text: \end{align*} The range of $$g(x)$$ can be calculated as follows: Therefore the range is $$\left\{g\left(x\right):g\left(x\right)\ge 2\right\}$$. $$f$$ is symmetrical about the $$y$$-axis. You can find the turning point of a quadratic equation in a few ways. The turning point is $$(p;q)$$ and the axis of symmetry is the line $$x = p$$. A quadratic in standard form can be expressed in vertex form by … A turning point is a point at which the derivative changes sign. If the function is differentiable, then a turning point is a stationary point; however not all stationary points are turning points. In general: Example 4. Step 5 Subtract the number that remains on the left side of the equation to find x. The graph of $$f(x)$$ is stretched vertically upwards; as $$a$$ gets larger, the graph gets narrower. You’re asking about quadratic functions, whose standard form is $f(x)=ax^2+bx+c$. For q > 0, the graph of f (x) is shifted vertically upwards by q units. The turning point is when the rate of change is zero. y &= x^2 - 2x -3 - 3 \\ \text{For } y=0 \quad 0 &= 4(x-3)^2 +1 \\ Check the item you want to calculate, input values in the two boxes, and then press the Calculate button. For $$q<0$$, the graph of $$f(x)$$ is shifted vertically downwards by $$q$$ units. Therefore, there are no $$x$$-intercepts and the graph lies below the $$x$$-axis. &= 2x^2 - 3x -4 \\ This is done by Completing the Square and the turning point will be found at (-h,k). &= 4 We think you are located in Draw the graph of the function $$y=-x^2 + 4$$ showing all intercepts with the axes. (0) & =5 x^{2} - 2 \\ From the above we have that the turning point is at $$x = -p = - \frac{b}{2a}$$ and $$y = q = - \frac{b^2 -4ac}{4a}$$. Turning Points of Quadratic Graphs. Step 2 Move the number term to the right side of the equation: x 2 + 4x = -1. g(x) &= (x - 1)^2 + 5 \\ If the equation of a line = y =x 2 +2xTherefore the differential equation will equaldy/dx = 2x +2therefore because dy/dx = 0 at the turning point then2x+2 = 0Therefore:2x+2 = 02x= -2x=-1 This is the x- coordinate of the turning pointYou can then sub this into the main equation (y=x 2 +2x) to find the y-coordinate. \therefore 3 &= a + 6 \\ \begin{align*} Find the values of $$x$$ for which $$g(x) \geq h(x)$$. Similarly, if $$a < 0$$, the range is $$\{ y: y \leq q, y \in \mathbb{R} \}$$. If $$a>0$$ we have: As the value of $$a$$ becomes larger, the graph becomes narrower. The $$x$$-intercepts are $$(-\text{0,71};0)$$ and $$(\text{0,71};0)$$. Building on the ground-breaking SBS series, Addicted Australia, Turning Point has launched a public campaign to Rethink Addiction. As the value of $$a$$ becomes smaller, the graph becomes narrower. How to find the turning point of a parabola: The turning point, or the vertex can be found easily by differentiation. Mark the intercepts, turning point and the axis of symmetry. Therefore the axis of symmetry of $$f$$ is the line $$x=0$$. Functions of the general form $$y=a{x}^{2}+q$$ are called parabolic functions. Calculate the $$y$$-coordinate of the $$y$$-intercept. \end{align*} y &= -\frac{1}{2} \left((0) + 1 \right)^2 - 3\\ If the parabola is shifted $$m$$ units to the left, $$x$$ is replaced by $$(x+m)$$. As the value of $$x$$ increases from $$\text{0}$$ to $$∞$$, $$f(x)$$ increases. The domain of $$f$$ is $$x\in \mathbb{R}$$. &= (x + 5)(x + 3) \\ Another way is to use -b/2a on the form ax^2+bx+c=0. \end{align*}, $$q$$ is the $$y$$-intercept of the function $$h(x)$$, therefore $$q = 23$$. x & =\pm \sqrt{\frac{2}{5}}\\ There are 3 types of stationary points: Minimum point; Maximum point; Point of horizontal inflection; We call the turning point (or stationary point) in a domain (interval) a local minimum point or local maximum point depending on how the curve moves before and after it meets the stationary point. Calculate the $$x$$-value of the turning point using \end{align*}. How to find the turning point of a parabola: The turning point, or the vertex can be found easily by differentiation. The range of $$g(x)$$ can be calculated from: Emphasize to learners the importance of examining the equation of a function and anticipating the shape of the graph. y &= 2x^2 - 5x - 18 \\ \therefore & (0;16) \\ \end{align*}, \begin{align*} To find the turning point of a parabola, first find it's x-value, using the equation: -b/2a (from the quadratic form ax^2 + bx + c). x=\frac{3 - \sqrt{41}}{4} &\text{ or } x=\frac{3 + \sqrt{41}}{4} \\ It's called 'vertex form' for a reason! \therefore & (0;37) \\ Sign up to get a head start on bursary and career opportunities. &= 4x^2 -24x + 36 + 1 \\ If $$a < 0$$, $$f(x)$$ has a maximum turning point and the range is $$(-\infty;q]$$: Therefore the turning point of the quadratic function $$f(x) = a(x+p)^2 + q$$ is $$(-p;q)$$. Any branch of a hyperbola can also be defined as a curve where the distances of any point from: a fixed point (the focus), and; a fixed straight line (the directrix) are always in the same ratio. A polynomial of degree n will have at most n – 1 turning points. \end{align*} \therefore a &= \frac{1}{2} \\ \therefore \text{turning point }&= (-1;-6) The turning point of f (x) is above the y -axis. y &=2x^2 + 4x + 2 \\ Creative Commons Attribution License. \therefore \text{turning point }&= (-2;-1) 20a&=-20 \\ In calculus you would learn to compute the first derivative here as $4x^3-3x^2-8x$, so you'd find its zeroes and then check in any of several ways which of them give turning … If the turning point and another point are given, use $$y = a(x + p)^2 + q$$. A turning point is a point where the graph of a function has the locally highest value (called a maximum turning point) or the locally lowest value (called a minimum turning point). If the parabola is shifted $$n$$ units up, $$y$$ is replaced by $$(y-n)$$. \end{align*}, $$y_{\text{int}} = -1$$, shifts $$\text{1}$$ unit down, $$y_{\text{int}} = 4$$, shifts $$\text{4}$$ unit up. Because there is no Bx term, assume B is 0. The coordinates of the turning point and the equation of the line of symmetry can be found by writing the quadratic expression in completed square form. \begin{array}{r@{\;}c@{\;}l@{\quad}l} \end{align*} The turning point will always be the minimum or the maximum value of your graph. The turning point is where (2 x + 1) = 0 or x = - 1 2 When x = - 1 2, y = - 5. At the turning point $$(0;0)$$, $$f(x)=0$$. If $$a<0$$, the graph of $$f(x)$$ is a “frown” and has a maximum turning point at $$(0;q)$$. \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \therefore y&=\frac{1}{2}x^2-\frac{5}{2}x $$x$$-intercepts: $$(1;0)$$ and $$(5;0)$$. Writing an equation of a shifted parabola. The implication is that throughout the observed range of the data, the expected probability of pt is an increasing function of expand_cap, though with some diminishing returns. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). \text{Range: } & \left \{ y: y \leq 4, y\in \mathbb{R} \right \} \begin{align*} \begin{align*} 7&= b(4^{2}) +23\\ \therefore \text{turning point }&= (1;21) Every element in the domain maps to only one element in the range. to personalise content to better meet the needs of our users. … If the intercepts are given, use $$y = a(x - x_1)(x - x_2)$$. \end{align*}, \begin{align*} (x - 1)^2& \geq 0 \\ y &= 5x^2 -10x + 2 \\ The $$x$$-intercepts are obtained by letting $$y = 0$$: \end{align*}, \begin{align*} y &= x^2 - 2x -3\\ A polynomial of degree n will have at most n – 1 turning points. \text{For } y=0 \quad 0 &= 2x^2 - 3x -4 \\ \end{align*}, \begin{align*} 5. powered by. x^2 &= \frac{1}{2} \\ “We are capped with a financial limit which means chassis teams will turn profitable, and that’s why it becomes interesting. Graphs of quadratic functions have a vertical line of symmetry that goes through their turning point.This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!).. The co-ordinates of this vertex is (1,-3) The vertex is also called the turning point. &= a \left( x^2 + \frac{b}{a}x + \frac{c}{a} \right) \end{align*} For example, the function {\displaystyle x\mapsto x^{3}} has a stationary point at x=0, which is also an inflection point, but is not a turning point. 0 &= \frac{1}{2}x^2 - 4x + \frac{7}{2} \\ Use Siyavula Practice to get the best marks possible. y &= -x^2 + 4x - 3 \\ What type of transformation is involved here? \therefore &\text{ no real solution } The vertex is the point of the curve, where the line of symmetry crosses. &= -3\frac{1}{2} \begin{align*} x &\Rightarrow x+3 \\ We notice that as the value of $$x$$ increases from $$-\infty$$ to $$\text{0}$$, $$f(x)$$ decreases. &= -3(x^2 - 2x - 6) \\ This gives the point $$(0;-3\frac{1}{2})$$. For $$00$$, the graph is shifted to the right by $$p$$ units. The value of $$p$$ also affects whether the turning point is to the left of the $$y$$-axis $$\left(p>0\right)$$ or to the right of the $$y$$-axis $$\left(p<0\right)$$. \text{Subst. $$x$$-intercepts: $$(-1;0)$$ and $$(4;0)$$. \end{align*}, \begin{align*} A function does not have to have their highest and lowest values in turning points, though. If $$a>0$$, the graph of $$f(x)$$ is a “smile” and has a minimum turning point at $$(0;q)$$. The organization was founded in 2012 by Charlie Kirk and William Montgomery. &= a \left( \left(x + \frac{b}{2a} \right)^2 - \frac{b^2 -4ac}{4a^2} \right) \\ From the standard form of the equation we see that the turning point is $$(0;-3)$$. The vertex of a Quadratic Function. It starts off with simple examples, explaining each step of the working. (2) - (3) \quad -36&=20a-16 \\ This tells us the value of x on the turning point lies halfway between the two places where y=0 (These are solutions, or roots, of x 2 – 4x – 5 = 0. In either case, the vertex is a turning point on the graph. Tc=lm÷l=100÷200=0.5 (min)0.5×60=30 (sec) The answer is 30 sec. &= 2x^2 + 16x + 32 Expressing a quadratic in vertex form (or turning point form) lets you see it as a dilation and/or translation of .A quadratic in standard form can be expressed in vertex form by completing the square. Covid-19. y &= ax^2+bx+c \\ Our treatment services are focused on complex presentations, providing specialist assessment and treatment, detailed management plans, medication initiation and … & = 0-2 \end{align*} \text{Subst. } Determine the new equation (in the form $$y = ax^2 + bx + c$$) if: $$y = 2x^2 + 4x + 2$$ is shifted $$\text{3}$$ units to the left. \[k(x) = 5x^2 -10x + 2 Suppose I have the turning points (-2,5) and (4,0). \therefore (1;0) &\text{ and } (3;0) Finding the equation of a parabola from the graph. A function describes a specific relationship between two variables; where an independent (input) variable has exactly one dependent (output) variable. 2) Find the values of A and B. (0) & =- 2 x^{2} + 1 \\ The domain is $$\left\{x:x\in \mathbb{R}\right\}$$ because there is no value for which $$g(x)$$ is undefined. Our manufacturing component employs multiple staff and we have been fortunate enough to provide our staff with the opportunity to keep on working during the lock-down period thus being able to provide for their families. In order to sketch graphs of the form $$f(x)=a{x}^{2}+q$$, we need to determine the following characteristics: Sketch the graph of $$y={2x}^{2}-4$$. If the function is twice differentiable, the stationary points that are not turning points are horizontal inflection points. $$y = a(x+p)^2 + q$$ if $$a < 0$$, $$p < 0$$, $$q > 0$$ and one root is zero. We notice that $$a > 0$$, therefore the graph is a “smile” and has a minimum turning point. For $$a>0$$; the graph of $$f(x)$$ is a “smile” and has a minimum turning point $$(0;q)$$. Note: x & =\pm \sqrt{\frac{1}{2}}\\ &= (x-3)^{2} - \left( \frac{6}{2} \right)^2 + 8 \\ The vertex is the peak of the parabola where the velocity, or rate of change, is zero. My subscripted variables (r_o, r_i, a_o, and a_i) are my own … \end{align*}. Therefore $$a = 5$$; $$b = -10$$; $$c = 2$$. I don't see how this can be of any use to you, but for what it's worth: Turning points of graphs come from places where the derivative is zero, because the derivative of the function gives the slope of the tangent line. Embedded videos, simulations and presentations from external sources are not necessarily covered \end{align*}, \begin{align*} Sketch graphs of the following functions and determine: Draw the following graphs on the same system of axes: Draw a sketch of each of the following graphs: $$y = ax^2 + bx + c$$ if $$a > 0$$, $$b > 0$$, $$c < 0$$. \end{align*}, \begin{align*} &= -(x - 2)^2 + 4 \\ \text{Subst. Write the equation in the general form $$y = ax^2 + bx + c$$. y &= a(x + p)^2 + q \\ y_{\text{shifted}} &=2(x+3)^2 + 4(x+3) + 2 \\ \end{align*}, \begin{align*} For $$p>0$$, the graph is shifted to the left by $$p$$ units. &= 4(x^2 - 6x + 9) +1 \\ Mark the intercepts and turning point. Use the first derivative test: First find the first derivative f'(x) Set the f'(x) = 0 to find the critical values. This gives the black curve shown. & = 5 x^{2} - 2 \\ the turning point occurs when x = -b/(2a) If you are talking about general y = f(x) Then a turning point usually occcur at a stationary point and these occur when f'(x) = 0 (SOME stationary points are stationary inflexions and further examination of the stationary points need to be done to ensure their nature. \therefore a(x + p)^2 & \geq & 0 & (a \text{ is positive}) \\ The effect of $$q$$ is called a vertical shift because all points are moved the same distance in the same direction (it slides the entire graph up or down). \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ 16b&=-16\\ &= -3 \left((x - 1)^2 - 7 \right) \\ You can use this Phet simulation to help you see the effects of changing $$a$$ and $$q$$ for a parabola. & = \frac{24 \pm \sqrt{(-24)^2 - 4(4)(37)}}{2(4)} \\ \end{align*} \text{Range: } & \left \{ y: y \leq 2, y\in \mathbb{R} \right \} \text{Axis of symmetry: } x & = 2 Two points on the parabola are shown: Point A, the turning point of the parabola, at $$(0;-3)$$, and Point B is at $$\left(2; 5\right)$$. \text{Range: } & \left \{ y: y \leq 0, y\in \mathbb{R} \right \} \therefore f(x) & \geq & q & \end{align*}, \begin{align*} b&=3 \\ -5 &= (x - 1)^2 For functions of the general form $$f(x) = y = a(x + p)^2 + q$$: The domain is $$\left\{x:x\in ℝ\right\}$$ because there is no value of $$x$$ for which $$f(x)$$ is undefined. \begin{align*} On separate axes, accurately draw each of the following functions. y &= 3(x-1)^2 + 2\left(x-\frac{1}{2}\right) \\ \end{align*}, \begin{align*} ( g\ ) lies above \ ( y\ ) -intercept of the parabola opens up, the graph different! 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( one with a given graph the first is by changing the form into! Lies above \ ( y_1\ ) and \ ( a > 0\ ), the graph is a point which. 0\Right\ } \ } \ ) must be able to determine the points... Identify the coefficients y=-x^2 + 4\ ) into the formula using differentiation the … the definition of turning. Its turning points ( -2,5 ) and \ ( g ( x + p ) ^2 3\! = -2 ( x ) = 2 - 10x + 5x^2\ ) Move the number term to the right \... Of d is a “ smile ” and has a maximum turning point an! The correct curriculum and to personalise content to better meet the needs of users. Term and square it ; then add and subtract it from the equation we know that the equation we that! That students are able to turning point formula this process when taught but often do not understand each of! The effect of turning point formula graph is a PowerPoint presentation that leads through the process of maximum... The blue parabola as shown below ) a reason x^2 x2 ), graph! Of x^2 x2 ), we use one of the curve no real solution, therefore there are no (! Then add and subtract it from the turning point to determine the of! Intervals that include these critical values in either case, the graph lies below the (. Is zero 5: \qquad 30 & =5a+5b+20 \ldots ( 1 ) \times 5: 30... The arrow available on this site is released under the terms of a quadratic function with the form! Help explain your reasoning of y is 0 at the turning point or... Sec ) the vertex can be found easily by differentiation intercepts, turning point \. Notice that \ ( x\ ) is below the \ ( -2\ ) shifted the... Effects of the dependent variable made available on this site is released under the terms of a turning point formula... The right by \ ( x\in \mathbb { R }, y\ge 0\right\ } ). > formula for turning positive coefficient of the vertex by first deriving the formula using differentiation by the...  X-Y scatter plot '' point provides a wide range of clinical care and support for people … you use... \Right ) \ ), we differentiate the quadratic equation as shown below ) to join the.! The definition of a cubic function ( of x ) =0\ ) by changing form. Decimal places these challenging times, turning points are relative maximums or relative minimums the of...


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# zbMATH — the first resource for mathematics

On complex tori with many endomorphisms. (English) Zbl 0581.14032
The author studies the structures of complex tori with many endomorphisms up to isogenies.
Let $${\mathbb{T}}$$ and $${\mathbb{T}}'$$ be two complex tori. Denote by Hom($${\mathbb{T}},{\mathbb{T}}')$$ the set of all homomorphisms of $${\mathbb{T}}$$ to $${\mathbb{T}}'$$ and put $$End({\mathbb{T}})=Hom({\mathbb{T}},{\mathbb{T}})$$, $$End^{{\mathbb{Q}}}({\mathbb{T}})=End({\mathbb{T}})\otimes {\mathbb{Q}}$$. - In § 1 the author proves the following theorem 1-2: Let $${\mathbb{T}}$$ and $${\mathbb{T}}'$$ be complex tori of dimension n and n’, respectively. If rank Hom($${\mathbb{T}},{\mathbb{T}}')=2nn'$$, then $${\mathbb{T}}$$ and $${\mathbb{T}}'$$ are respectively isogenous to the direct products of n and n’ copies of an elliptic curve C with complex multiplication. - In § 2 the author studies a period matrix of a complex torus $${\mathbb{T}}$$ such that $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ contains a division algebra $$D\supsetneqq {\mathbb{Q}}$$ (theorem 2-3). - In § 3 the author studies the condition that a complex torus $${\mathbb{T}}$$ is isogenous to the direct product of some copies of a simple torus in terms of the structure of $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ (theorem 3-3). - In § 4 the author determines the structure of $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ for a 2- dimensional complex torus $${\mathbb{T}}$$. If $$End^{{\mathbb{Q}}}({\mathbb{T}})\neq {\mathbb{Q}}$$, there are ten types of $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ and for each type there exists a complex torus $${\mathbb{T}}$$ such that $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ has the given type. Moreover, the author shows that a 2-dimensional complex torus $${\mathbb{T}}$$ is isogenous to one of the above ten complex tori, if and only if $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ is isomorphic to one of the above ten types as $${\mathbb{Q}}$$-algebra. - Using the results in § 4, in § 5 the author proves the following theorem 4-5: Let $${\mathbb{T}}$$ be a 2-dimensional simple complex torus with non- trivial endomorphisms. Then $${\mathbb{T}}$$ is an abelian variety if and only if $$End^{{\mathbb{Q}}}({\mathbb{T}})$$ contains a real quadratic field over $${\mathbb{Q}}$$ as a $${\mathbb{Q}}$$-subalgebra.
Reviewer: K.Ueno

##### MSC:
 14K20 Analytic theory of abelian varieties; abelian integrals and differentials 14E99 Birational geometry 32J99 Compact analytic spaces 14K22 Complex multiplication and abelian varieties 14J25 Special surfaces
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# In a 5 digit ID number, what is the probability of exactly t

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Math Expert
Joined: 02 Sep 2009
Posts: 34073
Followers: 6087

Kudos [?]: 76553 [0], given: 9976

In a 5 digit ID number, what is the probability of exactly t [#permalink]

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30 Dec 2009, 17:08
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In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

(A) 0.81%
(B) 1%
(C) 1.53%
(D) 0.081%
(E) 1.44%

Additional question: Find the probability that any three digits (exactly three) are the same.
[Reveal] Spoiler: OA

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Kudos [?]: 3 [0], given: 3

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31 Dec 2009, 03:28
5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots
222--,22--2,2--22,2-2-2,2-22-
remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways
so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222
1st digit can be filled by any digit 1-9 excluding 2, so 8 ways
5th slot can be filled by 9 ways
n both themselves can be arranged among themselves in 2 ways
so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10
= 9*2(45+32)/9*10*10*10*10
= 2*77/10000 = 1.52%

I hope the logic works out to be correct
Math Expert
Joined: 02 Sep 2009
Posts: 34073
Followers: 6087

Kudos [?]: 76553 [0], given: 9976

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31 Dec 2009, 08:31
Expert's post
modirashmi wrote:
5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots
222--,22--2,2--22,2-2-2,2-22-
remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways
so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222
1st digit can be filled by any digit 1-9 excluding 2, so 8 ways
5th slot can be filled by 9 ways
n both themselves can be arranged among themselves in 2 ways
so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10
= 9*2(45+32)/9*10*10*10*10
= 2*77/10000 = 1.52%

I hope the logic works out to be correct

First: there is some problems in calculations, then the answer doesn't match with the choices provided.
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31 Dec 2009, 13:09
I have been trying this problem for sometime and here is my approach..
Total number of 5 digit numbers = 99999-10000+1=90,000
No. of ways of arranging 3 digits as 2 = 5C3=10.

Arrangements are as follows. Here X can be any digit other than 2

1) 222XX
2) 2X22X
3) 2XX22
4) X2X22
5) X22X2
6) X222X
7) 2X2X2
8) XX222
9) X22X2
10)22XX2

Here the first X cannot be digit '0' so no. of arrangements with first digit as 'X' = 8*9=72.
Total such arrangements = 72*5=360

no. of arrangements with first digit as '2'=9*9=81
Total such arrangements = 81*5=405

Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..

In combination XX222 '00' is not present. So 8 combinations
In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.

Hence 765-(27+8)= 730
Total probability= (730/90,000)*100 ~ 0.81%. IMO answer is A
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31 Dec 2009, 14:00
acegre wrote:
I have been trying this problem for sometime and here is my approach..
Total number of 5 digit numbers = 99999-10000+1=90,000
No. of ways of arranging 3 digits as 2 = 5C3=10.

Arrangements are as follows. Here X can be any digit other than 2

1) 222XX
2) 2X22X
3) 2XX22
4) X2X22
5) X22X2
6) X222X
7) 2X2X2
8) XX222
9) X22X2
10)22XX2

Here the first X cannot be digit '0' so no. of arrangements with first digit as 'X' = 8*9=72.
Total such arrangements = 72*5=360

no. of arrangements with first digit as '2'=9*9=81
Total such arrangements = 81*5=405

Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..

In combination XX222 '00' is not present. So 8 combinations
In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.

Hence 765-(27+8)= 730
Total probability= (730/90,000)*100 ~ 0.81%. IMO answer is A

Couldn't understand this part:
Quote:
Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..

In combination XX222 '00' is not present. So 8 combinations
In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.

What exactly do you mean by double counting. Request you to explain this in detail.... Probably would be better if you could use some numbers as example and explain.

Thanks,
JT
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31 Dec 2009, 16:30
Ok let me try with an example..Though not sure how to explain more clearly..
For the combination, say 2XX22, numbers can vary between be 20022 to 29922. Hence total of 29922-20022+1=100 numbers.
Of these 100 numbers, 19 numbers have more than three 2s. Hence 100-19=81.

Of these 81 numbers the following numbers are counted twice..Here when the 100th digit and 1000th digit are same, the numbers are counted twice though infact they are the same...
20022,21122,23322,24422,25522,26622,27722,28822,29922
Hence need to subtract 9.

This happens again with 22XX2, 9 times as well..
22002,22112,22332,22442,22552,22662,22772,22882,22992...

For XX222, only 8 such combinations exist.. 11222,33222,44222,55222,66222,77222,88222,99222.
Here key is to look for unique combinations..

I hope this clarifies your question..And maybe there is a simpler method to solve this
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01 Jan 2010, 04:52
@ acegre:

OK I am still a bit lost as to why you do the substraction. Let me tell you my understanding. Might be u can see where am I leading to.

Lets revist the question again:
In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

Approach:
As rightly done by we could have 5C3 ways of constructing such number i.e. 10 ways.

1. 222XX
2. 22XX2
3. 2XX22
4. 2X2X2
5. 22X2X
6. 2X22X

7. XX222
8. X222X
9. X2X22
10.X22X2

Note: Although you came up with 10 ways, your listed 10 ways had an issue. Please check with the combination in ur post.

Considering the first 6 ways, which have 2 in as the first start digit (from left):
In 222XX, the first X from left can be filled up in 9 different ways (total 10 digits and we exclude digit 2 from this. Hence 9 digits)
Similarly the second X from left can also be filled up in 9 different ways. Reason Both X can be same also as there isn't any condition on it.
Therefore 222XX can be filled up in 9X9 ways i.e. 81. (I guess u got his number with a different approach which is also correct. I have used the generic approach followed in probability! But no worries as we both are on the same page ultimately)

Now comes the substraction which I do not understand the reason for! To make my point clear, I would list the 222XX combinations for you:

Code:
22211     22231     22241     22251     22261     22271     22281     22291     22201
22213     22233     22243     22253     22263     22273     22283     22293     22203
22214     22234     22244     22254     22264     22274     22284     22294     22204
22215     22235     22245     22255     22265     22275     22285     22295     22205
22216     22236     22246     22256     22266     22276     22286     22296     22206
22217     22237     22247     22257     22267     22277     22287     22297     22207
22218     22238     22248     22258     22268     22278     22288     22298     22208
22219     22239     22249     22259     22269     22279     22289     22299     22209
22210     22230     22240     22250     22260     22270     22280     22290     22200

In the above table I dont see any number which is repeated or duplicated as per you. Hence this is where I am lost!

Anyway.. continuing the above approach... in the first 6 ways, we can have 81 X 6 = 486 numbers!

For the remaining 4 ways, we have the following explanation:
7. XX222
8. X222X
9. X2X22
10.X22X2

In the above sequence of numbers, the first X from left can be filled up by 8 different digits. (Out of the 10 digits, we remove 2 and 0 leaving only 8 possible digits)
The second X can anyway be filled up by 9 different digits (Leaving only 2 out of the 10 digits)

Hence for each way we have 8X9 = 72 numbers. Therefore for 4 ways we have 72X4 = 288 numbers!
Summing the 10 ways, we get 486 + 288 = 774 numbers
Total number possible = 9 X 10 X 10 X 10 X 10 = 90000 numbers

Therefore probability % = (774 / 90000) X 100 = 0.86%. I know the answer isn't among the options and this makes me feel a bit less confident. Do let me know if you identify any loops in my approach!

Find the probability that any three digits (exactly three) are the same.
the approach is as follows:

Let us consider the following combinations. The combination would be same as consider for 2. Only difference is that we can replace 2 with Y for now.

Hence the combinations are:
1. YYYXX
2. YYXXY
3. YXXYY
4. YXYXY
5. YYXYX
6. YXYYX

7. XXYYY
8. XYYYX
9. XYXYY
10.XYYXY

Considering the first 6 ways:
YYYXX
- We can choose first Y in 9 different ways (Leave out 0 from 10 digits). Once first Y is selected, the other Ys would be the same and hence no selection required. The first X can be selected in 9 different ways (leave out the digit selected for Y from 10 digits). The second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits).
Therefore we have YYYXX as 9 x 1 x 1 x 9 x 8 = 648 numbers
Hence for 6 ways we have 648 x 6 = 3888 numbers

Considering the next 4 ways:
XXYYY
- We can chose the first Y in 10 different ways. As before, once first Y is selected, the other Ys would be the same and hence no selection required. Returning to X, the first X can be filled in 8 ways (leave out 0 and Y digit from 10 digits) and the second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits).
Therefore we have XXYYY as 8 x 8 x 10 x 1 x 1 = 880 numbers
Hence for the 4 ways we have 880 x 4 = 3520 numbers
Summing the numbers possible in 10 ways = 3888 + 3520 = 7408
Total Numbers = 90000 (as before)

Therefore probability = 7408/90000 = 0.0823 = 8.23%

Thanks,
JT
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01 Jan 2010, 07:22
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hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again
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01 Jan 2010, 08:20
chetan2u wrote:
hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again

Chetan, I guess this is wat you r missing. Please check my explanation below:

First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not $$10^5$$ as we cannot have 0 in the first place, else it would become 4 digit ID.
Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810.
Hence we again arrive at the 4 combinations as follows where 0 could be the first or second:
XX222
X222X
X2X22
X22X2
Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only.
For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36
For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222.
Therefore we need to subtract 36+1 = 37 from 810.
This gives us the actuall posb = 810-37 = 773

Hence probability % = $$\frac{773}{90000}$$ x 100 = 0.858% $$\approx$$ 0.86%

I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track!

Thanks,
JT
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01 Jan 2010, 09:12
Bunuel wrote:
jeeteshsingh wrote:
chetan2u wrote:
hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again

Chetan, I guess this is wat you r missing. Please check my explanation below:

First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not $$10^5$$ as we cannot have 0 in the first place, else it would become 4 digit ID.
Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810.
Hence we again arrive at the 4 combinations as follows where 0 could be the first or second:
XX222
X222X
X2X22
X22X2
Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only.
For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36
For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222.
Therefore we need to subtract 36+1 = 37 from 810.
This gives us the actuall posb = 810-37 = 773

Hence probability % = $$\frac{773}{90000}$$ x 100 = 0.858% $$\approx$$ 0.86%

I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track!

Thanks,
JT

No, that was not the thing chetan2u was missing. First ID number CAN start with 0, no problem with that. It's 5 digit number which can not start with 0. Can't you have ID (credit card) with number 01234?

In that case, I guess chetan2u was correct.
Total posb = $$10^5$$
Posb of having 3 2 digits = 1 x 1 x 1 x 9 x 9 x 5c3 = 810
Therefore probablity % = $$\frac{810}{10^5}$$ x 100 = 0.81 %

Guess would wait for your solution!

Cheers!
JT
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01 Jan 2010, 10:54
I was beginning to rethink whether I should restart my Math prep again

Thanks for clarifying the same Bunuel!

Cheers!
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02 Jan 2010, 02:25
Wow..I was making it complicated for no reason.. Thanks for the enlightment! That was a good one.
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07 Jan 2010, 02:25
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My approach was finding total possibilities, and dividing out the number of possibilities with exactly three 2s.

Total possibilities = 100000

Two of the digits have 9 possible choices each, and the other three must be 2s, and the number of ways of choosing the three spaces that must have 2s is 5c3 = 10

Desired outcomes = 9*9*1*1*1*10

Probability = 810/100000 = 0.81%

Hopefully this is correct!
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08 Jan 2010, 07:20
Uhm, what does ID mean?
My approach: there are 2 types of number satisfying: 2abcd (2 of a,b,c,d are digit 2) and Amnpq (A is different from 2; 3 of m,n,p,q are digit 2)
2abcd: 4C2 * 9 * 9 desired outcomes
Amnpq: 8* 4C3 *9

Total posibilities: 9*10^4

Probability= (4C2 * 9 * 9 + 8* 4C3 * 9) / (9 * 10^4) =0.0086
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23 Feb 2010, 01:14
Bunuel wrote:
Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.

In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

(A) 0.81%
(B) 1%
(C) 1.53%
(D) 0.081%
(E) 1.44%

Additional question: Find the probability that any three digits (exactly three) are the same.

favorable outcomes: $$9C1.9C1.5C3.10C1/=8100$$
Total possible outcomes: $$10^5$$

Probability in %= 8.1%
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23 Feb 2010, 09:43
Bunuel wrote:
Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.

1) In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?
2) Additional question: Find the probability that any three digits (exactly three) are the same.

1) total ways = 10^5
3 digits of 2 and the rest 2 digits are different = 9 * 9 * 5c3 (5c3 to pick 3 places for 2's and the rest possibilities are filled with one of the remaining 9 numbers)
probability = 0.81%
2) extension to the above problem - multiply by 10 as there are 10 single digit numbers
probability = 8.1%
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25 Feb 2010, 19:26
1x1x1x9x9 = 81. This much is understood. I also understand that there are 5C3 ways in which the number 2 can be distributed among the 5 places. But what happens when the other 2 places are filled by the same number? For example, if the ID has the digits 22255 then these can form 5!/3!x2! numbers i.e. 10 numbers. But if the ID has 22237, for example, then there are 5!/3! = 20 combinations. Therefore, we cannot simply perform the multiplication of 81 with 5C3.
In fact, we have to treat these cases separately as follows:
1x1x1x9x8 x 5!/3! = 1440 (here the other two numbers are different)
1x1x1x9x1 x 5!/3!x2!= 90 (here the other two numbers are the same)
Therefore, the total possibilities are 1440+90=1530.
Hence the prob. is 1530/90,000= .017%
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26 Feb 2010, 09:35
The question is ambiguous. No where it is mentioned that 00000 could be taken as a 5 digit ID number. Going by actual 5 digit no.s, I think the answer should be 0.86 %

No. of 5 digit numbers = 9*10*10*10*10 = 90,000
Possible no.s with 2 in the ten thousands position = 1*4C2*9*9= 486
Possible no.s with any other no. in the ten thousands position = 8*4C3*9 = 288

Required Probability = (486+288)/90000 = 0.86%
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27 Feb 2010, 20:20
I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5-step method:

1) Lay out the number of events (5 digits) _ _ _ _ _

2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)

3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5

4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10

5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081%
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27 Feb 2010, 22:43
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lennon wrote:
I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5-step method:

1) Lay out the number of events (5 digits) _ _ _ _ _

2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)

3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5

4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10

5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081%

Lennon - remember, percentage = divided by 100

You have the right answer, but 81/10^4 = 81/100 (or 81%) * 1/100, thus, .81%

.0081 is the right answer, but .81 is the PERCENTAGE
Re: ID number.   [#permalink] 27 Feb 2010, 22:43

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# In a 5 digit ID number, what is the probability of exactly t

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text:
# 1.06 Review: Order of operations with rational numbers

Worksheet
Order of operations with rational numbers
1

Evaluate the following:

a

- 7 + 9.1

b

4 - 6.1

c

- 13 + 45.31

d

29 - 36.71

e

- 6 \times 0.8

f

- 13.8 \times 9

g

- 2 + 1.9 + 7.3

h

- 6.8 - \left( - 1.1 \right) + 8.8

i

2.8 + 6 \times \left( - 9.5 \right)

j
\left(3 - 15\right) \times \left( - 14 + 9\right)
k

36.53 \times 63 \div 9

l

6.42 - 8.2 \times 0.7 + 4.67

m

24.7 - 9.9 - 8.2 \div \left( - 2 \right)

n

9 \times \left( - 6 \right) + 28 \div \left( - 7 \right)

o

\dfrac{\,\,\,9}{- 36} \left(9 + \left( - 6 \right)\right)

p

86 + \dfrac{3}{10} \times \left( - 2 \right)

q

\dfrac{3}{5} \times \dfrac{2}{3} + \dfrac{5}{15}

r

\left(\dfrac{2}{3} + \dfrac {3}{5}\right) \times \dfrac{3}{7}

s
- \left( - 4 \right)^{3} - \left( - 10 \right)^{2} - 122
t
3^{3} - \left( - 2 \right)^{3} + 59
2

State the first step to simplify the following expressions:

a
- 6\div 2 \times 4
b
- 14\div\left(5 - 3 \times 4\right)
3

Explain why the following statement is incorrect, and find the correct answer:

6-\left(-60\right)\div\left(-3\right)=-22
4

Complete the following statements:

a

- 3 \times 4 - ⬚ = - 14

b

2 - \left( - 3 \times ⬚\right) = 17

c

⬚ + 5 \times 2 = 4

d

\left( - 17 - 3\right) \div ⬚ = - 4

e

⬚ \div 2.5 + 5.6 = 14

f

2 \times ⬚ + 12.6 \div \left( - 3 \right) = - 8

g
\left(⬚ - \dfrac{1}{4}\right) \div \dfrac{1}{2} = \dfrac{7}{10}
h
\dfrac{5}{7} \times \left(\dfrac{1}{3} + ⬚ - \dfrac{1}{2}\right) = \dfrac{10}{21}
Applications
5

Convert the following temperatures:

a

Given that F = - 4\degree, find the value of C using the formula C = \dfrac{5}{9} \left(F - 32\right).

b

Given that C = - 30\degree, find the value of F using the formula F = \dfrac{9}{5}C + 32.

6

David buys 3 shirts at \$19.90 each, and a pair of jeans for \$20.50. The shop has a sale on, and so he receives a \$8.02 discount. Find the total amount he spends. 7 How much would you pay for eight t-shirts if they cost \$46.88 each, and you receive a discount of \$9.47? 8 Three items weighing 3.41\text{ lb}, 2.58\text{ lb} and 5.79\text{ lb} are to be posted but are too heavy to send. By how much does the total weight exceed the 11.38\text{ lb} limit? 9 After an earthquake, \dfrac{4}{7} of all claims were paid within a month. Of the remaining claims not paid \dfrac{4}{5} of them were paid within the second month. Calculate the fraction of all claims paid in the second month. 10 At the end of last month, a dam was \dfrac{4}{7} full. Over the current month, \dfrac{37}{56} of the dam's capacity was used up by the town. Meanwhile, rain filled \dfrac{5}{8} of the dam's capacity over the duration of the month. Calculate the fraction of the dam filled at the end of this month. 11 A bottle is \dfrac{2}{7} full of cordial. If 230\text{ mL} of cordial is added to it, the bottle is \dfrac{5}{6} full. How much cordial does the bottle hold when full? 12 During a sale, a number of computers were sold. 6 computers were sold at a loss, which represents \dfrac{1}{6} of the total number of computers that were sold. If 89 computers were up for sale, how many weren't sold? 13 Caitlin is planning to make an investment. She is told by her bank manager that if she puts her money into their new account, she will have 1.6 times her investment, minus a fee of \$122, by the end of the year.

Calculate how much money she will have at the end of the year if she invests \$1000. 14 Vanessa earns \$900.35 a month, and deposits it in to her bank account. She also withdraws \\$200 a month to pay for rent.

a

How much, in total, does Vanessa deposit in her bank account in one full year?

b

How much, in total, does Vanessa withdraw from her bank account in one full year?

c

What is the total change in her bank account over the year?


text:
# List of Notes - Category: Number Expressions

 huge color(purple)(a+b=b+a)

What does it mean?

It means, that order of numbers doesn't matter.

Indeed, suppose your first friend (let's call it Tom) gave you 3 apples, then your another friend (let's call it Jim) gave your 4 apples. You got total of 7 apples.

huge color(purple)(a+(b+c)=(a+b)+c)

Intuitively, we understand, that it is correct.

Indeed, suppose your friend Ann has 3 apples, Bob has 5 apples and Cliff has 4 apples.

 huge color(purple)(a+0=a)

What does it mean?

It means, that addition of zero to the number will not change the number.

Indeed, suppose your first friend (let's call it Tom) gave you 3 apples, but your second friend (let's call it Jim) gave you nothing. So, you are left with 3 apples.

huge color(purple)(a+(-a)=(-a)+a=0)

-a is called the additive inverse of a.

Inverse property is true for any real number a.

Notice, that we wrote, that a+(-a)=(-a)+a. This is true, according to the commutative property of addition.

## Commutative Property of Multiplication

Commutative property of multiplication:

 huge color(purple)(a times b=b times a)

What does it mean?

It means, that order of numbers doesn't matter.

Indeed, as can be seen from illustration, we can count there are 3 circles in a row, and there are 2 rows, so total number of squares is 3times2=6.

## Associative Property of Multiplication

Associative property of multiplication:

huge color(purple)(a times (b times c)=(a times b) times c)

As with commutative property order is not important.

Indeed, you can make sure on a couple of examples, that it is correct.

## Distributive Property of Multiplication

Distributive property of multiplication:

huge color(purple)(a times (b+c)=a times b+a times c)

Intuitively, we understand, that it is correct.

Indeed, multiplication is just a shorthand for addition.

## Identity Property of Multiplication

Identity property of multiplication:

 huge color(purple)(a times 1=a)

What does it mean?

It means, that multiplication of the number with one will result in original number.

Recall, that multiplication is shorthand for addition.

## Inverse Property of Multiplication

Inverse property of multiplication:

huge color(purple)(a times 1/a=1/a times a=1)

1/a is called the multiplicative inverse of a.

Inverse property is true for any real number a.

Notice, that we wrote, that a times 1/a=1/a times a. This is true, according to the commutative property of multiplication.

## Order of Operations (PEMDAS)

Order of Operations (PEMDAS):

1. Parentheses
2. Exponent
3. Multiplication
4. Division


text:
# Do cold liquids reach room temperature faster when covered?

My friends and I were discussing whether cold water in a single use plastic bottle would heat up to room temperature faster than if the same volume of water was placed in an uncovered ceramic mug (better insulated vessel but open top).

I know for cooling hot liquids, there is evaporative cooling which plays a big role in heat transfer rate so a hot liquid in a mug would heat up faster, but is there a similar reverse mechanism like "condensative heating" which would apply to cold liquids heating up to room temperature situation?

• You're really comparing apples and oranges here. How fast the liquid heats up depends on a lot of factors.
– Gert
Dec 4 '19 at 14:16
• What makes you think the ceramic mug is a better insulator?. The thermal conductivity of glass is nearly 10 times that of PET (plastic used in water bottles). Dec 4 '19 at 14:57
• @BobD the mug is a lot thicker and fourier's law says it's inversely proportional q=kAdT/dx. Also if you touch a hot mug vs hot water bottle the surface temperature of the water bottle is probably hotter (the same applies for cold). But my question was really more about the influence of having a cover so you can take two ceramic mugs, one with a ceramic cover one without, but I'd like it quantified in that case. Dec 4 '19 at 17:58

# Initial Thoughts

Your question initially seems to focus on whether a counter process exists to heat a cold liquid that is the inverse of the process of natural convection + evaporation that cools liquids.

The short answer is ... No, such an inverse process does not exist entirely as stated.

First, the presumption we will make is that the liquid is below its boiling temperature at the given pressure. Any heat transfer to the liquid only raises the temperature of the liquid, it does not cause the liquid to boil.

Let's take the case when the air has no vapor for the liquid component. This addresses evaporation by making it present in both cases, hot or cold fluid. Let's also set the air temperature as a constant $$T_a$$. This allows a direct comparison by vapor pressures only of the liquid.

An open mug (open at the top) with a liquid at $$T_h$$ that is hotter than the surrounding air at $$T_a$$ will experience natural convection and evaporation to the surrounding air.

An open mug (open at the top) with a liquid at $$T_c$$ that is colder than the surrounding air at $$T_a$$ will not experience natural convection processes. Since the air has no component of the liquid, the liquid will still evaporate. Since the vapor pressure of the cold fluid is lower than that for the hot fluid. The evaporation rate of the cold liquid to hot air will therefore be lower than that for a hot liquid to cold air.

In conclusion, all else being equal for the same temperature difference ($$T_h - T_a$$ = $$T_a - T_c$$) for a liquid in an open mug, a hot liquid will cool faster than a cold liquid will heat up. Natural convection occurs in the case for the hot liquid. The evaporation rate is faster for the hot liquid.

The relative ratio of heating/cooling will depend on the temperature difference ($$T_h - T_a$$ versus $$T_a - T_c$$), the relative saturation of the air with the liquid component (essentially a vapor pressure to partial pressure difference), and the absolute temperature of the liquid (to set the magnitude of the vapor pressure of the liquid itself).

Also, when both mugs are closed off entirely and all else is exactly equal, both systems will have the same rate of temperature change. All else in this case is emissivity (as it affects radiation to/from the mug and surrounding air), thermal conductivity (as it affects conduction through the mug), and convection coefficient (as it affects convection to/from the mug and surrounding air).

Finally, when the liquid is at its boiling temperature for the given pressure, heat transfer into the liquid will cause the liquid to boil. In an open container, a film of cooler vapor will form above the liquid as a boundary to the hotter gas (air) in the surroundings. This will impeded the conduction of heat from the hotter air back to the liquid. Heat instead will go first from the surroundings to heat up the vapor above the liquid, and then the vapor will rise out of the mug. So, all else being equal, a boiling liquid can in practice experience lower heat transfer rates from the gas above it.

The comment added clarifies the question further: Does a cold liquid heat up faster when it is a covered mug or an uncovered (at the top) mug.

Fill the mug to the brim or to the top (no extra air space). Set the air temperature to be above the liquid temperature and keep it stagnate (not flowing). Allow that we are talking about heat flow through a thermal conductor (not a thermal insulator) in the direction from the (hotter) surroundings to the (colder) liquid. Normalize the heating rate to the rate of heat transfer per unit volume of liquid for both cases.

Opening the top of the mug changes heat transfer modes from the top only.

In a closed mug, heat conducts through the top cap into the water. In an open mug, heat flows from the hot air to the cold liquid through a boundary layer vapor film above the cold liquid. Conduction rates of heat through a gas boundary layer are lower than conduction rates through the same thickness of a (thermally conductive) solid. As the solid cap is thicker, the conduction rate across it becomes smaller. As an order of magnitude, the thermal conductivity of stagnate air is perhaps a factor of 10x lower than glass. For a 100 micron air boundary layer above the liquid, a 10 micron ceramic plate has the same conduction rate, and a 1 mm thick ceramic plate as 100x smaller conduction rate.

--> The jury is out here. When the solid cap is thicker than the boundary layer using the ratio of thermal conductivities of the air to the cap material, the open mug will have more heat conduction from the top air.

Evaporation occurs from the open mug (and not at all from the closed mug). The evaporating liquid builds a colder vapor layer above the liquid. This slows down the conduction rate to the liquid below because heat that would go to heat the liquid is instead taken to heat the cold vapor.

Finally, in the open mug, the surrounding gas radiates heat to the colder liquid. In the case that the cap is only as thick as the boundary layer on the open mug and when we neglect evaporation as a process to limit heat transfer to the open mug liquid, radiation can be a deciding factor to make the open mug heat faster.

# Conclusion

The jury is out in practice. You can set the restrictions on the system configuration, apply first principles, and reach a conclusion either way. Otherwise, the arguments become "what-if" cases that go on forever.

• This addresses most of it, but OP does potentially have a point about "condensative heating" that it might be helpful to address. This could be especially important when the mugs are closed off. Assuming the mugs aren't ideal insulators, you might be able to make the case that there is a slight heating effect when the ambient air condenses on the cold mug; and that effect would still exist when the mugs are sealed; while it would prevent the reverse (evaporation) in the hot mug.
– JMac
Dec 4 '19 at 15:22
• Doesn't this really depend on the humidity of the air, though? If the partial pressure of vapor in the warm air is higher than the vapor pressure of the cold liquid, wouldn't there be a net absorption of vapor into the liquid? Dec 4 '19 at 15:28
• In fact, isn't the statement "an open mug of liquid colder than the air will not evaporate" disproved by everyday experience? If what you're saying was true, puddles wouldn't evaporate after a rainstorm, right? Dec 4 '19 at 15:30
• Puddles are generally warmer than the surrounding air. Yes humidity is to be considered and I will address it. Dec 4 '19 at 16:59
• @JeffreyJWeimer Actually, here's a really obvious counterexample: liquid nitrogen boils at 77 K under 1 atm of pressure. It cannot be higher than this temperature in the liquid state at that pressure. This is much, much colder than the temperature of the surrounding air (which is at 270-300 K). And yet an open container of liquid nitrogen eventually evaporates away, despite being colder than the air it's surrounded by. Dec 4 '19 at 18:02

It's quite in reverse. If you want a colder liquid than it's surroundings (room temperature) to heat-up faster, then you need to spit liquid onto some vessel which would have biggest uncovered area, so to maximize liquid's direct contact area with air molecules. So that air molecules could transfer their kinetic energy to liquid faster. Otherwise air molecules have to overcome vessel walls material thermal conductivity. Of course, I assume that thermal conductivity of vessel material is lower than air's, otherwise we will get the reverse situation (there may exists special materials so to maximize heat transfer). Another way to understand this phenomenon, just think why you need to keep Vacuum flask closed if you want to keep longer it's content temperature different (colder/hotter) than it's surroundings. (Hint: vacuum flask has very low thermal conductivity)


text:
## CryptoDB

### Paper: Cryptanalysis of Stream Ciphers Based on Arrays and Modular Addition

Authors: Souradyuti Paul URL: http://eprint.iacr.org/2007/054 Search ePrint Search Google In modern cryptography, stream ciphers are most useful in applications where information needs to be encrypted/decrypted at high speed (e.g. high resolution streaming video data) or when low footprint (gates/memory) encryption is required. In the literature, there exist plenty of stream ciphers whose internal states are based on arrays and that they use modular additions to generate output streams. The abundance of array-based stream ciphers with modular additions can be attributed to the fact that, when implemented in software skillfully, they are able to produce outputs at a very high speed. The main contribution of this thesis is a unified analysis of stream ciphers based on arrays and modular addition. During the process, we detect cryptographic weaknesses in the designs of 9 widely known stream ciphers or pseudorandom bit generators (PRBGs). At first, we show some theoretical results on solving an important class of equations known as \emph{differential equations of addition} (DEA) that combine modular additions over two different algebraic groups such as GF(2) and GF($2^{32}$). The results include, \bite \item proof of the fact that the satisfiability of an arbitrary set of DEA is in the complexity class \pP,\item deriving all the solutions of an arbitrary set of DEA. \eite Next, we apply these results to attack a practical stream cipher named Helix (designed by Ferguson \emph{et al.}) with both chosen plaintexts and adaptive chosen plaintexts. In the second phase, the thesis closely scrutinizes a number of array-based stream ciphers (or PRBGs) in order to estimate their resistance against distinguishing attacks. We eventually discover, counter-intuitively, that the correlations between the array-indices and their associated array-elements, which apparently seem to be useful from the point of view of implementation purposes, can be exploited to mount distinguishing attacks on such type of ciphers if adequate precautions are not taken. In support of our theoretical findings, we point out distinguishing attacks on 8 practical array-based stream ciphers (or PRBGs), namely RC4 (designed by Rivest), RC4A (designed by Paul and Preneel), Py, Py6 (designed by Biham and Seberry), IA, ISAAC (designed by Jenkins Jr.), GGHN, NGG (by Gong \emph{et al.}); our attacks are based on the dependence of array-elements on array-indices. In all the cases we work under the assumption that the key-setup algorithms of the ciphers produce uniformly distributed internal states. We detect flaws in the mixing of bits in the keystream generation algorithms. Our analysis can be explained as the extension, development, adaptation and deeper characterization of the \ti{fortuitous states attacks} on the RC4 cipher by Fluhrer and McGrew in 2000.
##### BibTeX
@misc{eprint-2007-13336,
title={Cryptanalysis of Stream Ciphers Based on Arrays and Modular Addition},
booktitle={IACR Eprint archive},
keywords={secret-key cryptography /},
url={http://eprint.iacr.org/2007/054},
note={Ph.D. thesis, Katholieke Universiteit Leuven, B. Preneel (supervisor), 145+xxiv pages, November 2006. Souradyuti.Paul@esat.kuleuven.be 13560 received 15 Feb 2007, last revised 15 Feb 2007},


text:
# American Institute of Mathematical Sciences

November  2014, 34(11): 4459-4486. doi: 10.3934/dcds.2014.34.4459

## Topological and ergodic properties of symmetric sub-shifts

 1 School of Mathematics, The University of Manchester, Oxford Road, Manchester, M13 9PL, United Kingdom

Received  July 2013 Revised  February 2014 Published  May 2014

The family of symmetric one sided sub-shifts in two symbols given by a sequence $a$ is studied. We analyse some of their topological properties such as transitivity, the specification property and intrinsic ergodicity. It is shown that almost every member of this family admits only one measure of maximal entropy. It is shown that the same results hold for attractors of the family of open dynamical systems arising from the doubling map with a centred symmetric hole depending on one parameter, and for the set of points that have unique $\beta$-expansion for $\beta \in (\varphi,2)$ where $\varphi$ is the Golden Ratio.
Citation: Rafael Alcaraz Barrera. Topological and ergodic properties of symmetric sub-shifts. Discrete & Continuous Dynamical Systems - A, 2014, 34 (11) : 4459-4486. doi: 10.3934/dcds.2014.34.4459
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Stat. Phys., 135 (2009), 519. doi: 10.1007/s10955-009-9747-8. Google Scholar [18] D. Lind and B. Marcus, An Introduction to Symbolic Dynamics and Coding,, Cambridge University Press, (1995). doi: 10.1017/CBO9780511626302. Google Scholar [19] R. Mañé, Introdução à teoria ergódica, volume 14 of Projeto Euclides [Euclid Project],, Instituto de Matemática Pura e Aplicada (IMPA), (1983). Google Scholar [20] J. Nilsson, On numbers badly approximable by dyadic rationals,, Israel J. Math., 171 (2009), 93. doi: 10.1007/s11856-009-0042-9. Google Scholar [21] J. Nilsson, The fine structure of Dyadically badly approximable numbers,, ArXiv e-prints, (2010). Google Scholar [22] W. Parry, On the \beta -expansions of real numbers,, Acta Math. Acad. Sci. Hungar., 11 (1960), 401. doi: 10.1007/BF02020954. Google Scholar [23] W. Parry, Intrinsic Markov chains,, Trans. Amer. Math. Soc., 112 (1964), 55. doi: 10.1090/S0002-9947-1964-0161372-1. Google Scholar [24] K. Petersen, Chains, entropy, coding,, Ergodic Theory Dynam. Systems, 6 (1986), 415. doi: 10.1017/S014338570000359X. Google Scholar [25] N. Sidorov, Arithmetic dynamics,, In Topics in dynamics and ergodic theory, (2003), 145. doi: 10.1017/CBO9780511546716.010. Google Scholar [26] N. Sidorov, Supercritical holes for the doubling map,, Acta Mathematica Hungarica, (2014), 1. Google Scholar [27] M. Urbański, Invariant subsets of expanding mappings of the circle,, Ergodic Theory Dynam. Systems, 7 (1987), 627. doi: 10.1017/S0143385700004247. Google Scholar [28] P. Walters, An Introduction to Ergodic Theory, volume 79 of Graduate Texts in Mathematics,, Springer-Verlag, (1982). Google Scholar [29] B. Weiss, Intrinsically ergodic systems,, Bull. Amer. Math. Soc., 76 (1970), 1266. doi: 10.1090/S0002-9904-1970-12632-5. Google Scholar [30] B. Weiss, Subshifts of finite type and sofic systems,, Monatsh. Math., 77 (1973), 462. doi: 10.1007/BF01295322. Google Scholar show all references ##### References:  [1] S. Akiyama and K. Scheicher, Symmetric shift radix systems and finite expansions,, Math. Pannon., 18 (2007), 101. Google Scholar [2] J. Allouche, M. Clarke and N. Sidorov, Periodic unique beta-expansions: The Sharkovskiĭ ordering,, Ergodic Theory Dynam. Systems, 29 (2009), 1055. doi: 10.1017/S0143385708000746. Google Scholar [3] S. Baker, Generalised golden ratios over integer alphabets,, Integers, 14 (2014), 1. Google Scholar [4] R. Bowen, Periodic points and measures for Axiom A diffeomorphisms,, Trans. Amer. Math. Soc., 154 (1971), 377. Google Scholar [5] R. Bowen, Some systems with unique equilibrium states,, Math. Systems Theory, 8 (): 193. doi: 10.1007/BF01762666. Google Scholar [6] M. Boyle, Algebraic aspects of symbolic dynamics,, In Topics in symbolic dynamics and applications (Temuco, (1997), 57. Google Scholar [7] M. Brin and G. Stuck, Introduction to Dynamical Systems,, Cambridge University Press, (2002). doi: 10.1017/CBO9780511755316. Google Scholar [8] S. Bundfuss, T. Krüger and S. Troubetzkoy, Topological and symbolic dynamics for hyperbolic systems with holes,, Ergodic Theory Dynam. Systems, 31 (2011), 1305. doi: 10.1017/S0143385710000556. Google Scholar [9] V. Climenhaga and D. J. Thompson, Intrinsic ergodicity beyond specification: \beta-shifts, S-gap shifts, and their factors,, Israel J. Math., 192 (2012), 785. doi: 10.1007/s11856-012-0052-x. Google Scholar [10] M. F. Demers and P. Wright, Behaviour of the escape rate function in hyperbolic dynamical systems,, Nonlinearity, 25 (2012), 2133. doi: 10.1088/0951-7715/25/7/2133. Google Scholar [11] C. Dettmann, Open circle maps: Small hole asymptotics,, Nonlinearity, 26 (2013), 307. doi: 10.1088/0951-7715/26/1/307. Google Scholar [12] P. Erdös, I. Joó and V. Komornik, Characterization of the unique expansions 1=\sum$^\infty_{i=1}q^{-n_i}$ and related problems,, Bull. Soc. Math. France, 118 (1990), 377.   Google Scholar [13] P. Glendinning and N. Sidorov, Unique representations of real numbers in non-integer bases,, Math. Res. Lett., 8 (2001), 535.  doi: 10.4310/MRL.2001.v8.n4.a12.  Google Scholar [14] P. Glendinning and N. Sidorov, The doubling map with asymmetrical holes,, Ergodic Theory and Dynamical Systems Ergodic Theory and Dynamical Systems, (2013), 1.  doi: 10.1017/etds.2013.98.  Google Scholar [15] B. M. Gurevič, Uniqueness of the measure with maximal entropy for symbolic dynamical systems that are close to Markov ones,, Dokl. Akad. Nauk SSSR, 204 (1972), 15.   Google Scholar [16] N. T. A. Haydn, Phase transition in one-dimensional subshifts,, Discrete Contin. Dyn. Syst., 33 (2013), 1965.  doi: 10.3934/dcds.2013.33.1965.  Google Scholar [17] G. Keller and C. Liverani, Rare events, escape rates and quasistationarity: Some exact formulae,, J. Stat. Phys., 135 (2009), 519.  doi: 10.1007/s10955-009-9747-8.  Google Scholar [18] D. Lind and B. Marcus, An Introduction to Symbolic Dynamics and Coding,, Cambridge University Press, (1995).  doi: 10.1017/CBO9780511626302.  Google Scholar [19] R. Mañé, Introdução à teoria ergódica, volume 14 of Projeto Euclides [Euclid Project],, Instituto de Matemática Pura e Aplicada (IMPA), (1983).   Google Scholar [20] J. Nilsson, On numbers badly approximable by dyadic rationals,, Israel J. Math., 171 (2009), 93.  doi: 10.1007/s11856-009-0042-9.  Google Scholar [21] J. Nilsson, The fine structure of Dyadically badly approximable numbers,, ArXiv e-prints, (2010).   Google Scholar [22] W. Parry, On the $\beta$-expansions of real numbers,, Acta Math. Acad. Sci. Hungar., 11 (1960), 401.  doi: 10.1007/BF02020954.  Google Scholar [23] W. Parry, Intrinsic Markov chains,, Trans. Amer. Math. Soc., 112 (1964), 55.  doi: 10.1090/S0002-9947-1964-0161372-1.  Google Scholar [24] K. Petersen, Chains, entropy, coding,, Ergodic Theory Dynam. Systems, 6 (1986), 415.  doi: 10.1017/S014338570000359X.  Google Scholar [25] N. Sidorov, Arithmetic dynamics,, In Topics in dynamics and ergodic theory, (2003), 145.  doi: 10.1017/CBO9780511546716.010.  Google Scholar [26] N. Sidorov, Supercritical holes for the doubling map,, Acta Mathematica Hungarica, (2014), 1.   Google Scholar [27] M. Urbański, Invariant subsets of expanding mappings of the circle,, Ergodic Theory Dynam. Systems, 7 (1987), 627.  doi: 10.1017/S0143385700004247.  Google Scholar [28] P. Walters, An Introduction to Ergodic Theory, volume 79 of Graduate Texts in Mathematics,, Springer-Verlag, (1982).   Google Scholar [29] B. Weiss, Intrinsically ergodic systems,, Bull. Amer. Math. Soc., 76 (1970), 1266.  doi: 10.1090/S0002-9904-1970-12632-5.  Google Scholar [30] B. Weiss, Subshifts of finite type and sofic systems,, Monatsh. Math., 77 (1973), 462.  doi: 10.1007/BF01295322.  Google Scholar
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# Difference between revisions of "L4D2 Level Design/Multi-Gamemode Support"

## Contents

info_gamemode entity

In L4D2, You can use the new info_gamemode entity to trigger different events depending on the game mode in which a map is played. This includes variant map appearance, path arrangement, item placement, or any other logic.

The most direct benefit is that you no longer have to ship multiple versions of the same level. A single .BSP can support all gamemodes, thus dramatically reducing the size of your add-on.

If there is a drawback, it is certainly in the increased complexity of your VMF file. However, as we shall see, this can be easily managed using visgroups and the new .VMF instance functionality described previously.

## Learning by example

Let's look at how gamemode logic can be used to add scavenge in an existing map, in this case one from the Deadline add-on campaign.

First, open the source file for the second map of Deadline 2:

Steam\steamapps\common\left 4 dead 2\sdk_content\DeadLine2\mapsrc\deadline02.vmf

## Overall structure

The scavenge-specific entities are in their own instance, which also lives in its own visgroup.

One of the features of this map is a lift in the center that must be triggered. Fictionally, the "noise" also creates a horde panic event. This is used to create two halves of the map and limit gameplay to each area. While this is desirable in co-op, where maps are very linear, scavenge gameplay is much more arena-oriented. So we'll need to have the map behave differently when launched in scavenge mode.

Let's look at how this is set up:

In the visgroup panel, click on the checkbox next to the "+SCAVENGE" visgroup to turn it on and off. You'll see all the scavenge-specific content activate and deactivate. There is just one func_instance in this visgroup, and it references the following file:

"instance/l4d_deadline02_scavenge.vmf"

The idea here is to have all the elements that are common to all gamemodes in the main map source file: "deadline02.vmf" and to have elements that are gamemode-specific each live in their own separate .VMF files, and have them be referenced into the main map as instances. The L4D2 level designers find it easier to manage everything this way, and depending on the complexity of your map, you may find it easier as well.

## Examining the instance

Editing the Scavenge instance.

Click on the Scavenge instance with the selection tool and open up its properties dialog. At the bottom left, click the Edit Instance button.

Hammer opens up the scavenge .VMF for editing. Now we can see just the scavenge-related items: The starting areas, the generator, weapons, and some obstacles that alter the map and navigation.

It's a little more complicated, though. Some elements need to be spawned only in scavenge mode, and others must be killed if the map is run anything but scavenge mode. Remember that visgroups and instances are only meaningful in Hammer. Once you run your map, the VBSP process collapses all the instances, which means everything gets lumped together, and disabled visgroups are ignored. So we need a mechanism for controlling these in game. This is where the info_gamemode entity comes in.

## info_gamemode settings

The gamemode logic entities.

The icon looks like a set of playing cards.

Tip: You can also use Map > Entity Report to find it. Enable filter by class, and type in "info_gamemode".
Outputs that fire depending on gamemode.

Select it and click on the Properties button. The class info is pretty sparse, but let's look at the Outputs tab. That's where all the action is.

Here you can see that if the map loads in any gamemode other than scavenge, it fires off a logic_relay named "relay_kill_scavenge" that kills all of the scavenge-specific items in this map.

That takes care of removing everything, which is the easy scenario. Adding things, however, needs to be done in stages, so we have broken things out into their own logic_relay entities.

When the map is first loaded in scavenge mode, it fires off the "OnScavenge" event. At this point, the Director is still getting ready, and it's important not to spawn or kill entities until it is finished initializing.

After the map has been set up and the Director has been initialized, it fires off "OnScavengePostIO". We do several things here:

1. First, we do any point_template spawning that we need to do. This example map has templates for scavenge-specific weapons, items, and the generator with the fill point.
2. It also actuates the lift automatically, thus raising the ladder and disarming the call button.
3. The last thing we do in post IO is adjust navigation. We've blocked some paths to balance the map, and the nav blockers are set to block.

"OnScavengeMatchStart" is, as I'm sure you've guessed, fired off for each match.

One of the nice things about the info_gamemode is that you can have several of them in a map. Instead of relying on a single one to do everything, the Deadline 2 map uses one in each game mode instance file. Each one is responsible for spawning and killing stuff related to that particular gamemode alone. This takes a bit more effort to set up, but it helps keep things a little simpler when you're editing. And keeping things in their own "buckets" really helps when you resume editing a map that you haven't touched for a while.

## Using info_director

info_director outputs.

In some cases, you will not be able to fully compartmentalize a gamemode. For example, scavenge maps require that the info_director, which is in the main map because it is used by all game modes, have certain outputs mapped as shown to the right.

You'll notice that survival has its own instance as well, with coop and versus modes are combined together into their own. Of course, you may organize your map and support gamemodes as you see fit, with or without instances.


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# RSA Attack, known public key and ciphertext. Can decrypt the message m?

I'm stuck with this problem: Assume Alice and Bob are using RSA to communicate each other. Given Bob's public key $(217,7)$, Alice send the ciphertext $c=53$ to Bob through RSA encryption. Now, Eve can read the ciphertext $c$, how can Eve decrypt the ciphertext and read plaintext $m$?

• @user30025 Is this (let’s just call it) “homework”? What research have you done? What have you tried? – e-sushi Dec 22 '15 at 21:07

Some examples of answers to similar questions:

Hint: $N = 217 =\ ?\ \times\ ?$

In the public key, $e$ is the exponent and $N$ is modulus, therefore $e < N$.
If $m$ is your message, $m^e \mod N = c$ is the cipher.
To decipher you need to find $d$, such as: $d \times e = 1 \mod \phi(N)$.
You will then be able to find $m$: $c^d \mod N = m$

• N = 217, public key = 7. You can find them, you know that p and q are prime. – Biv Dec 22 '15 at 20:55
• public key is represented as (217, 7), can you please explain me what values are e and n? – user30025 Dec 22 '15 at 21:00
• 1 is not prime. it is N = p * q. Any method to find them is fine, you can even ask google if 217 is prime. :) – Biv Dec 22 '15 at 21:56
• You need to factor $N$ not $e$. 7 is prime so it can not be a product of two primes. 217 ist not prime, but consists of two factors (excluding 1 and 217 itself). – Kevin__ Dec 22 '15 at 21:58
• If factoring $N$ was forbidden by the rule of the homework, of If $N$ was too large to factor, there is the option to try every possible plaintext, encipher it using the public key, and check against the ciphertext. – fgrieu Dec 28 '15 at 10:37


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# Tetration Forum

Full Version: Operations with fractional index between + and * ?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
This is only a sketch, yet; I didn't compute values. Also there is one conceptional problem.

Consider the binary operations add(x,b) = x+b , mult(x,b)=x*b , pow(x,b) = x^b (the version with left associativity) and their iterable notations add(x,b,h)=x+b*h, mult(x,b,h) = x*b^h , pow(x,b,h) = x^b^h

It may look overartificial to find the schröder-functions for fractional iterates of add(), mult() and pow(), since it is trivial to insert fractional heights h into the formulae.

But for more intense study let's try to express this with the use of schröder-functions.
Always we have the general form of composition
g ° b^h ° f
or
g( b^h * f(x))
where f is the appropriate schröder-function and g its inverse.

logb ( b^(b*h) * (b^x))

where the schröder-function is b^x and its inverse the logb (=log to base b)
Explicitely we have by this

logb( b^(b*h) * (b^x)) = logb( b^(x+b*h)) = x + b*h

Now we look at the mult-function. The schröder-function is simply the id-function

id(b^h*id(x)) = x*b^h

The pow-function has the logb-function as schröder-function:

b^( b^h * logb(x)) = b^(logb( x^b^h)) = x^(b^h)

Now we see, that the schröder-functions for the three basic operations are the iterates of the exp-function:

mul() --> exp_b°0(x) = id(x)
pow() --> exp_b°-1(x) = logb(x)

where the iteration-heights of the exp-function may serve as integer index with an unlucky offset. But let's ignore that unusual offset.
What we immediately see is, that we could study which sense it makes to define operations fractionally indexed between add() and mul() based on fractional iterates of the exp-/log-function.

However, for one inconsistency I didn't find a smooth workaround yet: for add() the cofactor to the schröder-function is b^(b*h), while for mult() and pow() it is b^h only.
I also have not yet numerical examples, and think I'll need some more time for this (I'm short, courses begin next week), but thought, perhaps someone other could already look at this.

What do you think?
Hmm, couldn't resist....
I just tried this for some numerical data without further checking, whether this all can eventually give a meaningful concept.
Anyway.
I've considered the ten fractionally indexed oprations between mul() and pow(). Base is sqrt(2), initial value x0 is 2.0 because with x0=1 the power-operation would give trivial result. The iteration-height h is 1,2 and 3. The (fractional) operation-index is k; k=0 means mul(x0,sqrt(2),h), k=1 means pow(x0,sqrt(2),h)

Here are the results:

Code:
´          h=1                h=2             h=3 ´    |  2.0* b^1.0      2.0* b^2.0      2.0* b^3.0                             k  | 2.82842712475   4.00000000000   5.65685424949  <--- mul(2,sqrt(2),h)  , direct computation     -----------------------------------------------------                           0.0  2.82842712475   4.00000000000   5.65685424949  = mul()                   0.1  2.81156857833   4.00000000000   5.75322682321                         0.2  2.79480389679   4.00000000000   5.85853861394                         0.3  2.77814247936   4.00000000000   5.97272443872                         0.4  2.76159351333   4.00000000000   6.09569942066                         0.5  2.74516595391   4.00000000000   6.22787081195                         0.6  2.72886850525   4.00000000000   6.37054764165                         0.7  2.71270960247   4.00000000000   6.52607412279                         0.8  2.69669739492   4.00000000000   6.69762735731                         0.9  2.68083973075   4.00000000000   6.88877250162                         1.0  2.66514414269   4.00000000000   7.10299330132  = pow()             -----------------------------------------------------                          | 2.66514414269   4.00000000000   7.10299330132    <--- pow(2,sqrt(2),h)  , direct computation      |  2.0^ b^1.0       2.0^ b^2.0      2.0^ b^3.0

Gottfried
Gottfried, this is really an interesting approach.
I need to ponder about in some quiet hour, if I get some time.
Let me rewrite this so I understand it better.

(1) $({+}b)^h(x) = x + bh = \mathcal{S}[{+}b]^{-1}(b^h\mathcal{S}[{+}b](x))$
(2) $({\times}b)^h(x) = xb^h = \mathcal{S}[{\times}b]^{-1}(b^h\mathcal{S}[{\times}b](x))$
(3) $(\!{\uparrow}b)^h(x) = x^{b^h} = \mathcal{S}[\!{\uparrow}b]^{-1}(b^h\mathcal{S}[\!{\uparrow}b](x))$
and
(4) $\mathcal{S}[{+}b](x) = a^{x/b}$ for almost any a.
(5) $\mathcal{S}[{\times}b](x) = x$
(6) $\mathcal{S}[\!{\uparrow}b](x) = \log_a(x)$ for almost any a.

so what I think you mean is that you can set $a=b^b$ in (4), and $a=b$ in (6) so that there is a common base? That is quite interesting. Honestly, I had never seen that before. This effectively reduces the problem of "nice" noninteger lower hyperoperations to the problem of "nice" noninteger tetration.

I am well aware of another similar situation with the commutative hyperoperations, where the general form can be written:
$H_{n+2}(a, b) = \exp^{n}\left(\log^{n}(a)\log^{n}(b)\right)$
this can also be considered a way of reducing noninteger hyperoperations to noninteger tetration.
@henryk - well, take your time
@andrew - yes, i think you got this right.

@all...

well, if the multiplicative coefficient to the schröder-function, which is of the form b^h, would be consistent, I would have made "a thing of it". But since their forms for the three operations are not constant and also not "linear" (in some sense) I doubt this all is eventually much helpful.

Just try to continue that sequence to higher integer indexes.. it looks to me as some well known try to extend the sequence of operations with some symmetric approach, as Andy has referred to.
So my first impression was, that this all might be -besides its nice appeal from the fractional indexes between mul() and pow() - somehow trivial.

Actually, if I don't get an intriguing and practicable idea how to "linearize" the structure of the cofactor of the schröder-function, then I feel this all as somehow "suboptimal" (though extremely interesting & animating...). But may be, my notorious trend to perfectionisnm keeps me away from recognizing some present useful aspect...

You see: I'm not having the feeling of a proper standing here...

<G>
[... including some updating...]
Just had another thought on this matter.

In order for this interpolation to work, you should also check to see if it applies for ranks above 3, for example (to use a combination Knuth-Bromer-Mueller arrow system), we have the first 5 lower hyperoperations are: $\{a + b, ab, a{\uparrow}b, a{\downarrow}{\uparrow}b, a{\downarrow}{\downarrow}{\uparrow}b\}$, and so we should check that:
$\mathcal{S}[{\downarrow}{\uparrow}b](x) = \log_a(\log_a(x))$, and
$\mathcal{S}[{\downarrow}{\downarrow}{\uparrow}b](x) = \log_a(\log_a(\log_a(x)))$
and I have a feeling that this is not the case.
So I checked up on it, and no, it is not the case. Also, I have some bad news. If we assume the Schroder functions (accordingly, the Abel functions) are iterated exponentials, then that means that the associated hyperoperation hierarchy would have the recursion rule:

$HE_n(x, y) := \exp(HE_{n-1}(\log(x), y))$

which (if n=2 is multiplication) requires that (n=1) is $HE_1(x, y) = x + \log(y)$, which is not addition, so this scheme does not form a hyperoperation hierarchy. However, if I might draw your attention to the Tetration Ref. section 2.3.4, the recursion rule:

$HR_n(x, y) := \exp(HR_{n-1}(\log(x), \exp^{n-3}(y)))$

has many of the same characteristics, and satisfies addition (n=1), mul (n=2), and powers (n=3). The only difference is in the heighest exponent.

$HR_n(x, y) = HE_n(x, \exp^{\circ\left({n-2 \atop 2}\right)}(y))$