 5/3 0 0 Cj-Zj -1/3 0 4 -5/3 0 0 5 X2 8/3 2/3 1 0 1/3 0 0 4 4 X3 14/15 -4/15 0 1 -2/15 1/5 0 -ve 0 S3 89/15 41/15 0 0 -2/15 -4/5 1 89/41 Zj 34/15 5 4 17/15 4/5 0 Cj-Zj 11/15 0 0 - -4/5 0 17/15 5 X2 50/41 0 1 0 15/41 8/41 - 10/41 4 X3 62/41 0 0 1 -6/41 5/41 4/41 3 X1 89/41 1 0 0 -2/41 - 15/41 12/41 Zj 3 5 4 45/41 24/41 11/41 Zj-Cj 0 0 0 -ve -ve -veAll Zj –Cj < 0 for non-basic variable. Therefore the optimal solution is reached. X1=89/41, X2=50/41, X3=62/41 Z = 3*89/41+5*50/41+4*62/41 = 765/41Example 2:(unbounded) Max Z=4x1+x2+3x3+5x4 Subject to 4x1-6x2-5x3-4x4 -20 -3x1-2x2+4x3+x4 10 -8x1-3x2+3x3+2x4 20 x1 , x 2 , x3 , x4 0.Solution:Since the RHS of the first constraints is negative , first it will be made positive by multiplying byDepartment of Mechanical Engineering, SJBIT Page 21
23. 23. Operations Research [06ME74]–1, -4x1+6x2+5x3+4x4 20Introducing non-negative slack variable s1, s2 & s3 to convert inequality constraint to equality then theLP problem becomes. Max Z=4x1+x2+3x3+5x4+0s1+0s2+0s3 Subject to -4x1+6x2+5x3+4x4+0s1=20 -3x1-2x1+4x3+x4+s3=10 -8x1-3x2+3x3+2x4+s4=20 x1,x2,x3,x4,s1,s2,s3 0 Cj 4 1 3 5 0 0 0 Cb Variables Solution X1 X2 X3 X4 S1 S2 S3 Ratio 0 S1 20 -4 6 5 4 1 0 0 5 0 S2 10 -3 -2 4 1 0 1 0 10 0 S3 20 -8 -3 3 2 0 0 1 10 Zj 0 0 0 0 0 0 0 CJ-Zj 4 1 3 5 0 0 0 5 X4 5 -1 3/2 5/4 1 ¼ 0 0 -ve 0 S2 5 -2 -7/2 11/4 0 -1/4 1 0 -ve 0 S3 10 -6 -6 ½ 0 -1/2 0 1 -ve Zj -5 15/2 25/4 5 5/4 0 0 Cj-Zj 9 -13/2 -13/4 0 -5/4 0 0Since all the ratio is negative, the value of incoming non-basic variable x1 can be made as large as welike without violating condition. Therefore, the problem has an unbounded solution.Example 3:(infinite solution) Max Z=4x1+10x2 Subject to 2x1+x2 10, 2x1+5x2 20, 2x1+3x2 18. x1 , x 2 0.Solution:Department of Mechanical Engineering, SJBIT Page 22
24. 24. Operations Research [06ME74]Introduce the non-negative slack variables to convert inequality constraint to equality, then the LPproblem becomes, Max Z=4x1+10x2+0s1+0s2+0s3 Subject to 2x1+x2+s1=10, 2x1+5x2+s2=20, 2x1+3x2+s3=18. x1, x2, s1, s2, s3 0. Cj 4 10 0 0 0 Cb Variables Soln X1 X2 S1 S2 S3 Ratio 0 S1 10 2 1 1 0 0 10 0 S2 20 2 5 0 1 0 4 0 S3 18 2 3 0 0 1 6 Zj 0 0 0 0 0 Cj-Zj 4 10 0 0 0 0 S1 6 8/5 0 1 -1/5 0 15/4 10 X2 4 2/5 1 0 1/5 0 10 0 S3 6 4/5 0 0 -3/5 1 15/2 Zj 4 10 0 2 0 *x1=0 Cj-Zj 0 0 0 -2 0 X2=4,Z=40 4 X1 15/4 1 0 5/8 -1/8 0 10 X2 5/2 0 1 -1/4 ¼ 0 0 S3 3 0 0 -1/2 -1/2 1 Zj 4 10 0 0 0 Cj-Zj 0 0 0 0 0*All Cj-Zj is either 0 or negative, it gives the optimal basic feasible solution.But one of non-basic variable (x1) is 0. it indicates the existence of an alternative optimal basic feasiblesolution.If 2 basic feasible optimal solution are known, an infinite number of non-basic feasible optimal solutioncan be derived by taking any weighted average of these 2 solutions.Department of Mechanical Engineering, SJBIT Page 23
25. 25. Operations Research [06ME74] Variables 1 2 General solution X1 0 15/4 X1=0 +15/4 (1- ) X2 4 5/2 X2=4 +5/2 (1- ) Minimization CaseIn certain situations it is difficult to obtain an initial basic feasible solution (a) When the constraints are of the form E (aj.xj) bj. Xj 0. But some RHS constraints are negative. Then in this case after adding the non-negative slack variables Si the initial solution so obtained will be si=bi. It violates the non-negative condition of slack variable. (b) When the constraints are of the form ‘ ’ E (aj.xj) bj Xj 0. In this case to convert the inequalities into equation, we are adding surplus variables, then we get the initial solution is -si=bi si=-bi which violates the non-negative condition of the variables.To solve these type of problems we are adding artificial variable. Thus the new solution to the given LPproblem does not constitute a solution to the original system of equations because the 2 system ofequation are not equivalent.Thus to get back to the original problem artificial variable must be driven to 0 in the optimal solution.There are 2 methods to eliminate these variables1) Two Phase method2) Big M method or Penalties.Big- M method or the method of penalties:In this method the artificial variables are assigned a large penalty (-M for max & +M for min. problems)in the objective function.Department of Mechanical Engineering, SJBIT Page 24
26. 26. Operations Research [06ME74]Example 4: Max Z =3x1-x2, Subject to 2x1+x2 2 x1+3x2 3 x2 4 x1,x2,x3 0Solution:Introduce slack, surplus & artificial variable to convert inequality into equality then the LP problembecomes Max Z =3x1-x2+0s1+0s2+0s3+0A1, Subject to 2x1+x2-s1+A1=2 x1+3x2+s2=3 x2+s3=4 x1,x2,x3,s1,s2,s3,A1 0 Cj 3 -1 0 0 0 -M Cb Variables Soln X1 X2 S1 S2 S3 A1 Ratio 1 A1 2 2 1 -1 0 0 1 1 0 S2 3 1 3 0 1 0 0 3 0 S3 4 0 1 0 0 1 0 - Zj -2M -M M 0 0 -M Cj-Zj 3+2M -1+M -M 0 0 0 3 X1 1 1 ½ -1/2 0 0 ½ - 0 S2 2 0 5/2 ½ 1 0 -1/2 2 0 S3 4 0 1 0 0 1 0 - Zj 3 3/2 -3/2 0 0 3/2 Cj-Zj 0 -5/2 3/2 0 - 3 X1 3 1 3 0 1 0 0 0 S1 4 0 5 1 2 0 -1 0 S3 4 0 1 0 0 1 0 Zj 3 9 0 3 0 0 Cj-Zj 0 -10 0 -3 0 4Department of Mechanical Engineering, SJBIT Page 25
27. 27. Operations Research [06ME74]Since the value of Cj-Zj is either negative or 0 under all columns the optimal solution has been obtained.Therefore x1=3 & x2=0, Z=3x1-x2 =3*3 =9Example 5 (A case of no feasible solution) Minimize Z=x1+2x2+x3, Subject to x1+1/2x2+1/2x3 1 3/2x1+2x2+x3>=8 x1, x2, x3 0Solution:Introduce stack, surplus & artificial variable to convert inequality into equality then the LP becomes, Max Z*=-x1-2x2-x3+0.s1+0s2-MA1 Where Z*=-Z Subject to x1+1/2x2+1/2x3+s1=1 3/2x1+2x2+x3-s2+A1=8 x1, x2, x3, s1, s2, A1 0 Cj -1 -2 -1 0 0 -M Cb Var Soln X1 X2 X3 S1 S2 A1 Ratio 0 S1 1 1 ½ ½ 1 0 0 2 -M A1 8 3/2 2 1 0 -1 1 4 Cj -3M/2 -2M -M 0 M -M Cj-Zj - - -1+M 0 -M 0 1+3M/2 2+2M -2 X2 2 2 1 1 2 0 0 -M A1 4 -5/2 0 -1 -4 -1 1 Cj -4+5m/2 -2 -2+M - M -M 4+4M Cj-Zj 3-5M/2 0 1-M 4-4M -M 0Since Cj - Zj is either negative or zero & the variable column contains artificial variable A1 is not at Zerolevel.In this method there is a possibility of many casesDepartment of Mechanical Engineering, SJBIT Page 26
28. 28. Operations Research [06ME74] 1. Column variable contains no artificial variable. In this case continue the iteration till an optimum solution is obtained. 2. Column variable contains at least one artificial variable AT Zero level & all Cj-Zj is either negative or Zero. In this case the current basic feasible solution is optimum through degenerate. 3. Column Variable contains at least one artificial variable not at Zero level. Also Cj - Zj<=0. In this case the current basic feasible solution is not optimal since the objective function will contain unknown quantity M, Such a solution is called pseudo-optimum solution. The Two Phase Method Phase I: Step 1: Ensure that all (bi) are non-negative. If some of them are negative, make them non-negative by multiplying both sides by –1. Step 2: Express the constraints in standard form. Step 3: Add non-negative artificial variables. Step 4: Formulate a new objective function, which consists of the sum of the artificial variables. This function is called infeasibility function. Step 5: Using simplex method minimize the new objective function s.t. the constraints of the original problem & obtain the optimum basic feasible function. Three cases arise: - 1. Min Z* & at least one artificial variable appears in column variable at positive level. In such a case, no feasible solution exists for the LPP & procedure is terminated. 2. Min W=0 & at least one artificial variable appears in column variable at Zero level. In such a case the optimum basic feasible solution to the infeasibility form may or may not be a basic feasible solution to the given original LPP. To obtain a basic feasible solution continue Phase I & try to drive all artificial variables out & then continue Phase II. 3. Min W=0 & no artificial variable appears in column variable. In such a case, a basic feasible solution to the original LPP has been found. Proceed to Phase II. Phase II: Use the optimum basic feasible solution of Phase I as a starting solution fir the original LPP. Using simplex method make iteration till an optimum basic feasible solution for it is obtained. Note: The new objective function is always of minimization type regardless of whether the given original LPP is of max or min type.Department of Mechanical Engineering, SJBIT Page 27
29. 29. Operations Research [06ME74] Example 6: Max Z=3x1+2x2+2x3, Subject to 5x1+7x2+4x3 7 -4x1+7x2+5x3 -2 3x1+4x2-6x3 29/7 Solution: Phase I Step 1: Since for the second constraint b2=-2, multiply both sides by –1 transform it to 4x1-7x2-5x3 2 Step 2: Introduce slack variables 5x1+7x2+4x3+s1=7 4x1-7x2-5x3+s2=2 3x1+4x3-6x3+s3=29/7 Step 3: Putting x1=x2=x3=0, we get s1=7,s2=2,s3=-29/7as initial basic feasible solution. However it is not the feasible solution as s1 is negative. Therefore introduce artificial variable A1 from the above Constraints can be written as 5x1+7x2+4x3+s1=7 4x1-7x2-5x3+s2=2 3x1+4x2-6x3-s3+A1=29/7 The new objective function is Z*=A1 Cj 0 0 0 0 0 0 1 CB Var Soln X1 X2 X3 S1 S2 S3 A1 Ratio 0 S1 7 5 7 4 1 0 0 0 1 0 S2 2 4 -7 -5 0 1 0 0 - 1 A1 29/7 3 4 -6 0 0 -1 1 29/28 Cj 3 4 -6 0 0 -1 1Department of Mechanical Engineering, SJBIT Page 28
30. 30. Operations Research [06ME74] Cj-Zj -3 -4 6 0 0 1 0 0 X2 1 5/7 1 4/7 1/7 0 0 0 7/5 0 S2 9 9 0 -1 1 1 0 0 1 1 A1 1/7 1/7 0 -58/7 -4/7 0 -1 1 1 Cj 1/7 0 -58/7 -4/7 0 -1 1 Cj-Zj -1/7 0 58/7 4/7 0 1 0 0 X2 2/7 0 1 294/7 3 0 5 -5 0 S2 0 0 0 521 37 1 63 -63 0 X1 1 1 0 -58 -4 0 -7 7 Cj 0 0 0 0 0 0 0 Cj-Zj 0 0 0 0 0 0 1Since all Cj - Zj 0 the objective function is 0 & no artificial variable appears in column variable thetable yields the basic feasible solution to the original problem.Phase II The original objective function is Max Z=3x1+2x2+2x3+0.s1+0.s2+0.s3 Cj 0 0 0 0 0 0 CB Var Soln X1 X2 X3 S1 S2 S3 Ratio 0 X2 2/7 0 1 42 3 0 5 1/147 0 S2 0 0 0 521 37 1 63 0 3 X3 1 1 0 -58 -4 0 -7 - Cj 3 2 -90 -6 0 -11 Cj-Zj 0 0 92 6 0 11 2 X2 2/7 0 1 0 59/521 - - 42/521 41/521 2 X3 0 0 0 1 37/521 1/521 63/521 3 X1 1 1 0 0 62/521 58/521 7/521 Cj 3 2 2 278/521 92/521 65/521 Cj-Zj 0 0 0 - - - 278/521 92 / 521 65/521Department of Mechanical Engineering, SJBIT Page 29
31. 31. Operations Research [06ME74] x1=1,x2=2/7,x3=0 z=25/7Example 7:(Unconstrained variables) Min Z=2x1+3x2 Subject to x1-2x2 0 -2x1+3x2 -6 x1,x2 unrestricted.Solution:The RHS of 2nd constraint id –ve so multiply both sides by –1 we get 2x1-3x2 6as variable x1 & x2 are unrestricted we express them as x1=y1-y2 x2= y3-y4 where yi 0 i=1,2,3,4thus the given problem is transformed to min Z=2y1-2y2+3y3-3y4 s.t. y1-y2- 2y3+2y4 0 -2y1-2y2+ 3y3-3y4 6introduce slack variables we get Min Z= 2y1-2y2+3y3-3y4+0s1+0s2 Subject to y1-y2- 2y3+2y4+0s1 0 -2y1-2y2+ 3y3-3y4+0s2 6 where all variables are 0 Cj 2 -2 3 -3 0 0 CB Var Soln Y1 Y2 Y3 Y4 S1 S2 Ratio 0 S1 0 1 -1 -2 2 1 0 0 0 S2 6 2 -2 -3 3 0 1 2 Cj 0 0 0 0 0 0Department of Mechanical Engineering, SJBIT Page 30
32. 32. Operations Research [06ME74] Cj-Zj 2 -2 3 -3 0 0 -3 Y4 0 ½ -1/2 -1 1 ½ 0 - 0 S2 6 ½ -1/2 0 0 -3/2 1 - Cj -3/2 3/2 3 -3 3/2 0 Cj-Zj 3/2 -7/2 0 0 -3/2 0All the ratios are –ve that the value of the incoming non-basic variable y2 can be made as large aspossible without violating the constraint. This problem has unbounded solution & the iteration stopshere.Note:If the minimum ratio is equal to for 2 or more rows, arbitrary selection of 1 of these variables may resultin 1 or more variable becoming 0 in the next iteration & the problem is said to degenerate.These difficulties maybe overcome by applying the following simple procedure called perturbationmethod. 1. Divide each element in the tied rows by the positive co-efficient of the key column in that row 2. Compare the resulting ratios, column-by-column 1st in the identity & then in the body from left to right. 3. The row which first contains the smallest algebraic ratio contains the outgoing variable. Linear programming - sensitivity analysisRecall the production planning problem concerned with four variants of the same product which weformulated before as an LP. To remind you of it we repeat below the problem and our formulation of it. Production planning problemA company manufactures four variants of the same product and in the final part of the manufacturingprocess there are assembly, polishing and packing operations. For each variant the time required forthese operations is shown below (in minutes) as is the profit per unit sold. Assembly Polish Pack Profit (£)Variant 1 2 3 2 1.50 2 4 2 3 2.50 3 3 3 2 3.00 4 7 4 5 4.50 Given the current state of the labour force the company estimate that, each year, they have 100000 minutes of assembly time, 50000 minutes of polishing time and 60000 minutes ofDepartment of Mechanical Engineering, SJBIT Page 31
33. 33. Operations Research [06ME74] packing time available. How many of each variant should the company make per year and what is the associated profit? Suppose now that the company is free to decide how much time to devote to each of the three operations (assembly, polishing and packing) within the total allowable time of 210000 (= 100000 + 50000 + 60000) minutes. How many of each variant should the company make per year and what is the associated profit? Production planning solution VariablesLet: xi be the number of units of variant i (i=1,2,3,4) made per yearTass be the number of minutes used in assembly per yearTpol be the number of minutes used in polishing per yearTpac be the number of minutes used in packing per yearwhere xi >= 0 i=1,2,3,4 and Tass, Tpol, Tpac >= 0 Constraints(a) operation time definitionTass = 2x1 + 4x2 + 3x3 + 7x4 (assembly)Tpol = 3x1 + 2x2 + 3x3 + 4x4 (polish)Tpac = 2x1 + 3x2 + 2x3 + 5x4 (pack)(b) operation time limitsThe operation time limits depend upon the situation being considered. In the first situation, where themaximum time that can be spent on each operation is specified, we simply have:Tass <= 100000 (assembly)Tpol <= 50000 (polish)Tpac <= 60000 (pack)In the second situation, where the only limitation is on the total time spent on all operations, we simplyhave:Tass + Tpol + Tpac <= 210000 (total time) ObjectivePresumably to maximise profit - hence we havemaximise 1.5x1 + 2.5x2 + 3.0x3 + 4.5x4which gives us the complete formulation of the problem.Department of Mechanical Engineering, SJBIT Page 32
34. 34. Operations Research [06ME74]A summary of the input to the computer package for the first situation considered in the question(maximum time that can be spent on each operation specified) is shown below.The solution to this problem is also shown below.We can see that the optimal solution to the LP has value 58000 (£) and that Tass=82000, Tpol=50000,Tpac=60000, X1=0, X2=16000, X3=6000 and X4=0.Department of Mechanical Engineering, SJBIT Page 33
35. 35. Operations Research [06ME74]This then is the LP solution - but it turns out that the simplex algorithm (as a by-product of solving theLP) gives some useful information. This information relates to: changing the objective function coefficient for a variable forcing a variable which is currently zero to be non-zero changing the right-hand side of a constraint.We deal with each of these in turn, and note here that the analysis presented below ONLY applies for asingle change, if two or more things change then we effectively need to resolve the LP. suppose we vary the coefficient of X2 in the objective function. How will the LP optimal solution change?Currently X1=0, X2=16000, X3=6000 and X4=0. The Allowable Min/Max c(i) columns above tell usthat, provided the coefficient of X2 in the objective function lies between 2.3571 and 4.50, the values ofthe variables in the optimal LP solution will remain unchanged. Note though that the actual optimalsolution value will change.In terms of the original problem we are effectively saying that the decision to produce 16000 of variant 2and 6000 of variant 3 remains optimal even if the profit per unit on variant 2 is not actually 2.5 (but liesin the range 2.3571 to 4.50).Similar conclusions can be drawn about X1, X3 and X4.In terms of the underlying simplex algorithm this arises because the current simplex basic solution(vertex of the feasible region) remains optimal provided the coefficient of X2 in the objective functionlies between 2.3571 and 4.50. for the variables, the Reduced Cost column gives us, for each variable which is currently zero (X1 and X4), an estimate of how much the objective function will change if we make that variable non-zero.Hence we have the tableVariable X1 X4Opportunity Cost 1.5 0.2New value (= or >=) X1=A X4=B or X1>=A X4>=BEstimated objective function change 1.5A 0.2BThe objective function will always get worse (go down if we have a maximisation problem, go up if wehave a minimisation problem) by at least this estimate. The larger A or B are the more inaccurate thisestimate is of the exact change that would occur if we were to resolve the LP with the correspondingconstraint for the new value of X1 or X4 added.Hence if exactly 100 of variant one were to be produced what would be your estimate of the newobjective function value?Department of Mechanical Engineering, SJBIT Page 34
36. 36. Operations Research [06ME74]Note here that the value in the Reduced Cost column for a variable is often called the "opportunity cost"for the variable.Note here than an alternative (and equally valid) interpretation of the reduced cost is the amountby which the objective function coefficient for a variable needs to change before that variable willbecome non-zero.Hence for variable X1 the objective function needs to change by 1.5 (increase since we are maximising)before that variable becomes non-zero. In other words, referring back to our original situation, the profitper unit on variant 1 would need to need to increase by 1.5 before it would be profitable to produce anyof variant 1. Similarly the profit per unit on variant 4 would need to increase by 0.2 before it would beprofitable to produce any of variant 4. for each constraint the column headed Shadow Price tells us exactly how much the objective function will change if we change the right-hand side of the corresponding constraint within the limits given in the Allowable Min/Max RHS column.Hence we can form the tableConstraint Assembly Polish PackOpportunity Cost (ignore sign) 0 0.80 0.30Change in right-hand side a b cObjective function change 0 0.80b 0.30cLower limit for right-hand side 82000 40000 33333.34Current value for right-hand side 100000 50000 60000Upper limit for right-hand side - 90000 75000For example for the polish constraint, provided the right-hand side of that constraint remains between40000 and 90000 the objective function change will be exactly 0.80[change in right-hand side from50000].The direction of the change in the objective function (up or down) depends upon the direction of thechange in the right-hand side of the constraint and the nature of the objective (maximise or minimise).To decide whether the objective function will go up or down use: constraint more (less) restrictive after change in right-hand side implies objective function worse (better) if objective is maximise (minimise) then worse means down (up), better means up (down)Hence if you had an extra 100 hours to which operation would you assign it? if you had to take 50 hours away from polishing or packing which one would you choose? what would the new objective function value be in these two cases?The value in the column headed Shadow Price for a constraint is often called the "marginal value" or"dual value" for that constraint.Department of Mechanical Engineering, SJBIT Page 35
37. 37. Operations Research [06ME74]Note that, as would seem logical, if the constraint is loose the shadow price is zero (as if the constraint isloose a small change in the right-hand side cannot alter the optimal solution). Comments Different LP packages have different formats for input/output but the same information as discussed above is still obtained. You may have found the above confusing. Essentially the interpretation of LP output is something that comes with practice. Much of the information obtainable (as discussed above) as a by-product of the solution of the LP problem can be useful to management in estimating the effect of changes (e.g. changes in costs, production capacities, etc) without going to the hassle/expense of resolving the LP. This sensitivity information gives us a measure of how robust the solution is i.e. how sensitive it is to changes in input data.Note here that, as mentioned above, the analysis given above relating to: changing the objective function coefficient for a variable; and forcing a variable which is currently zero to be non-zero; and changing the right-hand side of a constraintis only valid for a single change. If two (or more) changes are made the situation becomes morecomplex and it becomes advisable to resolve the LP. Linear programming sensitivity exampleConsider the linear program:maximise3x1 + 7x2 + 4x3 + 9x4subject tox1 + 4x2 + 5x3 + 8x4 <= 9 (1)x1 + 2x2 + 6x3 + 4x4 <= 7 (2)xi >= 0 i=1,2,3,4Solve this linear program using the computer package. what are the values of the variables in the optimal solution? what is the optimal objective function value? which constraints are tight? what would you estimate the objective function would change to if: o we change the right-hand side of constraint (1) to 10 o we change the right-hand side of constraint (2) to 6.5 o we add to the linear program the constraint x3 = 0.7Solving the problem using the package the solution is:Department of Mechanical Engineering, SJBIT Page 36
38. 38. Operations Research [06ME74]Reading from the printout given above we have: the variable values are X1=5, X2=1, X3=0, X4=0 the optimal objective function value is 22.0 both constraints are tight (have no slack or surplus). Note here that the (implicit) constraints ensuring that the variables are non-negative (xi>=0 i=1,2,3,4) are (by convention) not considered in deciding which constraints are tight. objective function change = (10-9) x 0.5 = 0.5. Since the constraint is less restrictive the objective function will get better. Hence as we have a maximisation problem it will increase. Referring to the Allowable Min/Max RHS column we see that the new value (10) of the right- hand side of constraint (1) is within the limits specified there so that the new value of the objective function will be exactly 22.0 + 0.5 = 22.5 objective function change = (7-6.5) x 2.5 = 1.25. Since we are making the constraint more restrictive the objective function will get worse. Hence as we have a maximisation problem it will decrease. As for (1) above the new value of the right-hand side of constraint (2) is within the limits in the Minimum/Maximum RHS column and so the new value of the objective function will be exactly 22.0 - 1.25 = 20.75 objective function change = 0.7 x 13.5 = 9.45. The objective function will get worse (decrease) since changing any variable which is zero at the linear programming optimum to a non-zero value always makes the objective function worse. We estimate that it will decrease to 22.0 - 9.45 = 12.55. Note that the value calculated here is only an estimate of the change in the objective function value. The actual change may be different from the estimate (but will always be >= this estimate).Note that we can, if we wish, explicitly enter the four constraints xi>=0 i=1,2,3,4. Although this isunnecessary (since the package automatically assumes that each variable is >=0) it is not incorrect.However, it may alter some of the solution figures - in particular the Reduced Cost figures may bedifferent. This illustrates that such figures are not necessarily uniquely defined at the linearprogramming optimal solution.Department of Mechanical Engineering, SJBIT Page 37
39. 39. Operations Research [06ME74] UNIT-III Transportation ModelFind the optimal solution for the following given TP model. Distribution center (to) 1 2 3 4 Supply Plants (From) 2 3 11 7 6 1 0 6 1 1 5 8 15 9 10 Requireme 7 5 3 2 ntsNote:If the supply & demand are equal then it is called balanced otherwise unbalanced.Non-Degenerate:A basic feasible solution to a (m x n) transportation problem that contains exactly (m+n-1) allocation inindependent position.Degenerate:A basic feasible solution that contains less than m+n-1 non-negative allocations Find Basic Feasible Solution1) North West Corner RuleStart in the northwest corner If D1<S1, then set x1 equal to D1 & proceed horizontally. If D1=S1, then set x1 equal to D1 & proceed diagonally. If D1>S1, then set x1 equal to S1 & proceed vertically. 2 3 11 7 6 0 6 1 0 6 1 1 1 0 5 8 15 9 5 3 2Department of Mechanical Engineering, SJBIT Page 38
40. 40. Operations Research [06ME74] 10 5 3 2 7 5 3 2 6 0 0 0 1it can be easily seen that the proposed solution is a feasible solution since all the supply & requirementconstraints are fully satisfied. In this method, allocations have been made without any consideration ofcost of transformation associated with them.Hence the solution obtained may not be feasible or the best solution.The transport cost associate with this solution is : Z=Rs (2*6+1*1+8*5+15*3+9*2) * 100 =Rs (12+1+40+45+18) * 100 =Rs 116002) Row Minima MethodThis method consists in allocating as much as possible in the lowest cost cell of the 1st row so that eitherthe capacity of the 1st plant is exhausted or the requirement at the jth distribution center is satisfied orbothThree cases arises: If the capacity of the 1st plant is completely exhausted, cross off the 1st row & proceed to the 2nd row. If the requirement of the jth distribution center is satisfied, cross off the jth column & reconsider the 1st row with the remaining capacity. If the capacity of the 1st plant as well as the requirement at jth distribution center are completely satisfied, make a 0 allocation in the 2nd lowest cost cell of the 1st row. Cross of the row as well as the jth column & move down to the 2nd row. 2 3 11 7 6 6 0 1 0 6 1 1 1 0 5 8 15 9 1 4 3 2 10 9 5 3 0 7 5 3 2Department of Mechanical Engineering, SJBIT Page 39
41. 41. Operations Research [06ME74] 1 0 0 0 0 Z=100 * (6*2+0*1+5*1+8*4+15*3+9*2) =100 * (12+0+5+32+45+18) =100 * 112 =112003) Column Minima Method 2 3 11 7 6 6 0 1 0 6 1 1 1 0 5 8 15 9 5 3 2 10 5 3 2 0 7 5 3 2 6 0 0 0 1 0Z= 2*6+1*1+5*0+5*8+15+18= 12+1+40+45+18= 1164) Least Cost Method This method consists of allocating as much as possible in the lowest cost cell/cells& then further allocation is done in the cell with the 2nd lowest cost.Department of Mechanical Engineering, SJBIT Page 40
42. 42. Operations Research [06ME74] 2 3 11 7 6 6 0 1 0 6 1 1 5 8 15 9 1 0 1 4 3 2 10 9 5 2 0 7 5 3 2 1 4 0 0 0 0 Z= 12+0+5+32+45+18 = 1125. Vogel’s Approximation Method2 3 11 7 6 1 0 [ 1] [ 1] [ 5] * 1 5 1 0 6 1 1 0 [ 1] * * * 15 8 15 9 10 7 8 0 [ 3] [ 3] [ 4 ] [ 4] 6 3 17 5 3 26 0 0 10 0[1] [3] [5] [6][3] [5] [4] [2][3] * [4] [2][5] [8] [4] [9]Department of Mechanical Engineering, SJBIT Page 41
43. 43. Operations Research [06ME74]z = 2 + 15 + 1 + 30 + 45 + 9 = 102Perform Optimality TestAn optimality test can, of course, be performed only on that feasible solution in which: (a) Number of allocations is m+n-1 (b) These m+n-1 allocations should be in independent positions.A simple rule for allocations to be in independent positions is that it is impossible to travel from anyallocation, back to itself by a series of horizontal & vertical jumps from one occupied cell to another,without a direct reversal of route.Now test procedure for optimality involves the examination of each vacant cell to find whether or notmaking an allocation in it reduces the total transportation cost.The 2 methods usually used are: (1) Stepping-Stone method (2) The modified distribution (MODI) method1. The Stepping-Stone MethodLet us start with any arbitrary empty cell (a cell without allocation), say (2,2) & allocate +1 unit to thiscell, in order to keep up the column 2 restriction (-1) must be allocated to the cell (1,2) and keep the row1 restriction, +1 must be allocated to cell (1,1) and consequently (-1) must be allocated to cell (2,1). 2 +1 3 -1 11 7 1 5 1 -1 0 +1 6 1 1 5 8 15 9 6 3 1The net change in the transportation cost is=0*1-3*1+2*1-1*1=-2Naturally, as a result of above perturbation, the transportation cost decreased by –2.Department of Mechanical Engineering, SJBIT Page 42
44. 44. Operations Research [06ME74]The total number of empty cell will be m.n- (m+n-1)=(m-1)(n-1)]Such cell evaluations that must be calculated. If anycell evaluation is negative, 0 1 10 4 the cost can be reduced. Sothat the solution under 2 12 6 consideration can beimproved. -3 -3 -2 7 5 62. The Modified Distribution (MODI) method or u-v methodStep 1: Set up the cost matrix containing the costs associated with the cells for which allocations have been made. V1=0 v2=0 v3=0 v4=0 u1=2 2 3 u2=3 1 u3=5 5 15 9Step 2: Enter a set of number Vj across the top of the matrix and a set of number Ui across the left side so that their sums equal to the costs entered in Step 1. Thus, u1+v1=2 u3+v1=5 u1+v2=3 u3+v3=15 u2+v4=1 2 u3+v4=9 3 Let v1=0 1 u1=2 ; u2=-3 ; u1=2 ; v2=1 ; v3=10 ; v4=4 5 15 9Step 3: Fill the vacant cells with the sum of ui & vjDepartment of Mechanical Engineering, SJBIT Page 43
45. 45. Operations Research [06ME74]Step 4: Subtract the cell values of the matrix of Step 3 from original cost matrix. 11-12 7-6 1+3 0+2 6-7 8-6The resulting matrix is called cell evaluation matrix. -1 1 4 2 -1step5: If any of the cell evaluations are negative the basic feasible 2solution is not optimal.Iterate towards optimal solutionSubstep1: From the cell evaluation matrix, identify the cell with the most negative entry. Let us choose cell (1,3).Substep2: write initial feasible solution. 1 5 + - 1 6 3 + - Check mark ( )the empty cell for which the cell evaluation is negative. This cell is chosen in substep1 & is called identified cell.Substep3: Trace a path in this matrix consisting of a series of alternately horizontal & vertical lines. Thepath begins & terminates in the identified cell. All corners of the path lie in the cells for whichallocations have been made. The path may skip over any number of occupied or vacant cells.Department of Mechanical Engineering, SJBIT Page 44
46. 46. Operations Research [06ME74]Substep4: Mark the identified cells as positive and each occupied cell at the corners of the pathalternatively negative & positive & so on.Note :In order to maintain feasibility locate the occupied cell with minus sign that has the smallestallocationSubstep5: Make a new allocation in the identified cell. 1-1=0 5 1 1 7 2 1 2 3 11 7 5 1 1 0 6 1 1 5 8 15 5 7 2 1The total cost of transportation for this 2nd feasible solution is=Rs (3*5+11*1+1*1+7*5+2*15+1*9)=Rs (15+11+1+35+30+9)=Rs 101Check for optimalityIn the above feasible solution a) number of allocations is (m+n-1) is 6 b) these (m+ n – 1)allocation are independent positions.Above conditions being satisfied, an optimality test can be performed.MODI methodStep1: Write down the cost matrix for which allocations have been made.Step2: Enter a set of number Vj across the top of the matrix and a set of number Ui across the left side sothat their sums equal to the costs entered in Step1. Thus, u1+v2=3 u3+v1=5Department of Mechanical Engineering, SJBIT Page 45
47. 47. Operations Research [06ME74] u1+v3=11 u3+v3=15 u2+v4=1 u3+v4=9 Let v1=0 u1=1 ; u2=-3 ; u3=5 ; v2=2 ; v3=10 ; v4=4 v1 v2 v3 v4 u1 3 11 u2 1 u3 5 15 9Step3: Fill the vacant cells with the sums of vj & ui v1=0 v2=2 v3=10 v4=4 U1=1 1 5 U2=-3 -3 -1 7 U3=5 7Step4: Subtract from the original matrix. 2-1 7-5 1 2 1+3 0+1 6-7 4 1 -1 8-7 -1Step5: Since one cell is negative, 2nd feasible solution is not optimal.Iterate towards an optimal solutionSubstep1: Identify the cell with most negative entry. It is the cell (2,3).Substep2: Write down the feasible solution. 5 1 + 1 - 7 2 - 1 +Substep3: Trace the path.Substep4: Mark the identified cell as positive & others as negative alternatively.Substep5: 5 1 Page 46 Third feasible solutionsDepartment of Mechanical Engineering, SJBIT
48. 48. Operations Research [06ME74] 1 7 1 2Z=(5*3+1*11+1*6+1*15+2*9+7*5)=100Test for optimalityIn the above feasible solutions a) number of allocation is (m+n-1) that is, 6 b) these (m+n-1) are independent.Step1: setup cost matrix 3 11 6 5 15 9Step2: Enter a set of number Vj across the top of the matrix and a set of number Ui across the left side sothat their sums equal to the costs entered in Step1. Thus, u1+v2=3 u3+v1=5 u1+v3=11 u3+v3=15 u2+v3=6 u3+v4=9let v1 = 0, u3 =5 u2=-4, u1=1, v2=2, v3=10, v4=4the resulting matrix is 0 2 10 4 1 1 5 -4 -4 -2 0 5 7Subtract from original cost matrix, we will get cell evaluation matrix 1 2 5 2 1Department of Mechanical Engineering, SJBIT Page 47
49. 49. Operations Research [06ME74] 1Since all the cells are positive, the third feasible solution is optimal solution. Assignment ProblemStep 1: Prepare a square matrix: Since the solution involves a square matrix, this step is not necessary.Step 2: Reduce the matrix: This involves the following substeps. Substep 1: In the effectiveness matrix, subtract the minimum element of each row from all theelements of the row. See if there is atleast one zero in each row and in each column. If it is so, stop here.If it is so, stop here. If not proceed to substep 2.Department of Mechanical Engineering, SJBIT Page 48
50. 50. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 49
51. 51. Operations Research [06ME74]SUBSTEP 2: Next examine columns for single unmarked zeroes and mark them suitably.SUBSTEP 3: In the present example, after following substeps 1 and 2 we find that their repetition isunnecessary and also row 3 and column 3 are without any assignments. Hence we proceed as follows tofind the minimum number of lines crossing all zeroes.SUBSTEP 4: Mark the rows for which assignment has not been made. In our problem it is the third row.SUBSTEP 5: Mark columns (not already marked) which have zeroes in marked rows. Thus column 1 ismarked.SUBSTEP 6: Mark rows(not already marked) which have assignmentsin marked columns. Thus row 1 ismarked.Department of Mechanical Engineering, SJBIT Page 50
52. 52. Operations Research [06ME74]SUBSTEP 7: Repeat steps 5 and 6 until no more marking is possible. In the present case this repetitionis not necessary.Department of Mechanical Engineering, SJBIT Page 51
53. 53. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 52
54. 54. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 53
55. 55. Operations Research [06ME74]Department of Mechanical Engineering, SJBIT Page 54
56. 56. Operations Research [06ME74] Cost = 61 THE TRAVELLING SALESMAN PROBLEMThe condition for TSP is that no city is visited twice before the tour of all the cities is completed. A B C D E A 0 2 5 7 1 B 6 0 3 8 2 C 8 7 0 4 7 D 12 4 6 0 5 E 1 3 2 8 0As going from A->A,B->B etc is not allowed, assign a large penalty to these cells in the cost matrixDepartment of Mechanical Engineering, SJBIT Page 55
57. 57. Operations Research [06ME74] A B C D E A B C D E � 1 4 6 0A A � 1 3 6 0 _ �B 4 1 6 0 _ � B 4 0 6 0 _ �C 4 3 0 3 _ � C 4 3 0 3D 8 0 2 � 1 D 8 0 1 � 1E 0 2 1 7 � E 0 2 0 7 � _ _ A B _ C D E _ A � 1 3 _ 6 _ B _ 4 � 0 6 0 0 _ 0 C 4 3 � 0 3 D 8 _ 1 � 1 E 0 2 0 0 _ 7 � _ 0 Which gives optimal for assignment problem but not for TSP because the path A->E, E-A, B->C->D->B does not satisfy the additional constraint of TSP The next minimum element is 1, so we shall try to bring element 1 into the solution. We have three cases. Case 1: Make assignment in cell (A,B) instead of (A,E). A B C D E A ∞ 1 3 6 0 B 4 ∞ 0 6 0 C 4 3 ∞ 0 3 D 8 0 1 ∞ 1 E 0 2 0 7 ∞ The resulting feasible solution is A->B,B-C, C->D,D->E,E->A and cost is 15 Now make assignment in cell (D,C) instead of (D,B) Department of Mechanical Engineering, SJBIT Page 56
58. 58. Operations Research [06ME74]Case 2: A B C D E A ∞ 1 3 6 0 B 4 ∞ 0 6 0 C 4 3 ∞ 0 3 D 8 0 1 ∞ 1 E 0 2 0 7 ∞Since second row does not have any assignment. We can choose minimum cost in that row and if anyassignment is there in that column, shift to next minimum cell. A B C D E A ∞ 1 3 6 0 B 4 ∞ 0 6 0 C 4 3 ∞ 0 3 D 8 0 1 ∞ 1 E 0 2 0 7 ∞Case 3: A B C D E A ∞ 1 3 6 0 B 4 ∞ 0 6 0 C 4 3 ∞ 0 3 D 8 0 1 ∞ 1 E 0 2 0 7 ∞Which is the same as case 1 . least cost route is given by a->b->c->d->e->aDepartment of Mechanical Engineering, SJBIT Page 57
59. 59. Operations Research [06ME74] UNIT-V Queuing theoryQueuing theory deals with problems which involve queuing (or waiting). Typical examples might be: banks/supermarkets - waiting for service computers - waiting for a response failure situations - waiting for a failure to occur e.g. in a piece of machinery public transport - waiting for a train or a busAs we know queues are a common every-day experience. Queues form because resources are limited. Infact it makes economic sense to have queues. For example how many supermarket tills you would needto avoid queuing? How many buses or trains would be needed if queues were to be avoided/eliminated?In designing queueing systems we need to aim for a balance between service to customers (short queuesimplying many servers) and economic considerations (not too many servers).In essence all queuing systems can be broken down into individual sub-systems consisting of entitiesqueuing for some activity (as shown below).Typically we can talk of this individual sub-system as dealing with customers queuing for service. Toanalyse this sub-system we need information relating to: arrival process: o how customers arrive e.g. singly or in groups (batch or bulk arrivals) o how the arrivals are distributed in time (e.g. what is the probability distribution of time between successive arrivals (the interarrival time distribution)) o whether there is a finite population of customers or (effectively) an infinite number The simplest arrival process is one where we have completely regular arrivals (i.e. the same constant time interval between successive arrivals). A Poisson stream of arrivals corresponds to arrivals at random. In a Poisson stream successive customers arrive after intervals which independently are exponentially distributed. The Poisson stream is important as it is a convenient mathematical model of many real life queuing systems and is described by a single parameter - the average arrival rate. Other important arrival processes are scheduled arrivals; batch arrivals; and time dependent arrival rates (i.e. the arrival rate varies according to the time of day). service mechanism: o a description of the resources needed for service to begin o how long the service will take (the service time distribution) o the number of servers availableDepartment of Mechanical Engineering, SJBIT Page 58
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## Summarize problem
Hi,
I've a table with the value of each costumer's order, an order may be divided in additional sub-orders or item according to the type of service required. All items from an order may have the same code like 1 for the 11234 order, or multiple codes like 1, 2, 3 for the 11234 order depending on the supplyer which can be the same or not. In any case each item code is unique to the order so, knowing the item code you can always recognize the order number.
In addition to the revenue, I've also a table with the associated expected costs for each items, constantly updated, and a table with the actual costs which is updated monthly.
What I would like to obtain, is a new table which uses the summarize function to summarize, for each order, the actual cost but since this table is uptaded less frequently, I'd also like to see the budget cost in the meantime:
SUMMARIZE('Actual Cost', 'Actual Cost [Order], "Direct Cost", IF(SUM('Actual Cost' [Amount]) = BLANK() || SUM(SUM('Actual Cost' [Amount]) < SUM('Cost Budget'[Amount]), SUM('Cost Budget'[Amount]) ,(SUM('Actual Cost' [Amount])))
The actual costs are always equal or higher than the expected ones because the budget is from the order confirmation to the supplier while the actual cost is from the purchase invoice which may have additional costs for certificates, assurance, customs duty.
My problem is that when I'tried that formula, the amount from the actual is correct while the budget is three times the real one, like Bi multiplies the value by the numbers of row, since the order is one but the item rows are three. Maybe is because the budget has items and the actual cost table has N. of Order.
I cannot use the RELATED function to create a new column with the order numbers in the budget table because of the many to many relationships.
Can someone help?
Thanks!
1 ACCEPTED SOLUTION
Solution Specialist
Hi, @Luca2020 ,
I suggest that you provide the model relationship of different tables, one to many or many to one, which is helpful to solve this problem.
Nevertheless, I have encountered similar problem before.
My solution is to create a new calculation column in the many-end table to calculate the number of rows that will be repeated after use RELATED() from one-end table, and then your measure can be divided by the value in this column to get a value that won't be repeated.
The formula for this calculate column is :
`column = countrows(RELATEDTABLE(<your main table>))`
Mark this post as solution if this helps, thanks!
Solution Specialist
Hi, @Luca2020 ,
I suggest that you provide the model relationship of different tables, one to many or many to one, which is helpful to solve this problem.
Nevertheless, I have encountered similar problem before.
My solution is to create a new calculation column in the many-end table to calculate the number of rows that will be repeated after use RELATED() from one-end table, and then your measure can be divided by the value in this column to get a value that won't be repeated.
The formula for this calculate column is :
`column = countrows(RELATEDTABLE(<your main table>))`
Mark this post as solution if this helps, thanks!
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# Questions tagged [propulsion]
Questions about the various methods used to propel aircraft.
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### How are Fan pressure of and Static pressure in the turbofan engine related?
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### Are subsonic nozzle fully expanded?
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• 101
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• 411
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### How does a partially covered air intake affect the thrust of a propeller ?
does a partially covered air intake of a ducted propeller significantly affect its thrust? If so, will it be like 10 or rather 90 percent less thrust ? Thanks for your answers! Simon
• 351
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I am currently doing research for a “white paper design” aircraft and I have always been interested in push pull aircraft as they seem to be rare and I wanted to see some challenges in making a very ...
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Cody
# Problem 640. Getting logical indexes
Solution 1825667
Submitted on 25 May 2019 by xr hu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
vec = [1 2 3 3 2 1]; v = 2; y_correct = [false true false false true false]; assert(isequal(binaryEqualsVector(vec,v),y_correct))
2 Pass
vec = [1 2 3 4 5 6]; v = 0; y_correct = [false false false false false false]; assert(isequal(binaryEqualsVector(vec,v),y_correct))
3 Pass
vec = [1 1 1 1 1]; v = 1; y_correct = [true true true true true]; assert(isequal(binaryEqualsVector(vec,v),y_correct))
4 Pass
vec = 'abcdef'; v = 'a'; y_correct = [true false false false false false]; assert(isequal(binaryEqualsVector(vec,v),y_correct))
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# Centrifugal Force: What Is It?
• alchemist
In summary, centrifugal force is the apparent outward force experienced by an object moving in a circular path, due to its inertia. It is not a real force. It is often confused with centripetal force, which is a real inward force that keeps an object in a circular path. Centrifugal force can be calculated using the formula F = m * v^2 / r. Some real-world examples include the feeling of being pushed outwards on a merry-go-round and the swinging motion of a thrown hammer. It is not a fundamental force, but can be explained by the laws of motion and gravity.
alchemist
is there such a thing as centrifugal force?
what is it?
If you swing a yoyo around your head, the centrifugal force is what your hand feels. The centripetal force is what the yoyo feels.
## 1. What is centrifugal force?
Centrifugal force is the apparent outward force experienced by an object that is moving in a circular path. It is a result of the object's inertia and is not actually a real force.
## 2. How is centrifugal force different from centripetal force?
Centrifugal force and centripetal force are often confused. While centrifugal force is an apparent outward force experienced by an object, centripetal force is a real inward force that keeps the object moving in a circular path.
## 3. How is centrifugal force calculated?
Centrifugal force can be calculated using the formula F = m * v^2 / r, where F is the centrifugal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.
## 4. What are some real-world examples of centrifugal force?
Some examples of centrifugal force include the feeling of being pushed outwards when riding a merry-go-round, the swinging motion of a hammer when it is thrown, and the outward bending of a spinning yo-yo string.
## 5. Is centrifugal force a fundamental force?
No, centrifugal force is not a fundamental force. It is an apparent force that can be explained by the laws of motion and gravity. It is also not included in the standard model of physics.
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Engineering Statics: Open and Interactive
Section2.32D Coordinate Systems & Vectors
A coordinate system gives us a frame of reference to describe a system that we would like to analyze. In statics we normally use orthogonal coordinate systems, where orthogonal means “perpendicular.” In an orthogonal coordinate system the coordinate direction are perpendicular to each other and thereby independent. The intersection of the coordinate axes is called the origin, and measurements are made from there. Both points and vectors are described with a set of numbers called the coordinates. For points in space, the coordinates specify the distance you must travel in each of the coordinate directions to get from the origin to the point in question. Together, the coordinates can be thought of as specifying a position vector, a vector from the origin directly to the point. The position vector gives the magnitude and direction needed to travel directly from the origin to the point.
In the case of force vectors, the coordinates are the scalar components of the force in each of the coordinate directions. These components locate the tip of the vector and they can be interpreted as the fraction of the total force which acts in each of the coordinate directions.
Three coordinate directions are needed to map our real three-dimensional world but in this section we will start with two, simpler, two-dimensional orthogonal systems: rectangular and polar coordinates, and the tools to convert from one to the other.
Subsection2.3.1Rectangular Coordinates
The most important coordinate system is the Cartesian system, which was named after the French mathematician René Descartes. In two dimensions the coordinate axes are straight lines rotated 90° apart named $$x\text{,}$$ and $$y\text{.}$$
In most cases, the $$x$$ axis is horizontal and points to the right, and the $$y$$ axis points vertically upward, however, we are free to rotate or translate this entire coordinate system if we like. It is usually mathematically advantageous to establish the origin at a convenient point to make measurements from, and to align one of the coordinate axes with a major feature of the problem.
Points are specified as an ordered pair of coordinate values separated by a comma and enclosed in parentheses, $$P = (x,y)\text{.}$$
Similarly, forces and other vectors will be specified with an ordered pair of scalar components enclosed by angle brackets,
\begin{equation*} \vec{F} = \langle F_x, F_y \rangle\text{.} \end{equation*}
Subsection2.3.2Polar Coordinates
The polar coordinate system is an alternate orthogonal system which is useful in some situations. In this system, a point is specified by giving its distance from the origin $$r\text{,}$$ and $$\theta\text{,}$$ an angle measured counter-clockwise from a reference direction – usually the positive $$x$$ axis.
In this text, points in polar coordinates will be specified as an ordered pair of values separated by a semicolon and enclosed in parentheses
\begin{equation*} P = (r\ ; \theta)\text{.} \end{equation*}
Angles can be measured in either radians or degrees, so be sure to include a degree sign on angle $$\theta$$ if that is what you intend.
Subsection2.3.3Coordinate Transformation
You should be able to translate points from one coordinate system to the other whenever necessary. The relation between $$(x,y$$) coordinates and $$(r;\theta)$$ coordinates are illustrated in the diagram and right-triangle trigonometry is all that is needed to convert from one representation to the other.
Rectangular To Polar for points (Given: $$x$$ and, $$y$$).
\begin{align} r \amp = \sqrt{x^2 + y^2}\tag{2.3.1}\\ \theta \amp = \tan^{-1}{\left(\frac{y}{x}\right)}\tag{2.3.2}\\ P \amp = (r \; ; \theta)\tag{2.3.3} \end{align}
Note2.3.4.
Take care when using the inverse tangent function on your calculator. Calculator angles are always in the first or fourth quadrant, and you may need to add or subtract 180° to the calculator angle to locate the point in the correct quadrant.
Polar to Rectangular for points (Given: $$r$$ and, $$\theta$$).
\begin{align} x \amp = r \cos \theta\tag{2.3.4}\\ y \amp = r \sin \theta\tag{2.3.5}\\ P \amp = (x,y)\tag{2.3.6} \end{align}
Rectangular To Polar for forces (Given: rectangular components).
If you are working with forces rather than distances, the process is exactly the same but triangle is labeled differently. The hypotenuse of the triangle is the magnitude of the vector, and sides of the right triangle are the scalar components of the force, so for vector $$\vec{A}$$
\begin{align} A \amp = \sqrt{A_x^2 + A_y^2}\tag{2.3.7}\\ \theta \amp = \tan^{-1}{\left(\frac{A_y}{A_x}\right)}\tag{2.3.8}\\ \vec{A} \amp = (A\; ; \theta)\tag{2.3.9} \end{align}
Polar to Rectangular for forces (Given: magnitude and direction).
\begin{align} A_x \amp = A \cos \theta\tag{2.3.10}\\ A_y \amp = A \sin \theta\tag{2.3.11}\\ \vec{A} \amp = \langle A_x, A_y \rangle = A \langle \cos \theta, \sin \theta \rangle \tag{2.3.12} \end{align}
Example2.3.5.Rectangular to Polar Representation.
Express point $$P = (-8.66, 5)$$ in polar coordinates.
$$P = (10\; ; 150°)$$
Solution 1.
Given: $$x = -8.66\text{,}$$ $$y = 5$$
\begin{align*} r \amp = \sqrt{x^2 +y^2} \amp \theta \amp =\tan^{-1} \left(\frac{y}{x}\right)\\ \amp = \sqrt{(-8.66)^2 + (5)^2 } \amp \amp =\tan^{-1} \left(\frac{5}{-8.66}\right)\\ \amp = 10 \amp \amp = \tan^{-1} (- 0.577)\\ \amp \amp \amp = -\ang{30} \end{align*}
You must be careful here and use some common sense. The $$\ang{-30}$$ angle your calculator gives you in this problem is incorrect because point $$P$$ is in the second quadrant, but your calculator doesn’t know this. It can’t tell whether the argument of $$\tan^{-1}(-0.577)$$ is negative because the $$x$$ was negative or because the $$y$$ was negative, so it must make an assumption and in this case it is wrong.
The arctan function on calculators will always return values in the first and fourth quadrant. If, by inspection of the $$x$$ and the $$y$$ coordinates, you see that the point is in the second or third quadrant, you must add or subtract $$\ang{180}$$ to the calculator’s answer.
So in this problem, $$\theta$$ is really $$\ang{-30} + \ang{180}\text{.}$$ After making this adjustment, the location of $$P$$ in polar coordinates is:
\begin{gather*} P = (10; \ang{150}) \end{gather*}
Solution 2.
Most scientific calculators include handy polar-to-rectangular and rectangular-to-polar functions that can save you time and help you avoid errors. Perhaps you should google your calculator model
1
google.com
to find out if yours does and learn how to use it?
Example2.3.6.Polar to Rectangular Representation.
Express $$\N{200}$$ force $$\vec{F}$$ as a pair of scalar components.
Given: The magnitude of force $$\vec{F} = \N{200}\text{,}$$ and from the diagram we see that the direction of $$\vec{F}$$ is $$\ang{30}$$ counter-clockwise from the negative $$x$$ axis.
Letting $$\theta = \ang{30}$$ we can find the components of $$\vec{F}$$ with right triangle trigonometry.
After making this adjustment, the location of $$\vec{F}$$ expressed in rectangular coordinates is:
If you would prefer not to apply the negative signs by hand, you can convert the $$\ang{30}$$ to an angle measured from the positive $$x$$ axis and let your calculator takes care of the signs. You may use either $$\theta = \ang{30} \pm \ang{180}\text{.}$$
For $$\theta = \ang{-150}$$
Although this approach is mathematically correct, experience has shown that it can lead to errors and we recommend that when you work with right triangles, use angles between zero and $$\ang{90}\text{,}$$ and apply signs manually as required by the physical situation.
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# Easy Ways to Learn Maths to Quickly Understand!
Mathematics is a subject that is always present in every curriculum, from elementary school to college. Not only in science and technology majors, even in social science departments mathematics is used to solve various quantitative problems.
On the other hand, many people find it difficult to learn mathematics. The mindset that mathematics is a scourge has spread everywhere. Many feel that Mathematics is a relatively difficult subject. Do you feel that way? But, if we understand the principles, then we will be able to learn mathematics well.
Because mathematics is a unique subject, it takes unique skills to conquer it. This time, we discuss about what we must understand to learn mathematics:
### 1. Need strong active learning
In learning mathematics, you really need to be active. Mathematics is not just a lesson to calculate something. Mathematics is a lesson that trains our logic for problem solving. Unlike other subjects, one chapter of mathematics lessons can cover various kinds of problems (questions) that we must solve.
Well, the way to get around this is to get our brains used to solving those problems. This is tantamount to solving life’s problems. We must actively practice our problem solving skills, so that if we meet any problem, we can solve it well.
### 2. Math is cumulative!
Mathematics is cumulative, starting from basic operations (+) and (-), increasing to (x) and (:). Learning math will be difficult if you jump around, from one material to another. Almost all branches of mathematics are related and influence each other.
In learning mathematics, what you learn today becomes an important part of what you will study tomorrow. If you miss something, you’ll have a hard time catching up on the next material, and if you don’t catch up soon, you’ll keep falling behind. This is why many people easily fail in learning mathematics.
### 3. Focus on understanding concepts, not memorizing
In some lessons, we will pass and get good grades by relying on rote memorization. We can memorize names, dates, body parts, etc. However, mathematics cannot be solved properly by relying on memorization.
Memorizing all the formulas is a bad idea. Understanding the concept is an absolute thing for us in learning mathematics. If you don’t understand the concept, by holding all the books in the world, or carrying formula notes, you still won’t be able to solve math problems well.
### 4. Make a list of vocabulary & math formulas
For many people, especially mathematicians, mathematics is a language in itself. Mathematics has so many terms, followed by so many formulas. To make it easier to learn mathematics, one trick that can be used is to make a list of vocabulary and formulas in mathematics, and read it often to understand and remember what it means and what it is used for.
### 5. Keep records well & clearly
If you have trouble learning math, taking notes with only certain points is obviously a waste of time. Make comprehensive notes, note important things that the teacher sometimes doesn’t write on the blackboard. Pay attention to the emphasis given by the teacher, take the core concepts and existing formulas, make yourself understand by rewriting them in your own language. But be careful, don’t get too busy taking notes until you miss the teacher’s explanation.
### 6. Do the exercises & homework well
Practice assignments or homework will often be given in learning mathematics. The tasks are given to encourage us to develop skills and familiarize ourselves with mathematical concepts and problem solving. If you want to master math, stop complaining about so many problems, and start working. Don’t cheat easily! The tip is to do this assignment while the lesson is still fresh in your mind, for example right after class or on the same day. Work in detail, showing every step of the problem solving you take.
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### Let D(t) represent the amount of drug in a patient (measured in grams) at time t
Question: Let D(t) represent the amount of drug in a patient (measured in grams) at time t (mea- sured in hour). The initial amount of drug is 90 grams. After 2 hours the amount is 70 grams. Assuming that the amount of drug decays exponentially, address the following problems. (a) Write down a differential equation that best describes the growth of the amount of the drug. (b) Find the general solution to the differential equation in part (a). (c) Use the additional provided information to determine D(t). (d) What is the half-life for D(t)?
Solution:
### How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g.
Question: How many grams of oxygen are present in 100.0 g Ca(NO3)2 is g. Use correct number of significant digits;
Solution: Mol weight of Ca(NO3)2 = 90 + 28 +96
= 164 g/mol
100 g of Ca(NO3)2 = 0.6097 mol of Ca(NO3)2
1 mol of Ca(NO3)2 contains 6 number of oxygen atoms.
so, 0.6097 moles of Ca(NO3)2 contains = 0.6097 * 6 = 3.6582 Now, mole of each ‘O’ atom is 16 g/mol So, 100 g of Ca(NO3)2 contains = 3.6582*16 = 58.5312 g of oxygen.
### A superconductor is a substance that is able to conduct electricity without resistance
Question: A superconductor is a substance that is able to conduct electricity without resistance, a property that is very desirable in the construction of large electromagnets. Metals have this property if cooled to temperatures a few degrees above absolute zero, but this requires the use of expensive liquid helium (boiling point 4 K). Scientists have discovered materials that become superconductors at higher temperatures, but they are ceramics. Their brittle nature has so far prevented them from being made into long wires. A recently discovered compound of magnesium and boron, which consists of 52.9% Mg and 47.1% B, shows special promise as a high-temperature superconductor because it is inexpensive to make and can be fabricated into wire relatively easily. What is the formula of this compound?
Solution:
Let us take 100 g of compound
Then, mass of Mg = 52.9g
and mass of boron = 47.1g
Moles of Mg = Mass/Molar Mass = 52.9g/24.31g/mol
= 2.18 mol
and moles of B = 47.1g/10.81g/mol
= 4.36 mol
So, the molar ratio of Mg and B:
nMg/nB = 2.18 mol/4.36 mol = 1/2
Formula of the compound is MgB2
### Assuming that all the blackbody radiation is emitted at this wavelength
Question: Consider a metal that is being welded
Assuming that all the blackbody radiation is emitted at this wavelength (a crude approximation), how many photons does a square centimeter of the metal emit per second?
Solution: Using stefan-boltzmann law, energy emitted by blackbody
E = σAtT^4
A = 1 cm^2 = 10^-4 m^2
t = 1s
but E =hcn/λ
σ = 5.67 * 10^-8
T = b/λ = 2.898 * 10^-3/λ
n = λσAtT^4/hc
n = 5.6710^-810^-41(2.89810^-3)^4/(6.6310^-34310^8(49010^-9)^3)
n= 1.71*10^22 photons
### Data Communications Networking Chapter 3, Problem 48E
Question: What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 μs and a processing time of 1 μs. The length of the link is 2000 Km. The speed of light inside the link is 2 × 108 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible?
Solution:
Given data:
Distance (length of the link) =2000 Km
=2000
103m
Propagation speed=2
108m/s
Frame size=5million bits
=5000000bits
The Bandwidth=5Mbps
=5
106 bps (1Mbps=106 bps)
Queuing time=10
2 (10 routers each having a queuing time of 2)
=20
Processing delay=10
1 (10 routers each having a processing time of 1)
=10
Total delay (Latency) = Propagation time + Transmission time + Queuing time
+ Processing delay.
Propagation time=
=
= 0.01s
Therefore, Propagation time=0.01s
Transmission time=
=
= 1s
Therefore, Transmission time=1s
Queuing time=10
2
=10
210-6s (1=10-6s)
=20
10-6s
= 0.00002s
Therefore, Queuing time=0.00002s
Processing delay=10
1
=10
10-6s (1=10-6s)
Therefore, Processing delay=0.000010s
Therefore, Total delay (Latency) =0.01+1+0.000020+0.000010
=1.01003s
### Drag the “orbitals with electrons” box to the energy scale shown.
Question: Drag the “orbitals with electrons” box to the energy scale shown. Your goal in this exercise is to show nitrogen in an sp3-hybridized state which is capable of forming 3 bonds with three other atoms. Fill answer fields from left to right. Place orbitals with two electrons on the left (if any).
Solution: Concepts and reason
An orbital is the distribution of the probability density that is associated with the given electron centered at a given point. If the point is the nucleus of an atom, an atomic orbital is the resultant. If the point is the center of mass of multiple nuclei within a molecule, a molecular orbital is the resultant.
When an atom is required to bond with the other atoms to form a molecule, the valence atomic orbitals with almost similar energies ‘hybridize’ to form hybridized orbitals. These orbitals of identical shapes have identical associated energies.
The number of hybrid orbitals formed is equal to the number of atomic orbital hybridizing. Fundamentals
The hybridization of an atom in a molecule could be correlated with the number of bonds formed by it. This is tabulated as follows:
The sum of the principal quantum number and the azimuthal quantum number informs about the energy associated with a given orbital.
$(n+l)\left( {{\rm{n + l}}} \right)(n+l)$
Here
$n−{\rm{n}}\,{\rm{ - }}\,n$− Principal quantum number
$l−{\rm{l}}\,{\rm{ - }}l$− Azimuthal quantum number
The electronic configuration of nitrogen is $1s22s22p3{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}}1s22s22p3$.
The $(n+l)\left( {{\rm{n + l}}} \right)(n+l)$ values associated with the atomic orbitals involved are as follows:
$1s−12s−22p−3\begin{array}{l}\\{\rm{1s}}\,{\rm{ - }}\,{\rm{1}}\\\\{\rm{2s}}\,{\rm{ - }}\,2\\\\{\rm{2p}}\,{\rm{ - }}\,{\rm{3}}\\\end{array}1s−12s−22p−3$
The atomic orbitals with lower $(n+l)\left( {{\rm{n + l}}} \right)(n+l)$ values have lower associated energies. Therefore, they are filled first. In nitrogen, the orbital with the lowest associated energy, $1s{\rm{1s}}1s$ , is filled foremost, followed by $2s{\rm{2s}}2s$ , and finally $2p{\rm{2p}}2p $.
The hybridized orbitals possess a blend of characteristics associated with their constituent atomic orbitals. Therefore, they could be placed at an intermediate position between the respective atomic orbitals (here, $2sand2p{\rm{2s}}\,{\rm{and}}\,{\rm{2p}}2sand2p$ ).
Hence, the arrangement is given below:
The arrangement is given below:
The $sp3{\rm{s}}{{\rm{p}}^{\rm{3}}}sp3$ orbitals possess $25%{\rm{25\% }}25%$ ‘s’ character and $75%{\rm{75\% }}75%$ ‘p’ character. Correspondingly, the stabilization or the effective energy of the hybridized orbitals is a little lesser than the un-hybridized $′2p′{\rm{'2p'}}′2p′$ orbitals, but quite greater than the un-hybridized $′2s′{\rm{'2s'}}′2s′$orbitals. The electronic population of the un-hybridized, valence orbitals is 666 . Correspondingly, the hybridized orbitals also have the same population.
### In C++ Implement a class 14. Inventory Bins p. Define the operators >, =.
Question: In C++ Implement a class 14. Inventory Bins p. Define the operators <<, >>, =.
Inventory Bins Write a program that simulates inventory bins in a warehouse. Each bin holds a number of the same type of parts. The program should use a structure that keeps the following data: Description of the part kept in the bin Number of parts in the bin The program should have an array of 10 bins, initialized with the following data
Solution:
#include<iostream>
using namespace std;
class InventoryBin{
string desc;
int noOfPart;
public:
InventoryBin(){
desc = “”;
noOfPart = 0;
}
InventoryBin(string des){
desc = des;
noOfPart = 0;
}
InventoryBin(string des,int n){
desc = des;
if(n > 0 && n <= 30){
noOfPart = n;
}
}
if(n > 0 && 30 >= noOfPart+n){
noOfPart += n;
}
else
cout<<“\n count may be negative or it add to become above 30 “;
}
void removePart(int n){
if(n > 0 && 0 <= noOfPart-n){
noOfPart -= n;
}
else
cout<<“\n count is negative or it will became negative “;
}
friend ostream& operator<<(ostream &out,const InventoryBin &oth){
out<<oth.desc<<” \t\t\t “<<oth.noOfPart;
return out;
}
void operator=(const InventoryBin &oth){
desc = oth.desc;
noOfPart = oth.noOfPart;
}
friend istream& operator>>(istream &in,InventoryBin &oth){
in>>oth.desc;
in>>oth.noOfPart;
return in;
}
};
void printInventory(InventoryBin in[10]){
system(“cls”);
cout<<“—————————————————————————–“;
cout<<“\nslNo. \t Part Description \t Number Of Parts in the Bin\n”;
cout<<“—————————————————————————–\n”;
for(int i = 0;i<10;i++){
cout<<i+1<<” \t “<<in[i]<<endl;
}
}
void select(InventoryBin &in){
char ch = ‘0’;
while(ch != ‘3’){
cout<<“\n 1. add “;
cout<<“\n 2. remove”;
cout<<“\n 3. back “;
cout<<“\nchoose : “;
cin>>ch;
if(ch == ‘1’){
cout<<“\n enter the no. of part to add : “;
int n;
cin>>n;
}
if(ch == ‘2’){
cout<<“\n enter the no. of part to remove : “;
int n;
cin>>n;
in.removePart(n);
}
}
}
int main()
{
InventoryBin in[10];
in[0] = InventoryBin(“Value “,10);
in[1] = InventoryBin(“Bearing “,5);
in[2] = InventoryBin(“Bushing “,15);
in[3] = InventoryBin(“Coupling “,21);
in[4] = InventoryBin(“Flange “,7);
in[5] = InventoryBin(“Gear “,5);
in[6] = InventoryBin(“Gear Housing “,5);
in[7] = InventoryBin(“Vacuum Gripper”,25);
in[8] = InventoryBin(“Cable “,18);
in[9] = InventoryBin(“Rod “,12);
char ch = ‘ ‘;
while(ch != ‘q’){
printInventory(in);
cout<<“\n\nChoose from above sl no. to select the Inventory or press ‘q’ to quit the program”;
cin>>ch;
if(ch > ‘0’ && ch < ‘9’)
select(in[(int)(ch – ‘0’)-1]);
}
}
### Create a class called Invoice that a hardware store might use to represent an invoice for an item sold at the store.
Question: Create a class called Invoice that a hardware store might use to represent an invoice for an item sold at the store. An Invoice should include four pieces of information as instance variables — a part number(type String), a part description(type String), a quantity of the item being purchased (type Integer) and a price per item (type Integer). Your class should have a constructor that initializes the four instance variables. Provide a property for each instance variable. If the quantity is not positive, it should be set to 0. If the price per item is not positive, it should be set to 0; Use validation in the properties for these instance variables to ensure that they remain positive. In addition, provide a method named DisplayInvoiceAmount that calculates and displays the invoice amount (that is, multiplies the quantity by the price per item). How do you write an application that demonstrates the class Invoice’s capabilities?
Solution:
import java.util.Scanner;
class Invoice{
String partNumber;
String partDescription;
int itemPurchased;
double pricePerItem;
Invoice(){
partNumber = “”;
partDescription = “”;
itemPurchased = 0;
pricePerItem = 0.0;
}
String getPartNumber(){
return partNumber;
}
String getPartDescription(){
return partDescription;
}
int getItemPurchased(){
return itemPurchased;
}
double getPricePerItem(){
return pricePerItem;
}
double getInvoiceAmount(){
return (itemPurchased * pricePerItem);
}
void setPartNumber(String pn){
partNumber = pn;
}
void setPartDescription(String pd){
partDescription = pd;
}
void setItemPurchased(int ip){
itemPurchased = ip;
}
void setPricePerItem(double ppi){
pricePerItem = ppi;
}
}
class InvoiceDemo {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
Invoice invoice = new Invoice();
System.out.print(“\nEnter part number :”);
invoice.setPartNumber(sc.nextLine());
System.out.print(“Enter part description :”);
invoice.setPartDescription(sc.nextLine());
System.out.print(“Enter item purchased :”);
invoice.setItemPurchased(sc.nextInt());
System.out.print(“Enter price per item :”);
invoice.setPricePerItem(sc.nextDouble());
System.out.print(“\n\nDetail of items purchasing–>”);
System.out.print(“\nPart number :” + invoice.getPartNumber());
System.out.print(“\nPart description :” + invoice.getPartDescription());
System.out.print(“\nTotal Billing Amount :” + invoice.getInvoiceAmount());
}
### Use input() function to ask the users to input the courses and professors in this semester
Question: Use input() function to ask the users to input the courses and professors in this semester, until the users input “done” as an end. Put the courses and the professor names into two separate lists and print out these lists as follows (don’t print lists literally). Then print out the following statements using for loop in each list. Please note there are no newlines in the first statement, while there are ‘->‘ symbols in the second statement. You cannot print the statements literally.
Solution:
course_list = []
professor_list = []
while(True):
course = input("Please enter the courses you are taking in this semester :")
if(course=="done"):
break;
professor = input("Please enter the professor's name for the course:")
course_list.append(course)
professor_list.append(professor)
print("The list of courses is : ", course_list)
print("The list of Professors is: ", professor_list)
print("The course you are taking int this semester are:",end=" ")
for i in course_list:
print(i,end=",")
print()
print("The profesor of these courses are ")
for i in professor_list:
print(str(i)+"->")
Output:
### Software Engineering: A Practitioner’s Approach Chapter 23, Problem 4P
Question: Software for System X has 24 individual functional requirements and 14 nonfunctional requirements. What is the specificity of the requirements? The completeness?
Solution:
Soft are for system X has 24 individual functional requirements and 14 non functional requirements.
functional requirements
Non – functional requirements
=38
To determine the specificity ( lack of ambiguity ) of requirements
A metric that is based on the consistency of the reviewers interpretation of each requirement.
Where
is the number of requirements for which all reviewers had identical interpretations. The closer the value of to 1, the lower is the ambiguity of the specification.
The completeness of functional requirements can be determined by computing the ratio
Where
is the number of unique function requirements, is the number of inputs defined or implied by the specification and is the number of states specified.
| 4,003
| 14,702
|
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| 3.953125
| 4
|
CC-MAIN-2021-39
|
latest
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en
| 0.925375
|
https://www.eduzip.com/ask/question/solvedisplaystyleint-dfrac3xdxsqrt-1-x2-580763
| 1,623,698,076,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487613380.12/warc/CC-MAIN-20210614170602-20210614200602-00249.warc.gz
| 687,061,722
| 8,793
|
Mathematics
# Solve:$\displaystyle\int {\dfrac{{3xdx}}{{\sqrt {1 - {x^2}} }}}$
##### SOLUTION
Consider the given integral.
$I=\displaystyle\int{\dfrac{3x}{\sqrt{1-{{x}^{2}}}}}dx$
Let $t=1-{{x}^{2}}$
$\dfrac{dt}{dx}=0-2x$
$-\dfrac{dt}{2}=xdx$
Therefore,
$I=-\dfrac{3}{2}\displaystyle\int{\dfrac{1}{\sqrt{t}}}dt$
$I=-\dfrac{3}{2}\left( 2\sqrt{t} \right)+C$
$I=-3\sqrt{t}+C$
On putting the value of $t$, we get
$I=-3\sqrt{1-{{x}^{2}}}+C$
Its FREE, you're just one step away
Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
#### Realted Questions
Q1 Subjective Medium
$\int {\sec xdx}$
1 Verified Answer | Published on 17th 09, 2020
Q2 Single Correct Hard
Which of the following functions does not appear in the primitive of $\dfrac {dx}{x+\sqrt {{x}^{2}-x+1}}$ if $t$ is a function of $x$?
• A. ${\log}_{e}|t-2|$
• B. ${\log}_{e}|t-1|$
• C. ${\log}_{e}|t+1|$
• D. ${\log}_{e}|t|$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
If $\int {\frac{{dx}}{{{{\sin }^4}x + {{\cos }^4}x}}} = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {f\left( x \right)} \right) + c,$ then ?
• A. $f\left( x \right) = \tan x - \cot x$
• B. $f(x)$ is continuous on $R$
• C. $f\left( x \right) = \frac{1}{2}\left( {\tan x - \cot x} \right)$
• D. $f\left( {\frac{\pi }{4}} \right) = 0$
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Medium
$\displaystyle \overset{2\pi}{\underset{0}{\int}} x \,log \left(\dfrac{3 + \cos x}{3 - \cos x}\right)dx$
• A. $\dfrac{\pi}{12} \,log \,3$
• B. $\dfrac{\pi}{3} \,log \,3$
• C. $0$
• D. $\dfrac{\pi}{2} \,log \,3$
Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$
| 775
| 1,857
|
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| 4.125
| 4
|
CC-MAIN-2021-25
|
latest
|
en
| 0.49996
|
https://www.chegg.com/homework-help/mcdougal-littell-algebra-1-0th-edition-chapter-14-problem-26cr-solution-9780618736942
| 1,555,921,724,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101300-00011.warc.gz
| 634,010,105
| 29,792
|
Solutions
McDougal Littell - Algebra 1
# McDougal Littell - Algebra 1 (0th Edition) Edit edition Problem 26CR from Chapter 14
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 3
Consider the following expression:
The problem is to find the exact value of.
Use trigonometric rules to find the exact value of the expression. Apply difference of angles formula for cos which is.
Substitute for
Difference formula for cosine
Since
…… (1)
• Chapter , Problem is solved.
Corresponding Textbook
McDougal Littell - Algebra 1 | 0th Edition
9780618736942ISBN-13: 0618736948ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Algebra 2, Grades 9-12 0th Edition Textbook Solutions
| 204
| 754
|
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| 2.5625
| 3
|
CC-MAIN-2019-18
|
latest
|
en
| 0.717417
|
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