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# Question: What Are Properties Of Proportions?
## What are the 3 kinds of proportion?
Types of ProportionsDirect Proportion.Indirect Proportion..
## What is difference between ratio and proportion?
A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours. A proportion on the other hand is an equation that says that two ratios are equivalent.
## What are the 3 types of ratios?
The three main categories of ratios include profitability, leverage and liquidity ratios. Knowing the individual ratios in each category and the role they plan can help you make beneficial financial decisions concerning your future.
## What are the different properties of proportion?
PROPERTIES OF PROPORTIONProperty 1 : An equality of two ratios is called a proportion. … Property 3 : In a proportion, … Property 4 : Three quantities a, b, c of the same kind (in same units) are said to be in continuous proportion. … Property 5 : … Property 6 : … Property 8 : … Property 10 : … Property 12 :More items…
## What are proportions used for?
Ratios and proportions are also used in business when dealing with money. For example, a business might have a ratio for the amount of profit earned per sale of a certain product such as \$2.50:1, which says that the business gains \$2.50 for each sale.
## What do proportions mean?
1 : harmonious relation of parts to each other or to the whole : balance, symmetry. 2a : proper or equal share each did her proportion of the work. b : quota, percentage. 3 : the relation of one part to another or to the whole with respect to magnitude, quantity, or degree : ratio.
## What are examples of proportions in real life?
Meanwhile, another car can fill up with a different amount of fuel than ours. The price per gallon stays the same, so the relationship between the gallons put in and the money paid is the same and therefore, filling each car’s tank with gas is proportional because they follow the same proportionality ratio.
## What are proportions in statistics?
A proportion refers to the fraction of the total that possesses a certain attribute. For example, suppose we have a sample of four pets – a bird, a fish, a dog, and a cat. Therefore, the proportion of pets with four legs is 2/4 or 0.50. …
## What is the property of ratio and proportion?
Important Properties of Proportion Dividendo – If a : b = c : d, then a – b : b = c – d : d. Componendo – If a : b = c : d, then a + b : b = c+d : d. Alternendo – If a : b = c : d, then a : c = b: d. Invertendo – If a : b = c : d, then b : a = d : c.
## What is the rule of proportion?
Fundamental rule of proportions means cross multiply. He explains that to arrive from an equation which has fractions into the one without equation multiply the top of left side with the bottom of right side and equal it with bottom of left side multiplied with top of the right side.
## What are examples of proportions?
Proportion says that two ratios (or fractions) are equal….Example: Rope40m of that rope weighs 2kg.200m of that rope weighs 10kg.etc.
## What are math proportions?
Proportion says that two ratios (or fractions) are equal. Example: 1/3 = 2/6. See: Equivalent Fractions. Proportions.
## What is the formula of mean proportion?
The mean proportional between the two terms of a ratio in a proportional is the square root of the product of these two. For example, in the proportion a:b :: c:d, we can define the mean proportional for the ratio a:b as the square root of the product of the two terms of the ratio or √ab.
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## How to solve equation: 25x+22y=488
Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
dgorack
Posts: 1
Joined: Tue May 21, 2013 3:38 am
Contact:
### How to solve equation: 25x+22y=488
Looking at the equation: 25x+22y=488, I can see that y=4 and x=16 but how to I write the steps to solve this equation?
25(16) + 22 (4) = 488
400 + 88 = 488
maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
Contact:
### Re: How to solve equation: 25x+22y=488
Looking at the equation: 25x+22y=488, I can see that y=4 and x=16 but how to I write the steps to solve this equation?
If you mean "what steps do I use to show that (x,y)=(16,4) is an answer?", then the steps you showed are just fine. If you mean "what steps do I use to find this and other answers?", then you can see how to do that here. The equation is for a straight line and every answer for the equation is a point on the line. You can pick whatever numbers you want for one of the letters and then you solve for the other letter. Like:
x=0: 25*0+22y=0+22y=22y, 22y=448, y=448/22=224/11
So (x,y)=(0,224/11) is also an answer!
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# Systems of Linear Equations
Markov Chains and
Dynamic Models
The refinery and economic supply-demand models of Section 1.2 were static
in the sense that we solved them once and that was it. There was one set of
production levels required, not a sequence of levels that would be needed
to describe an economy changing over time. A model that tries to predict
the behavior of a system over a period of time is called a dynamic model.
In this section we examine two dynamic linear models.
The first dynamic model we consider involves probability. This model,
called a Markov chain, will arise over and over in this book, so it is important
to understand the model well. The concepts of probability we need for this
model are simple and intuitive.
A Markov chain is a probabilistic model that describes the random
movement over time of some activity. At each period of time, the activity
is in one of several possible states. States might be amounts won in gam-
yield
heating oil = 520,
diesel oil = 890,
gasoline = 1040
This is getting quite close to our production goals. The overproduction
of 20 to 40 gallons in each product might be a reasonable safety margin
that is actually desirable.
To get closer, we should decrease x2 and x3 a little.
yield
heating oil = 504,
diesel oil = 852,
gasoline = 1006
This is an excellent fit. We have been a bit lucky. To do better, we
would probably have to use fractional values. (In the Exercises the
reader may need more tries to get this close.)
We next consider a slightly more complicated supply-demand model.
This model has the balancing advantage that trial-and-error calculations to
estimate a solution are easier. The reader is warned that it takes a little while
to get a feel for all the numbers in this model.
Example 2. A Model of General Economic
Supply-Demand
We present a linear model due to W. Leontief, a Nobel Prize-winning
economist. The model seeks to balance supply and demand throughout
a whole economy. For each industry, there will be one supply-demand
equation. In practical applications, Leontief economic models can have
hundreds or thousands of specific industries. We consider an example
with four industries.
The left-hand side of each equation is the supply, the amount
produced by the ith industry. Call this quantity xi; it is measured in
dollars. On the right-hand side, we have the demand for the product
of the ith industry. There are two parts to the demand. The first part
is demand for the output by other industries (to create other products
requires some of this product as input). The second part is consumer
demand for the product.
For a concrete instance, let us consider an economy of four general
industries: energy, construction, transportation, and steel. Suppose
that the supply-demand equations are
Industrial Demands Supply Transport. Energy Construct. Steel Consumer Demand Energy: x1= Construct.: x2= Transport. : x3= Steel: x4=
(6)
The first equation, for energy, has the supply of energy x1 on the
left. The terms on the right of this equation are the various demands
that this supply must meet. The first term on the right, 4x1 is the
input of energy required to produce our x1dollars of energy (A units
of energy input for one unit of energy output). Also, the second term
of .2x2 is the input of energy needed to make x2 dollars of construction.
Similarly, terms .2x3 and .2x4 are energy inputs required for transportation
and steel production. The final term of 100 is the fixed consumer
demand.
Each column gives the set of input demands of an industry. For
example, the third column tells us that to produce the x3 dollars of
transportation requires as input .2x3 dollars of energy, .2x3 dollars of
construction, and .1x3 dollars of steel. In the previous refinery model,
the demand for each product was a single constant quantity. In the
Leontief model, there are many unknown demands that each industry's
output must satisfy. There is an ultimate consumer demand for each
output, but to meet this demand industries generate input demands on
each other. Thus the demands are highly interrelated: Demand for
energy depends on the production levels of other industries. and these
production levels depend in turn on the demand for their outputs by
other industries, and so on.
When the levels of industrial output satisfy these supply-demand
equations, economists say that the economy is in equilibrium.
As in the refinery model, let us try to solve this system of equations
by trial-and-error. As a first guess, let us set the production levels
at twice the consumer demand (the doubling tries to account for the
interindustry demands). So , and
x4 = 0; these are our supplies. Given these production levels, we can
compute the demands from (6).
Supply Demand Energy: 200 Construct.: 100 Transport.: 200 Steel: 0
(7)
For our next approximations, let us try supply levels halfway
between the supply and demand values in (7). That is,
, and similarly, , and
Supply Demand Energy: 220 Construct.: 140 Steel: 165 Steel: 15
(8)
The second approximation is only moderately better. The interaction
effects between different industries are hard to predict. Adjusting production
levels was much easier in the refinery problem, where the
demand for each product was constant.
Let us stop trying to be clever and just use the simple-minded
approach of setting production levels (i.e., supply levels) equal to the
previous demand levels. So from (8), we try
Supply Demand Energy: 252 Construct. : 192 Transport. : 139 Steel: 30
(9)
The demand values here have been rounded to whole numbers. The
supplies and demands are getting a little closer together in (9).Repeating
the process of setting the new supply levels equal to the previous
demand levels (i.e., the demands on the right side in (9)) yields
Supply Demand Energy: 273 Construct. : 214 Transport. : 150 Steel: 33
(10)
Repeating this process again, we have
Supply Demand Energy: 289 Construct. : 229 Transport. : 155 Steel: 36
(11)
Observe that in successive rounds (9), (10), (11), supplies are
rising. This is because as we produce more, we need more input which
requires us to produce still more, and so on. It may be that this iteration
will go on forever, and no equilibrium exists. On the other hand, the
gap between supplies and demands is decreasing.
Leontief proposed a constraint on the input costs that we shall
show (in Section 3.4) guarantees that an equilibrium exists. The constraint
is
Input Constraint. Every industry is profitable: Every industry must
require less than \$1 of inputs to produce \$1 of output.
In mathematical terms, this means that the sum of the coefficients
in each column must be less than I. Our data in (6) satisfy this constraint,
so an equilibrium does exist for this four-industry economy.
Moreover, the iteration process of repeatedly setting production levels
equal to the previous demands will converge to this equilibrium. The
reader should check that the following numbers are equilibrium values
(rounded to the nearest integer).
Equilibrium: energy = 325, construction = 265,
transportation = 168, steel = 43
Note that any system of linear equations can be rewritten in the form
of supply-demand equations with xi appearing alone on the left side of the
ith equation, as in the Leontief supply-demand model (6). It is standard
practice to solve large systems of linear equations by some sort of iterative
method. The nature of the supply-demand equations suggested the iterative
scheme we used here, letting the demands from one round be the production
levels of the next round .
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Advertisement
## Calculate specific areas size in image
### Question
I'd like to calculate what area size share do the brown spots take within the area marked with pink lines. See the image here:
In case you wonder what that is - it's an E.coli bacteria colony (don't worry, nothing dangerous) :)
Anyway, I'd like to measure the area that brown spots take, I guess Photoshop will do (not very experienced with it though).
My line of thinking for doing this is:
• Crop the image to areas that are outlined with pink color (would get two images in this case). This is fairly easy and I would know how to do it (selection + crop image to selection).
• Make the image black&white, so the background is white and the spots are black (it should be easier then to calculate the size of the spots).
• Calculate/measure the size of the black spots. If I get this number, I can easily divide it by the size of the image and get the percentage of the area that these spots cover :)
What would be the steps to do that in Photoshop (if it's possible at all)? Or would you recommend some other program?
2015/05/09
1
2
5/9/2015 8:26:00 PM
A camera produces a image that loses any scale information. That is because the projection is flat. Objects at different distances appear sized differently. Likewise certain objects can appear the same.
Image 1: The projection loses sense of scale
This said its possible to know the size, if you have a pair of images (and multiple points in different depths to measure) and a reference measurement in the image.
It is also possible that you just want to compare sizes. In this case the petri dish size is known. Then it can be done if the camera is perpendicular to sample, or sample is perspective corrected right (but this incurs a hard to estimate error).
Image 2: Head on view of a petri dish, image source.
## Computing area by relation on perpendicular surface.
If you can photograph like head on like image 2. Then you can just measure the pixel area of the entire circle, and compare that to your marked area. Since the dimensions of the petri dish are known you get a simple relation.
Arp * (Apm /App ) = Arm
Where Arp is the real area of the petri dish (πr 2) in whatever units you want,
Apm is the pixel area of your region in pixels (pm = pixel measurement),
App is the pixel area of your petri dish floor (pm = pixel petri dish) and finally Am the are you wanted to measure.
You can get Photoshop to report number of selected pixels in the extended version of histogram view. This is the area of the selection in pixels.
Image 3: Dummy measurement, gets me an area of ~ 0.15*2800 = 420 mm2 for a petri dish with 60 mm diameter . No measurement confidence calculation done.
Please note due to scale loss effect above you can not use image are for measurement as the distance to target and size/off axis alignment of image is unknown. Measuring both areas ensures a measurement error that is better behaved even if the camera is slightly of axis for said sample.
## For fun
For a fun deepening understanding in aultomatic 3D correction see The Fundamental Matrix Song
PS: Not entirely Graphics design. After edit definitely no longer Graphics design.
2015/05/10
2
5/10/2015 10:42:00 AM
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# Roots of polynomial equations
• Dec 26th 2010, 01:38 PM
Femto
Roots of polynomial equations
I was happily answering some questions from my further pure textbook earlier until I encountered the last question; forgive me in advance for my lack of understanding, I'm only a sixth form student at the moment and I don't have as much knowledge as most people on this site!
Here's the question:
The roots of the equation $\displaystyle x^3 + ax + b$ are $\displaystyle \alpha, \beta, \gamma$. Find the equation with roots $\displaystyle \frac{\beta}{\gamma} + \frac{\gamma}{\beta}, \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}, \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Initially I tried to use substitution but that failed epicly. Overall, I'm really confused - please could somebody assist me?
Also, on a side note, I'm new here! Just thought you'd like to know (Smile)
• Dec 26th 2010, 01:44 PM
dwsmith
Since $\displaystyle \alpha, \ \beta, \ \mbox{and} \ \gamma$ are roots, $\displaystyle x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)$
Does this help?
• Dec 26th 2010, 01:46 PM
Femto
Quote:
Originally Posted by dwsmith
Since $\displaystyle \alpha, \ \beta, \ \mbox{and} \ \gamma$ are roots, $\displaystyle x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)$
Does this help?
Yes I kind of understand, but how do I obtain another equation with the new fractional roots presented in the question?
Would it perhaps make sense to write it out as this?
$\displaystyle (x - (\frac{\beta}{\gamma} + \frac{\gamma}{\beta}))(x - (\frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}))(x - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}))$
Really sorry, I'm getting a tad confused.
• Dec 26th 2010, 01:49 PM
dwsmith
$\displaystyle \displaystyle \left(x-\left(\frac{\beta}{\gamma}+\frac{\gamma}{\beta}\ri ght)\right)(\cdots)(\cdots)$
• Dec 26th 2010, 01:51 PM
dwsmith
Quote:
Originally Posted by Femto
Yes I kind of understand, but how do I obtain the new equation with the new fractional roots presented in the question?
Would it help to say:
$\displaystyle (x - (\frac{\beta}{\gamma} + \frac{\gamma}{\beta}))(x - (\frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}))(x - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}))$?
Now multiply it out.
I would leave it factored, but if you need it in x^3.... format, you need to multiply.
• Dec 26th 2010, 01:56 PM
Femto
Quote:
Originally Posted by dwsmith
Now multiply it out.
I would leave it factored, but if you need it in x^3.... format, you need to multiply.
Thanks; yikes this looks time consuming.
• Dec 26th 2010, 01:58 PM
dwsmith
$\displaystyle x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)=x^3-\alpha x^2-\gamma x^2-\beta x^2+\alpha\beta x+\beta\gamma x+\alpha\gamma x-\alpha\beta\gamma$
$\displaystyle b=-\alpha\beta\gamma$
$\displaystyle ax=x\alpha\beta +x\beta\gamma+x\alpha\gamma$
$\displaystyle 0x^2=-\alpha x^2-\gamma x^2-\beta x^2$
• Dec 26th 2010, 02:04 PM
Plato
Quote:
Originally Posted by Femto
Here's the question:
The roots of the equation $\displaystyle x^3 + ax + b$ are $\displaystyle \alpha, \beta, \gamma$. Find the equation with roots $\displaystyle \frac{\beta}{\gamma} + \frac{\gamma}{\beta}, \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}, \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
From the given we know that $\displaystyle \alpha+\beta+\gamma=0$,
$\displaystyle \alpha\beta+\alpha\gamma+\beta\gamma=a$ and $\displaystyle \alpha\beta\gamma=-b$.
You can use those and multiply out what you have setup.
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# 3/13/2012Data Streams: Lecture 161 CS 410/510 Data Streams Lecture 16: Data-Stream Sampling: Basic Techniques and Results Kristin Tufte, David Maier.
## Presentation on theme: "3/13/2012Data Streams: Lecture 161 CS 410/510 Data Streams Lecture 16: Data-Stream Sampling: Basic Techniques and Results Kristin Tufte, David Maier."— Presentation transcript:
3/13/2012Data Streams: Lecture 161 CS 410/510 Data Streams Lecture 16: Data-Stream Sampling: Basic Techniques and Results Kristin Tufte, David Maier
3/13/2012 Data Streams: Lecture 16 2 Data Stream Sampling Sampling provides a synopsis of a data stream Sample can serve as input for Answering queries “statistical inference about the contents of the stream” “variety of analytical procedures” Focus on: obtaining a sample from the window (sample size « window size)
3/13/2012 Data Streams: Lecture 16 3 Windows Stationary Window Endpoints of window fixed (think relation) Sliding Window Endpoints of window move What we’ve been talking about More complex than stationary window because elements must be removed from sample when they expire from window
3/13/2012 Data Streams: Lecture 16 4 Simple Random Sampling (SRS) What is a “representative” sample? SRS for a sample of k elements from a window with n elements Every possible sample (of size k) is equally likely, that is has probability: 1/ Every element is equally likely to be in sample Stratified Sampling Divide window into disjoint segments (strata) SRS over each stratum Advantageous when stream elements close together in stream have similar values nknk ( )
3/13/2012 Data Streams: Lecture 16 5 Bernoulli Sampling Includes each element in the sample with probability q The sample size is not fixed, sample size is binomially distributed Probability that sample contains k elements is: Expected sample size is nq ( ) q k (1-q) n-k nknk
3/13/2012 Data Streams: Lecture 16 6 Binomial Distribution - Example Expected Sample Size = 20*0.5 = 10 Binomial Distribution (n=20, q=0.5) Probability Sample Size
3/13/2012 Data Streams: Lecture 16 7 Binomial Distribution - Example Expected Sample Size = 20*1/3 ≈ 6.667 Binomial Distribution (n=20, q=1/3) Probability Sample Size
3/13/2012 Data Streams: Lecture 16 8 Bernoulli Sampling - Implementation Naïve: Elements inserted with probability q (ignored with probability 1-q) Use a sequence of pseudorandom numbers (U 1, U 2, U 3, …) U i [0,1] Element e i is included if U i ≤ q e1e1 Sample: e2e2 e6e6 e5e5 e4e4 e3e3 U 1 =0.5U 2 =0.1 e2e2 e5e5 U 3 =0.9 e7e7 U 4 =0.8 U 5 =0.2U 6 =0.3 e7e7 U 7 =0.0 Example q = 0.2
3/13/2012 Data Streams: Lecture 16 9 Bernoulli Sampling – Efficient Implementation Calculate number of elements to be skipped after an insertion (Δ i ) Pr {Δ i = j} = q(1-q) j If you skip zero elements, must get: U i ≤ q (pr: q) Skip one element, must get: U i > q, U i+1 ≤ q (pr: (1-q)q) Skip two elements: U i > q, U i+1 > q, U i+2 ≤ q (pr: (1-q) 2 q) Δ i has a geometric distribution
3/13/2012 Data Streams: Lecture 16 10 Geometric Distribution - Example Geometric Distribution q = 0.2 Probability Number of Skips (Δ i )
3/13/2012 Data Streams: Lecture 16 11 Bernoulli Sampling - Algorithm
3/13/2012 Data Streams: Lecture 16 12 Bernoulli Sampling Straightforward, SRS, easy to implement But… Sample size is not fixed! Look at algorithms with deterministic sample size Reservoir Sampling Stratified Sampling Biased Sampling Schemes
3/13/2012 Data Streams: Lecture 16 13 Reservoir Sampling Produces a SRS of size k from a window of length n (k is specified) Initialize a “reservoir” using first k elements For every following element, insert with probability p i (ignore with probability 1-p i ) p i = k/i for i>k (p i = 1 for i ≤ k) p i changes as i increases Remove one element from reservoir before insertion
3/13/2012 Data Streams: Lecture 16 14 Reservoir Sampling e1e1 Reservoir Sample: e2e2 e6e6 e5e5 e4e4 e3e3 Sample size 3 (k=3) Recall: p i = 1 i≤k, p i = i/k i>k p 1 =1p 2 =1 e1e1 e2e2 p 3 =1 e3e3 p 4 =3/4 p 5 =3/5p 6 =3/6 e7e7 p 7 =3/7 e8e8 p 8 =3/8 U 4 =0.5U 5 =0.1 U 6 =0.9U 4 =0.8 U 5 =0.2 e4e4 e5e5 e8e8
3/13/2012 Data Streams: Lecture 16 15 Reservoir Sampling - SRS Why set p i = k/i? Want S j to be a SRS from U j = {e 1, e 2, …, e j } Sj is the sample from Uj Recall SRS means every sample of size k is equally likely Intuition: Probability that e i is included in SRS from U i is k/i k is sample size, i is “window” size k/i = (#samples containing e i )/(#samples of size k) = ( ) i-1 k-1 ( ) ikik
3/13/2012 Data Streams: Lecture 16 16 Reservoir Sampling - Observations Insertion probability (p i = k/i i>k) decreases as i increases Also, opportunities for an element in the sample to be removed from the sample decrease as i increases These trends offset each other Probability of being in final sample is same for all elements in the window
3/13/2012 Data Streams: Lecture 16 17 Other Sampling Schemes Stratified Sampling Divide window into strata, SRS in each stratum Deterministic & Semi-Deterministic Schemes i.e. Sample every 10 th element Biased Sampling Schemes Bias sample towards recently-received elements Biased Reservoir Sampling Biased Sampling by Halving
3/13/2012 Data Streams: Lecture 16 18 Stratified Sampling
3/13/2012 Data Streams: Lecture 16 19 Stratified Sampling When elements close to each other in window have similar values, algorithms such as reservoir sampling can have bad luck Alternative: divide window into strata and do SRS in each strata If you know there is a correlation between data values (i.e. timestamp) and position in stream, you may wish to use stratified sampling
3/13/2012 Data Streams: Lecture 16 20 Deterministic Semi-deterministic Schemes Produce sample of size k by inserting every n/k th element into the sample Simple, but not random Can’t make statistical conclusions about window from sample Bad if data is periodic Can be good if data exhibits a trend Ensures sampled elements are spread throughout the window e1e1 e2e2 e6e6 e5e5 e4e4 e3e3 e7e7 e9e9 e8e8 e 11 e 10 e 12 e 13 e 17 e 16 e 15 e 14 e 18 n=18, k=6
3/13/2012 Data Streams: Lecture 16 21 Biased Reservoir Sampling Recall: Reservoir sampling – probability of inclusion decreased as we got further into the window (p i = i/k) What if p i was constant? (p i = p) Alternative: p i decreases more slowly than i/k Will favor recently-arrived elements Recently-arrived elements are more likely to be in sample than long-ago-arrived elements
3/13/2012 Data Streams: Lecture 16 22 ( ) Biased Reservoir Sampling For reservoir sampling, Probability that e i is included in sample S: If p i is fixed, that is set p i = p (0,1) Probability that e i is in final sample increases geometrically as i increases Pr {e i S} = p i j=max(i, k)+1 n k-p j k Pr {e i S} = p n - max(i, k) k-p k
3/13/2012 Data Streams: Lecture 16 23 Biased Reservoir Sampling Probability e i is included in final sample, p=0.2, k=10, n=40 Element index (i) Probability ( ).2 40 - max(i, 10) 10-.2 10
3/13/2012 Data Streams: Lecture 16 24 kk Biased Sampling by Halving Break into strata (Λ i ), Sample of size 2k Step 1: S = unbiased SRS samples of size k from Λ 1 and Λ 2 (i.e. use reservoir sampling) Step 2: Sub-sample S to produce a sample of size k, insert SRS of size k from Λ 3 into S Λ1Λ1 Λ2Λ2 Λ3Λ3 Λ4Λ4 kk kk
3/13/2012 Data Streams: Lecture 16 25 Sampling from Sliding Windows Harder than sampling from stationary window Must remove elements from sample as the elements expire from the window Difficult to maintain a sample of a fixed size Window Types: Sequence-based windows - contain n most recent elements (row-based window) Timestamp-based windows - contains all elements that arrived within past t time units (time-based windows) Unbiased sampling from within a window
3/13/2012 Data Streams: Lecture 16 26 Sequence-based Windows W j is a window of length n, j ≥ 1 W j = {e j, e j+1, … e j+n-1 } Want a SRS S j of k elements from W j Tradeoff between amount of memory required and degree of dependence between S j ’s
3/13/2012 Data Streams: Lecture 16 27 Complete Resampling Window size = 5, Sample size = 2 Maintain full window (W j ) Each time window changes, use reservoir sampling to create S j from W j Very expensive – memory, CPU O(n) (n=window-size) e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9 e 10 e 11 e 12 e 13 e 14 e 15 W1W1 W2W2 S 1 = {e2, e4} S 2 = {e3, e5}
3/13/2012 Data Streams: Lecture 16 28 Passive Algorithm Window size = 5, sample size = 2 When an element in the sample expires, insert the newly-arrived element into sample S j is a SRS from W j S j ’s are highly correlated If S 1 is a bad sample, S 2 will be also… Memory is O(k), k = sample size e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9 e 10 e 11 e 12 e 13 e 14 e 15 W1W1 W2W2 S 1 = {e 2, e 4 }S 2 = {e 2, e 4 } W3W3 S 3 = {e 7, e 4 }
3/13/2012 Data Streams: Lecture 16 29 Chain Sampling (Babcock, et al.) Improved independence properties compared to passive algorithm Expected memory usage: O(k) Basic algorithm – maintains sample of size 1 Get sample of size k, by running k chain- samplers
3/13/2012 Data Streams: Lecture 16 30 Chain Sampling - Issue Behaves as reservoir sampler for first n elements Insert additional elements into sample with probability 1/n e1e1 Sample: e2e2 e5e5 e4e4 e3e3 e1e1 W1W1 p 1 =1 p 2 =1/2p 3 =1/3 p 4 =1/3 e2e2 W2W2 W3W3 Now, what do we do?
3/13/2012 Data Streams: Lecture 16 31 Chain Sampling - Solution When e i is selected for inclusion in sample, select K from {i+1, i+2, … i+n}, e K will replace e i if e i expires while part of sample S Know e k will be in window when e i expires e1e1 Sample: e2e2 e5e5 e4e4 e3e3 e1e1 W1W1 p 2 =1/2p 3 =1/3 p 4 =1/3 e2e2 W2W2 W3W3 Choose K {3, 4, 5}, K=5 e5e5 Choose K {6, 7, 8}, K=7 e7e7 e5e5 e7e7
3/13/2012 Data Streams: Lecture 16 32 Chain Sampling - Summary Expected memory consumptin O(k) Chain sampling produces a SRS with replacement for each sliding window If we use k chain-samplers to get a sample of size k, may get duplicates in that sample Can over sample (use sample size k + α), then sub-sample to get a sample of size k
3/13/2012 Data Streams: Lecture 16 33 Stratified Sampling Divide window into strata and do SRS in each strata
3/13/2012 Data Streams: Lecture 16 34 Stratified Sampling – Sliding Window e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9 e 10 e 11 e 12 e 13 e 14 e 15 W1W1 ss 1 = {e 1,e 2 } Window size = 12 (n), stratum size 4 (m), stratum sample size = 2 (k) W j overlaps between 3 and 4 strata (l, l+1 strata) l = win_size/stratum_size = n/m (=3) Paper says sample size is between k(l-1) and k∙l, think should be k(l-1) – k(l+1) ss 2 = {e 6,e 7 }ss 3 = {e 9,e 11 } e 16 ss 2 = {e 14,e 16 } W2W2 W3W3
3/13/2012 Data Streams: Lecture 16 35 Timestamp-Based Windows Number of elements in window changes over time Multiple elements in sample expire at once Chain sampling relies on insertion probability = 1/n (n is window size) Stratified Sampling – wouldn’t be able to bound sample size
3/13/2012 Data Streams: Lecture 16 36 Priority Sampling (Babcock, et al.) Priority Sampler maintains a SRS of size 1, use k priority samplers to get SRS of size k Assign random, uniformly-distributed priority (0,1) to each element Current sample is element in window with highest priority Keep elements for which there is no other element with both higher priority and higher (later) timestamp
3/13/2012 Data Streams: Lecture 16 37 Priority Sampling - Example Keep elements for which there is no element with: higher priority and higher (later) timestamp e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9 e 10 e 11 e 12 e 13 e 14 e 15 W1W1 W2W2 W3W3.1.8.3 priority:.4.7.1.3.5.2.6.4.1.5.3 elt in sample elt stored in mem elt in window, not stored
3/13/2012 Data Streams: Lecture 16 38 Inference From a Sample What do we do with these samples? SRS samples can be used to estimate “population sums” If each element e i is a sales transaction and v(e i ) is dollar value of transaction v(e i ) = total sales of transactions in W Count: h(e i ) = 1 if v(e i ) > \$1000, h(e i ) = number of transactions in window for > \$1000 Can also do average e i W
3/13/2012 Data Streams: Lecture 16 39 SRS Sampling To estimate a population sum from a SRS of size k, expansion estimator: To estimate average, use sample average: α = Θ/n = (1/k) h(e i ) ^ eiSeiS ^ eiSeiS Θ = (n/k) h(e i ) ^ Also works for Stratified Sampling
3/13/2012 Data Streams: Lecture 16 40 Estimating Different Results SRS sampling is good for estimating population sums, statistics But, use different algorithms for different results Heavy Hitters algorithm Find elements (values) that occur commonly in the stream Min-Hash Computation set resemblance
3/13/2012 Data Streams: Lecture 16 41 Heavy Hitters Goal: Find all stream elements that occur in at least a fraction s of all transactions For example, find sourceIPs that occur in at least 1% of network flows sourceIPs from which we are getting a lot of traffic
3/13/2012 Data Streams: Lecture 16 42 Heavy Hitters Divide window into buckets of width w Current bucket id = N/w , N is current stream length Data structure D : (e, f, Δ) e - element f – estimated frequency Δ – maximum possible error in f If we are looking for common sourceIPs in a network stream D : (sourceIP, f, Δ)
3/13/2012 Data Streams: Lecture 16 43 Heavy Hitters Data structure D : (e, f, Δ) New element e: Check if e exists in D If so, f = f+1 If not, new entry (e, 1, b current -1) At bucket boundary (when b current changes) Delete all elements (e, f, Δ) if f + Δ b current If only one instance of f in bucket, entry for f deleted Deleting items that occur once per bucket For threshold s, output items: f (s-ε)N (w = 1/ε ) (N is stream size)
3/13/2012 Data Streams: Lecture 16 44 Min-Hash Resemblance, ρ, of two sets A, B = Min-hash signature is a representation of a set from which one can estimate the resemblance of two sets ρ(A,B) = | A B | / | A B | Let h 1, h 2, … h n be hash functions s i (A) = min(h i (a) | a A) (minimum hash value of h i over A) Signature of A: S(A) = (s 1 (A), s 2 (A), …, s n (A))
3/13/2012 Data Streams: Lecture 16 45 Min-Hash Resemblance estimator: ρ(A,B) = I(s i (A), s i (B)) I(x,y) = 1 if x=y, 0 otherwise ρ(A,B) = | A B | / | A B | h 1, h 2, … h n hash functions s i (A) = min(h i (a) | a A) S(A) = (s 1 (A), s 2 (A), …, s n (A)) i=1 n Count # times min hash value is equal Can substitute N minimum values of one hash function for minimum values of N hash functions ^
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Prerequisite: Graph and its representations
Examples:
Input: arr[][] = [ [0, 0, 1], [0, 0, 1], [1, 1, 0] ]
0 -> 2
1 -> 2
2 -> 0 -> 1
Input: arr[][] = [ [0, 1, 0, 0, 1], [1, 0, 1, 1, 1], [0, 1, 0, 1, 0], [0, 1, 1, 0, 1], [1, 1, 0, 1, 0] ]
0 -> 1 -> 4
1 -> 0 -> 2 -> 3 -> 4
2 -> 1 -> 3
3 -> 1 -> 2 -> 4
4 -> 0 -> 1 -> 3
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Adjacency Matrix: Adjacency Matrix is a 2D array of size V x V where V is the number of vertices in a graph. Let the 2D array be adj[][], a slot adj[i][j] = 1 indicates that there is an edge from vertex i to vertex j.
Adjacency List: An array of lists is used. The size of the array is equal to the number of vertices. Let the array be array[]. An entry array[i] represents the list of vertices adjacent to the ith vertex.
To convert an adjacency matrix to the adjacency list. Create an array of lists and traverse the adjacency matrix. If for any cell (i, j) in the matrix “mat[i][j] = 1“, it means there is an edge from i to j, so insert j in the list at i-th position in the array of lists.
Below is the implementation of the above approach:
## C++
`#include ` `using` `namespace` `std; ` ` ` `// CPP program to convert Adjacency matrix ` `// representation to Adjacency List ` ` ` `// converts from adjacency matrix to adjacency list ` `vector> convert( vector> a) ` `{ ` ` ``vector> adjList(a.size()); ` ` ``for` `(``int` `i = 0; i < a.size(); i++) ` ` ``{ ` ` ` ` ``for` `(``int` `j = 0; j < a[i].size(); j++) ` ` ``{ ` ` ``if` `(a[i][j] == 1) ` ` ``{ ` ` ``adjList[i].push_back(j); ` ` ``} ` ` ``} ` ` ``} ` ` ``return` `adjList; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``vector> a; ` ` ``vector<``int``> p({0, 0, 1}); ` ` ``vector<``int``> q({0, 0, 1}); ` ` ``vector<``int``> r({1, 1, 0}); ``// adjacency matrix ` ` ``a.push_back(p); ` ` ``a.push_back(q); ` ` ``a.push_back(r); ` ` ``vector> AdjList = convert(a); ` ` ``cout<<``"Adjacency List:"``< "` `<< AdjList[i][j] << endl; ` ` ``break``; ` ` ``} ` ` ``else` ` ``cout << ``" -> "` `<< AdjList[i][j]; ` ` ``} ` ` ``} ` `} ` ` ` `// This code is contributed by Surendra_Gangwar `
## Java
`// Java program to convert adjacency ` `// matrix representation to ` `// adjacency list ` `import` `java.util.*; ` ` ` `public` `class` `GFG { ` ` ` ` ``// Function to convert adjacency ` ` ``// list to adjacency matrix ` ` ``static` `ArrayList> convert(``int``[][] a) ` ` ``{ ` ` ``// no of vertices ` ` ``int` `l = a[``0``].length; ` ` ``ArrayList> adjListArray ` ` ``= ``new` `ArrayList>(l); ` ` ` ` ``// Create a new list for each ` ` ``// vertex such that adjacent ` ` ``// nodes can be stored ` ` ``for` `(``int` `i = ``0``; i < l; i++) { ` ` ``adjListArray.add(``new` `ArrayList()); ` ` ``} ` ` ` ` ``int` `i, j; ` ` ``for` `(i = ``0``; i < a[``0``].length; i++) { ` ` ``for` `(j = ``0``; j < a.length; j++) { ` ` ``if` `(a[i][j] == ``1``) { ` ` ``adjListArray.get(i).add(j); ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return` `adjListArray; ` ` ``} ` ` ` ` ``// Function to print the adjacency list ` ` ``static` `void` `printArrayList(ArrayList> ` ` ``adjListArray) ` ` ``{ ` ` ``// Print the adjacency list ` ` ``for` `(``int` `v = ``0``; v < adjListArray.size(); v++) { ` ` ``System.out.print(v); ` ` ``for` `(Integer u : adjListArray.get(v)) { ` ` ``System.out.print(``" -> "` `+ u); ` ` ``} ` ` ``System.out.println(); ` ` ``} ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``// Given Adjacency Matrix ` ` ``int``[][] a = { { ``0``, ``0``, ``1` `}, ` ` ``{ ``0``, ``0``, ``1` `}, ` ` ``{ ``1``, ``1``, ``0` `} }; ` ` ` ` ``// function to convert adjacency ` ` ``// list to adjacency matrix ` ` ``ArrayList> adjListArray = convert(a); ` ` ``System.out.println(``"Adjacency List: "``); ` ` ` ` ``printArrayList(adjListArray); ` ` ``} ` `} `
## Python
`# Python program to convert Adjacency matrix ` `# representation to Adjacency List ` ` ` `from` `collections ``import` `defaultdict ` `# converts from adjacency matrix to adjacency list ` `def` `convert(a): ` ` ``adjList ``=` `defaultdict(``list``) ` ` ``for` `i ``in` `range``(``len``(a)): ` ` ``for` `j ``in` `range``(``len``(a[i])): ` ` ``if` `a[i][j]``=``=` `1``: ` ` ``adjList[i].append(j) ` ` ``return` `adjList ` ` ` `# driver code ` `a ``=``[[``0``, ``0``, ``1``], [``0``, ``0``, ``1``], [``1``, ``1``, ``0``]] ``# adjacency matrix ` `AdjList ``=` `convert(a) ` `print``(``"Adjacency List:"``) ` `# print the adjacency list ` `for` `i ``in` `AdjList: ` ` ``print``(i, end ``=``"") ` ` ``for` `j ``in` `AdjList[i]: ` ` ``print``(``" -> {}"``.``format``(j), end ``=``"") ` ` ``print``() ` ` ` `# This code is contributed by Muskan Kalra. `
## C#
`// C# program to convert adjacency ` `// matrix representation to ` `// adjacency list ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Function to convert adjacency ` ` ``// list to adjacency matrix ` ` ``static` `List> convert(``int``[,] a) ` ` ``{ ` ` ``// no of vertices ` ` ``int` `l = a.GetLength(0); ` ` ``List> adjListArray = ``new` `List>(l); ` ` ``int` `i, j; ` ` ` ` ``// Create a new list for each ` ` ``// vertex such that adjacent ` ` ``// nodes can be stored ` ` ``for` `(i = 0; i < l; i++) ` ` ``{ ` ` ``adjListArray.Add(``new` `List<``int``>()); ` ` ``} ` ` ` ` ` ` ``for` `(i = 0; i < a.GetLength(0); i++) ` ` ``{ ` ` ``for` `(j = 0; j < a.GetLength(1); j++) ` ` ``{ ` ` ``if` `(a[i,j] == 1) ` ` ``{ ` ` ``adjListArray[i].Add(j); ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return` `adjListArray; ` ` ``} ` ` ` ` ``// Function to print the adjacency list ` ` ``static` `void` `printList(List> adjListArray) ` ` ``{ ` ` ` ` ``// Print the adjacency list ` ` ``for` `(``int` `v = 0; v < adjListArray.Count; v++) ` ` ``{ ` ` ``Console.Write(v); ` ` ``foreach` `(``int` `u ``in` `adjListArray[v]) ` ` ``{ ` ` ``Console.Write(``" -> "` `+ u); ` ` ``} ` ` ``Console.WriteLine(); ` ` ``} ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `Main(String[] args) ` ` ``{ ` ` ` ` ``// Given Adjacency Matrix ` ` ``int``[,] a = { { 0, 0, 1 }, { 0, 0, 1 }, { 1, 1, 0 } }; ` ` ` ` ``// function to convert adjacency ` ` ``// list to adjacency matrix ` ` ``List> adjListArray = convert(a); ` ` ``Console.WriteLine(``"Adjacency List: "``); ` ` ` ` ``printList(adjListArray); ` ` ``} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
Output:
```Adjacency List:
0 -> 2
1 -> 2
2 -> 0 -> 1
```
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Lesson Objectives
• Demonstrate an understanding of how to solve a system of linear equations in two variables using graphing
• Demonstrate an understanding of how to solve a system of linear equations in two variables using substitution
• Demonstrate an understanding of the addition property of equality
• Learn how to solve a system of linear equations in two variables using elimination
• Learn how to identify a system of linear equations in two variables with no solution
• Learn how to identify a system of linear equations in two variables with infinitely many solutions
## How to Solve a Linear System using Elimination
So far, we have learned how to solve a system of linear equations in two variables using graphing and substitution. In this lesson, we will learn another method known as "elimination". The goal of the elimination method is to eliminate one of the variables and obtain a linear equation in one variable. This allows us to get a solution for one of the variables. We can then use substitution to find the other unknown. The elimination method works best when one pair of variable terms are opposites, this means they will have opposite coefficients.
### Solving a Linear System using the Elimination Method
• Place both equations in standard form
• ax + by = c
• Transform one or both equations in such a way that one pair of variable terms are opposites
• This means the variables will have opposite coefficients
• Add the left sides of the equations and set this equal to the sum of the right sides, the result is a linear equation in one variable
• This is legal since we are adding equal quantities to both sides of an equation
• If a = b and c = d, then a + c = b + d
• Solve the equation and find the other unknown using substitution
• Check
Plug in for x and y in both original equations
Let's look at a few examples.
Example 1: Solve each linear system using elimination
-5x + 3y = 18
5x - 2y = -12
Step 1) Place both equations in standard form.
In this case, both equations are in standard form. We will label our equations as 1 and 2.
1) -5x + 3y = 18
2) 5x - 2y = -12
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we have -5x and 5x, which are opposites.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-5x + 3y = -12$$ $$\underline{+5x - 2y = -12}$$ The -5x and 5x will cancel: $$\require{cancel}\cancel{-5x} + 3y = 18$$ $$\underline{\cancel{+5x} - 2y = -12}$$ $$\hspace{4em}y = -24$$ On the left, we are left with (3y) + (-2y), which is just (y). On the right, we have (18) + (-12), which is (6).
Step 4) Solve the equation and find the other unknown.
In this case, we already know that y is 6. Let's plug this in for y in one of the original equations. Let's choose to plug in for y in equation 1.
-5x + 3y = 18
-5x + 3(6) = 18
-5x + 18 = 18
-5x = 0
x = 0
Our solution is the ordered pair (0,6).
Step 5) Check.
-5x + 3y = 18
-5(0) + 3(6) = 18
18 = 18
5x - 2y = -12
5(0) - 2(6) = -12
-12 = -12
Example 2: Solve each linear system using elimination
4x + 10y = -22
5y = 5x - 25
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 4x + 10y = -22
2) -5x + 5y = -25
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we can multiply both sides of equation 2 by (-2). This will give us (10y) in equation 1 and (-10y) in equation 2.
2) 10x - 10y = 50
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$\hspace{.5em}4x + 10y = -22$$ $$\underline{10x - 10y = 50}$$ The 10y and (-10y) will cancel: $$\hspace{.5em}4x + \cancel{10y} = -22$$ $$\underline{10x - \cancel{10y} = 50}$$ $$14x \hspace{3em} = 28$$ On the left, we are left with (10x) + (6x), which gives us (14x). On the right, we have (-22) + (50), which is (28).
This leaves us with the equation:
14x = 28
Step 4) Solve the equation and find the other unknown.
14x = 28
x = 2
Let's plug a (2) in for x in equation 1.
4x + 10y = -22
4(2) + 10y = -22
(8) + 10y = -22
10y = -22 - 8
10y = -30
y = -3
Our solution is the ordered pair (2,-3).
Step 5) Check.
4x + 10y = -22
4(2) + 10(-3) = -22
8 + (-30) = -22
-22 = -22
5y = 5x - 25
5(-3) = 5(2) - 25
-15 = 10 - 25
-15 = -15
Example 3: Solve each linear system using elimination
-9y = 2x - 16
-7x = 22 + 12y
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) -2x - 9y = -16
2) -7x - 12y = 22
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (7) and equation 2 by (-2). This will give us (-14x) in equation 1 and (14x) in equation 2.
1) -14x - 63y = -112
2) 14x + 24y = -44
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-14x - 63y = -112$$ $$\underline{+14x + 24y = -44}$$ The -14x and 14x will cancel: $$\cancel{-14x} - 63y = -112$$ $$\underline{\cancel{+14x} + 24y = -44}$$ $$\hspace{3.2em}-39y = -156$$ On the left, we are left with (-63y) + (24y), which is (-39y). On the right, we have (-112) + (-44), which is (-156).
This leaves us with the equation:
-39y = -156
Step 4) Solve the equation and find the other unknown.
-39y = -156
y = 4
Let's plug a 4 in for y in equation 1.
-2x - 9y = -16
-2x - 9(4) = -16
-2x - 36 = -16
-2x = 20
x = -10
Our solution is the ordered pair (-10,4).
Step 5) Check.
-9y = 2x - 16
-9(4) = 2(-10) - 16
-36 = -20 - 16
-36 = -36
-7x = 22 + 12y
-7(-10) = 22 + 12(4)
70 = 22 + 48
70 = 70
### Special Case Linear Systems
At this point, we have run across special case scenarios while using graphing and substitution methods. The majority of our problems with linear systems will have exactly one solution. This type of system is said to be "consistent", with equations that are "independent". When we have a system which contains two parallel lines, there will not be a solution. This type of system is said to be "inconsistent". Additionally, we will see systems that contain the same equation. These equations are said to be "dependent". For this type of system, there are infinitely many solutions.
### Linear Systems with No Solution
When using the elimination method, if both are variables are eliminated and we are left with a false statement, we know we have an inconsistent system. Our answer will be stated as "no solution" or ∅. Let's look at an example.
Example 4: Solve each linear system using elimination
16 = 55x + 10y
-22x - 4y = -4
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 55x + 10y = 16
2) -22x - 4y = -4
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (2) and equation 2 by (5). This will give us (20y) in equation 1 and (-20y) in equation 2.
1) 110x + 20y = 32
2) -110x - 20y = -20
We should be able to see a problem. Both variables are set up such that they have opposite coefficients. If we add the left sides together and set this equal to the sum of the right sides, we will get a false statement.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+110x + 20y = 32$$ $$\underline{-110x - 20y = -20}$$ The 110x and (-110x) will cancel and the 20y and (-20y) will also cancel: $$\cancel{+110x} + \cancel{+20y} = 32$$ $$\cancel{-110x} + \cancel{-20y} = -20$$ $$\hspace{6.65em}0 = 12$$ On the left, we are left with (0). On the right, we have (32) + (-20), which is (12).
0 = 12 (false)
We can stop and report our answer as "no solution".
### Solving Linear Systems with Infinitely Many Solutions
When using the elimination method, if both are variables are eliminated and we are left with a true statement, we know we have dependent equations. Our answer will be stated as "infinitely many solutions". Let's look at an example.
Example 5: Solve each linear system using elimination
15y + 3 = -24x
-150y - 30 = 240x
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 24x + 15y = -3
2) -240x - 150y = 30
At this point, we should be able to tell that multiplying equation 1 by (-10) will give us equation 2. Therefore, both equations of the system are the same and we will have infinitely many solutions. To show the full process, let's continue with the steps.
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (10). This will give us (240x) and (-240x).
1) 240x + 150y = -30
2) -240x - 150y = 30
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+240x + 150y = -30$$ $$\underline{-240x - 150y = 30}$$ The 240x and (-240x) will cancel and the 150y and (-150y) will also cancel: $$\cancel{+240x} + \cancel{+150y} = -30$$ $$\cancel{-240x} + \cancel{-150y} = 30$$ $$\hspace{7.25em}0 = 0$$ On the left, we are left with (0). On the right, we have (30) + (-30), which is (0).
0 = 0 (true)
We can stop and report our answer as "infinitely many solutions".
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Mathematics for Teaching Embedding the idea of functions in geometry lessons
## Embedding the idea of functions in geometry lessons
GeoGebra is a great tool to promote a way of thinking and reasoning about shapes. It provides an environment where students can observe and describe the relationships within and among geometric shapes, analyze what changes and what stays the same when shapes are transformed, and make generalizations.
When shapes or objects are transformed or moved, their properties such as location, length, angles, perimeters, and area changes. These properties are quantifiable and may vary with each other. It is therefore possible to design a lesson with GeoGebra which can be used to teach geometry concepts and the concepts of variables and functions. Noticing varying quantities is a pre-requisite skill towards understanding function and using it to model real life situations. Noticing varying quantities is as important as pattern recognition. Below is an example of such activity. I created this worksheet to model the movement of the structure of a collapsible chair which I describe in this Collapsible chair model.
Show angle CFB then move C. Express angle CFB in terms of δ, the measure of FCB. Show the next angle EFB then move C. Express EFB in terms of δ. Do the same for angle FBG.
[iframe https://math4teaching.com/wp-content/uploads/2011/07/locus_and_function.html 700 400]
Because CFB depends on FCB, the measure of CFB is a function of δ. That is f(δ) = 180-2δ. Note that the triangle formed is isosceles. Likewise, the measure of angle EFB is a function of δ. We can write this as g(δ) = 2δ. Let h be the function that defines the relationship between FCB and FBG. So, h(δ)=180-δ. Of course you would want the students to graph the function. Don’t forget to talk about domain and range. You may also ask students to find a function that relates f and g.
For the geometry use of this worksheet, read the post Problems about Perpendicular Segments. Note that you can also use this to help the students learn about exterior angle theorem.
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# Why are patterns and structure important in early math?
In mathematics, patterns are more than a beautiful design (though they are often that too), patterns follow a predictable rule and that rule allows us to predict what will come next. Mathematicians say that mathematics is the study of pattern—of patterns and structure in numbers, and patterns and structure in geometry. Seeing pattern and structure in the world around us is a key mathematical habit of mind and one that children are developing from the first days of life. Children are naturally attuned to patterns because it allows them to predict what will come next and make sense of their world. When we see patterns we are able to predict—to count on things happening—and feel more secure and confident. Noticing these routines and patterns in everyday life helps prepare children to notice other patterns. Many stories, dances, and chants follow a predictable pattern. “Five Little Monkeys” follow a pattern where the words repeat but the number of monkeys decreases by one each time. “Head, Shoulders, Knees, and Toes” follows a pattern that speeds up as you go. Dancing or movement patterns such as clap, clap, stomp, clap, clap, stomp also help children build an understanding of pattern that includes the kinesthetic. As you engage children in these everyday activities, help them notice the pattern and describe it in words.
### Why are patterns and structure important in early math? Sub-Topics
Repeating patterns are the ones we tend to think of first when we think of patterns. The stripes in the American flag are a repeating pattern: red, white, red, white, red, white. The repeating part, or unit, for the stripes is red, white. We can label this an AB pattern, where red is A and white Is B. The stripe pattern is ABABAB. Children may begin to understand that patterns are made up of repeating units, but it may take more time for them to be able to consistently identify the repeating unit or to create their own stable patterns. With time and experience, children will be able to see the underlying mathematical structure in patterns and can use symbols to represent the structure of the pattern. They can see that the pattern with fork, fork, spoon, spoon, fork, fork, spoon, spoon is related to the pattern blue, blue, yellow, yellow, blue, blue, yellow, yellow because they can both be generalized as AABB patterns.
Growing patterns keep increasing or decreasing by the same amount. In the example, the block towers increase in height by one block each time. It is a plus-one growing pattern. Our number system is a plus-one growing pattern too: 1, 2, 3, 4, 5. Each unit grows bigger by one.
Symmetrical patterns have segments that repeat but instead of repeating in a line, the segments are the same when flipped, folded, or rotated. Butterflies have mirror symmetry—the butterfly wings match when folded along a line through the middle of the butterfly. Snowflakes have mirror symmetry and rotational symmetry. The segments of a snowflake match when folded and the design looks the same when you turn or rotate it. There is symmetry in artwork, in buildings, in nature, and even in people and animals (our bodies are symmetrical if you draw a line down the middle—two ears, two arms, two legs, etc.)
Concentric patterns have circles or rings that grow from a common center like the ripples in the water or the circles that surround the bullseye on archery target.
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Search a number
11709 = 321301
BaseRepresentation
bin10110110111101
3121001200
42312331
5333314
6130113
746065
oct26675
917050
1011709
118885
126939
135439
1443a5
153709
hex2dbd
11709 has 6 divisors (see below), whose sum is σ = 16926. Its totient is φ = 7800.
The previous prime is 11701. The next prime is 11717. The reversal of 11709 is 90711.
It is an interprime number because it is at equal distance from previous prime (11701) and next prime (11717).
It can be written as a sum of positive squares in only one way, i.e., 6084 + 5625 = 78^2 + 75^2 .
It is not a de Polignac number, because 11709 - 23 = 11701 is a prime.
It is a fibodiv number, since the Fibonacci-like sequence with seeds 117 and 9 contains 11709 itself.
It is a nialpdrome in base 11.
It is a junction number, because it is equal to n+sod(n) for n = 11691 and 11700.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (11701) by changing a digit.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 642 + ... + 659.
It is an arithmetic number, because the mean of its divisors is an integer number (2821).
211709 is an apocalyptic number.
It is an amenable number.
11709 is a deficient number, since it is larger than the sum of its proper divisors (5217).
11709 is a wasteful number, since it uses less digits than its factorization.
11709 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1307 (or 1304 counting only the distinct ones).
The product of its (nonzero) digits is 63, while the sum is 18.
The square root of 11709 is about 108.2081327812. The cubic root of 11709 is about 22.7077063112.
The spelling of 11709 in words is "eleven thousand, seven hundred nine".
Divisors: 1 3 9 1301 3903 11709
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# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 156: 52
$\mu_k = 0.12$
#### Work Step by Step
Since the sled is moving at a constant speed, the net force on the sled is zero. Therefore, the force of kinetic friction is equal in magnitude to the horizontal component of the applied force. So; $F_f = (75~N)~cos(30^{\circ})$ $F_N~\mu_k = (75~N)~cos(30^{\circ})$ $[mg-(75~N)~sin(30^{\circ})]~\mu_k = (75~N)~cos(30^{\circ})$ $\mu_k = \frac{(75~N)~cos(30^{\circ})}{mg-(75~N)~sin(30^{\circ})}$ $\mu_k = \frac{(75~N)~cos(30^{\circ})}{(60~kg)(9.80~m/s^2)-(75~N)~sin(30^{\circ})}$ $\mu_k = 0.12$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2
Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2
Question 1.
Mention two branches of statistical inference?
Solution:
(i) Estimation (ii) Testing of Hypothesis
Question 2.
What is an estimator?
An estimator is a statistic that is used to infer the value of an unknown population parameter in a statistical model. The estimator is a function of the data arid so it is also a random variable.
Question 3.
What is an estimate?
Solution:
Any specific numerical value of the estimator is called an estimate. For example, sample means are used to estimate population means.
Question 4.
What is point estimation?
Solution:
When a single value is an estimate, the estimate is called a point estimate of the population parameter. In other words, an estimate of a population parameter given by a single number is called as point estimation.
Question 5.
What is interval estimation?
Solution:
Interval estimation is the use of sample data to calculate an interval of possible values of an unknown population parameter. For example the interval estimate for the population mean is (101.01, 102.63).This gives a range within which the population mean is most likely to be located.
Question 6.
What is confidence interval?
Solution:
A confidence interval L a type of interval estimate, computed from the statistics of the observed data, that might contain the true value of an unknown population parameter. The numbers at the upper and lower end of a confidence interval are called confidence limits. For example, if mean is 7.4 with confidence interval (5.4, 9.4), then the numbers 5.4 and 9.4 are the confidence limits.
Question 7.
What is null hypothesis? Give an example.
Solution:
Question 8.
Define the alternative hypothesis.
Solution:
The alternative hypothesis is the hypothesis that is contrary to the null hypothesis and it is denoted by H1.
For example if H1 : µ = 15, then the alternative hypothesis will be : H1 : µ ≠ 15, (or) H1 : µ < 15 (or) H1 : µ > 15.
Question 9.
Define the critical region.
Solution:
The critical region is the region of values that corresponds to the rejection of the null hypothesis at some chosen probability level. For the two-tailed test, the critical region is given below.
where α is the level of significance.
Question 10.
Define critical value.
Solution:
A critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis. It depends on the level of significance. For example, if the confidence level is 90% then the critical value is 1.645.
Question 11.
Define the level of significance
Solution:
The probability of type 1 error is known as level. of significance and it is denoted by The level of significance is usually employed in testing of hypothesis are 5% and 1%. The level of significance is always fixed in advanced before collecting the sample information.
Question 12.
What is a type I error?
Solution:
In statistical hypothesis testing, a Type f error is the rejection of a true null hypothesis. Example of Type I errors includes a test that shows a patient to have a disease when he does not have the disease, a fire alarm going on indicating a fire when there is no fire (or) an experiment indicating that medical treatment should cure a disease when in fact it does not.
Question 13.
What is the single-tailed test?
Solution:
A single-tailed test or a one-tailed test is a statistical test in which the critical area of a distribution is one-sided so that it is either greater than or less than a certain value, but not both. For the null hypothesis H0 : µ = 16.91, the alternative hypothesis H1 : µ > 16.91 or H1 : µ < 16.91 are one-tailed tests.
Question 14.
A sample of 100 items, draw from a universe with mean value 4 and S.D 3, has a mean value 3.5. Is the difference in the mean significant?
Solution:
Given Sample size n = 100
Sample mean $$\bar{x}$$ = 3.5
Population mean µ = 4
Population SD σ = 3
Now, null hypothesis H0 : µ = 4
Alternative hypothesis H1 : µ ≠ 4 (Two tail)
We take level of significance α = 5% = 0.05
The table value $$\mathrm{Z}_{\alpha / 2}$$ = 1.96
Test statistic
Since the alternative hypothesis is of the two-tailed test we can take |Z| = 1.667. We observe that 1.667 < 1.96 (i.e) |Z| < $$\mathrm{Z}_{\alpha / 2}$$. So at 5% level of significance, the null hypothesis H0 is accepted. Therefore, we conclude that there is no significant difference between the sample mean and the population mean.
Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with a mean height of 67.39 inches and standard deviation of 1.30 inches?
Solution:
Given Sample size n = 400
Sample mean $$\bar{x}$$ = 67.47
Population mean µ = 67.39
Population SD σ = 1.3
Null hypothesis H0 : µ = 67.39 inches
(the sample has been drawn from the population with mean heights 67.39 inches)
Alternative hypothesis H1 : µ ≠ 67.39 inches
(the sample has not been drawn from the population with mean height 67.39 inches)
The level of significance α = 5% = 0.05
Test statistic
The significant value or table value $$\mathrm{Z}_{\alpha / 2}$$ = 1.96. We see that 1.2308 < 1.96 (i.e) Z < $$\mathrm{Z}_{\alpha / 2}$$. Since the calculated value is less than the table value at 5% level of significance, the null hypothesis is accepted. Hence we conclude that the data does not provide us with any evidence against the null hypothesis. Thus, the sample has been drawn from a large population with a mean height of 67.39 inches and S.D 1.3 inches.
Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
Given Population mean µ = 76
Population SD σ = 8
Sample size n = 100
Sample mean $$\bar{x}$$ = 72
Significance level α = 0.05
Null hypothesis H0 : µ = 76
(i.e) there is no difference between the state scores and the national scores.
Alternative hypothesis H1 : µ ≠ 76
(i.e) there is a significant difference between the state scores and the nationals scores of the aptitude test.
Test statistic
The significant value or table value $$\mathrm{Z}_{\alpha / 2}$$ = 1.96. Comparing the calculated value and table value, we find that |Z| > $$\mathrm{Z}_{\alpha / 2}$$ (i.e) 5 > 1.96. So the null hypothesis is rejected and we accept the alternative hypothesis. So we conclude that at the significance level of 5%, there is a difference between the state scores and the national scores of the nationally administered amplitude test.
Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance?
Solution:
Given Population mean µ = 1800
Population SD σ = 100
Sample size n = 50
Sample mean $$\bar{x}$$ = 1850
Significance level α = 0.01
Null hypothesis H0 : µ = 1800
(i.e) the breaking strength of the cables has not increased, after the new technique in the manufacturing process.
Alternative hypothesis H1 : µ > 1800 (i.e) the new technique was successful.
Test statistic
The table value for the one-tailed test is Zα = 2.33.
Comparing the calculated value and table value, we find that Z > Zα (i.e.) 3.536 > 2.33.
Inference: Since the calculated value is greater than the table value at 1 % level of significance, the null hypothesis is rejected and we accept the alternative hypothesis. We conclude that by the new technique in the manufacturing process the breaking strength of the cables is increased. So the claim is supported at 0.01 level of significance.
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March 20, 2023
### How to convert Celsius to Fahrenheit?
The formula to convert Celsius to Fahrenheit is:
F = (9/5) * C + 32
where F is the temperature in Fahrenheit and C is the temperature in Celsius.
To use this formula, simply substitute the desired values for C and F. For example, if you want to convert 20 degrees Celsius to Fahrenheit, you would plug 20 into the formula like this:
F = (9/5) * 20 + 32
F = 36 + 32
F = 68 degrees Fahrenheit
To convert from Fahrenheit to Celsius, simply reverse the formula. So if you want to convert 68 degrees Fahrenheit to Celsius, you would plug 68 into the formula like this:
C = (9/5) * F – 32
C = (9/5) * 68 – 32
C = (9/5) * 68 – 32
C = (9/5) * 36 – 32
C = 20 degrees Celsius.
To perform this conversion using a calculator, you would simply use the formula that corresponds to the unit you want to convert from. This is because most calculators only have one button for converting between units of measurement (usually Celsius to Fahrenheit).
### what is 180 Celsius to Fahrenheit?
180 degrees Celsius is equivalent to 352 degrees Fahrenheit. This means that the temperature on both scales is equal, and it can be used interchangeably when measuring temperature. This can be useful for a variety of reasons, particularly when cooking or baking. To convert Celsius to Fahrenheit, you need to divide the number in Celsius by 9 and then multiply this number by 5. This will give you the desired Fahrenheit temperature.
### what is 36 Celsius in Fahrenheit ?
36 Celsius is equal to 96.8 Fahrenheit. To convert from Celsius to Fahrenheit, multiply by 1.8 and add 32.
### what is 18 Celsius in Fahrenheit ?
18 Celsius is equal to 64.4 Fahrenheit. To convert from Celsius to Fahrenheit, multiply by 1.8 and add 32.0.
### what is 180 Celsius in Fahrenheit?
To convert 180 Celsius to Fahrenheit, simply multiply by 1.8 and add 32. This gives you a result of 354 degrees Fahrenheit.
### what is 20 degrees Celsius in Fahrenheit?
To convert 20 degrees Celsius to Fahrenheit, simply multiply 20 by 1.8 and add 32. This gives you a final answer of 68 degrees
28 C = 82.4 F
40 C = 104 F
200 C = 392 F
30 C = 86 F
36.6 C = 98.1 F
36.7 C= 98.3 F
37 C = 99 F
38 C = 100.4 F
9 C = 48.2 F
25 C = 77 F
30 C = 86 F
35 C = 95 F
36.5 C = 97.7 F
36.9 C= 98.4 F
50 C = 122 F
10 C= 50 F
2 C= 35.6 F
22 C= 71.6 F
25 C= 77 F
### Convert the following from Celsius in Fahrenheit:
The celsius temperature scale is based on the freezing and boiling points of water, with 0 degrees representing the freezing point and 100 degrees representing the boiling point. To convert a celsius temperature to fahrenheit, multiply the celsius temperature by 1.8 and add 32 degrees.
15 C = (15 * 1.8) + 32 = 59 F
32 C = (32 * 1.8) + 32 = 89.6 F
37 C = (37 * 1.8) + 32 = 98.6 F
37.2 C = (37.2 * 1.8) + 32 = 99 F
37.5 C = (37.5 * 1.8) + 32 = 99.7 F
70 C = (70 * 1.8) + 32 = 158 F
40 C = (40 * 1.8) + 32 = 104 F
45 C = (45 * 1.8) + 32 = 113 F
80 C = (80 * 1.8) + 32 = 176 F
Happy
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DiscriminantSequence - Maple Help
RegularChains[ParametricSystemTools]
DiscriminantSequence
Compute the discriminant sequence of a polynomial
Calling Sequence DiscriminantSequence(p, v, R) DiscriminantSequence(p, q, v, R)
Parameters
R - polynomial ring p - polynomial of R q - polynomial of R v - variable of R
Description
• When input is only one polynomial p, the result of this function call is the list of polynomials in R which is the discriminant sequence of p regarded as a univariate polynomial in v; otherwise the discriminant sequence of p and q.
• For a univariate polynomial p of degree n, its discriminant sequence is a list of n polynomials in the coefficients of p. The signs of these polynomials determine the number of distinct complex (real) zeros of p. The discriminant sequence of two polynomials p and q, together with the discriminant sequence of p, can help determining the number of distinct real roots of p=0 such that q>0 or q<0. For the details, please see the reference listed below.
Examples
> $\mathrm{with}\left(\mathrm{RegularChains}\right):$
> $\mathrm{with}\left(\mathrm{ParametricSystemTools}\right):$
> $R≔\mathrm{PolynomialRing}\left(\left[x,y,t\right]\right)$
${R}{≔}{\mathrm{polynomial_ring}}$ (1)
> $p≔{x}^{2}+tx+y$
${p}{≔}{t}{}{x}{+}{{x}}^{{2}}{+}{y}$ (2)
> $q≔y{x}^{2}+ty$
${q}{≔}{y}{}{{x}}^{{2}}{+}{t}{}{y}$ (3)
> $\mathrm{lp1}≔\mathrm{DiscriminantSequence}\left(p,x,R\right)$
${\mathrm{lp1}}{≔}\left[{1}{,}{{t}}^{{2}}{-}{4}{}{y}\right]$ (4)
> $\mathrm{lp2}≔\mathrm{DiscriminantSequence}\left(p,q,x,R\right)$
${\mathrm{lp2}}{≔}\left[{1}{,}{y}{,}{-}{{t}}^{{2}}{}{{y}}^{{2}}{-}{2}{}{t}{}{{y}}^{{2}}{+}{2}{}{{y}}^{{3}}{,}{{t}}^{{5}}{}{{y}}^{{3}}{+}{{t}}^{{4}}{}{{y}}^{{3}}{-}{6}{}{{t}}^{{3}}{}{{y}}^{{4}}{+}{{t}}^{{2}}{}{{y}}^{{5}}{-}{4}{}{{t}}^{{2}}{}{{y}}^{{4}}{+}{8}{}{t}{}{{y}}^{{5}}{-}{4}{}{{y}}^{{6}}\right]$ (5)
References
Yang, L., "Recent advances in determining the number of real roots of parametric polynomials", J. Symb. Compt. vol. 28, pp. 225--242, 1999.
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# matlab
### Modern control: Solutions & state transition matrices
Monday, Sep 24, 2018 | 2 min read
Categories: Engineering,
Tags: Matlab, Control Systems,
The state equation for a linear time-invariant system: $$x’(t) = A x(t) + B u(t)$$ Can be solved for $x(t)$. We collect all terms in $x$: $$x’(t) - A x(t) = B u(t)$$ Multiply equation by $e^{-At}$ $$x’(t) e^{-At} - A x(t) e^{-At} = B u(t) e^{-At}$$ Using product rule $d(f;g) = f;dg + g;df$, where: To give: $$\frac{d}{dt} (e^{-At} x(t)) = B u(t) e^{-At}$$
### Modern control: State space equations
Monday, Sep 24, 2018 | 3 min read
Categories: Engineering,
Tags: Matlab, Control Systems,
In modern control approaches, systems are analyzed in time domain as a set of differential equations. Higher order differential equations are decomposed into sets of first order equations of state variables that represent the system internally. This produces three sets of variables: Input variables are stimuli given to the system. Denoted by $u$. Output variables are the result of the current system state and inputs. Denoted by $y$. State variables represent the internal state of a system which may be obscured in the output variables.
### Classical control: Transfer functions
Friday, Sep 21, 2018 | 4 min read
Categories: Engineering,
Tags: Matlab, Control Systems,
A transfer function relates the output of a system to its input when it is represented in the Laplace domain. An assumption is made that initial steady-state response is 0. If $Y(s)$ is the output of a system, $X(s)$ is the input, then the transfer function is: $$H(s) = \frac{Y(s)}{X(s)}$$ Example - A Car A car as a system: The input is the acceleration. The output is the total distance travelled.
### Control Systems: Overview
Monday, Sep 10, 2018 | 4 min read
Categories: Engineering,
Tags: Matlab,
A primer for classical control theory.
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https://www.aepochadvisors.com/covar/
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# COVAR
In this comprehensive guide, we will explore the COVAR formula in Excel, which is used to calculate the covariance between two sets of data. Covariance is a statistical measure that helps determine the degree to which two variables change together. A positive covariance indicates that the variables tend to increase or decrease together, while a negative covariance indicates that one variable tends to increase when the other decreases. Understanding covariance can be useful in various fields, such as finance, where it can help in portfolio optimization and risk management.
## COVAR Syntax
The syntax for the COVAR formula in Excel is as follows:
=COVAR(array1, array2)
Where:
• array1 is the first set of data points (required).
• array2 is the second set of data points (required).
Both arrays must have the same number of data points, and they should be numeric values. The COVAR formula will return the covariance between the two sets of data points.
## COVAR Examples
Let’s look at some examples of using the COVAR formula in Excel:
Example 1: Suppose we have two sets of data points representing the monthly returns of two stocks, A and B. We can use the COVAR formula to calculate the covariance between the returns of these two stocks:
=COVAR(A2:A13, B2:B13)
In this example, the formula calculates the covariance between the monthly returns of stock A (in cells A2 to A13) and stock B (in cells B2 to B13).
Example 2: If we have data on the daily temperatures and ice cream sales for a month, we can use the COVAR formula to determine the covariance between temperature and ice cream sales:
=COVAR(C2:C32, D2:D32)
This example calculates the covariance between daily temperatures (in cells C2 to C32) and daily ice cream sales (in cells D2 to D32). A positive covariance would indicate that higher temperatures are associated with higher ice cream sales, while a negative covariance would suggest the opposite.
## COVAR Tips & Tricks
Here are some tips and tricks to help you get the most out of the COVAR formula in Excel:
1. Remember that covariance is a measure of the degree to which two variables change together, but it does not indicate the strength of the relationship. To measure the strength of the relationship between two variables, consider using the correlation coefficient, which can be calculated using the CORREL or PEARSON formula in Excel.
2. If you need to calculate the covariance matrix for multiple variables, consider using the Data Analysis ToolPak add-in, which includes a covariance matrix tool.
3. Keep in mind that the COVAR formula is sensitive to the scale of the data. If the data points in your arrays have different units or scales, consider standardizing the data before calculating covariance.
## Common Mistakes When Using COVAR
Here are some common mistakes to avoid when using the COVAR formula in Excel:
1. Using arrays with different numbers of data points: Both arrays must have the same number of data points for the COVAR formula to work correctly. If the arrays have different numbers of data points, you may receive an error or incorrect results.
2. Using non-numeric data: The COVAR formula requires numeric data points. If your arrays contain non-numeric data, such as text or error values, the formula may return an error or incorrect results.
3. Interpreting covariance as correlation: While covariance measures the degree to which two variables change together, it does not indicate the strength of the relationship. Be cautious when interpreting the results of the COVAR formula, and consider using the correlation coefficient to measure the strength of the relationship between two variables.
## Why Isn’t My COVAR Working?
If you’re having trouble with the COVAR formula in Excel, consider the following troubleshooting tips:
1. Check that both arrays have the same number of data points: If the arrays have different numbers of data points, the COVAR formula may return an error or incorrect results.
2. Ensure that your arrays contain numeric data: The COVAR formula requires numeric data points. If your arrays contain non-numeric data, such as text or error values, the formula may return an error or incorrect results.
3. Verify that your formula is entered correctly: Double-check the syntax of your COVAR formula to ensure that it is entered correctly, including the correct cell references for your arrays.
## COVAR: Related Formulae
Here are some related formulae that you may find useful when working with the COVAR formula in Excel:
1. CORREL: Calculates the correlation coefficient between two sets of data points, which measures the strength and direction of the linear relationship between the variables. Syntax: =CORREL(array1, array2)
2. PEARSON: Calculates the Pearson correlation coefficient between two sets of data points, which is equivalent to the CORREL function. Syntax: =PEARSON(array1, array2)
3. SLOPE: Calculates the slope of the linear regression line for two sets of data points, which can help determine the relationship between the variables. Syntax: =SLOPE(known_y’s, known_x’s)
4. INTERCEPT: Calculates the intercept of the linear regression line for two sets of data points, which can help determine the relationship between the variables. Syntax: =INTERCEPT(known_y’s, known_x’s)
5. VAR.P and VAR.S: Calculate the population variance (VAR.P) or sample variance (VAR.S) of a set of data points, which can help measure the dispersion of the data. Syntax: =VAR.P(number1, [number2], …) or =VAR.S(number1, [number2], …)
By understanding the COVAR formula and its related functions, you can gain valuable insights into the relationships between variables in your data, helping you make more informed decisions in various fields, such as finance, marketing, and operations.
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We discussed about the registered capital contribution requirement under the New Company Law in our previous article. We got some inquiries asking for advice about
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I recently led a training session where we implemented the rules for scoring ten-pin bowling in Scala. It makes for a good case study. It’s small enough that you can pick up the rules in a few minutes, but the dependencies between frames makes calculating the score non-trivial. I decided to implement my own solution which turned into an interesting exercise in algebraic data and finite state machines. In this post I’ll describe my implementation and my process for developing it.
For my implementation I solely focused on scoring the game. I didn’t implement any parsing code, as that part of the problem didn’t interest me.
## The Data
The core of my approach is getting the data structure right. Once they’re in place the rest of the code is relatively straightforward. This approach relies on some foundational features of functional programming, namely algebraic data types and structural recursion. Lets have a quick diversion into these topics.
### Algebraic Data Types and Structural Recursion
“Algebraic data type” is a fancy phrase that functional programmers use to refer to data that is modelled in terms of logical ands and logical ors. Here are some examples:
• a User is a name and an email address and a password;
• a Result is a success or a failure;
• a List of A is:
• the empty list (conventionally called nil) or
• a pair (conventionally called cons) containing a head of type A and a tail of type List of A.
If a language has support for algebraic data types, once we have a description of data such as the examples above we can directly translate it into code. Let’s use the example of the list as it is the most complex.
Here’s how we can define it in Scala.
sealed trait List[A]
final case class Nil[A]() extends List[A]
final case class Cons(head: A, tail: List[A]) extends List[A]
Here’s the same thing in Typescript.
type Nil<A> = { kind: "nil" }
type Cons<A> = { kind: "cons", head: A, tail: List<A> }
type List<A> = Nil<A> | Cons<A>
const Nil = <A>(): List<A> =>
({ kind: "nil" });
const Cons = <A>(head: A, tail: List<A>): List<A> =>
Here’s Rust.
enum List<A>{
Nil,
Cons(A, Box<List<A>>)
}
Each language requires some language specific knowledge in the implementation. For example:
• in Scala we can choose between covariance and invariance;
• in Typescript we need to define constructors seperately;
• in Rust we must wrap recursion in a Box.
The general concept, however, applies to all these languages and we can transfer knowledge from one language to another. To avoid writing all the code three times for the rest of this post I’ll be sticking to Scala.
Given an algebraic data type we can implement any transformation on that type using structural recursion (also known as a fold or a catamorphism)[1]. The rules, informally, for structural recursion are:
• each case (logical or) in the data must have a case in the structural recursion; and
• if the data we are processing is recursive the function we are defining must be recursive at the same place.
Structural recursion cannot solve everything for us—we must add problem-specific code to fill out the implementation—but it gives us a substantial help.
Here’s one way we could write the structural recursion skeleton for a list in Scala.
def transform[A](list: List[A]): SomeResultType =
list match {
case Nil() =>
??? // Problem specific
case Cons(h, t) =>
??? // Problem specific but *must* include the recursion transform(t)
}
If we want to calculate the length of a list we can start with the skeleton
def length[A](list: List[A]): Int =
list match {
case Nil() =>
??? // Problem specific
case Cons(h, t) =>
??? // Problem specific but *must* include the recursion length(t)
}
and fill out the problem specific parts
def length[A](list: List[A]): Int =
list match {
case Nil() =>
0
case Cons(h, t) =>
1 + length(t)
}
Any (yes, really, any) other method we can write that transforms a list to something else (or even another list) is going to have the same skeleton. So in summary, all we have to do is work out how to model our data using logical ands and ors and then we immediately get for free:
• the representation of that data in code; and
• a generic template for transforming that data into anything.
Diversion over! Let’s get back to bowling.
## Bowling as an Algebraic Data Type
From reading the rules of bowling we can pull out a reasonably simple structure:
• A game consists of 10 frames
• Each frame can be a strike, a spare, or an open frame where
• an open frame is two rolls that sum to less than ten;
• a spare is one roll that is less than ten (the second rule is implied by the first roll); and
• a strike doesn’t need any additional information.
This is the model I started with but as I worked on it I realised the true model is more complicated because the final frame may have up to two bonus rolls. Hence I changed the model to
• A game consists of 9 frames and 1 final frame
• A frame can be a strike, a spare, or an open frame where
• an open frame is two rolls that sum to less than ten;
• a spare is one roll that is less than ten (the second roll is implied by the first roll); and
• a strike doesn’t need any additional information.
• A final frame can be a strike, a spare, or an open frame which have the same definition as above and also
• a spare final frame has one bonus roll; and
• a strike final frame has two bonus rolls.
These definitions fit the criteria for an algebraic data type (they consist of logical ands and ors) and therefore translate to code in a straightforward way. Rather than paste a big lump of code I’ll just link to Frame, FinalFrame, and Game in the code repository. Note that not all the invariants can be expressed in the type system. For example, we cannot express the criteria that the rolls in an open frame must sum to less than 10. The problem specification says we only need to consider valid data, but I put some dynamic checks in the “smart constructors” on the companion objects. This turned out to be useful as it caught some errors in my tests (which I’ll talk about in a bit.)
Now that we have defined the data we just need to write a structural recursion over the Game type. Well, not quite. The scoring rules have dependencies between frames. For example, if a frame is a strike the next two rolls are added to the score for that frame. We need to keep around the information about pending frames—frames that have yet to be scored—while we process the data.
Reading through the rules we can extract the following.
The pending frames can be
• a strike;
• a spare; or
• a strike and a strike .
This is another algebraic data type.
sealed trait Pending
case object Strike extends Pending
case object Spare extends Pending
case object StrikeAndStrike extends Pending
When we score a frame in a game we must calculate:
• the score for this frame if it is not pending futures frames;
• the score for any pending frames that are now complete; and
• the pending frames after this frame.
In this way the scoring algorithm is a finite state machine (FSM). The pending frames are the current state of the FSM, the current frame is the output, and we output the next state (the updated pending frames) in addition to a score.
It’s useful to wrap the Pending information up with the score calculated so far, which gives all the information to calculate the total score so far. I called this State. Note that Pending is wrapped in an Option; there may be no frames for which the score is pending.
final case class State(
score: Int,
pending: Option[Pending]
) {
def next(additionalScore: Int, newPending: Option[Pending]): State =
this.copy(
pending = newPending
)
}
With this definition the scoring function has type (State, Frame) => State, which is exactly the type of the transition function of a FSM. We can calculate the score of a List[Frame] by passing this function as the second argument to foldLeft, with the initial state forming the first argument. In code this is
frames.foldLeft(initialState)(transitionFunction)
The transition function, the scoring algorithm, is a structural recursion over the Frame as well as the State. The code is lengthy, but it isn’t hard to write and a good deal of it is generated by the IDE (in my case, Metals with Doom Emacs.)
### Testing
Testing was important. The scoring rules aren’t amenable to much support from the type system (though now I think about it I could have expressed the rules in a different way that would have given me more compiler support) which means testing is the next best way to ensure the code is correct. This is an excellent application for property-based testing, for which I used ScalaCheck.
I defined a few different generators for the various types of frames. For example here is how I generate open frames.
def genOpen: Gen[Frame] = {
for {
misses <- Gen.choose(1, 10)
hits = 10 - misses
roll1 <- Gen.choose(0, hits)
roll2 = hits - roll1
} yield Frame.open(roll1, roll2)
}
These generators enabled me to test both the examples given in the instructions and examples generated at random. I found quite a few errors with these tests, both in my scoring algorithm and in how I was generating data. Luckily they were all very easy to diagnose. As the scoring algorithm was very explicit it was easy to work out what I had done wrong (which was usually forgetting to include a roll somewhere).
## Conclusions
I hope this article has given an insight in how I approached this case study. In summary there are three important components:
• the core of my approach is to model the data correctly, as I know once I have the data model in place almost all of the rest of the code follows from it;
• recognising the scoring algorithm was a finite state machine was another insight I needed to model it cleanly; and
• using property-based testing allowed me to achieve a high degree of confidence in my implementation without a great deal of effort.
I have presented my process as if I moved straight from problem to implementation. This was not the case. It was a highly iterative process, and I changed the data model at least three times as I came to better understand the problem. I also interleaved developing the tests with the code under test.
Of course my approach is the only one. There is a write-up of a TDD approach in C# which may make an interesting contrast to mine.
1. Although this is well known in programming language theory I haven’t been able to find a reference that has a chance of being comprehensible to the average programmer. I think the first place to state this result is Data Structures and Program Transformation, but this uses the Bird-Meertens formalism which I find very hard to read. A tutorial on the universality and expressiveness of fold only considers folds on list, but uses Haskell and more standard mathematical notation. I imagine this is still quite obscure for most but it is an improvement! ↩︎
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# A certain baseball team has just completed its season. In
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A certain baseball team has just completed its season. In [#permalink]
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17 Apr 2008, 18:11
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A certain baseball team has just completed its season. In stadiums that seat 20,000 or fewer people, the team averaged 1 home run per game; in stadiums that seat between 20,000 and 40,000 people, the team averaged 2 home runs per game; and, in stadiums that seat 40,000 or more people, the team averaged 3 home runs per game. Obviously, the excitement of playing in front of large crowds motivated the team to hit more home runs.
Assuming that all stadiums during the season were filled to capacity, which of the following, if true, most undermines the argument above?
a) The team's leading home run hitter hit more home runs in mid-sized stadiums than in large stadiums.
b) The fans in the larger stadiums often cheered against the team.
c) The team averaged only 2 home runs per game when playing in the league’s largest stadium.
d) In order to create seating for the additional fans, the outfield walls in the larger stadiums were constructed closer to home base.
e) The team’s announcer cited crowd noise as a major motivator for the team.
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Re: CR - Home hitter [#permalink]
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17 Apr 2008, 18:30
i'll say D. Question asks us to basically find another reason for why the team hit more homers.
D gives that reason; if the walls of the outfield in the larger stadiums were closer to home, then the batters dont have to hit as much of a distance to get a home run. So it wasnt the excitement, but the fact that it was easier to hit home runs.
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Re: CR - Home hitter [#permalink]
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17 Apr 2008, 18:31
D for me.
moni77 wrote:
A certain baseball team has just completed its season. In stadiums that seat 20,000 or fewer people, the team averaged 1 home run per game; in stadiums that seat between 20,000 and 40,000 people, the team averaged 2 home runs per game; and, in stadiums that seat 40,000 or more people, the team averaged 3 home runs per game. Obviously, the excitement of playing in front of large crowds motivated the team to hit more home runs.
Assuming that all stadiums during the season were filled to capacity, which of the following, if true, most undermines the argument above?
a) The team's leading home run hitter hit more home runs in mid-sized stadiums than in large stadiums. [hmmm seems to be restating the story]
b) The fans in the larger stadiums often cheered against the team. [so what - argument is about having large crowds not who they cheer for]
c) The team averaged only 2 home runs per game when playing in the league’s largest stadium. [hmmm largest is not clear - does it mean larger size ? or more crowds ? this is not clear cut enough to be the answer]
d) In order to create seating for the additional fans, the outfield walls in the larger stadiums were constructed closer to home base. [ahhh shorter boundaries could cause more home runs - hence the answer]
e) The team’s announcer cited crowd noise as a major motivator for the team. [irrelevant]
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Re: CR - Home hitter [#permalink]
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17 Apr 2008, 21:28
i had literrally no idea of the game itself..
so the term ' home run' created a great amount of confusion in me..
if something similar appears on the test, i think it would not be fair from an general applicant's perpective.
iam not sure whether gmac already takes care of this.
regards,
Neo
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Re: CR - Home hitter [#permalink]
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17 Apr 2008, 21:51
They say that the questions are supposed to be "culturally sensitive" but during my actual GMAT test, the very first Math question had "nickels" and "dimes" and since I am not American, I really had no idea about that one.
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Re: CR - Home hitter [#permalink]
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18 Apr 2008, 06:49
I agree with D.
For those of you who watch cricket, think of a home run as a 6 and the outfield walls as the boundary.
The argument states that playing in front of large crowds is what caused the team to hit more HR. However, D provides an alternate explanation that the team hit more HR because the field was moved in.
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Re: CR - Home hitter [#permalink]
### Show Tags
18 Apr 2008, 07:07
Neochronic wrote:
i had literrally no idea of the game itself..
so the term ' home run' created a great amount of confusion in me..
if something similar appears on the test, i think it would not be fair from an general applicant's perpective.
iam not sure whether gmac already takes care of this.
regards,
Neo
That's the same thing I feel. I have no idea how this game works.
Anyway, OA is D.
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Re: CR - Home hitter [#permalink] 18 Apr 2008, 07:07
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# A certain baseball team has just completed its season. In
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# How is 438,500,000,000 written in scientific form?
Mar 8, 2018
4.385 $\times {10}^{11}$
#### Explanation:
In Scientific notation there is only one digit to the left of the decimal point. The other significant digits are written to the right of the decimal point. Then the digits are multiplied by a power of ten. The multiplication of the power of ten establishes the value of the digits and acts as the place holder.
In 4 38,500,000,000 The four is written to the left of the decimal point. 385 are significant numbers and are written to the right of the decimal point, 00,000,000 are zero's used as place holders and are not significant. The power of ten will replace the zero's as place holders.
In 438,500,000,000 there are 11 places between the four and the last zero. This is equal to 10 to the 11th power so
$4.385 \times {10}^{11}$
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# Cluster analysis using k-means method in R
K-means method for data clustering and its realization in R.
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### Cluster analysis using k-means method in R
1. 1. Cluster analysis using k-means method Vladimir Bakhrushin, Professor, D.Sc. (Phys. & Math.) [email protected]
2. 2. Formulation of the problem The task of cluster analysis is to divide the existing set of points on a certain number of groups (clusters) so that the sum of squares of points distances from cluster centers was minimal. At the point of minimum all cluster centers coincide with the centers of the corresponding areas of Voronoi diagram. Main algorithms: Hartigan and Wong Lloyd Lloyd-Forgy MacQueen
3. 3. The initial approximation First step is to set the initial approximation of cluster centers. To do this, such methods are most commonly used: to set the centers of clusters directly; to set the number of clusters k and take the first k points coordinates as centers; to set the number of clusters k and take the randomly selected k points coordinates as centers (it is appropriate to carry out calculations for several random runs of the algorithm).
4. 4. Iteration procedure 1. Placing of each point to the cluster center of which is the nearest to it. As a measure of closeness squared Euclidean distance is used most commonly, but other measures of distance also may be selected. 2. Recalculation of cluster centers coordinates. If the measure of closeness is the Euclidean distance (or its square), cluster centers are calculated as the arithmetic means of corresponding coordinates of points that belong to these clusters. The iterations are stopped when the specified maximum number of iterations is carried out, or if there is no longer change of the clusters composition.
5. 5. Limitation (shortcoming) Setting the number of clusters (initial approximation) Preliminary analysis of data Sensitivity to outliers Using of k-medians Limitations and shortcomings Using of random samples from arrays Slow work on large arrays
6. 6. Forming of data array a1 = matrix(c(rnorm(20, mean = 5, sd = 1), rnorm(20, mean = 5, sd = 1)), nrow=20, ncol = 2) a2 = matrix(c(rnorm(20, mean = 5, sd = 1), rnorm(20, mean = 13, sd = 1)), nrow=20, ncol = 2) a3 = matrix(c(rnorm(20, mean = 12, sd = 1), rnorm(20, mean = 6, sd = 1)), nrow=20, ncol = 2) a4 = matrix(c(rnorm(20, mean = 12, sd = 1), rnorm(20, mean = 12, sd = 1)), nrow=20, ncol = 2) a <- rbind(a1,a2,a3,a4) Function rbind() forms matrix a, in which the first 20 rows are the corresponding strings of matrix a1, next 20 – matrix a2 and so on.
7. 7. Group centers Next, we must calculate the matrix of values of formed group centers and display the results on a screen:
8. 8. Function kmeans() For forming the clusters by k-means method we can use the function: kmeans(x, centers, iter.max = 10, nstart = 1, algorithm = c("Hartigan-Wong", "Lloyd", "Forgy", "MacQueen") ) x – matrix of numerical data; centers – initial approximation of cluster centers or number of clusters (in the latter case, the appropriate number of randomly selected rows of the matrix will be taken as the initial approximation x); iter.max – maximum number of iterations; nstart – number of random sets which must be chosen if centers – is the number of clusters; algorithm – choice of clustering algorithm.
9. 9. Clustering results
10. 10. Clustering results
11. 11. Clustering results
12. 12. Comparison of centers Group (cluster) number xa ya xcl ycl a1 4,613619 5,169488 4,613619 5,169488 a2 4,570456 13,396202 4,570456 13,396202 a3 11,855793 5,936099 11,855793 5,936099 a4 12,197688 11,930728 12,197688 11,930728 b1 5,531175 5,405187 5,545309 5,527677 b2 5,340795 12,983168 5,472965 13,239925 b3 11,770917 6,725708 11,842934 6,916365
13. 13. Residues Using command sd(resid.a) we can calculate residues standard deviations. They are close to the given values of standard deviations of initial arrays. It confirms the adequacy of the clustering results.
14. 14. Results of the division on 3 clusters
15. 15. Results of the division on 5 clusters
16. 16. Within and between group variations
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K-means method for data clustering and its realization in R.
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What is APY (annual percentage yield)?
APY, a commonly used acronym for annual percentage yield, is the rate earned on an investment in a year, taking into account the effects of compounding interest. APY is calculated using this formula: APY= (1 + r/n )n – 1, where “r” is the stated annual interest rate and “n” is the number of compounding periods each year. APY is also sometimes called the effective annual rate, or EAR.
Deeper definition
When the APY is the same as the interest rate that is being paid on a person’s investment, he is earning simple interest. When the APY is higher than the interest rate, however, the interest is being compounded, which means he is earning interest on his accumulating interest.
People sometimes confuse APY with APR. APR refers to the annual interest rate without taking compounding interest into account. APY, on the other hand, does take into account the effects of compounding within a year. The difference between the two can have important implications for borrowers and investors.
When banks or other financial institutions are looking for clients for interest-bearing investments, such as money market accounts and certificates of deposit, it is in their best interests to promote their best APY, not their APR. APY is higher than APR, so it looks like a better investment for the client.
The more frequent the compounding periods, the higher the APY. Thus, people who save money in their bank accounts should check how often the money is compounded. Typically, daily or quarterly is better than annual compounding, but make sure to check the quoted APY for each option beforehand.
APY example
If an individual deposits \$1,000 into a savings account that pays 5 percent interest annually, he will make \$1,050 at the end of year.
However, the bank may calculate and pay interest every month, in which case he would end the year with \$1,051.16. In the latter case, he would have earned an APY of more than 5 percent. The difference may not be huge, but after several years (or with larger deposits), the difference is significant. In this example, APY is calculated like this:
Annual percentage yield = (1+0.5/12)^12-1= 5.116 percent
APY can show investors exactly how much interest they will earn. With this information, they can compare options. They will be able to decide which bank is the best, and whether or not they want to go for a higher rate.
Use this calculator to see if you’re on track to meet your investment goal.
Other Investing Terms
Prudent investor rule
Prudent investor rule is a term every investor should understand. Bankrate explains it.
Fiduciary rule
The fiduciary rule describes what a financial advisor can do with your money.
Repurchase agreement (repo loan)
A repurchase agreement is a short-term loan to raise quick cash. Bankrate explains.
Derivative
Investing
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# Homework Help: Friction problem involving 2 blocks sliding in 2 directions with 2 frictions
1. Nov 9, 2008
### 2FAST4U8
1. The problem statement, all variables and given/known data
http://streetrodjohn.home.comcast.net/~streetrodjohn/physics.jpg [Broken]
A 1-kg block is pushed against a 4-kg block on a horizontal surface of coefficient of friction 0.25, as shown in the figure. Determine the minimum force needed to ensure that the 1-kg block does not slip down. Assume that the coefficient of friction at the interface between the block is 0.4. Hint: The two blocks exert equal and opposite forces on each other.
2. Relevant equations
F=(A+B)a
Ag=fs
N=Ba
Ag=$$\mu$$Ba
3. The attempt at a solution
In class we have done similiar problems, only with a frictionless horizontal surface. I don't know how to account for the horizontal coefficient of friction of .25 in this problem. Using the above equations and ignoring the horizontal coefficient of friction, I get this:
a=(Ag)/($$\mu$$B
F=(((A+B)Ag)/($$\mu$$B))
F=(((4+1)(1)(9.8))/((.4)(4)))
F=30.62N
Now how do I account for the horizontal coefficient of friction of .25? Thanks in advance for any help you can provide.
Last edited by a moderator: May 3, 2017
2. Nov 9, 2008
### asleight
Well, examining block one, we notice that we want $$\sum\vec{F}=m\vec{a}=0=\vec{F}_f+\vec{F}_g\rightarrow\vec{F}_f=-\vec{F}_g$$. From this, we can examine the frictional force, specifically:
$$\vec{F}_f=\mu_i\vec{N}\rightarrow m\vec{g}/\mu_i=\vec{N}$$.
What next? :)
Last edited by a moderator: May 3, 2017
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Breaking News
# Square Root Of .35
Square Root Of .35. (1 24 51 10), the tablet also gives an example where one side of. The inverse operation of squaring 35.72 is extracting the square root of 35.72. N √ a = b b n = a. Free online fraction to decimal calculator to convert fraction 35/50 to decimal form 0.8367 i.e.
A square root of a number 'x' is a number y such that y 2 = x, in other words, a number y whose square is y. Besides the real values of \sqrt[2]{45.35}. Given square root is 35 9 √ 35 9 can be written as 35 1/2 9 1/2 or √35 √9 finding the numerator square root and denominator square root we get the value as 5.9161 3.
## To find the value, we should know the prime factorization of 1225.
Square root of 35 = 5.9161 updated edited (completed) by jayden caldwell 1 35 2 the symbol √ is called radix, or radical sign the number below the radix is the radicand is 35 a perfect. To find the value, we should know the prime factorization of 1225. In the next section we elaborate on the term “square.” what is 35.72 squared?
## A Square Root Of A Number 'X' Is A Number Y Such That Y 2 = X, In Other Words, A Number Y Whose Square Is Y.
Now click the button “find square root” to get the output step 3:. The square root of 35 is 5.91607978309962 √ 35 = 5.91607978309962 proof that the square root of 35 is 5.91607978309962 the square root of 35 is defined as the only positive real number.
### Square Roots Is A Specialized Form Of Our Common Roots Calculator.
By trial and error method, we can see that, there does not exist any integer whose square is 35.
### Kesimpulan dari Square Root Of .35.
It is is denoted by √, known as the radical sign or radix. A square root of a number 'x' is a number y such that y 2 = x, in other words, a number y whose square is y. A square is a flat.
See also Optic Fibres Are Mainly Used For Which Of The Following
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0
# what is the answer for this inequalities 4r-9>7
This is solving inequalites
### 1 Answer by Expert Tutors
Aaron M. | Experienced Math to English TranslatorExperienced Math to English Translator
5.0 5.0 (7 lesson ratings) (7)
0
Eugena,
When you are working with an equation involving inequalities (<,>) you can treat it just like an equation with an = sign. So we can use the same rules, and we will be trying to get the r "by itself". In order to do this we have to do the opposite of whatever is happening to r to both sides of our equation.
Currently we are subtracting 9 from the r side of the equation, so we will add 9 to each side giving us,
4r-9+9>7+9
4r>16.
Now we are multiplying r by 4, so we will divide both sides by 4. This gives us,
(4r)/4>16/4
r>4.
I hope this helps!
While this is a correct answer for solving this particular problem, be very careful with this. I agree that it is helpful to think of inequalities as being like equations. But they are actually two distinct things, and you want to get your terminology correct.
Next while it is true that you can solve this inequality by treating it just like an equation, that is not true for all inequalities. If you have to multiply or divide by a negative number, then you have to change the direction of the inequality. For example, if this were -4r - 9 > 7 instead, we would solve it slighting differently:
-4r - 9 + 9 > 7 + 9
-4r > 16
(-4r)/-4 < 16/-4 (notice we changed from > to < on this step!)
r < -4
So while this solution works, you can't just say that inequalities can always be treated just like equations.
Thank you for your comment Brian. I was simply explaining how to work through this problem, but it is good to point out that in solving different problems a different method is needed.
Eugena, I hope our tag-teamed response helps!
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Worksheet: Applications of Systems of Equations
Q1:
Two cousins were born 8 years apart. How old will the elder be when the sum of their ages is equal to 20 years?
• A 6 years
• B 12 years
• C 8 years
• D 14 years
• E 16 years
Q2:
I am thinking of two numbers. Use the clues to determine what the numbers are.
• The product of the two numbers is 48.
• The difference between the two numbers is 2.
• A 9 and 7
• B 6 and 4
• C 5 and 3
• D 8 and 6
• E7 and 5
Q3:
A concert venue sells single tickets for \$40 each and couple tickets for \$65. If the total revenue was and 321 tickets were sold, how many of each type of ticket did they sell?
• A100 single tickets, 221 couple tickets
• B210 single tickets, 111 couple tickets
• C121 single tickets, 200 couple tickets
• D111 single tickets, 210 couple tickets
Q4:
Jacob has to invest. His intent is to earn interest on his investment. He can invest part of his money at interest and part at interest. How much does Jacob need to invest in each option to get a total return on his ?
• A at , at
• B at , at
• C at , at
• D at , at
• E at , at
Q5:
Two planes flying in opposite directions pass each other. One is flying at 450 mph and the other at 550 mph. How long will it take before they are 4000 miles apart?
Q6:
Sophia invested 1.1 million dollars in two land investments. Her return on the first investment, Swan Peak, was a increase on the money she had invested. On her second investment, Riverside Community, she earned over what she had invested. Given that she earned \$1 million in profit, how much did she invest in each of the land deals?
• ASwan Peak: \$750 000, Riverside: \$250 000
• BSwan Peak: \$350 000, Riverside: \$750 000
• CSwan Peak: \$650 000, Riverside: \$350 000
• DSwan Peak: \$750 000, Riverside: \$350 000
Q7:
A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was \$12.50, and the price for an adult ticket was \$16.00. The register confirms that were taken in. How many student tickets and adult tickets were sold?
• Astudent tickets: 250, adult tickets: 100
• Bstudent tickets: 200, adult tickets: 150
• Cstudent tickets: 300, adult tickets: 50
• Dstudent tickets: 150, adult tickets: 200
Q8:
A Jeep and a BMW entered a highway at the same interchange, heading in opposite directions. The Jeep entered the highway 30 minutes before the BMW and traveled 7 mph slower than the BMW. Two hours after the BMW entered the highway, the two cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control.
• Athe Jeep: 72 mph, the BMW: 80 mph
• Bthe Jeep: 72 mph, the BMW: 65 mph
• Cthe Jeep: 57 mph, the BMW: 61 mph
• Dthe Jeep: 65 mph, the BMW: 72 mph
Q9:
The difference between the measures of the acute angles in a right-angled triangle is . Find the measure of each acute angle.
• A ,
• B ,
• C ,
• D ,
• E ,
Q10:
If , find the values of and .
• A ,
• B ,
• C ,
• D ,
Q11:
You invested in two accounts: in account A and in account B. Both accounts earn simple interest and, after one year, you have earned in interest. Given that account B earns half a percent less than twice the interest account A earns, what are the interest rates for your accounts?
• Aaccount A: , account B:
• Baccount A: , account B:
• Caccount A: , account B:
• Daccount A: , account B:
Q12:
Madison starts riding her bike at 20 mph. After a while, she slows down to 12 mph and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mph?
Q13:
A chemistry teacher needs to mix a salt solution with a salt solution to make 20 qt of a salt solution. How many quarts of each solution should the teacher mix to get the desired result?
• A 12 qt of the solution, 8 qt of the solution
• B 5 qt of the solution, 15 qt of the solution
• C 8 qt of the solution, 12 qt of the solution
• D 15 qt of the solution, 5 qt of the solution
• E 10 qt of the solution, 10 qt of the solution
Q14:
At the beginning of a year, you invested \$10 000 into two accounts, A and B, which receive simple interest and simple interest respectively. At the end of that year, you had \$10 710 in your combined accounts. How much was invested in each account?
• Aaccount A: \$5 000, account B: \$5 000
• Baccount A: \$3 000, account B: \$7 000
• Caccount A: \$8 000, account B: \$2 000
• Daccount A: \$7 000, account B: \$3 000
Q15:
Noah buys 5 apples and 3 bananas from a grocery store and pays \$3.40. Madison buys 3 apples and 2 bananas from the same store and pays \$2.10. Work out the price of a single apple and a single banana.
• A An apple is 25¢ and a banana is 15¢.
• B An apple is 26¢ and a banana is 14¢.
• C An apple is 47¢ and a banana is 6¢.
• DAn apple is 50¢ and a banana is 30¢.
• E An apple is 45¢ and a banana is 35¢.
Q16:
A rectangle’s length is 16 cm less than four times its width. Given that its perimeter is equal to that of a square of side 12 cm, find the dimensions of the rectangle.
• A 3 cm, 22 cm
• B 4 cm, 8 cm
• C 3 cm, 13 cm
• D 8 cm, 16 cm
Q17:
I am thinking of two numbers. Use the clues to determine what the numbers are.
• When you divide one by the other, the quotient is 8.
• The sum of the two numbers is 81.
• A 71 and 10
• B 70 and 11
• C 4 and 8
• D 72 and 9
• E 75 and 6
Q18:
Four times the weight of Michael is 150 pounds more than the weight of Victoria. Four times the weight of Victoria is 660 pounds less than seventeen times the weight of Michael. Four times the weight of Michael plus the weight of Benjamin equals 290 pounds. Liam would balance all three of the others. Find the weights of the four people.
• AMichael’s weight = 60 lb, Victoria’s weight = 90 lb, Liam’s weight = 50 lb, Benjamin’s weight = 200 lb
• BMichael’s weight = 90 lb, Victoria’s weight = 60 lb, Liam’s weight = 200 lb, Benjamin’s weight = 50 lb
• CMichael’s weight = 90 lb, Victoria’s weight = 60 lb, Liam’s weight = 50 lb, Benjamin’s weight = 200 lb
• DMichael’s weight = 60 lb, Victoria’s weight = 90 lb, Liam’s weight = 200 lb, Benjamin’s weight = 50 lb
• EMichael’s weight = 50 lb, Victoria’s weight = 60 lb, Liam’s weight = 90 lb, Benjamin’s weight = 200 lb
Q19:
A man’s age is 4 times his son’s age. In five years, the sum of their ages will be 105. How old are they now?
• A 20 years, 62 years
• B 23 years, 60 years
• C 25 years, 43 years
• D 19 years, 76 years
• E 21 years, 84 years
Q20:
In a test with 20 questions, marks are awarded for each correct answer and marks are deducted for each incorrect answer. Benjamin answered 12 questions correctly and 8 questions incorrectly, and he scored 44 points. Emma answered 14 questions correctly and 6 questions incorrectly, and she scored 58 points. How many points were deducted for each incorrect answer?
• A 6 points
• B 5 points
• C 3 points
• D 2 points
• E 4 points
Q21:
A library decides to put in place a new fine system as a stronger incentive for borrowers to return items on time. Instead of issuing a fine of per day and per item, the fine is for one day after the return date, for two days, for three days, and so on. How much more will a borrower pay for an item returned 7 days after the return date?
• A
• B
• C
• D
• E
Q22:
A bank offers two savings account options. In the first one, a deposit of is received every year for an initial capital of . In the second option, an annual interest rate of is offered and it’s compounded monthly. Which option is more profitable after 30 years?
• Aoption 2
• Boption 1
Q23:
A farmer finds there is a linear relationship between the number of bean stalks, , she plants and the yield, , each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form that gives the yield when stalks are planted.
• A
• B
• C
• D
• E
Q24:
A clothing business finds there is a linear relationship between the number of shirts, , it sells and the price, , that it charges per shirt. In particular, historical data shows that, when the price was \$30, they sold shirts, whereas, when the price was \$22, they sold . Find a linear equation in the form that gives the price they can charge if they want to sell shirts.
• A
• B
• C
• D
• E
Q25:
William buys 4 burgers and 3 hot dogs for \$16 from a fast-food vendor. Matthew buys 3 burgers and 4 hot dogs for \$15.50 from the same vendor. Work out the price of a single burger and a single hot dog.
• A A burger is \$4.42 and a hot dog is \$0.56.
• B A burger is \$2 and a hot dog is \$2.50.
• C A burger is \$0.56 and a hot dog is \$4.42.
• DA burger is \$2.50 and a hot dog is \$2.
• E A burger is \$3 and a hot dog is \$2.50.
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# Prime Number Arithmetic Progression
• FeDeX_LaTeX
In summary: , in summary, the least possible value of the largest term in an arithmetic progression of seven distinct primes is 97.
FeDeX_LaTeX
Gold Member
"Determine the least possible value of the largest term in an arithmetic progression of seven distinct primes."
I really have no clue what to do here. Is there a general tactic that you can use to do this, other than trial and error? Some experimenting gives you these of arithmetic progressions:
5, 11, 17, 23, 29
5, 17, 29, 41, 53
7, 19, 31, 43
3, 7, 11
41, 47, 53, 59
61, 67, 73, 79
7, 37, 67, 97, 127, 157
107, 137, 167, 197, 227, 257
53, 113, 173, 233, 293, 353
I haven't found one that gives me a string of 7 primes yet and I've just been looking at primes under 100.
Of course there are general rules to follow when finding the strings that I spotted (shouldn't add a number to a prime which will land you on a multiple of 5, such as adding 12 to a prime excluding 2).
EDIT: Okay, I think I know how to solve this problem. If I had 4, 6 or 8, I'll never get a streak longer than 5, but if I choose a difference of +10 (or a multiple of 10), I might find one more easily.
Last edited:
Remember, Prime Numbers except 2 or 3 can be expressed as 6k±1.
AGNuke said:
Remember, Prime Numbers except 2 or 3 can be expressed as 6k±1.
Thanks for this. So would it be worthwhile to only consider the primes that are 1 and 5 (mod 6)?
I assume that it would better help you to determine the prime number and thus the relevant progression.
Start from k=1, we get 5 and 7. Both are Primes.
k=2, we get 11 and 13.
k=3, we get 17 and 19.
k=4, we get 23 and 25, not a prime. And so on...
It may help you to get a proper listing, like it is probable that AP can be formed with common difference of 6, 12...
You can also look for Prime Generating Programs, if you want I can give you one.
Last edited:
But I thought that the required arithmetic progression here had to have a common difference that was a multiple of 10? All primes (except 2) are odd, and if you add a common difference that is a multiple of 6 to an odd prime, you will quickly end up with a multiple 5 and you will be unable to get an AP longer than 5 terms.
If the difference is 10 which is 1 mod 3 you get a number divisible by 3 at every third step. So you need to include 3 into the difference, it must be at least 2*3*5=30. That is 3 mod 7 and you will bump into a multiple of 7 in every 7th step...
ehild
## What is a Prime Number Arithmetic Progression?
A Prime Number Arithmetic Progression is a sequence of prime numbers in which the difference between any two consecutive numbers is constant. For example, 3, 7, 11, 15 is a Prime Number Arithmetic Progression with a common difference of 4.
## How do you find the next number in a Prime Number Arithmetic Progression?
To find the next number in a Prime Number Arithmetic Progression, you can add the common difference to the previous number in the sequence. For example, if the previous number is 11 and the common difference is 4, the next number would be 15.
## What is the formula for the nth term in a Prime Number Arithmetic Progression?
The formula for the nth term in a Prime Number Arithmetic Progression is n x d + a, where n is the term number, d is the common difference, and a is the first term in the sequence. For example, the 5th term in the sequence 3, 7, 11, 15 would be 5 x 4 + 3 = 23.
## What is the significance of Prime Number Arithmetic Progressions in mathematics?
Prime Number Arithmetic Progressions have been studied for centuries and have many applications in mathematics, including in number theory, algebra, and cryptography. They also help us better understand the distribution of prime numbers and their properties.
## Are there any famous unsolved problems related to Prime Number Arithmetic Progressions?
Yes, there are several famous unsolved problems related to Prime Number Arithmetic Progressions, including the Green-Tao theorem, which states that there are arbitrarily long arithmetic progressions of prime numbers, and the Twin Primes Conjecture, which suggests that there are infinitely many pairs of prime numbers that differ by 2.
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## How is emf produced in alternator?
Typically, a rotating magnet, called the rotor turns within a stationary set of conductors wound in coils on an iron core, called the stator. The field cuts across the conductors, generating an induced EMF (electromotive force), as the mechanical input causes the rotor to turn.
## What are the 3 main requirements for electromagnetic induction?
What is this electromagnetic induction of which you speak?
• The size of the magnetic field. The more flux lines there are, the more flux lines there are for the conductor to cut.
• The active length of the conductor.
• The speed at which the conductor passes through the field.
What are the 3 requirements for voltage production by magnetism?
Strength of the electromagnet (the stronger the magnetic field, the greater the induced voltage) Speed of rotation of the conductor or magnet (the faster the rotation, the greater the induced voltage)
### What are the three methods of producing induced emf?
The induced emf can be produced by changing :
• (i) the magnetic induction (B),
• (ii) area enclosed by the coil (A) and.
• (iii) the orientation of the coil (θ) with respect to the magnetic field.
### What is the formula for emf of an alternator?
The emf induced by the alternator or synchronous generator is three-phase alternating in nature. Z = 2T, where T is the number of coils or turns per phase. One turn has two coil sides or conductor as shown in the below diagram.
What is the equation for alternating emf?
The equation of an alternating emf is V=120sin(100πt)cos(100πt) Volt.
#### What conditions must happen for electricity to be generated?
To produce an electric current, three things are needed: a supply of electric charges (electrons) which are free to flow, some form of push to move the charges through the circuit and a pathway to carry the charges. The pathway to carry the charges is usually a copper wire.
#### What is Fleming’s left hand rule?
Fleming’s left – hand rule states that if we stretch the thumb, middle finger and the index finger of the left hand in such a way that they make an angle of 90 degrees(Perpendicular to each other) and the conductor placed in the magnetic field experiences Magnetic force.
What are 3 ways to create a voltage difference?
If there is a surplus of electrons at one end of a conductor and a deficiency at the other end, a current flows. Certain devices create this difference in charge so current flows….Electromotive Force (EMF) and Potential Difference
• Friction.
• Chemical.
• Pressure.
• Heat.
• Light.
• Magnetism.
## What is the difference between voltage and EMF?
Voltage is the potential difference between two points which causes current to flow. It is the amount of energy per unit charge while moving between two points. EMF or electromotive force is the amount of energy supply to the charge by battery cell. The intensity of voltage is lower than EMF and non-constant.
## Which method is used to induce an emf?
1. Emf induced by changing the magnetic induction. The magnetic induction can be changed by moving a magnet either towards or away from a coil and thus an induced emf is produced in the coil.
What is the basic cause of induced emf?
The most basic cause of an induced EMF is change in magnetic flux. Placing a current carrying coil that is moving constantly in a stable and static magnetic field. This will cause a change in the area vector and hence, EMF will be generated.
### How is EMF induced in an alternator?
How is emf induced in an alternator? EMF or electromotive force, i.e.Voltage, is generated when the magnets inside the alternator turn within the wire coil of the alternator. What is emf equation of three phase transformers?
### Where can I find the induced EMF equation?
This induced EMF can be found by the EMF equation of the alternator which as follow: Lets, P = No. of poles. Z = No. of Conductors or Coil sides in series/phase i.e. Z = 2T…Where T is the number of coils or turns per phase (Note that one turn or coil has two ends or sides)
What kind of armature winding is used in alternator?
Types of Armature Winding of Alternator. There are different types of armature winding used in alternator. The windings can be classified as. Single phase and poly phase armature winding. Concentrated winding and distributed winding. Half coiled and whole coiled winding. Single layer and double layer winding.
#### How does the flux of an alternator relate to the winding?
The useful flux which links with both windings is due to combined mmf of the armature winding and field winding. When the armature winding of an alternator carries current then an mmf sets in armature. This armature mmf reacts with field mmf producing the resultant flux, which differs from flux of field winding alone.
#### How is voltage regulation of alternator by EMF method?
The voltage regulation of alternator by EMF method involves the EMF quantities of all the armature parameters (armature resistance, Armature leakage reactance, armature reaction). The drop due to armature reaction is not considered, because it does not occur due to any of the physical element but due to interaction of armature flux with main flux.
How is alternating emf generated in a coil?
EMF is generated in a coil when there is a relative movement between the coil and the magnetic field. Alternating emf is generated if the change in flux-linkage of the coil is cyclic. Since electromechanical energy conversion requires relative motion between the field and armature winding, either of these could be placed on the stator or rotor.
## Where are the conductors located in an alternator?
In Alternator, we are having a large number of conductors which are systematically placed over the armature to obtain a smooth curve. In actual construction, the armature conductors (source of emf) form the stationary part (stator) and the field windings (for producing flux) are placed on the rotating part (rotor).
## What kind of field system does an alternator have?
Alternators consists of a dc heteropolar field system as in a dc machine and a three phase armature winding whose coil arrangement is quite different from that of a d.c. machine. In this unit, first we will consider the constructional features and EMF equation of alternator.
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# Thursday April 17, 2025
## Calculating 28 days before Thursday May 15, 2025 by hand
This page helps you figure out the date that is 28 days before Thursday May 15, 2025. We've made a calculator to find the date before a certain number of days before a specific date. If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculatorto type in a new question or days from a specific date if you want to add 28 days.
But for all you time sickos out there who want to calculate 28 days before Thursday May 15, 2025 - here's how you do it:
1. Start with the Input Date (Thursday May 15, 2025): Write it down! I can't stress this enough
2. Count in Weeks: Recognize that 28 days is approximately 4.0 weeks. Count forward 4.0 weeks (5.6 work weeks) from the input date. This takes you to .
3. Add Remaining Days: Since you've counted 4.0 weeks, you only need to add the remaining days to reach Thursday April 17, 2025
4. Use Mental Math: If Thursday May 15, 2025 is a Thursday, then compare that to if 28 is divisible by 7. That way, you can double-check if April 17 matches that Thursday.
## Thursday April 17, 2025 Stats
• Day of the week: Thursday
• Month: April
• Day of the year: 107
## Counting 28 days backward from Thursday May 15, 2025
Counting backward from today, Thursday April 17, 2025 is 28 before now using our current calendar. 28 days is equivalent to:
28 days is also 672 hours. Thursday April 17, 2025 is 29% of the year completed.
## Within 28 days there are 672 hours, 40320 minutes, or 2419200 seconds
Thursday Thursday April 17, 2025 is day number 107 of the year. At that time, we will be 29% through 2025.
## In 28 days, the Average Person Spent...
• 6014.4 hours Sleeping
• 799.68 hours Eating and drinking
• 1310.4 hours Household activities
• 389.76 hours Housework
• 430.08 hours Food preparation and cleanup
• 134.4 hours Lawn and garden care
• 2352.0 hours Working and work-related activities
• 2163.84 hours Working
• 3541.44 hours Leisure and sports
• 1921.92 hours Watching television
## Famous Sporting and Music Events on April 17
• 2011 "Game of Thrones", based on the fantasy novels by George R. R. Martin premieres on HBO
• 1976 NL greatest comeback: trailing 12-1, Philadelphia Phillies beat Chicago Cubs, 18-16 in 10 innings at Wrigley Field; Mike Schmidt hits 4 consecutive HRs
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# Mercator¶
The Mercator projection is a cylindrical map projection that origins from the 15th century. It is widely recognized as the first regularly used map projection. The projection is conformal which makes it suitable for navigational purposes.
Classification Conformal cylindrical Available forms Forward and inverse, spherical and elliptical projection Defined area Global, but best used near the equator Implemented by Gerald I. Evenden Options +lat_ts Latitude of true scale. Defaults to 0.0 +k_0 Scaling factor. Defaults to 1.0
## Usage¶
Applications should be limited to equatorial regions, but is frequently used for navigational charts with latitude of true scale (+lat_ts) specified within or near chart’s boundaries. Often inappropriately used for world maps since the regions near the poles cannot be shown [Evenden1995].
Example using latitude of true scale:
\$ echo 56.35 12.32 | proj +proj=merc +lat_ts=56.5
3470306.37 759599.90
Example using scaling factor:
echo 56.35 12.32 | proj +proj=merc +k_0=2
12545706.61 2746073.80
Note that +lat_ts and +k_0 are mutually exclusive. If used together, +lat_ts takes precedence over +k_0.
## Mathematical definition¶
The formulas describing the Mercator projection are all taken from G. Evenden’s libproj manuals [Evenden2005].
### Spherical form¶
For the spherical form of the projection we introduce the scaling factor:
$k_0 = \cos \phi_{ts}$
#### Forward projection¶
$x = k_0 \lambda$
$y = k_0 \ln \left[ \tan \left(\frac{\pi}{4} + \frac{\phi}{2} \right) \right]$
#### Inverse projection¶
$\lambda = \frac{x}{k_0}$
$\phi = \frac{\pi}{2} - 2 \arctan \left[ e^{-y/k_0} \right]$
### Elliptical form¶
For the elliptical form of the projection we introduce the scaling factor:
$k_0 = m\left( \phi_ts \right)$
where $$m\left(\phi\right)$$ is the parallel radius at latitude $$\phi$$.
We also use the Isometric Latitude kernel function $$t()$$.
Note
m() and t() should be described properly on a separate page about the theory of projections on the ellipsoid.
#### Forward projection¶
$x = k_0 \lambda$
$y = k_0 \ln t \left( \phi \right)$
#### Inverse projection¶
$\lambda = \frac{x}{k_0}$
$\phi = t^{-1}\left[ e^{ -y/k_0 } \right]$
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## What is Calculation Of Mean?
Example:
The following data relate to the age of a group of Govt. employees. Calculate the arithmetic mean and standard deviation:
Age 50—55 45—50 40—45 35-40 30—35 25—30 20—25
No. of 25 30 40 45 80 110 170
Employees.
Solution: Calculation of Mean and S.D. by step-deviation method
Age Mid- No. of Dev. From Step Devia- (X) (mv) (f) Dx dx2 Fdx fdx2 S0-5S 52.5 25 15 +3 9 +75 225 45—50 47.5 30 +10 +2 +60 120 40-45 42.5 40 +5 +1 +40 40 35—40 37.5 45 0 0 +0 0 30—35 32.5 80 — 5 —1 —80 80 25—30 27.5 110 —10 —2 —220 440 20-25 22.5 170 —15 —3 —510 1530 ∑f = 500 ∑fdx = -635 ∑fdx2 = 2435
Arithmetic Mean or X = A + (∑fdx)/N) × i = 37.5 + (-635)/500 × 5 = 31.15 years
Standard-Deviation σ = √(∑fdX2))/N- ((∑fdx)/N )2 ) × i = √(2435/500- ((-635)/500 )2 ) × 5
= √(3.27 )×5
## Email Based, Online Homework Assignment Help in Calculation Of Mean
Transtutors is the best place to get answers to all your doubts regarding calculation of mean. Transtutors has a vast panel of experienced Statistics tutorswho can explain the different concepts to you effectively. You can submit your school, college or university level homework or assignment to us and we will make sure that you get the answers related to calculation of mean.
## Related Topics
All Statistics Topics
More Q&A
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Home » Problems (Page 2)
# Category Archives: Problems
## A light year is the distance light travels in one year . How many meters are there in one light year?
The speed of light in vacuum is 3 x 10^8 m/s. That means in one second it travels 300000000 metres.
As we know, in one year, there are 365 days; in each day there are 4 hours; in each hour there are 60 minutes and in each minute there are 60 seconds.
Therefore, the distance traveled by light in one year = 3 x 10^8 x 365 x 24 x 60 x 60 = 1 light year = 9.4605284 × 1015 meters
## Learn Integral Calculus from Scratch
“Where can I learn integration fundamentals”
Just find below one of the best choice for you.
http://www.intmath.com/integration/integration-intro.php
Visit http://www.intmath.com/ for learning the fundamentals of Mathematics
## Factors affecting frequency of sound produced by a stretched string
Study how the frequency of sound produced will change in each case with the following strings of length 15cms when the strings are tied between 2 ends-
• aluminium string
• copper string
• cotton string
• metallic string
• jute string
Also study how the pitch changes when the strings are made taught and loose. Study how the frequency of sound changes with thickness of the following strings
• cotton strings
• copper strings
This seems to be a homework question or a project question. Therefore I am not giving a detailed answer so as not to tamper the basic aim of assigning a project.
The frequency of sound produced by a stretched string depends on the following factors:
1. the length of the string
2. the linear mass density (i.e; the mass per unit length) of the string
3. the tension in the string
When you are using strings of different materials, the factor which changes is the mass per unit length and the same is true when you are changing the thickness.
When you make the string more taut, the tension increases and vice versa.
The question is given for a constant length. Therefore the case of effect of changing length does not come into picture.
The formula showing the relationship is $\large \fn_jvn f=\frac{1}{2L}\sqrt{\frac{T}{m}}$
it is evident from the formula that the frequency of sound is
• inversely proportional to the length
• directly proportional to the square root of tension in the string and
• inversely proportional to the square root of linear density of the string.
on proper substitution, the formula can be recast as
$\large \fn_jvn f=\frac{1}{Ld}\sqrt{\frac{T}{\pi \rho }}$
and this will be more convenient for you to answer the questions.
I recommend that you try to explore by actually performing the experiments.
## Change in Resistance due to stretching a wire
“A piece of wire is redrawn without any change in volume so that its radius become half the original. Compare the new resistance with the original value.”
When we redraw the wire, the volume remains constant and the resistivity also remains constant. SO, the variables are
1. area of cross section and
2. the length
When radius becomes half, the area of cross section increases in such a way that A1l1 = A2l2. This implies that when length is halved, area of cross section is doubled.
We know that $\large \inline \fn_cs R=\frac{\rho l}{a}$
Therefore, the new resistance becomes (1/4) th the original value.
So, R2:R1 = 1:4
## More Problems from Kinematics
1. “On a 120 km track, a train moves the first 30 km at a uniform speed of 30 km/h. How fast must the train in the next 90 km so as to have average speed 60 km/h for the entire trip?”
2. “A person goes form one place to another on bike
with uniform speed 40 km/h and returns back at
uniform speed of 60 km/h. Find average speed
during the journey.”
## A numerical from Centre of mass
Two objects, of mass 1 kg and 2 kg, are moving with velocities equal to +2 m/sec and -3 m/sec. The center of mass of the two objects is moving at velocity
=(1×2+2x(-3)/(1+2)=(2-6)/3=-4/3=-1.33 m/s (-ve sign shows that the centre of mass will be moving in the direction of the second body)
## A numerical Problem from optics
When an object is kept at a distance of 60cm from a concave mirror, the magnification is 1/2. Where should the object be placed to get magnification of 1/3??
Ans:
In the first part, u=-60, m=(-)1/2 implies v=-30 cm (Since the image must be real as concave lens form virtual image of bigger size only)
Substituting in eqn
or
we get f = –20 cm
In II case
m=-1/3
v=u/3
f=-20cm
substituting in
we get u=4f=4 x (-20)=-80 cm
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# Probability (part 5) | Summary and Q&A
234.8K views
May 3, 2008
by
Probability (part 5)
## TL;DR
The video explains the probability of rolling a 7 in Monopoly using two six-sided dice.
## Install to Summarize YouTube Videos and Get Transcripts
### Q: How can the probability of rolling a 7 in Monopoly be calculated?
The probability can be calculated by finding the number of favorable outcomes (sum of two dice equal to 7) and dividing it by the total number of possible outcomes (6 x 6 = 36).
### Q: What are the favorable outcomes of rolling a 7?
The favorable outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) - a total of six outcomes.
### Q: What is the probability of rolling a 7 in Monopoly?
The probability is calculated as 6 favorable outcomes divided by 36 total outcomes, which equals 1/6 or approximately 0.167.
### Q: Can the grid be used to calculate probabilities of other outcomes?
Yes, the grid can be used to calculate the probabilities of various outcomes by counting the number of favorable outcomes and dividing by the total number of possible outcomes.
## Summary & Key Takeaways
• The content discusses the probability of rolling a 7 in Monopoly using two six-sided dice.
• A grid is used to represent all the possible outcomes of the two dice rolls.
• The top row represents the outcomes of the first dice, while the second row represents the outcomes of the second dice.
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# New Pattern Coding Decoding Quiz
Directions (Q. 1- 5): Study the following information to answer the given questions:
In a certain code,
“RUNG MADLY BUN JAR” is written as ‘24#G, 7%Y, 16\$N, [email protected]
“JAM RAPID BUT MONKEY” is written as ‘[email protected], 28#D, 10\$T, 8%Y’
“MACAROON BIRR RACISM JEEP” is written as ‘21%N, 13\$R, 20#M, [email protected]
“RATIO JUKE BAILEE MOD” is written as ‘17#O, [email protected], 28\$E, 26%D’
We Recommend Testbook APP 100+ Free Mocks For RRB NTPC & Group D Exam Attempt Free Mock Test 100+ Free Mocks for IBPS & SBI Clerk Exam Attempt Free Mock Test 100+ Free Mocks for SSC CGL 2021 Exam Attempt Free Mock Test 100+ Free Mocks for Defence Police SI 2021 Exam Attempt Free Mock Test 100+ Free Mocks for UPSSSC 2021 Exam Attempt Free Mock Test
1. The code for the word ‘RAN’ is
1.8%N
2.8*N
3.17#N
4.16#N
5.None of these
Explanation :
RAN – 16#N
R – #
No of letters – 3+(Reverse Alphabet order of N = 13) = 16
Last letter – N
2. The code ‘26#E’ denotes which of the following word?
1.ROSE
2.REVERSE
3.RUPEE
4.RANDOM
5.None of these
Explanation :
“26#E”– ROSE
R – #
Last letter – E
No of letters – 4+(Reverse Alphabet order of E = 22) = 26
3. Which of the following is the code for “MONK”?
1.20%K
2.26%K
3.28!K
4.24!K
5.None of these
Explanation :
M – %
Last letter – R
No of letters – 4+(Reverse Alphabet order of K = 16) = 20
4. Which of the following denotes @ symbol?
1.A
2.M
3.B
4.R
5.J
Explanation :
[email protected], M=%, R=#, B=\$
5. Which of the following is the code for ‘MUSIC BITE RACE JARL’ ?
[email protected], 13#C, 5?E, 13%L
2.21%E, 6\$L, 18#E, [email protected]
[email protected], 17#L, 7?E, 18%C
[email protected], 29%C, 26#E, [email protected]
5.None of these
Answer – [email protected], 29%C, 26#E, [email protected]
Explanation :
MUSIC BITE RACE JARL = [email protected], 29%C, 26#E, [email protected]
Q(6 –10) Study the information below and answer the following question: –
In a certain code language,
‘COT DAINTY FUN EPOCH’ is written as “&23T, \$31Y, *17N, #13H”
‘COW DOLL FLUENT EMBOW’ is written as “&26W, \$16L, *26T, #28W”
‘CON DOORS FAIR ELEVEN’ is written as “&17N, \$24S, *22R, #20N”
‘FLOPPY COR ENTER ENG’ is written as “*31Y, &21R, #23R, #10G”
1. Which of the following is the code for “FEAR”?
1.*22R
2.*13R
3.#15R
4.\$13R
5.None of these
Explanation :
First Position – F = *
Last Position – R
Middle Position – 18 + 4 = 22
2. Which of the following denotes # symbol?
1.C
2.D
3.E
4.F
5.None of these
Explanation :
# – E.
3. Which of the following denotes \$ symbol?
1.C
2.D
3.E
4.F
5.None of these
Explanation :
D = \$
4. The code ‘#24T’ denotes which of the following word?
1.Egg
2.Eagle
3.Elephant
4.Edit
5.None of these
Explanation :
First Position – symbol denotes first letter = (E-#)
Middle Position – 20 + 4 = 24; Last Position – T
5. By using the given code word, find the code word for ‘FAN COAL EBONY DACTYL’?
1. *18N, &15L, #32Y, \$18L
2. *17N, &16L, #31Y, \$18L
3. *17N, &16L, #23Y, \$18L
4. *18N, &16L, #33Y, \$18L
5. *17N, &16L, #30Y, \$18L
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## 20049
20,049 (twenty thousand forty-nine) is an odd five-digits composite number following 20048 and preceding 20050. In scientific notation, it is written as 2.0049 × 104. The sum of its digits is 15. It has a total of 3 prime factors and 8 positive divisors. There are 12,960 positive integers (up to 20049) that are relatively prime to 20049.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 15
• Digital Root 6
## Name
Short name 20 thousand 49 twenty thousand forty-nine
## Notation
Scientific notation 2.0049 × 104 20.049 × 103
## Prime Factorization of 20049
Prime Factorization 3 × 41 × 163
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 20049 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 20,049 is 3 × 41 × 163. Since it has a total of 3 prime factors, 20,049 is a composite number.
## Divisors of 20049
1, 3, 41, 123, 163, 489, 6683, 20049
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 27552 Sum of all the positive divisors of n s(n) 7503 Sum of the proper positive divisors of n A(n) 3444 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 141.594 Returns the nth root of the product of n divisors H(n) 5.82143 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 20,049 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 20,049) is 27,552, the average is 3,444.
## Other Arithmetic Functions (n = 20049)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 12960 Total number of positive integers not greater than n that are coprime to n λ(n) 3240 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2271 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 12,960 positive integers (less than 20,049) that are coprime with 20,049. And there are approximately 2,271 prime numbers less than or equal to 20,049.
## Divisibility of 20049
m n mod m 2 3 4 5 6 7 8 9 1 0 1 4 3 1 1 6
The number 20,049 is divisible by 3.
## Classification of 20049
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (20049)
Base System Value
2 Binary 100111001010001
3 Ternary 1000111120
4 Quaternary 10321101
5 Quinary 1120144
6 Senary 232453
8 Octal 47121
10 Decimal 20049
12 Duodecimal b729
20 Vigesimal 2a29
36 Base36 fgx
## Basic calculations (n = 20049)
### Multiplication
n×i
n×2 40098 60147 80196 100245
### Division
ni
n⁄2 10024.5 6683 5012.25 4009.8
### Exponentiation
ni
n2 401962401 8058944177649 161573771817684801 3239392551172762575249
### Nth Root
i√n
2√n 141.594 27.1663 11.8993 7.25134
## 20049 as geometric shapes
### Circle
Diameter 40098 125972 1.2628e+09
### Sphere
Volume 3.37572e+13 5.05121e+09 125972
### Square
Length = n
Perimeter 80196 4.01962e+08 28353.6
### Cube
Length = n
Surface area 2.41177e+09 8.05894e+12 34725.9
### Equilateral Triangle
Length = n
Perimeter 60147 1.74055e+08 17362.9
### Triangular Pyramid
Length = n
Surface area 6.96219e+08 9.49756e+11 16369.9
## Cryptographic Hash Functions
md5 f9d15850234291f8df6f03c468fb5cc7 b2e9e9dd348ff37a681303126d62eac2149e73c8 3b6bb6bdbfe2de241f81901cb5c52b61fcbd8effbe5e90d5393725e5c00d88f5 4a53cfdda58d3087d993919a62ba7748ddb7977f0a357635219ca12857215bc3b08656342403282c97525d4a97327ff1052e58d4e7efc94fa6408ca59f7076dd fe9fa5122e3eda6f4aeac20bb0d964a01c812a2f
| 1,453 | 4,107 |
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JEE > Important Formulas: Applications of Derivatives
# Important Formulas: Applications of Derivatives - Mathematics (Maths) for JEE Main & Advanced
``` Page 1
Page # 11
APPLICATION OF DERIVATIVES
1. Equation of tangent and normal
Tangent at (x
1
, y
1
) is given by (y – y
1
) = f ?(x
1
) (x – x
1
) ; when, f ?(x
1
) is real.
And normal at (x
1
, y
1
) is (y – y
1
) = –
) x ( f
1
1
?
(x – x
1
), when f ?(x
1
) is nonzero
real.
2. Tangent from an external point
Given a point P(a, b) which does not lie on the curve y = f(x), then the
equation of possible tangents to the curve y = f(x), passing through (a, b)
can be found by solving for the point of contact Q.
f ?(h) =
a h
b ) h ( f
?
?
And equation of tangent is y – b =
h a
f(h) b
?
?
(x – a)
3. Length of tangent, normal, subtangent, subnormal
(i) PT =
2
m
1
| k | 1 ? = Length of Tangent
p p( (h h, ,k k) )
N N M M T T
(ii) PN =
2
| k | 1 ? m = Length of Normal
(iii) TM =
m
k
= Length of subtangent
(iv) MN = |km| = Length of subnormal.
Page 2
Page # 11
APPLICATION OF DERIVATIVES
1. Equation of tangent and normal
Tangent at (x
1
, y
1
) is given by (y – y
1
) = f ?(x
1
) (x – x
1
) ; when, f ?(x
1
) is real.
And normal at (x
1
, y
1
) is (y – y
1
) = –
) x ( f
1
1
?
(x – x
1
), when f ?(x
1
) is nonzero
real.
2. Tangent from an external point
Given a point P(a, b) which does not lie on the curve y = f(x), then the
equation of possible tangents to the curve y = f(x), passing through (a, b)
can be found by solving for the point of contact Q.
f ?(h) =
a h
b ) h ( f
?
?
And equation of tangent is y – b =
h a
f(h) b
?
?
(x – a)
3. Length of tangent, normal, subtangent, subnormal
(i) PT =
2
m
1
| k | 1 ? = Length of Tangent
p p( (h h, ,k k) )
N N M M T T
(ii) PN =
2
| k | 1 ? m = Length of Normal
(iii) TM =
m
k
= Length of subtangent
(iv) MN = |km| = Length of subnormal.
Page # 12
4. Angle between the curves
Angle between two intersecting curves is defined as the acute angle between
their tangents (or normals) at the point of intersection of two curves (as shown
in figure).
tan ? =
2 1
2 1
m m 1
m m
?
?
5. Shortest distance between two curves
Shortest distance between two non-intersecting differentiable curves is always
along their common normal.
(Wherever defined)
6. Rolle’s Theorem :
If a function f defined on [a, b] is
(i) continuous on [a, b]
(ii) derivable on (a, b) and
(iii) f(a) = f(b),
then there exists at least one real number c between a and b (a < c < b) such
that f ?(c) = 0
7. Lagrange’s Mean Value Theorem (LMVT) :
If a function f defined on [a, b] is
(i) continuous on [a, b] and (ii) derivable on (a, b)
then there exists at least one real numbers between a and b (a < c < b) such
that
b a
f( )b f( )a
?
?
= f ?(c)
Page 3
Page # 11
APPLICATION OF DERIVATIVES
1. Equation of tangent and normal
Tangent at (x
1
, y
1
) is given by (y – y
1
) = f ?(x
1
) (x – x
1
) ; when, f ?(x
1
) is real.
And normal at (x
1
, y
1
) is (y – y
1
) = –
) x ( f
1
1
?
(x – x
1
), when f ?(x
1
) is nonzero
real.
2. Tangent from an external point
Given a point P(a, b) which does not lie on the curve y = f(x), then the
equation of possible tangents to the curve y = f(x), passing through (a, b)
can be found by solving for the point of contact Q.
f ?(h) =
a h
b ) h ( f
?
?
And equation of tangent is y – b =
h a
f(h) b
?
?
(x – a)
3. Length of tangent, normal, subtangent, subnormal
(i) PT =
2
m
1
| k | 1 ? = Length of Tangent
p p( (h h, ,k k) )
N N M M T T
(ii) PN =
2
| k | 1 ? m = Length of Normal
(iii) TM =
m
k
= Length of subtangent
(iv) MN = |km| = Length of subnormal.
Page # 12
4. Angle between the curves
Angle between two intersecting curves is defined as the acute angle between
their tangents (or normals) at the point of intersection of two curves (as shown
in figure).
tan ? =
2 1
2 1
m m 1
m m
?
?
5. Shortest distance between two curves
Shortest distance between two non-intersecting differentiable curves is always
along their common normal.
(Wherever defined)
6. Rolle’s Theorem :
If a function f defined on [a, b] is
(i) continuous on [a, b]
(ii) derivable on (a, b) and
(iii) f(a) = f(b),
then there exists at least one real number c between a and b (a < c < b) such
that f ?(c) = 0
7. Lagrange’s Mean Value Theorem (LMVT) :
If a function f defined on [a, b] is
(i) continuous on [a, b] and (ii) derivable on (a, b)
then there exists at least one real numbers between a and b (a < c < b) such
that
b a
f( )b f( )a
?
?
= f ?(c)
Page # 13
8. Useful Formulae of Mensuration to Remember :
1. Volume of a cuboid = ?bh.
2. Surface area of cuboid = 2( ?b + bh + h ?).
3. Volume of cube = a
3
4. Surface area of cube = 6a
2
5. Volume of a cone =
3
1
? ?r
2
h.
6. Curved surface area of cone = ?r ? ( ? = slant height)
7. Curved surface area of a cylinder = 2 ?rh.
8. Total surface area of a cylinder = 2 ?rh + 2 ?r
2
.
9. Volume of a sphere =
3
4
?r
3
.
10. Surface area of a sphere = 4 ?r
2
.
11. Area of a circular sector =
2
1
r
2
?, when ? is in radians.
12. Volume of a prism = (area of the base) × (height).
13. Lateral surface area of a prism = (perimeter of the base) × (height).
14. Total surface area of a prism = (lateral surface area) + 2 (area of
the base)
(Note that lateral surfaces of a prism are all rectangle).
15. Volume of a pyramid =
3
1
(area of the base) × (height).
16. Curved surface area of a pyramid =
2
1
(perimeter of the base) ×
(slant height).
(Note that slant surfaces of a pyramid are triangles).
```
## Mathematics (Maths) for JEE Main & Advanced
129 videos|359 docs|306 tests
## Mathematics (Maths) for JEE Main & Advanced
129 videos|359 docs|306 tests
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| 4.5 | 4 |
CC-MAIN-2023-40
|
latest
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| 0.842723 |
https://matt-rickard.com/zero-knowledge-proofs?ref=matt-rickard.com
| 1,725,731,264,000,000,000 |
text/html
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crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00509.warc.gz
| 381,047,928 | 7,379 |
# Zero Knowledge Proofs
Apr 4, 2022
How do you prove that you know something secret without revealing the secret?
Let's say that Alice wants to prove to her red-green color blind friend Bob that two otherwise indistinguishable balls, one green, and one red, are different colors – without revealing which is green and which is red.
Alice gives the balls to Bob. Next, Bob puts the balls behind his back, randomly switching hands or keeping them in the same hand 50% of the time.
Alice has to then "guess" whether Bob switched or didn't switch hands. If they are different colors, Alice should be able to guess correctly 100% of the time. If they are the same, she can't do better than 50% in the long run. The more guesses ("proofs") Alice performs, the higher the probability that Alice can tell the balls apart (2-t).
Alice has proved that she can tell the balls apart without revealing to Bob which ball is green and which is red.
| 212 | 939 |
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| 3.71875 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.958714 |
https://en.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/tangent-planes-and-local-linearization/v/local-linearization
| 1,721,065,529,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00675.warc.gz
| 195,862,084 | 104,896 |
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### Course: Multivariable calculus>Unit 3
Lesson 1: Tangent planes and local linearization
# Local linearization
A "local linearization" is the generalization of tangent plane functions; one that can apply to multivariable functions with any number of inputs. Created by Grant Sanderson.
## Want to join the conversation?
• The formula for local linearization reminded me of Taylor Polynomial (Taylor Approximation) in single variable functions. Are these two related in some way?
• Yes, when you take a Taylor polynomial and discard everything with larger than 1st order derivative, you get a local linearization for your single variable function - a line approximating your function at a given point. You can do this at multivariable calculus too - here you get a plane instead of a line. And off course, you can approximate your surfaces further with larger order partial derivatives - for examle quadratic approximations of multivariable functions are analogous to approximating single variable functions with 2nd degree Taylor polynomials.
• why did Grant say that it is not technically a linear function?
• I think he meant the linear function definition which requires that the graph of the function goes through the origin. The tangent planes in this video do not (in general) pass through the origin, therefore, they are not represented by linear functions, strictly speaking.
• If it is in 3D, can I randomly take two directional derivative to get the function of the plane? For example:
We have a function f(x, y), and we can find its gradient.
I randomly use a vector, say [1, 0], and find its directional derivative at a specific point P(a, b, c), and let's say the result is 5. Then we have a vector [1, 0, 5].
I do this again with a different vector, and get another vector, say [0, 1, -2].
Therefore the equation for the plane will be:
[x, y, z] = [a, b, c] + t[1, 0, 5] + s[0, 1, -2]
Is this method correct? If it is, is there a way to make this more general, to make it work in higher dimentions, or in other words, to vectorize it?
• Yes, this works perfectly fine. The simplest way is to always use the coordinate vectors, (1, 0) and (0, 1). If the plane is z = ax + by + c, then the gradient is (a, b) everywhere. Then taking the directional derivative in the x direction, we get a. In the y direction, it's b. So two vectors are (1, 0, a) and (0, 1, b), and we shift them by (0, 0, c). The parameterization is x = s, y = t, z = as + bt + c. But, as you might see, this isn't a really useful trick since you've just obtained a relatively obvious result.
• What's really cool to me is that this formula echoes exactly the equation for tangent lines in single variable calculus.
• ∇f( x₀ ) ⋅ ( x - x₀ ) + f( x₀ ) is the 3D equivalent of:
f’( xₒ ) ⋅ ( x - x₀ ) + f( xₒ )
• should it be del f(xo,yo) in the vector form?
• So in the multivariate case this is just the Jacobian matrix of f evaluated at a point p0, the best linear approximation to f at a point p0. In this univariate case, can we see this as the directional derivative in the direction of a "nudge vector" comprised of a nudge in x (dx) and a nudge in y (dy) directions, from x0 to x, plus an offset (f evaluated at x0) ? I'm trying to see how we generalize this from a regular single variable scalar functions, to multivariable scalar functions and multivariable vector functions, and if this applies to higher dimensions, i.e, tensors of 2, and above rank.
• At , why is the gradient expressed as f(x0) + ∇f(x0) dotted by (x-x0), instead of f(x0, y0) + ∇f(x0, y0) dotted by (x-x0, y-y0)?
(1 vote)
• The bold x represents the vector <x, y>
the bold xo is the vector <xo, yo>
• What does Grant mean when he says "variable multiplied by a constant"? It will still be a variable so why does he mention it?
(1 vote)
• For higher dimensional inputs would we have tangent n-spaces?
(1 vote)
• No, a tangent plane of a graph of a function requires 2 partial derivatives and a point of reference in space. For a function with n number of inputs, the condition still applies. However, if you have to find approximations or direction of steepest ascent in higher dimensional graphs, there are different tools for the same.
(1 vote)
| 1,131 | 4,435 |
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| 4.3125 | 4 |
CC-MAIN-2024-30
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latest
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http://www.ck12.org/algebra/percent-problems/lesson/user%3AemFwcHlzMDAxMGRAZ21haWwuY29t/Percent-Equations/
| 1,462,352,061,000,000,000 |
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| 424,908,318 | 28,514 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Percent Problems
## Find percents by using decimals.
Estimated12 minsto complete
%
Progress
Practice Percent Problems
Progress
Estimated12 minsto complete
%
Percent Equations
Suppose that there was an article in your school newspaper that said that 80% of the students in your school plan on attending the prom. It also said that 500 students in your school plan on attending the prom. Would you be able to tell from this information how many students there are in your school? In this Concept, you'll learn how to solve equations involving percentages so that you can determine information such as this.
### Watch This
For more help with percent equations, watch this 4-minute video recorded by Ken’s MathWorld. How to Solve Percent Equations (4:10)
### Guidance
Now that you remember how to convert between decimals and percents, you are ready for the percent equation:
The key words in a percent equation will help you translate it into a correct algebraic equation. Remember the equal sign symbolizes the word “is” and the multiplication symbol symbolizes the word “of.”
#### Example A
Find 30% of 85.
Solution: You are asked to find the part of 85 that is 30%. First, translate into an equation:
Convert the percent to a decimal and simplify:
#### Example B
A dime is worth what percent of a dollar?
Solution:
Since a dime is 10 cents and a dollar is 100 cents, we can set up the following equation:
#### Example C
50 is 15% of what number?
Solution: Translate into an equation:
Rewrite the percent as a decimal and solve:
### Vocabulary
Percent equation:
### Guided Practice
6 is 2% of what number?
Solution:
First, use the percent equation:
We can also use the fractional form of a percentage. Substitute in for 2%, since they are equivalent expressions:
6 is 2% of 300.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Percent Problems (14:15)
2. 50% of $9.00 is what number? 3. of 16 is what number? 4. 9.2% of 500 is what number 5. 8 is 20% of what number? 6. 99 is 180% of what number? 7. What percent of 7.2 is 45? 8. What percent of 150 is 5? 9. What percent of 50 is 2500? 10.$3.50 is 25% of what number?
| 643 | 2,601 |
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| 4.8125 | 5 |
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# Subsets of sets containing empty set [duplicate]
Why is $\{\emptyset\}$ not a subset of $\{\{\emptyset\}\}$?
It contains this element, but why is it not a subset?
-
## marked as duplicate by Rory Daulton, Claude Leibovici, Eric Wofsey, Joel Reyes Noche, Jack's wasted lifeFeb 5 at 12:19
You're trying to interpret the word "contain" too broadly. For example, a library may be said to contain letters, words, sentences, paragraphs, and chapters; but when we think of a library as a set, we generally think of it as a set of books, so only the books are members of the set, and not the piecemeal contents/components of the books. – Scott Feb 4 at 18:25
In addition to the excellent answers below, look at math.stackexchange.com/questions/1281436/… -- especially part 3 of the question. – David K Feb 4 at 22:41
$\in\ne\subset$. – Martín-Blas Pérez Pinilla Feb 5 at 7:26
Incidentally, there is a word for a set of which every element is also a subset, and that word is "transitive". $\{\{\emptyset\}\}$ is not a transitive set, but $\{\{\emptyset\},\emptyset\}$ is. – Steve Jessop Feb 5 at 11:05
It is not a subset because its element $\emptyset$ does not belong to the set $\{\{\emptyset\}\}$.
-
The element $\{\emptyset\}$ and the element $\emptyset$ are different.
$\{\emptyset\}$ is an element of $\{\{\emptyset\}\}$, whereas $\emptyset$ is not.
A subset is a set whose every element is also a part of the given set.
Thus, the subsets of $\{\{\emptyset\}\}$ are $\{\{\emptyset\}\}$ and the empty set $\{\}$, also denoted by $\emptyset$.
EDIT:
In one sentence, (Thanks to @Henry)
$\{\{∅\}\}$ has a single element $\{∅\}$ and two subsets $\{\{∅\}\}$ and $∅$, while $\{∅\}$ has a single element $∅$ and two subsets $\{∅\}$ and $∅$.
-
You could say that $\{\{\emptyset\}\}$ has a single element $\{\emptyset\}$ and two subsets $\{\{\emptyset\}\}$ and $\emptyset$. Meanwhile $\{\emptyset\}$ has a single element $\emptyset$ and two subsets $\{\emptyset\}$ and $\emptyset$. – Henry Feb 4 at 22:49
@Henry That's a nice way of saying it, have added it to the answer. Thank you. – GoodDeeds Feb 5 at 10:43
You can think of sets like plastic bags if you want; the empty set is just a plastic bag with nothing in it, $\{\emptyset\}$ is a plastic bag with another plastic bag in it, and $\{\{\emptyset\}\}$ is three layers of plastic bags.
The element relation $A\in B$ means that you could open up bag $B$ and take out $A$.
The subset relation $A\subset B$ means that every object that you could directly take out of $A$ can also be directly taken out of $B$.
So, look at $\{\emptyset\}$. You can "open it up and" take out $\emptyset$, but you can't do that with $\{\{\emptyset\}\}$. Therefore, $\{\emptyset\}\not\subset \{\{\emptyset\}\}$.
-
You didn't define B, which makes this answer a bit hard to follow. – djechlin Feb 4 at 22:08
@djechlin Fixed, thanks! – Deusovi Feb 4 at 22:24
sure you can, just open the bag and take the plastic bag with nothing inside out. – djechlin Feb 4 at 22:42
@153330 Then you have fun with whatever visualization or lack of visualization you want. – Deusovi Feb 5 at 3:02
The bag layers' count would be more obvious if you replace $\emptyset$ with$\{\,\}$ after the first use. – CiaPan Feb 5 at 10:16
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| 3.84375 | 4 |
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https://mulloverthings.com/are-convolutions-linear/
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| 420,176,224 | 7,223 |
MullOverThings
Useful tips for everyday
# Are convolutions linear?
## Are convolutions linear?
, Convolution is a linear operator and, therefore, has a number of important properties including the commutative, associative, and distributive properties.
## Is 2D convolution linear?
Convolution is a linear operation. It then follows that the multidimensional convolution of separable signals can be expressed as the product of many one-dimensional convolutions.
## Is convolution function linear operator?
Short answer, Convolution is a linear operator (check here) but what you are defining in context of CNN is not convolution, it is cross-correlation which is also linear in case of images (dot product). where each A(ejω) is the Fourier Transform of a(t) So this is the basic idea of discrete convolution.
## What is a 2D convolution?
The 2D convolution is a fairly simple operation at heart: you start with a kernel, which is simply a small matrix of weights. This kernel “slides” over the 2D input data, performing an elementwise multiplication with the part of the input it is currently on, and then summing up the results into a single output pixel.
## Which is better linear or circular convolution?
Linear convolution is the basic operation to calculate the output for any linear time invariant system given its input and its impulse response. Circular convolution is the same thing but considering that the support of the signal is periodic (as in a circle, hence the name).
## Why do we use linear convolution?
Linear convolution is a mathematical operation done to calculate the output of any Linear-Time Invariant (LTI) system given its input and impulse response. Circular convolution is essentially the same process as linear convolution. Circular convolution is also applicable for both continuous and discrete-time signals.
## What is a 2D convolution layer?
The 2D Convolution Layer A filter or a kernel in a conv2D layer “slides” over the 2D input data, performing an elementwise multiplication. The kernel will perform the same operation for every location it slides over, transforming a 2D matrix of features into a different 2D matrix of features.
## What is importance of 2D convolution?
Convolution is the most important and fundamental concept in signal processing and analysis. By using convolution, we can construct the output of system for any arbitrary input signal, if we know the impulse response of system.
## What is the difference between linear and circular convolution?
6 Answers. Linear convolution is the basic operation to calculate the output for any linear time invariant system given its input and its impulse response. Circular convolution is the same thing but considering that the support of the signal is periodic (as in a circle, hence the name).
## What are the two properties in linear operator?
A function f is called a linear operator if it has the two properties: f(x+y)=f(x)+f(y) for all x and y; f(cx)=cf(x) for all x and all constants c.
## What is the advantage of circular convolution?
Although DTFTs are usually continuous functions of frequency, the concepts of periodic and circular convolution are also directly applicable to discrete sequences of data. In that context, circular convolution plays an important role in maximizing the efficiency of a certain kind of common filtering operation.
## What are the applications of FFT algorithm?
It covers FFTs, frequency domain filtering, and applications to video and audio signal processing. As fields like communications, speech and image processing, and related areas are rapidly developing, the FFT as one of the essential parts in digital signal processing has been widely used.
## How to define discrete time convolution in 4.3?
Discrete time convolution is an operation on two discrete time signals defined by the integral (4.3.1) (f ∗ g) [ n] = ∑ k = − ∞ ∞ f [ k] g [ n − k] for all signals f, g defined on Z. It is important to note that the operation of convolution is commutative, meaning that
## How is multidimensional discrete convolution used in signal processing?
In signal processing, multidimensional discrete convolution refers to the mathematical operation between two functions f and g on an n -dimensional lattice that produces a third function, also of n -dimensions. Multidimensional discrete convolution is the discrete analog of the multidimensional convolution of functions on Euclidean space.
## Which is the result of a linear convolution?
The linear convolution result of two arbitrary M × N and P × Q image functions will generally be ( M + P − 1) × ( N + Q − 1), hence we would like the DFT G ˆ ˜ to have these dimensions. Therefore, the M × N function f and the P × Q function h must both be zero-padded to size ( M + P − 1) × ( N + Q − 1).
## Is the convolution of separable signals a linear operation?
Convolution is a linear operation. It then follows that the multidimensional convolution of separable signals can be expressed as the product of many one-dimensional convolutions.
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• https://me.yahoo.com
# Multiple Continuous Beams
The deflection of Continuous Beams with more than one span.
## Continuous Beams
When a Beam is carried on three or more supports it is said to be Continuous. It is possible to use an extension of the Moment-Area method ( See "Bending of Beams Part 3") to obtain a relationship between the Bending Moments at three points (Usually Supports.)
A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.
On the drawing the areas $\inline&space;A_1$ and $\inline&space;A_2$ are the Free Bending Moment areas obtained by treating the Beam as over two separate spans $\inline&space;l_1$ and $\inline&space;l_2$ . If the actual Bending Moments at these points are $\inline&space;M_1$, $\inline&space;M_2$ and $\inline&space;M_3$. Then a Fixing Moment diagram consisting of two trapezia can be introduced and the actual Bending Moment will be the Algebraic sum of the two diagrams.
In the lower figure the Elastic Line of the deflected Beam is shown.
The deflections $\inline&space;\delta&space;_1$ and $\inline&space;\delta&space;_2$ are measure relative to the left hand support and are positive upwards. $\inline&space;\displaystyle&space;\theta$ is the slope of the beam over the central support and $\inline&space;Z_1$ and $\inline&space;Z_2$ are the intercepts for $\inline&space;l_1$ and $\inline&space;l_2$
$\therefore\;\;\;\;\;\;\theta&space;&space;=&space;\frac{Z_1&space;+&space;\delta&space;_1}{l_1}&space;=&space;\frac{Z_2\;+\[\delta&space;_2&space;-&space;\delta&space;_1]}{l_2}$
Note. This assumes that the slopes everywhere are small.
$\frac{A_1\bar{x}_1&space;-&space;(\displaystyle\frac{M_1\;l_1}{2})(\displaystyle\frac{2\;l_1}{3})&space;-&space;(\displaystyle\frac{M_2l_1}{2})(\displaystyle\frac{2\;l_1}{3})}{E\;I}&space;+&space;\frac{\delta&space;_1}{l_1}$
$\;\;\;\;\;\;\;=&space;-&space;\frac{A_2\bar{x}_2&space;+&space;(\displaystyle\frac{M_3&space;l_2}{2})(\displaystyle\frac{2\;l_2}{3})&space;+&space;(\displaystyle\frac{M_2l_2}{2})(\displaystyle\frac{2\;l_2}{3})}{E\;I}&space;+&space;\frac{\delta&space;_2&space;-&space;\delta&space;_1}{l_1}$
Note that $\inline&space;\displaystyle&space;Z_2$ is a negative intercept.
The above equation can be written as:
$\frac{M_1\;l_1}{I_1}&space;+&space;2M_2\left(\frac{l_1}{I_1}&space;+&space;\frac{l_2}{I_2}&space;\right)&space;+&space;\frac{M+3\;l_2}{I_2}\;&space;=&space;6\left(\frac{A_1\bar{x}_1}{I_1\;l_1}&space;+&space;\frac{A_2\;\bar{x}_2}{I_2\;l_2}&space;\right)&space;+&space;6E\;\left[\left(\frac{\delta&space;_1}{l_1}\right)&space;+&space;\left(\frac{\delta&space;_1&space;-&space;\delta&space;_2}{l_2}&space;\right)&space;\right]$
If $\inline&space;\displaystyle&space;I_1&space;=&space;I_2$
$M_1\;l_1&space;+&space;2M_2(l_1&space;+&space;l_2)&space;+&space;M_3\;l_2&space;=&space;6\left(\frac{A_1\bar{x}_1}{l_2}&space;\right)&space;+&space;6\;E\;I\left[&space;\frac{\delta&space;_1}{l_1}&space;+&space;\frac{\delta&space;_1&space;-&space;\delta&space;_2}{l_2}\right]$
If the supports are at the same level:
$M_1\;l_1&space;+&space;2M_2(&space;l_1&space;&space;+&space;l_2)&space;+&space;M_3\;l_2&space;=&space;6\left(\frac{A_1\bar{x}_1}{l_2}&space;\right)&space;+&space;6\;E\;I\left(\frac{A_1\bar{x}_1}{l_1}&space;+&space;\frac{A_2\;\bar{x}_2}{l_2}&space;\right)$
If the Ends are Simply Supported then $\inline&space;\displaystyle&space;M_1&space;=&space;M_3&space;=&space;0$
$M_2(l_1&space;+&space;l_2)&space;=&space;3\;\left(\frac{A_1\;\bar{x}_1}{l_1}&space;+&space;\frac{A_2\;\bar{x}_2}{l_2}&space;\right)$
## Clapeyron's Equation Or The Equation Of Three Moments
Span is the distance between two intermediate supports for a structure, e.g. a beam or a bridge.
Equation (1) is the most general form of The Equation of Three Moments. Equations (2) (3) and (4) are simplifications to meet particular needs. Of these Equation (3) is the form most frequently required.
Example:
[imperial]
##### Example - Example 1
Problem
A Beam Ad 60 ft. long rests on supports at $\inline&space;A$, $\inline&space;B$, and $\inline&space;C$ which are at the same level. $\inline&space;AB&space;=&space;24&space;ft.$ and $\inline&space;BC&space;=&space;30&space;ft.$ The loading is 1 ton/ft. throughout and in addition a concentrated load of 5 tons acts at the mid-point of $\inline&space;AB$ and a load of 2 tons acts at $\inline&space;D$
Draw the Shear Force and Bending Moment diagrams.
Workings
$M_a&space;=&space;0$
$M_c&space;=&space;2\times6&space;+&space;6\times3&space;=&space;30\;tons-ft.$
Applying Equation (3) to the span $\inline&space;A\;B\;C$
$2M_b\times&space;54&space;+&space;30\times30&space;=$
$=&space;6\left[\left(\frac{1}{2}\times\frac{5\times24}{4}\times24&space;\right)\times\frac{12}{24}&space;+&space;\left(\frac{2}{3}\times\frac{24^2}{8}\times24&space;\right)\times\frac{12}{24}\;&space;+\;\left(\frac{2}{3}\times\frac{30^2}{8}\times30&space;\right)\times\frac{15}{30}&space;\right]$
$=&space;6\times1881$
$\therefore\;\;\;\;\;\;\;M_b&space;=&space;96.2\;tons-ft.$
The Bending Moment at mid-point of $\inline&space;AB$
$=&space;5\times&space;\frac{24}{4}&space;+&space;\frac{24^2}{8}&space;-&space;\frac{M_b}{2}&space;=&space;53.9\;tons-ft.$
The Bending Moment at the mid-point of $\inline&space;BC$
$=&space;\frac{30^2}{8}&space;-&space;\frac{1}{2}\left(M_b&space;+&space;30&space;\right)&space;=&space;49.4\;tons-ft.$
To find the reactions at the supports:
$\inline&space;M_b\;=&space;-&space;R_a\times&space;24&space;+&space;24\times&space;12&space;+&space;5\times&space;12$ for $\inline&space;A\;B$
$\inline&space;M_b\;=&space;-&space;R_c\times&space;30&space;+&space;36\times&space;18&space;+&space;2\times&space;36$ for $\inline&space;B\;C\;D$
$\therefore\;\;\;\;\;\;R_a&space;=&space;\left(\frac{288&space;+&space;60&space;-&space;96.2}{24}&space;\right)&space;=&space;10.49\;tons\;\;\;\;say&space;\;10.5\;tons$
And $\inline&space;R_c&space;=&space;\left(\displaystyle\frac{540&space;+&space;72&space;-&space;96.2}{30}&space;\right)&space;=&space;21.5\;tons$
By difference,
$R_b&space;=&space;60&space;+&space;5\;+2&space;-&space;10.5&space;-&space;21.5&space;=&space;35\;tons$
From the Shear Force diagram it can be seen that the maximum Bending Moment occurs either at a distance of 13.5 ft. from $\inline&space;C$ where:
$M&space;=&space;21.5\times&space;13.5&space;-&space;\frac{19.5^2}{2}&space;-&space;2\times19.5&space;=&space;62\;tons-ft.$
Or at a distance of 10.5 ft. from $\inline&space;A$ where:
$M&space;=&space;10,5\times&space;10.5&space;-&space;\frac{10.5^2}{2}&space;=&space;55.2\;tons-ft.$
The combined Bending Moment diagram is shown at the bottom of the sketch.
## Beams With More Than Two Spans.
Where a Beam extends over more than three Supports the Equation of Three Moments is applied to each group of three in turns. In general if there are $\inline&space;n$ Supports there will be $\inline&space;n&space;-&space;2$ unknown Bending Moments ( excluding the Ends) and $\inline&space;n&space;-&space;2$ equations to solve simultaneously.
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# Parallel Resistors Explained
Calculation of parallel resistors for basic electronic circuits – Parallel Resistors are resistors connected in parallel to other resistors.
Two resistors, three resistors and so on.
Well, previously we have studied the Explanation and function of series resistors. Now we will learn about Parallel Resistors.
The basic circuit of a parallel resistor is as follows:
Parallel Resistor Functions are:
• Decrease a specific resistance value.
• Current Divider
## A. Reducing Specific Resistance Values
Why should we reduce the resistance value?
Resistors have a standard value in their production. Resistor manufacturing companies, do not print resistors with many values.
Consider the following pdf, which contains 168 lists of resistor values with a tolerance of 5% that are commonly produced:
For example, in our circuit design, a resistor with a value of 2819 ohms or 2.8K Ohm is needed.
However, only 2.7K and 3K Ohm resistors are available on the market.
then, how can we make a resistor with a value of 2.8K ohms?
The trick is to make a parallel resistor using a resistor of 3.6K ohms and 13K ohms = 2.8K ohms.
The formula for parallel resistor circuit is:
Rtotal = 1/R1 + 1/R2 + 1/R3 + ..... + 1/Rn
Ket:
R1 = Resistor 1
R2 = Resistor 2
R3 = Resistor 3
Rn = Resistor n
*n = next value
## B. Parallel Resistor As Current Divider
Things to remember from this series are:
The voltage of each resistor is the same, while the current of each resistor is different“.
By their nature, the resistor will cause a voltage drop or voltage drop across the series resistor.
Notice in the image below, there are three resistors connected in series.
The voltage applied to points A and B is 10V.
Each resistor has a resistance value of 3K, 2K, and 5K Ohm connected in parallel.
When measuring the respective currents in the resistor using a multimeter, the results are 3V, 2V and 5V.
1. Find Rtotal
Rtotal = 1/R1 + 1/R2 + 1/R3
= 967.74 Ohm
2. Find Current (I1)
I1 = V / R1
= 10V / 3K
= 0.0033333333333333 A
= 3.3333333333333 mA
3. Find Current (I2)
I2 = V / R2
= 10V / 2K
= 0.005 A
= 5mA
4. Find Current (I3)
I3 = V / R3
= 10 / 5K
= 0.002 A
= 2mA
5. Find Total Current Itotal
Itotal = I1 + I2 + I3
= 3.3mA + 5mA + 2mA
= 10.3333333333333 mA
6. Check it
R = V / I
= 10V / 10.3333333333333 mA
= 10V / 0.0103333333333333 A
= 967.7419 Ohm
At this point, you understand, right?
Thank you for visiting the Chip Piko website. Hopefully this article can add to our insight in a series of series.
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# What is the relationship between corresponding sides, altitudes, and medians in similar triangles?
Dec 6, 2015
The ratio of their lengths is the same.
#### Explanation:
Similarity can be defined through a concept of scaling (see Unizor - "Geometry - Similarity").
Accordingly, all linear elements (sides, altitudes, medians, radiuses of inscribed and circumscribed circles etc.) of one triangle are scaled by the same scaling factor to be congruent to corresponding elements of another triangle.
This scaling factor is the ratio between the lengths of all corresponding elements and is the same for all elements.
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Law of Syllogism
Law of Syllogism
Imagine you are at the ticket counter of your favorite gaming store.
You are third in the queue.
Each transaction with a customer takes about 2 minutes.
You have only 10 minutes to get your ticket before the counter closes.
What can you infer from the above situation? Can you draw any conclusion?
You may be very happy that there is a fair chance that you can get your ticket.
While drawing this conclusion, you must have used the "Law of syllogism."
Those logical inferences that you draw are outcomes of the law of syllogism.
Even in a game of chess, you use several logical conclusions for taking the next steps in the game.
For example, if I defend my king what happens to the knight right up there!
Let us look at the law of syllogism in more detail and understand the definition of the law of syllogism and its examples in real life.
Lesson Plan
1 Law of Syllogism in Geometry 2 Thinking Out of the Box! 3 Solved Examples on Law Of Syllogism 4 Challenging Questions on Law Of Syllogism 5 Interactive Questions on Law Of Syllogism
Law of Syllogism in Geometry
Law of Syllogism Definition
The word "Syllogism" has a Greek origin and it means deduction or inference.
Syllogism refers to drawing inferences from given prepositions or sentences.
The Law of Syllogism is actually a part of deductive reasoning where we arrive at conclusions by logical reasoning.
It is similar to the transitive property: if a = b and b= c, then a=c.
It is like a chain rule.
Law of Syllogism Example
Statement 1: If it is a Monday, I have school.
Statement 2: If I have school, I have my math class.
The conclusion that we can draw from the above two statements is, "If it is a Monday, then I have math class."
This law of syllogism is a wonderful tool for proving many mathematical statements, especially in geometry.
Structure of a Syllogism
In the rule of syllogism, there are three parts involved.
Each of these parts is called a conditional argument.
The hypothesis is the conditional statement that follows after the word if.
The inference follows after the word then
To represent each phrase of the conditional statement, a letter is used.
The pattern looks like this:
Statement 1: If P, then Q.
Statement 2: If Q, then R.
Statement 3: If P, then R.
Statements 1 and 2 are called the premises of the given argument.
If they are true, then the correct inference must be statement 3.
Using the Law of Syllogism to Draw a Conclusion
Let us look at this geometry problem.
Draw a conclusion from the following true statements using the Law of Syllogism.
P: If a quadrilateral is a square, then it has four right angles.
Q: If a quadrilateral has four right angles, then it is a rectangle.
Here, statement P is true but statement Q is not true.
So, even though there is an immediate inference that a square is a rectangle, it is not valid as P is true, but not Q.
What Are the 3 Types Of Syllogisms?
Syllogisms are arguments usually with two statements (or premises).
Major premise: a point in general.
Minor premise: a particular argument.
The conclusion is based on both statements.
There are 3 main types of syllogisms. They are as follows.
Conditional Syllogism: If A is true, then B is true (If A, then B).
Categorical Syllogism: If A is in C, then B is in C.
Disjunctive Syllogism: If A is true, B is not true (A or B).
Now that we know what syllogism is about, let us try some examples to understand it better.
Think Tank
Choose the correct option using the law of syllogism:
Statement 1: All bats are mammals.
Statement 2: No birds are bats.
Conclusions:
a) No birds are mammals.
b) Some birds are not mammals.
c) No bats are birds.
d) All mammals are bats.
Solved Examples
Example 1
Help John draw a conclusion using the law of syllogism.
Statement 1: If a number ends in 0, then it is divisible by 10.
Statement 2: If a number is divisible by 10, then it is divisible by 5.
Solution
Let P be the statement "The number ends in 0"; let Q be the statement "It is divisible by 10"; and let R be the statement "It is divisible by 5."
Then (1) and (2) can be re-written as:
1) If P, then Q.
2) If Q, then R.
Thus, by the Law of Syllogism, we can infer:
3) If P, then R.
That means, if a number ends in 0, then it is divisible by 5.
$$\therefore$$ If a number ends in 0, then it is divisible by 5.
Example 2
Draw a conclusion using the law of syllogism.
All bikes have wheels. I ride a bike.
Solution
This scenario belongs to a categorical syllogism.
Major Premise: All bikes have wheels.
Minor Premise: I ride a bike.
So, by the law of syllogism, we can conclude that my bike has wheels.
$$\therefore$$ Conclusion is - My bike has wheels.
Example 3
Noah had a conversation with his friend.
Noah: All bunnies are cute.
Friend: My cousin's pet is also cute.
Therefore, the pet is a bunny.
Do you think the conclusion is valid?
Solution
These statements do not fall in any specific category of syllogism, so there is a high chance that we may end up in fallacy.
Major Premise: All bunnies are cute.
Minor Premise: My cousin's pet is also cute.
Conclusion: The pet is a bunny.
Not every cute pet is a bunny.
$$\therefore$$ The conclusion is not valid.
Example 4
Help Harry draw a conclusion using the law of syllogism.
$$\angle A$$ and $$\angle C$$ are equal.
$$\angle B$$ and $$\angle C$$ are equal.
Solution
So, the statements can be considered as:
P: $$\angle A = \angle C$$
Q: $$\angle B = \angle C$$
R: $$\angle A = \angle B$$
Then (1) and (2) can be re-written as:
1) If P, then Q.
2) If Q, then R.
So, by the Law of Syllogism, we can infer:
3) If P, then R.
Thus, $$\angle A = \angle B = \angle C$$
$$\therefore \angle A = \angle B = \angle C$$
Challenging Questions
1. Choose the valid conclusion using the law of syllogism:
Statements:
• All technicians are villagers.
• No villager is a doctor.
• All doctors are managers.
Conclusions:
1. No technician is a manager.
2. All villagers being managers is a possibility.
Interactive Questions
Here are a few activities for you to practice.
Let's Summarize
The mini-lesson targeted the fascinating concept of the law of syllogism. The math journey around the law of syllogism starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
1. What is the law of detachment?
The law of detachment helps to arrive at a new valid conclusion from the given statements.
1) If P, then Q
2) P
By the Law of Detachment, we can conclude that Q is valid.
For example,
1) If you are a bird, then you live in a nest.
2) You are a bird.
Let P be the statement, "You are a bird"; let Q be the statement, "You live in a nest".
By the Law of Detachment, we can conclude that Q is valid.
$$\therefore$$ You live in a nest.
2. What is the Law of Contrapositive?
The negation and inversion of the original statement which conveys the same meaning is called the contrapositive.
Interchange the hypothesis and the conclusion of the inverse statement to form the contrapositive of the given statement.
The law of contraposition states that the given statement is valid if and only if its contrapositive is true.
3. What is the Law of Converse?
Interchange the hypothesis and the conclusion in order to form the converse of the given statement.
For example,
Statement: If two triangles are congruent, then their corresponding angles are equal.
Converse: If the corresponding angles of the two triangles are equal, then the triangles are congruent.
Note that, the converse of a statement need not hold good in every case.
4. What are the three types of syllogism?
There are 3 main types of syllogisms. They are as follows.
Conditional Syllogism: If A is true, then B is true (If A, then B).
Categorical Syllogism: If A is in C, then B is in C.
Disjunctive Syllogism: If A is true, B is not true (A or B).
5. What is the pattern of a syllogism?
In the rule of syllogism, there are three conditional arguments.
The hypothesis is the conditional statement that follows after the word if.
The inference follows after the word then.
The pattern looks like this:
Major Premise: If P, then Q.
Minor Premise: If Q, then R.
Conclusion: If P, then R.
6. What is the purpose of syllogism?
The law of syllogism is especially used in proving geometrical statements.
It is also used to derive logical conclusions from the given statements (or premises).
More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus
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What process could you use to solve equations
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What process could you use to solve equations of the form (x + a)(x + b) = 0? In this case there are only two factors (values being multiplied) that are being set equal to 0. How would you solve the equation if there were more than two factors? Explain.
Dec 18, 2018
#1
0
could i use quadratic equation ?
Dec 18, 2018
#2
+109524
+1
What process could you use to solve equations of the form (x + a)(x + b) = 0? In this case there are only two factors (values being multiplied) that are being set equal to 0. How would you solve the equation if there were more than two factors? Explain.
What you have here is
(x + a) * (x + b) = 0
The ONLY way that two things can multiply together to give zero is if ONE or BOTH of them IS ZERO.
so
x+a=0 or x+b=0
Now you can 'digest' that properly and then finish the question.
Dec 18, 2018
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The number of different $n \times n$ symmetric matrices with each element being either 0 or 1 is: (Note: $\text{power} \left(2, X\right)$ is same as $2^X$)
1. $\text{power} \left(2, n\right)$
2. $\text{power} \left(2, n^2\right)$
3. $\text{power} \left(2,\frac{ \left(n^2+ n \right) }{2}\right)$
4. $\text{power} \left(2, \frac{\left(n^2 - n\right)}{2}\right)$
edited | 3.2k views
0
This might help ...
0
In symmetric matrix, $A[i][j] = A[j][i]$. So, we have choice only for either the upper triangular elements or the lower triangular elements. Number of such elements will be $n + (n-1) + (n-2) + \cdots + 1 = n\frac{(n+1)}{2} = \frac{(n^2+n)}{2}$. Now, each element being either 0 or 1 means, we have 2 choices for each element and thus for $\frac{(n^2+n)}{2}$ elements we have $2^{\frac{(n^2+n)}{2}}$ possibilities.
Choice is $C.$
edited
+14
The problem is similar to finding the number of symmetric relations possible on an n element set.
Now if we consider the n pairs of reflexive relation, those n pairs also satisfy the property of symmetric relation and each of these n pair may or may not be present in a symmetric relation.
Moreover there are (n2 - n)/2 diagonal pairs in an nxn matrix each of which should be present to have relation symmetric.
Means if we have pair (x,y) we need to have pair (y,x)
So, total symmetric relations possible on an n element set :
2n . 2 (n2-n)/2 = 2 n(n+1)/2
0
why not twice of this ?How about if matrix diagonal is from upper right to lower left.Would not that be one more case?
0
Didn't get you :(
0
what about counting case (ii) ?
0
Your second matrix is an invalid lower triangular matrix, as some elements below the main diagonal are zero.
A triangular matrix is upper or lower by the fact that the elements above or below the main diagonal respectively, are non-zero.
0
@ Ayush Upadhyaya The got the concept of no. of symmetric relation possible but i cannot understand how you relate it with matrix.
Matrices with 0 or 1 can be viewed as adjacency matrix of an undirected graph. There can be at max $n(n-1)/2$ edges in such graph.
So each edge can either be present or not. That gives total of $2^{(n^{2}-n)/2}$ , but this does not include n combination of self loops. so the total number of combincation is $2^{(n^{2}+n)/2}$.
for diagnol elements we have 2 choices either 0 or 1 So , 2^n for rest of the elements we can pair which will have 2 choices either 0 or 1 total pairs =(n^2-n)/2
2^n * 2^((n^2-n)/2) = Option C
consider a 5*5 square matrix, where entries in each cell represent the number of ways we can fill it. Since only 0 or 1 can be inserted therefore 2 ways.
$\bigl(\begin{smallmatrix} 2& 1 & 1& 1 &1\\ 2& 2 &1 &1 &1 \\ 2&2 &2 &1 & 1\\ 2&2 & 2 &2 &1 \\ 2&2 &2 & 2 & 2 \end{smallmatrix}\bigr)$
Now diagonal entries can be filled by any of 0 or 1 , so 2 ways for diagonal elements
but for non-diagonal we need to fix the elements in either upper/lower triangle of matrix, Because its a symmetric matrix hence it will be mirror image about diagonal.
for e.g. if A[2,1]= 1, then A[1,2]=1 OR if A[2,1]= 0, then A[1,2]=0
fixing lower triangle: number of ways to fill A[2,1] is 2 , so number of ways to fill A[1,2] is 1
otherwise fixing upper triangle: number of ways to fill A[1,2] is 2, so number of ways to fill A[2,1] is 1
number of n*n symmetric = $2^{n}*2^{n-1}*2^{n-1}$............$2^{1}$
=$2^{\frac{n(n+1)}{2}}$
1
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# Solution to Sum of Exponential Squared Series
• m00se
In summary, the "Solution to Sum of Exponential Squared Series" is a mathematical formula used to find the sum of a series of exponential squared terms. It is calculated by plugging in the values for "a" and "r" and using the formula for finding the sum of an infinite geometric series. This formula has various real-life applications in fields such as finance, physics, and biology. However, it is limited in its use as it only works for convergent series. The "Solution to Sum of Exponential Squared Series" can be used to solve real-world problems by calculating values and probabilities of events occurring over time and making predictions based on past data.
m00se
I know:
$$\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$$
However, is there a similar solution for:
$$\sum_{n=0}^\infty \left(\frac{x^n}{n!}\right)^2$$Thanks in advance; I'm not very good at this kind of maths (I teach statistics ), and I've been struggling with this one for a while.
Last edited:
## 1. What is the "Solution to Sum of Exponential Squared Series"?
The "Solution to Sum of Exponential Squared Series" is a mathematical formula used to find the sum of a series of exponential squared terms. It is often used in statistics and probability to calculate the probabilities of events occurring over a period of time.
## 2. How is the "Solution to Sum of Exponential Squared Series" calculated?
The formula for the "Solution to Sum of Exponential Squared Series" is: S = a + ar + ar^2 + ar^3 + ..., where "a" is the initial term, "r" is the common ratio, and the series continues until infinity. It is calculated by plugging in the values for "a" and "r" and then using the formula for finding the sum of an infinite geometric series.
## 3. What are some real-life applications of the "Solution to Sum of Exponential Squared Series"?
The "Solution to Sum of Exponential Squared Series" can be applied in various fields such as finance, physics, and biology. For example, it can be used to calculate compound interest in financial investments, decay rates of radioactive materials in physics, and growth rates of populations in biology.
## 4. Are there any limitations to using the "Solution to Sum of Exponential Squared Series"?
One limitation of using the "Solution to Sum of Exponential Squared Series" is that it assumes that the series is convergent, meaning that the sum of the terms approaches a finite value. If the series is divergent, meaning that the sum of the terms does not approach a finite value, then the formula cannot be used.
## 5. How can the "Solution to Sum of Exponential Squared Series" be used to solve real-world problems?
The "Solution to Sum of Exponential Squared Series" can be used to solve real-world problems by providing a way to calculate the total value or probability of a series of events occurring over time. It can also be used to make predictions and projections based on past data, such as in financial forecasting or population growth studies.
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q.1) find number of 5 digit number in which digit are in ascending order.
+5
for ascending, we need to avoid $0$, we cannot place it anywhere, so we're left with $9$ options,
So, we'll be choosing $5$ digits out of $9$,i.e. $^9C_5$
and after choosing there is only $1$ way to arrange them.
∴ Answer will be $^9C_5 \times 1 = ^{9}C_5 = 126$
0
0
repetition not allowed
0
in question it is mention that we have to arrange in ascending order ,have you any such number in which digit is repeated but it's still in ascending order???
+1
@BASANT
then it is non-decreasing order...
@srestha mam
i didn't ask for this particular question, i asked for generally, due to it's look more tough.. May anyone have any simple method?
0
@Shaik
in that case ans also be same
because , take any example
like 2235
arrangement done in 1 way only
0
no mam,
check this question https://gateoverflow.in/8399/gate2015-3-5
and should check the first link provided on the comment on the question
+1 vote
out of 1,2,3,4,5,6,7,8,9 pick any 5 digits. only one way to arrange them(ascending order). So, 9C5.
1
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# Math.Atan(Double) Method
## Definition
Returns the angle whose tangent is the specified number.
``````public:
static double Atan(double d);``````
``public static double Atan (double d);``
``static member Atan : double -> double``
``Public Shared Function Atan (d As Double) As Double``
#### Parameters
d
Double
A number representing a tangent.
#### Returns
Double
An angle, θ, measured in radians, such that -π/2 ≤ θ ≤ π/2.
-or-
NaN if `d` equals NaN, -π/2 rounded to double precision (-1.5707963267949) if `d` equals NegativeInfinity, or π/2 rounded to double precision (1.5707963267949) if `d` equals PositiveInfinity.
## Examples
The following example demonstrates how to calculate the arctangent of a value and display it to the console.
``````// This example demonstrates Math.Atan()
// Math.Atan2()
// Math.Tan()
using namespace System;
int main()
{
double x = 1.0;
double y = 2.0;
double angle;
double result;
// Calculate the tangent of 30 degrees.
angle = 30;
radians = angle * (Math::PI / 180);
Console::WriteLine( "The tangent of 30 degrees is {0}.", result );
// Calculate the arctangent of the previous tangent.
angle = radians * (180 / Math::PI);
Console::WriteLine( "The previous tangent is equivalent to {0} degrees.", angle );
// Calculate the arctangent of an angle.
String^ line1 = "{0}The arctangent of the angle formed by the x-axis and ";
String^ line2 = "a vector to point ({0},{1}) is {2}, ";
String^ line3 = "which is equivalent to {0} degrees.";
radians = Math::Atan2( y, x );
angle = radians * (180 / Math::PI);
Console::WriteLine( line1, Environment::NewLine );
Console::WriteLine( line2, x, y, radians );
Console::WriteLine( line3, angle );
}
/*
This example produces the following results:
The tangent of 30 degrees is 0.577350269189626.
The previous tangent is equivalent to 30 degrees.
The arctangent of the angle formed by the x-axis and
a vector to point (1,2) is 1.10714871779409,
which is equivalent to 63.434948822922 degrees.
*/
``````
``````// This example demonstrates Math.Atan()
// Math.Atan2()
// Math.Tan()
using System;
class Sample
{
public static void Main()
{
double x = 1.0;
double y = 2.0;
double angle;
double result;
// Calculate the tangent of 30 degrees.
angle = 30;
Console.WriteLine("The tangent of 30 degrees is {0}.", result);
// Calculate the arctangent of the previous tangent.
Console.WriteLine("The previous tangent is equivalent to {0} degrees.", angle);
// Calculate the arctangent of an angle.
String line1 = "{0}The arctangent of the angle formed by the x-axis and ";
String line2 = "a vector to point ({0},{1}) is {2}, ";
String line3 = "which is equivalent to {0} degrees.";
Console.WriteLine(line1, Environment.NewLine);
Console.WriteLine(line3, angle);
}
}
/*
This example produces the following results:
The tangent of 30 degrees is 0.577350269189626.
The previous tangent is equivalent to 30 degrees.
The arctangent of the angle formed by the x-axis and
a vector to point (1,2) is 1.10714871779409,
which is equivalent to 63.434948822922 degrees.
*/
``````
``````' This example demonstrates Math.Atan()
' Math.Atan2()
' Math.Tan()
Class Sample
Public Shared Sub Main()
Dim x As Double = 1.0
Dim y As Double = 2.0
Dim angle As Double
Dim result As Double
' Calculate the tangent of 30 degrees.
angle = 30
radians = angle *(Math.PI / 180)
Console.WriteLine("The tangent of 30 degrees is {0}.", result)
' Calculate the arctangent of the previous tangent.
angle = radians *(180 / Math.PI)
Console.WriteLine("The previous tangent is equivalent to {0} degrees.", angle)
' Calculate the arctangent of an angle.
Dim line1 As [String] = "{0}The arctangent of the angle formed by the x-axis and "
Dim line2 As [String] = "a vector to point ({0},{1}) is {2}, "
Dim line3 As [String] = "which is equivalent to {0} degrees."
angle = radians *(180 / Math.PI)
Console.WriteLine(line1, Environment.NewLine)
Console.WriteLine(line3, angle)
End Sub
End Class
'
'This example produces the following results:
'
'The tangent of 30 degrees is 0.577350269189626.
'The previous tangent is equivalent to 30 degrees.
'
'The arctangent of the angle formed by the x-axis and
'a vector to point (1,2) is 1.10714871779409,
'which is equivalent to 63.434948822922 degrees.
'
``````
## Remarks
A positive return value represents a counterclockwise angle from the x-axis; a negative return value represents a clockwise angle.
Multiply the return value by 180/Math.PI to convert from radians to degrees.
This method calls into the underlying C runtime, and the exact result or valid input range may differ between different operating systems or architectures.
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# Resources tagged with: Working systematically
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### There are 342 results
Broad Topics > Thinking Mathematically > Working systematically
### Jigsaw Pieces
##### Age 7 to 11 Challenge Level:
How will you go about finding all the jigsaw pieces that have one peg and one hole?
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Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes?
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Design an arrangement of display boards in the school hall which fits the requirements of different people.
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Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
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Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line?
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What is the best way to shunt these carriages so that each train can continue its journey?
### Tetrahedra Tester
##### Age 11 to 14 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### Factor Lines
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Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
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Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?
### LOGO Challenge - the Logic of LOGO
##### Age 11 to 16 Challenge Level:
Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?
### Map Folding
##### Age 7 to 11 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Waiting for Blast Off
##### Age 7 to 11 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### Maths Trails
##### Age 7 to 14
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Square Corners
##### Age 7 to 11 Challenge Level:
What is the greatest number of counters you can place on the grid below without four of them lying at the corners of a square?
### Four Triangles Puzzle
##### Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Window Frames
##### Age 5 to 14 Challenge Level:
This task encourages you to investigate the number of edging pieces and panes in different sized windows.
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### LOGO Challenge - Triangles-squares-stars
##### Age 11 to 16 Challenge Level:
Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.
### Isosceles Triangles
##### Age 11 to 14 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### First Connect Three
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Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for?
### Putting Two and Two Together
##### Age 7 to 11 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Cuboids
##### Age 11 to 14 Challenge Level:
Can you find a cuboid that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Number Sandwiches
##### Age 7 to 14 Challenge Level:
Can you arrange the digits 1, 1, 2, 2, 3 and 3 to make a Number Sandwich?
### Bochap Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku combines all four arithmetic operations.
### Making Maths: Double-sided Magic Square
##### Age 7 to 14 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
##### Age 11 to 14 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
### Broken Toaster
##### Age 7 to 11 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Rectangle Outline Sudoku
##### Age 11 to 16 Challenge Level:
Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.
### Twin Corresponding Sudoku III
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Twinkle Twinkle
##### Age 7 to 14 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Intersection Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
##### Age 7 to 11 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Number Daisy
##### Age 11 to 14 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Three Sets of Cubes, Two Surfaces
##### Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
### Painting Possibilities
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
### Knight's Swap
##### Age 7 to 11 Challenge Level:
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Making Squares
##### Age 7 to 11
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### Button-up Some More
##### Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
##### Age 7 to 11 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
##### Age 11 to 14 Challenge Level:
A few extra challenges set by some young NRICH members.
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Ratio Sudoku 3
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios or fractions.
### Ratio Sudoku 1
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios.
### Coded Hundred Square
##### Age 7 to 11 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Wallpaper Sudoku
##### Age 11 to 16 Challenge Level:
A Sudoku that uses transformations as supporting clues.
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Quick Answer: Is .2 A Real Number?
How do you know if a number is a real number?
Points to the right are positive, and points to the left are negative.
Any point on the line is a Real Number: The numbers could be whole (like 7).
Is 3 a real number?
The real numbers include natural numbers or counting numbers, whole numbers, integers, rational numbers (fractions and repeating or terminating decimals), and irrational numbers. The set of real numbers is all the numbers that have a location on the number line. Integers …, −3, −2, −1, 0, 1, 2, 3, …
What does R mean in math?
real numbersList of Mathematical Symbols • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Page 1. List of Mathematical Symbols. • R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers.
Is infinity a real number?
Infinity is a “real” and useful concept. However, infinity is not a member of the mathematically defined set of “real numbers” and, therefore, it is not a number on the real number line.
Is irrational number is a real number?
In mathematics, the irrational numbers are all the real numbers which are not rational numbers. That is, irrational numbers cannot be expressed as the ratio of two integers.
What are not real numbers?
A non-real, or imaginary, number is any number that, when multiplied by itself, produces a negative number. Mathematicians use the letter “i” to symbolize the square root of -1. An imaginary number is any real number multiplied by i. For example, 5i is imaginary; the square of 5i is -25.
Is a decimal considered a real number?
Therefore, all of these rational and irrational numbers, including fractions, are considered real numbers. Real numbers that include decimal points are known as floating point numbers because the decimal floats within the numbers.
What kind of number is zero?
0 is a rational, whole, integer and real number. Some definitions include it as a natural number and some don’t (starting at 1 instead).
Which are the real numbers?
All the natural numbers are integers but not all the integers are natural numbers. … Real numbers are the numbers which include both rational and irrational numbers. Rational numbers such as integers (-2, 0, 1), fractions(1/2, 2.5) and irrational numbers such as √3, π(22/7), etc., are all real numbers.
Is zero a number Yes or no?
Zero is not positive or negative. Even though zero is not a positive number, it’s still considered a whole number. … So, to answer the question is zero a natural number – yes it is on a number line and when identifying numbers in a set; but also no, because it’s not used to count objects.
Can you Google search a phone number?
There are many ways of looking up a cell phone online, including using Google. You can easily do a free reverse phone number lookup with Google online. The search engine actually used to have a service, Phonebook Search Operator, that was specifically for those who wanted to look up a phone number.
Is 5 a whole number?
In mathematics, whole numbers are the basic counting numbers 0, 1, 2, 3, 4, 5, 6, … and so on. 17, 99, 267, 8107 and 999999999 are examples of whole numbers. Whole numbers include natural numbers that begin from 1 onwards.
Is 3.14 a real number?
Rational Numbers and Pi The number pi, denoted π, is a well-known irrational number that is generally known to have value 3.14. However, 3.14 is actually the number π rounded to two decimal places, and is not the true value of π.
What are the 5 subsets of real numbers?
The real numbers have the following important subsets: rational numbers, irrational numbers, integers, whole numbers, and natural numbers.
Is .1 a real number?
What Are Real Numbers? Edit. Real numbers consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers.
Is 12 a real number?
Integers: {… … As you can see, −12 is an integer, but it is also a rational number because it can be made into a fraction: −121 and it is real because it can be found on the number line.
Is 1 a prime number?
Proof: The definition of a prime number is a positive integer that has exactly two positive divisors. However, 1 only has one positive divisor (1 itself), so it is not prime. … A prime number is a positive integer whose positive divisors are exactly 1 and itself.
What is the difference between natural and real numbers?
): The counting numbers {1, 2, 3, …} are commonly called natural numbers; however, other definitions include 0, so that the non-negative integers {0, 1, 2, 3, …} are also called natural numbers. Natural numbers including 0 are also called whole numbers. … All rational numbers are real, but the converse is not true.
Is √ 3 an irrational number?
The square root of 3 is the positive real number that, when multiplied by itself, gives the number 3. The square root of 3 is an irrational number. … It is also known as Theodorus’ constant, named after Theodorus of Cyrene, who proved its irrationality.
Is 0.4 A irrational number?
Answer. since the number is non terminating and non repeating therefore it is irrational number.
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# Network Theory Questions and Answers – Dot Convention in Magnetically Coupled Circuits
«
»
This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Dot Convention in Magnetically Coupled Circuits”.
1. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________
a) 25
b) 50
c) 100
d) 200
Explanation: Q = $$\frac{f_0}{BW}$$
And f0 = 1/2π (LC)0.5
BW = R/L
Or, Q = $$\frac{1}{R} (\frac{L}{C})^{0.5}$$
When R, L and C are doubled, Q’ = 50.
2. In the circuit given below, the input impedance ZIN of the circuit is _________
a) 0.52 – j4.30 Ω
b) 0.52 + j15.70 Ω
c) 64.73 + j17.77 Ω
d) 0.3 – j33.66 Ω
Explanation: ZIN = (-6j) || (ZA)
ZA = j10 + $$\frac{12^2}{(j30+j6-j2+4)}$$
= 0.49 + j5.82
ZIN = $$\frac{(-j6)(0.49+j5.82)}{(-j6+0.49+j5.82)}$$
= 64.73 + j17.77 Ω.
3. The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I(t) for t>0 is ________
a) 0.2e-125tu(t) mA
b) 20e-1250tu(t) mA
c) 0.2e-1250tu(t) mA
d) 20e-1000tu(t) mA
Explanation: CEQ = $$\frac{0.8 × 0.2}{0.8+0.2}$$ = 0.16
VC (t=0) = 100 V
At t≥0,
The discharging current I (t) = $$\frac{V_O}{R} e^{-\frac{t}{RC}}$$
= $$\frac{100}{5000} e^{- \frac{t}{5×10^3×0.16×10^{-6}}}$$
= 0.2e-1250tu(t) mA.
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4. In the circuit shown, the voltage source supplies power which is _____________
a) Zero
b) 5 W
c) 10 W
d) 100 W
Explanation: Let the current supplied by a voltage source.
Applying KVL in outer loop,
10 – (I+3) × (I+1) – (I+2) × 2 = 0
10 – 2(I+3) – 2(I-2) = 0
Or, I = 0
∴ Power VI = 0.
5. In the circuit shown below the current I(t) for t≥0+ (assuming zero initial conditions) is ___________
a) 0.5-0.125e-1000t A
b) 1.5-0.125e-1000t A
c) 0.5-0.5e-1000t A
d) 0.375e-1000t A
Explanation: I (t) = $$\frac{1.5}{3}$$ = 0.5
LEQ = 15 mH
REQ = 5+10 = 15Ω
L = $$\frac{L_{EQ}}{R_{EQ}}$$
= $$\frac{15 × 10^3}{15} = \frac{1}{1000}$$
I (t) A – (A – B) e-t = 0.5 – (0.5-B) e-1000t
= 0.5(0.5 – 0.375) e-1000t
= 0.5 – 0.125 e-1000t
I (t) = 0.5-0.125e-1000t.
6. Initial voltage on capacitor VO as marked |VO| = 5 V, VS = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0+ is _____________
a) 1 V
b) -1 V
c) $$\frac{13}{3}$$ V
d) –$$\frac{13}{3}$$ V
Explanation: Applying voltage divider method, we get,
I = $$\frac{V}{R_{EQ}}$$
= $$\frac{8}{1+1||1}$$
= $$\frac{8}{1+\frac{1}{2}} = \frac{16}{3}$$ A
I1 = $$\frac{16}{3} × \frac{1}{2} = \frac{8}{3}$$ A
And $$I’_2 = \frac{V}{R_{EQ}} = \frac{5}{1+1||1}$$
= $$\frac{5}{1+\frac{1}{2}} = \frac{10}{3}$$ A
Now, $$I’_1 = I’_2 × \frac{1}{1+1}$$
= $$\frac{10}{3} × \frac{1}{2} = \frac{5}{3}$$ A
Hence, the net current in 1Ω resistance = I1 + $$I’_1$$
= $$\frac{8}{3} + \frac{5}{3} = \frac{13}{3}$$ A
∴ Voltage drop across 1Ω = $$\frac{13}{3} × 1 = \frac{13}{3}$$ V.
7. For a unit step signal u (t), the response is V1 (t) = (1-e-3t) for t>0. If a signal 3u (t) + δ(t) is applied, the response will be (considering zero initial conditions)?
a) (3-6e-3t)u(t)
b) (3-3e-3t)u(t)
c) 3u(t)
d) (3+3e-3t)u(t)
Explanation: For u (t) = 1, t>0
V1 (t) = (1-e-3t)
Or, V1 (s) = $$\left(\frac{1}{s} + \frac{1}{s+3}\right) = \frac{3}{s(s+3)}$$
And T(s) = $$\frac{V_1 (S)}{u(S)} = \frac{3}{s+3}$$
Now, for R(s) = ($$\frac{3}{s}$$ + 1)
Response, H(s) = R(s) T(s) = $$(\frac{3+s}{s}) (\frac{3}{s+3}) = \frac{3}{s}$$
Or, h (t) = 3 u (t).
8. In the circuit given below, for time t<0, S1 remained closed and S2 open., S1 is initially opened and S2 is initially closed. If the voltage V2 across the capacitor C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ will be ____________
a) 1 V
b) 2 V
c) 1.5 V
d) 3 V
Explanation: When S1 is closed and S2 is open,
VC1 (0) = VC1 (0+) = 3V
When S1 is opened and S2 is closed, VC2 (0+) = VC2 (0+) = 3V.
9. In the circuit given below, the switch S1 is initially closed and S2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through C during t=0+ is _____________
a) 55 A
b) 5.5 A
c) 45 A
d) 4.5 A
Explanation: By KCL, we get,
$$\frac{V_L}{10} – 10 + \frac{V_L-10}{10}$$ = 0
Hence, 2 VL = 110
∴ VL = 55 V
Or, IC = $$\frac{55-10}{10}$$ = 4.5 A.
10. In the circuit given below, the switch S1 is initially closed and S2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0+ is _____________
a) 55 A
b) 5.5 A
c) 45 A
d) 4.5 A
Explanation: By KCL, we get,
$$\frac{V_L}{10} – 10 + \frac{V_L-10}{10}$$ = 0
Hence, 2 VL = 110
∴ VL = 55 V.
11. An ideal capacitor is charged to a voltage VO and connected at t=0 across an ideal inductor L. If ω = $$\frac{1}{\sqrt{LC}}$$, the voltage across the capacitor at time t>0 is ____________
a) VO
b) VO cos(ωt)
c) VO sin(ωt)
d) VO e-ωtcos(ωt)
Explanation: Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.
Voltage across capacitor will be decreasing from VO and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t>0 is VO cos(ωt).
12. In the figure given below, what is the RMS value of the periodic waveform?
a) 2$$\sqrt{6}$$ A
b) 6$$\sqrt{2}$$ A
c) $$\sqrt{\frac{4}{3}}$$ A
d) 1.5 A
Explanation: The rms value for any waveform is = $$\sqrt{\frac{1}{T} \int_0^T f^2 (t)dt)}$$
= $$[\frac{1}{T}(\int_0^{\frac{T}{2}}(mt)^2 dt + \int_{\frac{T}{2}}^T 6^2 dt]^{1/2}$$
= $$[\frac{1}{T}(\frac{144}{T^2} × \frac{T^3}{3×8} + 36 × \frac{T}{2})]^{\frac{1}{2}}$$
= $$[6+18]^{\frac{1}{2}} = \sqrt{24} = 2\sqrt{6}$$ A.
13. In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________
a) 14.7 A
b) 18.5 A
c) 40 A
d) 50 A
Explanation: Using KVL, 100 = R$$\frac{dq}{dt} + \frac{q}{C}$$
Or, 100 C = RC$$\frac{dq}{dt}$$ + q
Now, $$\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t dt$$
Or, 100C – q = (100C – qo) e-t/RC
I = $$\frac{dq}{dt} = \frac{(100C – q_0)}{RC} e^{-1/1}$$
= 40e-1 = 14.7 A.
14. In the circuit given below, the switch is closed at time t=0. The voltage across the inductance just at t=0+ is ____________
a) 2 V
b) 4 V
c) -6 V
d) 8 V
Explanation: A t=0+,
I (0+) = $$\frac{10}{4||4+3} = \frac{10}{5}$$ = 2A
∴ I2(0+) = $$\frac{2}{2}$$ = 1A
VL(0+) = 1 × 4 = 4 V.
15. A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________
a) A[1 – exp(-$$\frac{B}{RC}$$)]
b) $$\frac{AB}{RC}$$
c) A
d) A exp(-$$\frac{B}{RC}$$)
Explanation: VC = $$\frac{1}{C} ∫Idt$$
= $$\frac{1}{C} ∫_0^B \frac{A}{R} e^{-\frac{t}{RC}}$$ dt
VC = A[1 – exp(-$$\frac{B}{RC}$$)]
Hence, maximum voltage = V [1 – exp (-$$\frac{B}{RC}$$)].
Sanfoundry Global Education & Learning Series – Network Theory.
To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.
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# Katie Robertson
(423) 623-3811 ext. 319
# Bio
I am one of the first grade teachers here at Newport Grammar School. I went to NGS K-8 and Cocke County High School 9-12. I graduated from Carson-Newman University with my Bachelors in Education (K-6). I have been teaching for 12 years. This is my fifth year teaching at Newport Grammar School. My husband's name is Chris and we have two sons, Shane (age 7) and Theo (age 5). I graduated from Cumberland University in May 2022 with my Masters of Arts in Education.
### Watch my "ALL ABOUT ME" video!
Mrs. Robertson's First Grade Class (2023-24)
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Ethan H.
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Lux J.
Blake K.
Alyssa N.
Leighanna R.
Boston R.
Dylan W.
Rylee W.
### Here's what we're learning right now...
These are the TN standards we are currently covering in our curriculum:
1st Quarter Math Essential Questions:
iReady Unit 2: Numbers within 20, Addition and Subtraction and Representing Data
1.OA.A.1 Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g. by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
1.OA.B.3 Apply properties of operations as strategies to add and subtract.
1.OA.C.6 Add and subtract with 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on, making ten; decomposing a number leading to a ten using the relationship between addition and subtraction; and creating equivalent but easier or known sums.
Represent and interpret data.
1.MD.C.5 Organize, represent, and interpret data with up to three categories. Ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
2nd Quarter ELA Essential Questions:
WRITING FOCUS: Informational Writings
SS UNIT: Citizenship Unit and Change Over Time Unit
Unit 2 ESSENTIAL QUESTIONS:
Week of 10/23/23
Unit 2 Week 3 Story → At a Pond
Where do animals live together?
What are possessive nouns?
Reading Skills: Main Idea and Key Details
Spelling Pattern: nd, sk, nk word families
Week of 10/30/23
Unit 2 Week 4 Story → Nell’s Books
How do people help out in a community?
What are the elements of fiction, especially fantasies?
What are nouns, esp. common and proper nouns?
What are syllables?
Reading Skills: Visualize, Key Details, Plot
Spelling Pattern: th, sh, ng word families
Week of 11/6/23
Unit 2 Week 5 Story → Follow the Map
How can you find your way around?
What are plural nouns? How should you add -s or -es to a word to make it plural?
Spelling Pattern: ch, tch, wh, ph word families
Wonders Unit 3 ESSENTIAL QUESTIONS:
Week of 11/13/23
Unit 3 Week 1 Story → On My Way to School
How do we measure time?
What is a fantasy?
What are onomatopoeias?
How do I identify character, setting, and plot?
How do I use commas in a series?
How do I use past and present tense verbs?
Spelling Pattern: Long a: a, a_e
Week of 11/27/23
Unit 3 Week 2 Story → The Big Yuca Plant
How do plants change as they grow?
How do I identify the plot and sequence of a story?
What is fantasy?
How do I make and confirm predictions?
How should I use present and past tense verbs?
How do I distinguish between singular and plural nouns?
Spelling Pattern: Long i: i_e
Week of 12/4/23
Unit 3 Week 3 Story → The Gingerbread Man
What is a folktale?
What is a verb? What is subject-verb agreement?
What is past tense? What is future tense?
How can I use commas correctly?
Reading Skills: Plot, Cause/Effect, Dictionary Skills
Spelling Pattern: Soft c and g
Week of 1/8/24
Unit 3 Week 4 Story → Long Ago and Now
How is life different than it was long ago?
How can I use commas in dates?
When should I use is? are?
Reading Skills: Compare and Contrast within Text; Fact or Opinion; Nonfiction
Spelling Pattern: Long o: o_e; Long u: u_e; Long e: e_e
RLA Standards/Objectives
Click on the document below for a detailed list of Wonders Unit 1-6 ELA standards.
### Starting 1st Grade- What You Need to Know
Click on this week's PDF file to see a list of important dates and events as well as this week's homework assignments.
### SPELLING CHOICE BOARD
Please remember that the Newport Grammar School's homepage has TONS of digital resources for your everyday use.
Just scroll to the VERY top of this webpage, look to the far right, and click on the yellow rectangle that says "Digital Resources" for all sorts of FABULOUS links and icons. Or...you can simply click here:
https://www.newportgrammar.org/digital
"Useful Links" is also another great yellow rectangle you can click on at the top right side of this webpage.
### Specials Schedule
ATTENTION PARENTS: Please be sure to answer my parent survey so that I have your up-to-date contact info. Click on the purple Google form icon below to get started.
HOMEWORK
New assignments come home every Monday night and are not be due back until Friday. I always assign the following to be practiced nightly:
-That week's reading passage & "Fast & Fluent" page
-Spelling words list
-Sight word flash cards
-Writing Journal- one response to the reading each week with a matching illustration
-Math Fast Facts flash cards
Homework Hotline (615) 298-6636
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Question Video: Finding the Range of a Function from Its Graph | Nagwa Question Video: Finding the Range of a Function from Its Graph | Nagwa
# Question Video: Finding the Range of a Function from Its Graph Mathematics
Which of the following is the range of the function π(π₯) = |π₯ + 8| β 7? [A] β β {β7} [B] β [C] (β7, β) [D] [β7, β) [E] β β {β8}
03:40
### Video Transcript
Which of the following is the range of the function π of π₯ is equal to the absolute value of π₯ plus eight minus seven? Is it option (A) the set of real numbers minus the set containing negative seven? Option (B) the set of real numbers. Option (C) the open interval from negative seven to β. Is it option (D) the left-closed, right-open interval from negative seven to β or option (E) the set of real numbers minus the set containing negative eight?
In this question, weβre asked to determine the range of a given function. And we can see the given function π of π₯ contains the absolute value function, and weβre given a graph of this function. This means we can answer this question algebraically by looking at the equation of the function. Or we can answer the question graphically by considering the meaning of the range with respect to the graph of the function. Letβs use the latter method. Itβs the set of all output values of the function given its domain, which is the set of possible input values.
Letβs attempt to determine the range of this function from its graph. To do this, we know the π₯-coordinate of a point on the graph tells us the input value of the function and the corresponding π¦-coordinate tells us the output of the function. For example, we can see the graph of this function passes through the point with coordinates negative 30, 15. Therefore, π evaluated at negative 30 must be equal to 15. 15 is an element of the range of this function. Itβs a possible output. And we can, of course, verify this by substituting negative 30 into the function π of π₯.
But the range of our function is the set of all possible output values of the function. Since the π¦-coordinates of points on the graph tells us the possible outputs of the function, the range of the function is the set of all π¦-coordinates of points on its graph. So letβs try and determine the π¦-coordinates of points which lie on the graph. To do this, we can note that there is a point with lowest π¦-coordinate. Itβs the point which lies right on the corner. However, we cannot determine the π¦-coordinate of this point just from the diagram, so weβll need to determine the coordinates of this point by using the function.
To do this, we recall π of π₯ is the absolute value of π₯ plus eight, and then we subtract seven. And the point with lowest π¦-coordinate will be when the output value of this function is the lowest. So we need to make the absolute value of π₯ plus eight minus seven as small as possible. To do this, we note that the absolute value of π₯ plus eight will always be greater than or equal to zero. And we canβt affect the value of negative seven since itβs a constant. Therefore, the smallest output of this function will be when the absolute value of π₯ plus eight is equal to zero. This occurs when π₯ is negative eight. Therefore, weβve shown that negative eight is the input value of the lowest output of our function. Negative eight is the π₯-coordinate of the corner.
And we can use this to determine the π¦-coordinate. The π¦-coordinate will be π evaluated at negative eight. And we can evaluate our function at negative eight by substituting negative eight into the function. Itβs the absolute value of negative eight plus eight minus seven. And this is equal to negative seven. Therefore, negative seven is an element of the range of our function and no value smaller than negative seven is an element of the range. Itβs the smallest element of the range.
To determine the rest of the range of this function, we need to recall that the graph of this function continues indefinitely in both directions. And in particular, this means for any π¦-value greater than or equal to negative seven, there is a point on the graph of the function with this as a π¦-coordinate. In other words, itβs a possible output of the function. Therefore, the range of this function includes negative seven and is unbounded. It goes all the way to β. We write this as the left-closed, right-open interval from negative seven to β. Hence, we were able to show the range of the function π of π₯ is equal to the absolute value of π₯ plus eight minus seven is option (D): the left-closed, right-open interval from negative seven to β.
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# If x and y are positive integers, what is the remainder when y^x is
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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]
Hi All,
We're told that X and Y are positive integers. We're asked for the remainder when Y^X is divided by 2.
This question can be solved by TESTing VALUES and/or by using Number Properties. It's worth noting that when dividing an integer by 2, the only possible remainders are 0 and 1.
1) Y^2 is an ODD integer.
Fact 1 tells us that Y^2 is an ODD integer - and we already know that X and Y are both POSITIVE INTEGERS.
(Even)^2 = Even
(Odd)^2 = Odd
This means that Y MUST be ODD. By extension, an ODD number raised to an INTEGER power will ALWAYS be ODD. Fact 1 essentially tells us that Y^X will ALWAYS be an ODD number. Dividing ANY odd number by 2 will ALWAYS give us a remainder of 1.
Fact 1 is SUFFICIENT
2) XY is an EVEN integer.
The information in Fact 2 means that one - or both - of the two integers are EVEN.
IF....
X=2, Y=1, then Y^X = 1 and the answer to the question is 1.
X=1, Y=2, then Y^X = 2 and the answer to the question is 0.
Fact 2 is INSUFFICIENT
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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]
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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]
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mathleaks.com mathleaks.com Start chapters home Start History history History expand_more
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Writing and Graphing One-Variable Inequalities
Exercise 1.12 - Solution
a
The number of single tickets is and we know that a single ticket costs Thus, we can express the cost of single tickets as
Now we can write an inequality to represent when this expression is less than the cost of a monthly pass, which costs This gives us
b
To determine which option is less expensive after rides, we can substitute into the inequality. If the it holds, the single ticket option is less expensive. If not, the monthly option is less expensive.
We can see that it will cost to buy single tickets. Thus, this is less expensive than buying the monthly pass.
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### Quantitative Aptitude Topic of the day: Mixture and Alligation
1. Alligation
It is the process to find the solution of problems concerned with compounding/mixing of ingredients which differ in price or quality.
Simply put it is the rule used to find the ratio in which two or more ingredients at a given value must be mixed to produce a mixture of desired value
2. Mean Price
It is the cost price of the unit quantity of the mixture.
3. Useful Formulae
If two ingredients A and B of price x and y are mixed and the price of resultant mixture is M (mean price)then the rule of alligation gives the ratio (R) in which these ingredients are mixed
R=(M−y)/(x−M)
This formula can be visualised as:
Thus the required ratio is,
R=(M−y) / (x−M)= (y−M) / (M−x)
4. Replacement of Part of Solution Formula:
Many questions of Mixture and alligation section are based on the following formula:
Lets Assume a container contains a solution composed of different ingredients . Some quantity of solution is then taken out and it is replaced with one of the ingredients. This process is repeated n times then,
Final Amount of ingredient of solution that is not replaced
= Initial Amount×(Vol. after removal / Vol. after replacing)^n
This formula can also be applied to ratios.
Final ratio of ingredient not replaced to total
= Initial ratio *(Vol. after removal / Vol. after replacing)^n
Attempt the following questions to check your level of preparedness for ratio proportion questions in competitive exams. You can answer the questions in comments section. Also make sure you checkout tomorrow’s (September 11) Winner’s Curry Lunch update for solutions to these questions along with the formulae used in VIDEO format by our Subject Expert.
Question 1 : In what ratio must water be mixed with milk to gain 25% on selling the mixture at cost price?
A) 4:01
B) 1:04
C) 3:02
D) 2:03
Question 2: Two vessel contains milk and water mixed in the ratio of 2:3 and 3:4. Find the ratio in which these two are to be mixed to get a new mixture in which ratio of milk to water is 7:10 respectively.
A) 10:07
B) 7:03
C) 5:03
D) 3:05
Question 3: A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A) 1/5
B) 1/10
C) 2/5
D) 2/10
Question 4: A shopkeeper has a mixture of 24 kg of sugar of worth Rs. 35 per/kg with another quality of sugar of Rs.42 a kg. By selling the mixture at Rs. 41.04 per kg, he may gain 8%, how much sugar of second quality, had added in the mixture.
A)18 kg
B) 32 kg
C) 16 kg
D) 28 kg
Question 5: In what ratio must wheat at Rs.3.20 per kg be mixed with wheat at Rs.2.90 per kg so that the mixture be worth Rs.3.08 per kg?
A)3:04
B) 3:03
C) 3 : 02
D) 4:05
VIDEO SOLUTIONS FOR SEPTEMBER 9 Ratio and Proportion QUESTIONS:
Share these profit loss questions with everyone you know and check their answers. Also let us know your comments on video solutions.
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03-02ChapGere.0003
# 03-02ChapGere.0003 - f between the ends of the bar Solution...
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Solution 3.4-11 Tapered tube 206 CHAPTER 3 Torsion t thickness (constant) d A , d B average diameters at the ends d B 2 d A (approximate formula) I P d 3 t 4 T T A B L A NGLE OF TWIST Take the origin of coordinates at point O . I P ( x ) [ d ( x ) ] 3 t 4 td A 3 4 L 3 x 3 d ( x ) x 2 L ( d B ) x L d A B L d B = 2 d A d A O x d (x) d x L For element of length dx : For entire bar: f 2 L L d f 4 TL 3 Gtd A 3 2 L L dx x 3 3 TL 2 Gtd A 3 d f Tdx GI P ( x ) Tdx G ¢ td 3 A 4 L 3 x 3 4 TL 3 Gtd A 3 # dx x 3 Problem 3.4-12 A prismatic bar AB of length L and solid circular cross section (diameter d ) is loaded by a distributed torque of constant intensity t per unit distance (see figure). (a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist
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Unformatted text preview: f between the ends of the bar. Solution 3.4-12 Bar with distributed torque A B t L t 5 intensity of distributed torque d 5 diameter G 5 shear modulus of elasticity (a) M AXIMUM SHEAR STRESS (b) A NGLE OF TWIST f 5 # L d f 5 32 t p Gd 4 # L x dx 5 16 tL 2 p Gd 4 d f 5 T ( x ) dx GI p 5 32 tx dx p Gd 4 T ( x ) 5 tx I P 5 p d 4 32 T max 5 tL t max 5 16 T max p d 3 5 16 tL p d 3 A B t L dx x A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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# Thread: Can someone run me through how to do this, my book is not helping at all
1. ## Can someone run me through how to do this, my book is not helping at all
that being said i have gave it a go but would be helpful if someone could run through this practise question with me
1. A company manufactures widgets and knows that when they set the price of
one widget to £25 then their customers demand 30 widgets. However, if they
reduce the price of a widget to £20, demand then increases to 35 widgets. In
order to produce widgets the company faces fixed costs of £300 and variable
costs of £15 per unit.
a) Use the data regarding price and demand to determine the linear
demand function (recall the demand function will be of the form
p = aQ + b where p is price and Q is quantity) (9)
2. ## Re: Can someone run me through how to do this, my book is not helping at all
Substituting two given pairs (price, quantity) into the equation p = aQ + b, you get two equations on a and b. Then you need to solve the system of these two equations.
3. ## Re: Can someone run me through how to do this, my book is not helping at all
thankyou but i still do not understand 100%, i am very new to all of this
4. ## Re: Can someone run me through how to do this, my book is not helping at all
Originally Posted by yoshimuru
thankyou but i still do not understand 100%, i am very new to all of this
Substituting two given pairs (price, quantity) into the equation p = aQ + b, you get two equations on a and b. Then you need to solve the system of these two equations.
25 = 30a + b
20 = 35a + b
solve the system of equations for a and b ...
5. ## Re: Can someone run me through how to do this, my book is not helping at all
a=-1
b=55
thats what i got, sound right to you guys?, think im getting this tho not sure
6. ## Re: Can someone run me through how to do this, my book is not helping at all
Originally Posted by yoshimuru
a=-1
b=55
thats what i got, sound right to you guys?, think im getting this tho not sure
Substitute your values in one of the original equations:
30a + b = 25
30(-1) + 55 = -30 + 55 = 25
So you're correct...OK?
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# What is the probability of rolling a pair of dice and doubles or a sum of 6 is rolled?
Contents
The number of different outcomes on two dice is the number of different outcomes on the first die (6) TIMES the number of different outcomes on the second (6), or 36. The probability of rolling doubles is 1/6, because there are 6 ways to roll doubles. 6/36 = 36.
## When two dice are rolled what is the probability of rolling doubles or a sum of 8?
Probabilities for the two dice
Total Number of combinations Probability
8 5 13.89%
9 4 11.11%
10 3 8.33%
11 2 5.56%
## What is the probability of rolling a pair of dice and getting a sum of two?
We have a probability of 1/6 that the first die rolls 2, and a probability of 1/6 that the second die rolls 2, thus making a combination (2,2) with the probability 1/36.
## What is the probability of rolling a sum of 10 or doubles?
If you consider the chart on the webpage given, there are 36 combination of rolls you can get from two dice. When you consider the sum being 10, there are only 3 combinations. So, the probability of getting a 10 would be 3/36 = 1/12.
## How many ways are there to roll a sum of 5 or doubles on two dice?
To calculate your chance of rolling doubles, add up all the possible ways to roll doubles (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice.
## What is probability of getting a sum of 20 when rolling to dice?
Step-by-step explanation: The maximum sum that we can get when we roll 2 dice is 12. So, the probability of getting 20 is obviously .
## What is the probability of getting a total of 7 when rolling two dice?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is the probability of rolling a sum of 6 on the two number cubes?
There are 36 possible outcomes in rolling two six-sided cubes. Of those 36 possibilities, five of them result in a sum of 6 .
## What is the probability of rolling a 7 or 11 with two dice?
What about 7 OR 11? There are 6 x 6 or 36 options, all are equally likely, 7 occurs 6 times, so the chances are 6/36 or 1/6. 11 occurs 2 times so chances are 2/36 or 1/18. 7 or 11 are 8 of the 36 options so 8/36 or 2/9.
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## When two sided dice are rolled There are 36 possible outcomes?
Every time you add an additional die, the number of possible outcomes is multiplied by 6: 2 dice 36, 3 dice 36*6 = 216 possible outcomes.
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Courses
# RD Sharma Solutions (Part - 2)- Ex-22.1, Data Handling I Collection Organisation Data, Class 7 Class 7 Notes | EduRev
## Class 7 : RD Sharma Solutions (Part - 2)- Ex-22.1, Data Handling I Collection Organisation Data, Class 7 Class 7 Notes | EduRev
The document RD Sharma Solutions (Part - 2)- Ex-22.1, Data Handling I Collection Organisation Data, Class 7 Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7
#### Question 7:
In a study of number of accidents per day, the observations for 30 days were obtained as follows:
Prepare a frequency distribution table.
Required frequency-distribution table:
Number of Accidents Number of Days 0 2 1 3 2 6 3 3 4 4 5 6 6 6
#### Question 8:
Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13, 14, 13, 12, 14, 13, 14, 15, 13, 14, 13, 14, 16, 12, 14 13, 14, 15, 16, 13, 14, 13, 12, 17, 13, 12, 13, 13, 13, 14
Frequency Distribution Table is :
Ages (in years) Number of Students 12 4 13 12 14 9 15 2 16 2 17 1
#### Question 9:
Following figures relate the weekly wages (in Rs.) of 15 workers in a factory:
300, 250, 200, 250, 200, 150, 350, 200, 250, 200, 150, 300, 150, 200, 250
Prepare a frequency table.
(i) What is the range in wages (in Rs)?
(ii) How many Workers are getting Rs 350?
(iii) How many workers are getting the minimum wages?
Frequency Distribution Table is
Wages (in Rs.) No. of Workers 150 3 200 5 250 4 300 2 350 1
(i) The range in wages (in Rs.) = 350 -150 = 200.
(ii) Only 1 worker is getting Rs. 350.
(iii) 3 workers are getting the minimum wages, i.e, Rs. 150.
#### Question 10:
Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class VI of a school:
9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9, 12, 17, 19, 19, 20, 9
(i) What is the range of marks?
(ii) What is the highest mark?
(iii) Which mark is occurring more frequently?
Required frequency-distribution table:
Marks Frequency 9 6 12 4 17 4 18 2 19 4 20 3 25 2
(i) Range of marks: 25-9=16.
(ii) The highest mark is 25.
(iii) 9 is occurring most frequently.
#### Question 11:
In a mathematis test following marks were obtained by 40 students of class VI. Arrange these marks in a table using, tally marks.
(i) Find how many students obtained marks equal to or more than 7?
(ii) How many students obtaned marks below 4?
The Frequency Distribution Table is :
Marks Tally Marks Frequency 1 II 2 2 III 3 3 III 3 4 7 5 6 6 7 7 5 8 IIII 4 9 III 3
(i) 12 students obtained marks equal to or more than 7.
(ii) Only 8 students obtained marks below 4.
#### Question 12:
(i) Arrange the names of sweets in a table using tally marks.
(ii) Which sweet is preferred by most of the students.
(ii) Ladoo is preferred by most of the students, 12 students.
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## Intro to Statistics, Statistics 1034. Chapter 4 Notes
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# Intro to Statistics, Statistics 1034. Chapter 4 Notes Stat 1034
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## About this Document
Information about probability!
COURSE
Elementary Statistics I
PROF.
Sarah Myers
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Probability, Statistics, Statistics 1, Introduction To Statistics
KARMA
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Chapter 4 Elementary Probability Theory Material Extracted From Textbook (Brase, Charles Henry., and Corrinne Pellillo. Brase. Understandable Statistics: Concepts and Methods . 11th ed. N.p.: Cengage Learning, n.d. Print.) 4.1 What is Probability? Probability is a numerical measure between 0 and 1 that describes the likelihood that an event will occur. Probabilities closer to 1 indicate that the event is more likely to occur. Probabilities closer to 0 indicate that the event is less likely to occur. Probability Assignments 1. A probability assignment based on intuition incorporates past experience, judgement, or opinion to estimate the likelihood of an event . 2. A probability assignment based on relative frequency uses the formula: Probability of event = relative frequency = f/n Where f is the frequency of the event occurrence in a sample of n observations. 3. A probability assignment based on equally likely outcomes uses the formula: Probability of event = # of outcomes favorable to event / total # of outcomes Law of Large Numbers: In the long run, as the sample size increases and increases, the relative frequencies of outcomes get closer and closer to the theoretical (or actual) probability value. ● The underlying assumption we make is that if events occurred a certain percentage of times in the past, they will occur about the same percentage of times in the future. Statistical Experiment/Statistical Observation: Can be thought of as any random activity that results in a definite outcome. Event: Is a collection of one or more outcomes of a statistical experiment or observation. Simple Event: Is one particular outcome of a statistical experiment. Sample Space: The set of all simple events. ● The sum of the probabilities of all simple events in a sample space must equal 1. Interpreting Probabilities: ● The closer the probability is to 1, the more likely the event is to occur. ○ Just because the event of a probability is high, it is not certainty that the event will occur. ● Similarly, if the likelihood of an event is low, it is possible that the event might occur. Events with low probability but big consequences are of special concern. ● Some of people’s biggest mistakes in a person’s life can result from either misjudging: a) the size of an event’s impact. b) the likelihood the event will occur. ● An event of great importance cannot be ignored even if it has a low probability of occurrence. What Does the Probability of an Event Tell Us? ● The probability of an event A tells us the likelihood that event A will occur. If the probability is 1, the event A is certain to occur. If the probability is 0, the event A will not occur. ● The probability of event A applies only in the context of conditions surrounding the sample space containing event A. ● If we know the probability of event A, then we can easily compute the probability of event not A in the context of the same sample space. P(notA)= 1P(A). Probability Related to Statistics ● If probability did not exist, then inferential statistics would not exist. ● Probability you know the overall description of the population. The central problem is to compute the likelihood of a specific outcome. ● Statistics you know only the result of a sample drawn from a statistic. 4.2 Some Probability Rules Compound Events Conditional Probability and Multiplication Rules: Independent Events: Two events are independent if the occurrence or nonoccurrence of one event does not change the probability that the other event will occur. Dependent Events: Two events are dependent if the occurrence or nonoccurrence of one event changes the probability that the other event will occur. Why Does the Independence or Dependence Matter? ● The type of the events determines the way we compute the probability of the two events happening together. Multiplication for Independent Events P(A and B) = P(A) x P(B) Multiplication for Dependent Events P(A and B) = P(A) x P(B|A) P(A and B) = P(B) x P(A|B) Conditional Probability: The notation P(A, given B) denotes the probability that event A will occur given that event B has occurred. Insert Conditional Probability Rule How to Use the Multiplication Rules 1. First determine whether A and B are independent events. If P(A) = P(A|B), then the events are independent. 2. If A and B are independent events: P(A and B) = P(A) x P(B). 3. If A and B are any events, P(A and B) = P(A) x P(B|A) or P(A and B) = P(B) x P(A|B). What does Conditional Probability Tell Us? Conditional probability of two events A and B tell us: ● The probability that event A will happen under the assumption that event B has happened (or is guaranteed to happen in the future). This probability is designated P(A|B) and is read “probability of A given event B.” Note that P(A|B) might be larger or smaller than P(A). ● The probability that event B will happen under the assumption that event A has happened. This probability is designated P(B|A). Note that P(A|B) and P(B|A) are not necessarily equal. ● If P(A|B) = P(A) or P(B|A) = P(B), then events A and B are independent. This means the occurrence of one of the events does not change the probability that the other event will occur. ● Conditional probabilities enter into the calculations that two events A and B will both happen together. P(A and B) = P(A) x P(B|A) also P(A and B) = P(A)x P(B) In the case that events A and B are independent, then the formulas for P(A and B) simplify to. P(A and B) = P(A) x P(B). ● If we know the values of P(A and B) and P(B), then we can calculate the value of P(A|B). Mutually Exclusive/ Disjoint: Two events are mutually exclusive or disjoint if they cannot occur together. In particular, events A and B are mutually exclusive if P(A and B) = 0. Addition Rule for Mutually Exclusive Events A and B P(A or B) = P(A) + P(B) General Addition Rule for any Events A and B (Not Mutually Exclusive) P(A or B) = P(A) + P(B) P(A and B) How to Use the Addition Rules 1. First determine whether A and B are mutually exclusive events. If P(A and B) = 0, then the events are mutually exclusive. 2. If A and B are mutually exclusive events, P(A or B) = P(A) + P(B). 3. If A and B are any events, P(A or B) = P(A) + P(B) P(A and B). What Does the Fact that Two Events are Mutually Exclusive Tell Us? If two events A and B are mutually exclusive, then we know the occurrence of one of the events means that the other event will not happen. In terms of calculations, this tells us: ● P(A and B) = 0 for mutually exclusive events. ● P(A or B) = P(A) =P(B) for mutually exclusive events. ● P(A|B) =0 and P(B|A) = 0 for mutually exclusive events. That is, if event B occurs, then event A will not occur, and vice versa. 4.3 Tree and Counting Techniques ● The probability formula requires that we be able to determine the number of outcomes in the sample space. ● When an outcome of an experiment is composed of a series of events, the multiplication rule gives us the total number of outcomes . Tree Diagram: A visual display of the total number of outcomes of an experiment consisting of a series of events. Helps determine the total number of outcomes and individual outcomes. Factorial Notation: Procedure: What Do Counting Rules Tell Us? Counting rules tell us the total number of outcomes created by combining a sequence of events in specified ways. ● The multiplication rule tells us the total number of possible outcomes for a sequence of events. Tree diagrams provide a visual display of all the resulting outcomes. ● The permutation rule tells us the total number of ways we can arrange in order n distinct objects into a group of size r. ● The combination rule tells us how many ways we can form n distinct objects into a group of size r. The order of the objects is irrelevant.
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# 「学习笔记」伯努利数
### 定义
$B_n = [n = 0] - \frac 1{n + 1} \sum_{i=0}^{n-1} \binom {n + 1} i B_i$
$\hat{B}(x) = \sum_{i \geq 0} B_i \frac {x^i} {i!} = \frac x{e^x - 1}$
### 性质
$S_k(n) = \frac 1{k + 1} \sum_{i=0}^k \binom{k + 1}i B_i n^{k - i + 1}$
$$\hat{S}_n(x) = \sum_{i \geq 0} \frac {x^i} {i!} S_i(n)$$，那么：
\begin{aligned} \hat{S}_n(x) &= \sum_{i \geq 0} \frac {x^i} {i!} \sum_{j=0}^{n-1} j^i\\ &= \sum_{j=0}^{n-1} \sum_{i \geq 0} \frac {(jx)^i} {i!}\\ &= \sum_{j=0}^{n-1} e^{jx}\\ &= \frac {e^{nx} - 1}{e^x - 1} \end{aligned}
\begin{aligned} S_k(n) &= k! [x^k] \frac{e^{nx} - 1}{e^x - 1}\\ &= k! \sum_{i=0}^k \frac{B_i}{i!} [x^{k-i+1}] (e^{nx} - 1)\\ &= \frac 1{k + 1} \sum_{i=0}^k \binom{k+1}i B_i n^{k - i + 1} \end{aligned}
### 例子
#### 洛谷P3711 仓鼠的数学题
\begin{aligned} g(x) &= \sum_{k=0}^n \frac {a_k} {k + 1} \sum_{i=1}^{k + 1} \binom{k + 1}i x^i B_{k - i + 1}\\ &= \sum_{k=0}^n a_k k! \sum_{i = 1}^{k + 1} \frac {x^i} {i!} \frac {B_{k - i + 1}}{(k - i + 1)!}\\ &= \sum_{i=0}^n \frac {x^{i + 1}}{(i + 1)!} \sum_{k=i}^n a_k k! \frac{B_{k - i}}{(k - i)!} \end{aligned}
posted @ 2021-02-23 18:55 xgzc 阅读(266) 评论(0编辑 收藏 举报
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https://www.physicsforums.com/threads/f-frequency-of-small-diameter-organ-pipe.104461/
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# F frequency of small diameter organ pipe
1. Dec 17, 2005
### msimard8
This is a multiple choice question
The fundamental frequency of a small diameter organ pipe is
a) directly proportional to its length
b) inversely proportional to its length
c) independent of its length
d) inversely proportional to its diameter
e) directly proportional to its diameter
well this is what i know (assume)
the organ pipe is closed at one end
what formulas should i consider solving this problem
2. Dec 17, 2005
### mukundpa
go to text
3. Dec 17, 2005
### msimard8
whats that suppose to mean
4. Dec 17, 2005
### mukundpa
in your text book i think they have derived formulae for fundamental frequency and their overtones for open and cloced organ pipes, with the diagrams.
5. Dec 17, 2005
### msimard8
the only formula i see is
Ln=(2n=1) wavlength/4
which doesnt describe the question
Right now I am thinking that the diameter has no effect because, all diameter does is increase or decrease the amplitude of the wavelength which effects the loudness or intensity of the sound.
So therefore d and e are eliminated
The length of the pipe determines how many waves can fit in the pipe.
umm so confused
6. Dec 18, 2005
### mukundpa
dont get confused,
The formula is about the of a closed tube rasonating with nth harmonic of a given frequency f, and is
Ln=(2n-1) wavlength/4
( if it is + it is for nth overtone)
(n = 1) gives first hormonic or fundamentalfrequency,length is wavlength/4
(n = 2) gives first overtone frequency, length is 3*wavlength/4
(n = 3) gives second overtone frequency, length of tube is 5*wavlength/4 and so on
forget this here
Now on to the question
The frequency is given by
f = c/lembda = wave velocity / wavelength
in a colsed organ pipe standing waves will be produces with wavelength
4L(fundamental)= f0
4L/3(first overtone) = f1 = 3f0
5L/4(second overtone) =f2 =5f0 and so on
so the fundamental frequency of a close orgon pipe is
f0 = c/(4*L)
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0
# What is a factorial function in Visual Basic?
Updated: 8/10/2023
Wiki User
16y ago
' Iterative solution
Function iterativeFactorial(ByVal n As Long) As Long
Dim factorial As Long = 1
For i As Long = 1 To n
factorial *= i
Next
Return factorial
End Function
' Recursive solution
Function recursiveFactorial(ByVal n As Long) As Long
If n <= 1 Then
Return n
End If
Return n * recursiveFactorial(n - 1)
End Function
Wiki User
15y ago
Wiki User
16y ago
Factorial is a Mathematical Function.
Factorial returns the product of all numbers from 1 to itself
e.g. Factorial 5 = 5*4*3*2*1 = 120
It is expressed as n! = factorial of n
To implement it in Visual Basic, there are two methods-
Function factorial(ByVal n as Integer) as Integer
If n =< 1 Then factorial = 1:Exit Function
factorial = n * factorial(n-1)
End Function
Function factorial(ByVal n as Integer) as Integer
factorial = 1
Dim a as Integer
For a = 1 to n
factorial = factorial * a
Next 'a
End Function
Wiki User
11y ago
dim num as integer, factorial as single
num=inputbox("enter a number")
factorial = 1
for x = 1 to num
factorial = factorial * x
next x
print"factorial is" ; factorial
or By Recursive Method
Private Function FindFactorial(number As Integer)
If number < 1 Then
FindFactorial = 1
Else
FindFactorial = number * FindFactorial(number - 1)
End If
End Function
' recursive is faster and simpler for finding factorial
Earn +20 pts
Q: What is a factorial function in Visual Basic?
Submit
Still have questions?
Related questions
### What is a function procedure in Visual Basic?
A function is essentially a subroutine that is ment to be used by other subroutines.
### What function in Visual Basic will return a date from a computer?
(it used to be) DATE\$
### What is rounding in Visual Basic?
Rounding in Visual Basic is the method of rounding an integer up, or flooring an integer, which is rounding down. To round up, you use the System.Math.Round function. To round down, or floor, you use the System.Math.Floor function.
### What is visual basic control?
Visual Basic Controls work on Visual Studio for Visual Basic and Applications that made by Visual Basic.
### How do you get square root in Microsoft visual basic?
The function is Sqr() in VB6 and Math.Sqr() in .NET.
### Write a C-like program fragment that calculate the factorial function for argment 12 with do while loop?
#!/usr/bin/perl print factorial(\$ARGV[11]); sub factorial { my(\$num) = @_; if(\$num == 1) { return 1; # stop at 1, factorial doesn't multiply times zero } else { return \$num * factorial(\$num - 1); # call factorial function recursively } }
### The year of invention of visual basic?
Visual Basic was started in 1991.
### Who developed visual basic?
Microsoft is the developer of visual basic
### Who invented the visual basic?
Visual Basic was created by a team at Microsoft.
### What is the difference between visual basic and visual c?
The programming language: Visual Basic is a BASIC-like (or BASIC-derived) language, Visual C is... well C.
No.
### Who is the developer of Visual Basic?
Microsoft is the developer of Visual Basic.
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# What is Place Value?
Here at Whizz Education, we know the importance of having parental figures involved in a child’s education. If you, as parents, can show that maths is fun and relevant in everyday life, you’ll be a dramatic help to teachers and educators in the classroom!
In this – the first of our ‘how to’ blogs – we’ll guide you through how to teach your child about place value with a few, fun and easy-to-execute games and exercises whether your child is in Reception or Year 8.
But first, let’s remind ourselves of just what exactly ‘place value’ is…
## What is Place Value?
Learning about place value allows a student to know the ‘value’ of a digit within a number by looking at its ‘place’ in relation to a decimal point. It is important that children understand that whilst a digit can be the same, its value depends on where it is in the number.
Take, for example, the number 725.1.
• The digit ‘7’ in the number ‘725.1’ has a value of 700, or 7 hundreds
• The digit ‘2’ has a value of 20, or 2 tens
• The digit ‘5’ has a value of 5, or 5 ones
• The digit ‘1’ has a value of 0.1, or 1 tenths
## Why is it important to learn about place value?
This kind of information can seem self-explanatory to adults. However, learning the place value of different digits within a number is an important skill for many, more complicated mathematical operations.
An understanding of place value helps children, at different stages, to:
• Determine which of two numbers is larger in value and get them to explain why.
e.g. £7501 > £7.501
• Partition the numbers ( to break up the numbers into tens and ones to find the answer)for addition and subtraction
e.g. 70 + 6 = 76
e.g. 78 – 56 = (70 + 8) – (50 + 6) = (70 – 50) + (8 – 6) = 22
• Discover the factors of a number for multiplication and division
e.g. the place value of the digit ‘5’ in 500 is five hundreds, or 5 x 100
• Read’ numbers – to be able to point to a digit within a number and know its value
• Grasp the language of statistics and know the meaning of terms such as one tenth of the population or five hundred times the size
• Line up numbers correctly for any mathematical operation (addition, subtraction, multiplication, or division)
e.g.
10 345
6 237 –
=
• Round numbers to the nearest ten, one hundred, or one thousand
As you can see, place value is important across the board for handling money, weighing goods, and reading data in the news, among other things.
## How can I help my child to understand place value within the home?
Maths-Whizz provides access to a number of free interactive lessons on place value from Reception through to Year 8 students. They provide a good starting point for different age groups.
To make this clearer to children, teachers like to represent this in the form of place value columns, where the constituent numbers are placed according to whether they’re 100s, 10s or 1s and so on. For further fun, produce a grid like the following (on paper, on a whiteboard, or using other objects from your home). Depending on their age, see if your child can:
• Tell which of two numbers is larger
• Write a two-, or three- or four-digit number in the correct place on the grid
• Move the numbers to make it 10 or 100 times bigger or smaller
• State how many ‘ones’ or ‘tens’ etc. are in the number
• Identify the quantity value of any digit (e.g.10 000) or the column value e.g. 10 thousands
Throughout primary school, concrete maths resources are used to help make place value easy to understand for children. As a further activity, why not use objects (blocks, beans, spaghetti hoops and marbles) or drawn shapes (dots, squares, circles and triangles) to represent ‘hundredths ’, ‘tenths’, ‘ones’, or ‘tens’.
Write a number and see if your child can pile/draw the objects in the correct quantities. For example, the number 75 would need 7 hair bands representing ‘tens’ and 5 coins representing ‘ones’ (and as an extension, try exchanging a hairband for 10 coins!).
As your child gets older, the size of the numbers they are expected to deal with grows. For example, year 1’s will be taught to count to 100. Year 2’s will learn the place value of each digit in a two-digit number (tens, ones). Year 3’s will be able to count to 1000 and partition two-digit numbers e.g. 75 = 70 + 5. By year 4 students can recognise the place value of any digit in a four-digit number.
Place value is intrinsically linked to many other areas of maths; a solid understanding of place value provides the essential number knowledge needed to complete calculations, including addition, subtraction, multiplication, division, and fractions.
Why not send us pictures of you learning about place value this week on our Facebook page or Twitter account?
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# Double Angle Formula Worksheet
## Double Angle And Half Angle Identities Worksheets
Recapitulate The Application Of Double And Half Angle Formulas With These Printable High School Worksheets Observe The Angle Measure Check If It Can Be Expressed In Double Or Half Angle And Then Apply The Appropriate Formula To Simplify Download The Set 3 Worksheets Express As A Single Trigonometric Function Simplifying Complex Trigonometric Expressions Becomes Easy With Double And Half
### Double Angle And Half Angle Formulas Worksheet Dsoftschools
Some Of The Worksheets Below Are Double Angle And Half Angle Formulas Worksheet Understanding The Double Angle Formulas Understanding The Power Reduction Formulas Using Double Angle Formulas To Find Exact Values Using The Double Angle Formula For Tangent To Find An Exact Value
#### Double Angle And Half Angle Identities
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##### Double Angle Formula Worksheets Kiddy Math
Double Angle Formula Displaying Top 8 Worksheets Found For This Concept Some Of The Worksheets For This Concept Are Double And Half Angle Identities Date Period The Double Angle Formulae 22 More Trigonometric Identities Work 1 Of 2 Using Double And Half Angle Formulas Angle Sumdifference Identities Double Angle Power Reducing And Half Angle Formulas W Trigonometry Chapter 4 Name 4
###### The Double Angle Formulae Mathcentre
The Double Angle Formulae Forsin2a Cos2a Andtan2a We Start By Recalling The Addition Formulae Which Have Already Been Described In The Unit Of The Same Name Sin A B Sinacosb Cosasinb Cos A B Cosacosb Sinasinb Tan A B Tana Tanb 1 Tanatanb We Consider What Happens If We Let B Equal To A
Double And Half Angle Formulas Worksheets Teaching
In This Worksheet The Student Will Be1 Using Double Angle Formulas Half Angle Formulas Law Of Sines2 Decoding The Pun I Would Not Buy Anything With Velcro Nocalc Precalc Mathacrostics By Value Added Publishing Will Be The Fifth Book Written In Their Mathacrostics Series An Entire Digital V Subjects Precalculus Trigonometry Grades 11 Th 12 Th Staff Types Worksheets
Quiz Worksheet Double Angle Formula Study
Check Your Understanding Of The Double Angle Formula By Completing Practice Problems From The Interactive Quiz Or Printable Worksheet Get The
Double And Half Angle Formulas Practice
Double And Half Angle Formulas Practice Use A Double Angle Identity To Find The Exact Value Of Each Expression 1 Tan 3 4 And 3 2 Find Tan 2 2 Tan 5 12 And 3 2 Find Cos 2 3 Tan 5 12 And 3 2 2 Find Cos 2 4 Sin 7 25 And 3 2 2 Find Cos 2 5 Sin 2 2 3 And 2 Find Tan 2
Double And Half Angle Identities Date Period
Use A Double Angle Or Half Angle Identity To Find The Exact Value Of Each Expression 9 Cot 3 3 3 10 Cot 2 3 3 3 11 Sec 5 12 6 2 12 Cot 60 3 3 13 Cot 240 3 3 14 Cot 5 3 3 3 1 W J2 U0y1 X2a 2kqugtxa T Zs9o3fatrwia Jr Qee 4lfltcr W 8 Oasl Nl U 9ruisg Qhctest Nrkeis3e9rnvie Wdi T M Omra1drem Ewpiytih I Si Yn Nfyiknni Ot8ec La Jldgne1bgrea X Z26 Z Worksheet By
Double And Half Angle Formulae Teaching Resources
Including Triple Angle Formulae This Website And Its Content Is Subject To Our Terms And Conditions
Double Angle Formula Worksheet. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
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Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused.
Double Angle Formula Worksheet. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
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#### Career
Hi all, I am looking for quant trading jobs.
Can anyone share some of quant interview questions?
here some questions for reference
1. Find x if xxx... = 2
2. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
3. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?
4. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
5. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again?
6. Calculate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$
Original Posted by - b'Terminator':
here some questions for reference
1. Find x if xxx... = 2
2. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
3. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?
4. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
5. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again?
6. Calculate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$
How much time did you have to solve these questions?
Original Posted by - b'Terminator':
here some questions for reference
1. Find x if xxx... = 2
2. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
3. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?
4. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
5. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again?
6. Calculate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$
hey can you share the answers as well?
• I got x = sqrt(2) for question 1
• and my answer for question 6 is 2
is it correct?
Original Posted by - b'Gary Fung': How much time did you have to solve these questions?
i think 20-30 min. That is 5 min each
Original Posted by - b'Nic C':
hey can you share the answers as well?
• I got x = sqrt(2) for question 1
• and my answer for question 6 is 2
is it correct?
yes i got the same answers
Original Posted by - b'Terminator':
here some questions for reference
1. Find x if xxx... = 2
2. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
3. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?
4. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
5. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again?
6. Calculate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$
For question 5, i got t= 3927.2727...=1 hour 5 minutes 27.27 seconds
Original Posted by - b'\xe2\x9a\xa1\xe6\xaf\x94\xe5\x8d\xa1\xe8\xb6\x85\xe2\x9a\xa1': For question 5, i got t= 3927.2727...=1 hour 5 minutes 27.27 seconds
can you explain how to get this answer?
Original Posted by - b'Terminator':
here some questions for reference
1. Find x if xxx... = 2
2. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
3. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?
4. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
5. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again?
6. Calculate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}$
Could you give the answers as well?
Do you have more interesting questions?
Here another set of interview questions! Enjoy guys!
Questions
1. For a 3 sets tennis game, would you bet on it finishing in 2 sets or 3 sets?
2. I have a square, and place three dots along the 4 edges at random. What is the probability that the dots lie on distinct edges?
3. You have 10 people in a room. How many total handshakes if they all shake hands?
4. Two decks of cards. One deck has 52 cards, the other has 104. You pick two cards separately from a same pack. If both of two cards are red, you win. Which pack will you choose?
5. What is 39*41?
6. A group of people wants to determine their average salary on the condition that no individual would be able to find out anyone else's salary. Can they accomplish this, and, if so, how?
7. How many digits are in 99 to the 99th power?
8. A line of 100 passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the nth passenger in line has a ticket for the seat number n.) Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
9. What is the sum of the numbers one to 100?
10. You have a 3 gallon jug and 5 gallon jug, how do you measure out exactly 4 gallons? Is this possible?
11. You have 17 coins and I have 16 coins, we flip all coins at the same time. If you have more heads then you win, if we have the same number of heads or if you have less then I win. What's your probability of winning?
12. What is the probability you draw two cards of the same color from a standard 52-card deck? You are drawing without replacement.
13. You’re in a room with three light switches, each of which controls one of three light bulbs in the next room. You need to determine which switch controls which bulb. All lights are off to begin, and you can’t see into one room from the other. You can inspect the other room only once. How can you find out which switches are connected to which bulbs? Is this possible?
14. In world series, what are the odds it goes 7 games if each team equal chance of winning?
15. Given 100 coin flips, what is the probability that you get an even number of heads?
16. There are 5 balls, 3 red, and 2 black. What is the probability that a random ordering of the 5 balls does not have the 2 black balls next to each other?
17. What is the least multiple of 15 whose digits consist only of 1's and 0's?
18. Is 1027 a prime number?
19. Does the price of a call option increase when volatility increases?
20. 2 blue and 2 red balls, in a box, no replacing. Guess the color of the ball, you receive a dollar if you are correct. What is the dollar amount you would pay to play this game?
21. What is the singles digit for 2^230?
1. Two sets - Let p=prob team 1 wins and q=prob team 2 wins. p^2 + q^2 = probability finish in two sets. 2pq = probability finish in three sets. p^2 + q^2 always >= 2pq, so the answer is two sets.
2. 3/8 - Given the edge the first dot is on, the probability the other two dots are on distinct edges is (3/4)*(2/4)
3. 45 - (10 choose 2) = 45 -- this is the total number of ways two people can shake hands.
4. 104 card pack - (52/104)(51/103) > (26/52)(25/51), or 51/103 > 25/51
5. 1599 - 3941 = (40-1)(40+1) = 40*40 - 1 = 1599
6. Yes, it’s possible - The first person thinks of a random number, say X. This person adds this number to her salary. The rest of the group simply adds their salary to the initial number. Then, the first person subtracts the random number X and divides the total salary sum by the size of the group to obtain the average.
7. 198 - 99^99 = (100)^(99) * (.99)^99 = (10)^(198) * (.99)^99. You can convince yourself 10^198 has 199 digits, and 0.99^.99 approaches 1/e. Thus, (10)^(198) * (.99)^99 has 198 digits.
8. 0.5 - The fate of the last passenger is determined the second either the first or last seat on the plane is taken. This statement is true because the last person will either get the first seat or the last seat. All other seats will necessarily be taken by the time the last passenger gets to pick his/her seat. Since at each choice step, the first or last seat has an equal probability of being taken, the last person will get either the first or last with equal probability: 0.5.
9. 5050 - Sum of numbers from 1,2....n = (n)(n+1)/2. You can also think about this problem by pairing off numbers - 1 and 100, 2 and 99, 3 and 98, 4 and 97, etc. We have 50 of these pairs, and each pair sums up to 101, so the final sum = 50101 = 5050.
10. Yes, it’s possible - Fill up the 3 gallon jug. Then, pour the liquid into the 5 gallon jug. Fill the 3 gallon jug again, and then fill the 5 gallon jug until it is full. We now have 1 gallon remaining in the 3 gallon jug. We empty the five gallon jug and pour the remaining 1 gallon into our 5 gallon jug. Finally, we fill the 3 gallon jug and add this to the 5 gallon jug (which already had 1 gallon). We are left with 4 gallons in the 5 gallon jug.
11. 0.5 - Use recursion - The initial 16 flips have the same probability of everything. Thus, the game completely depends on if the last coin flip is tails or head (50/50 chance of H vs. T).
12. 25/51 - You either draw a black or a red card first. Then, there are 51 cards left in the deck and 25 of these cards have the same color. Thus, the probability is 25/51.
13. Yes, it’s possible - Leave switch 1 off. Then, turn switch 2 on for ten minutes. After the ten minutes, turn it off and quickly turn on switch 3. Now, go into the room. The currently lit up bulb connects to switch 3. The bulb that off but still warm is from switch 2, and the remaining bulb is from switch 1.
14. 20/64 - Out of the first three games, each team needs to win three. Thus, (6 choose 3)*(.5^6) = 20/64, as each team has a 1/2 probability of winning each game.
15. 1/2 - Whether there is an odd or even number of heads is ultimately determined by the final flip (50/50 chance of being heads vs. tails), for any number of flips.
16. 0.6 - Because of repeats of black/red balls, there are 10 combinations of red/black balls: (5 choose 2) or (5 choose 3) spots to put the black or red balls, respectively. There are 4 places that 2 black balls can be next to each other, so the other 6 combinations do NOT have two black balls next to each other.
17. 1110 - The last digit must be zero (30, 45, 60, 75, etc.). Multiples of 15 never end in 1. Then, starting checking numbers. 10, 100, 110, 1000, 1100, 1110. You will quickly arrive at the answer if you are good with your mental math.
18. No - 1027 = 1000 + 27 = 10^3 + 3^3. We know a^3 + b^3 can be factored, so 1027 is NOT prime.
19. Yes - sometimes a rare finance question is included in these interviews; remember that both time and volatility increase the prices of both calls and puts
20. 17/6 dollars - You’ll always get the last ball right as your sampling w/o replacement. The first ball you have a 50% chance of getting right. The second ball you have a 2/3 chance of getting right.
21. 4 - Repeating patterns -- 2,4,8,6,2 -- follow the pattern.
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## Why You've Probably Been Measuring Your Targets Wrong
Before TargetView, we used to measure the group-size of our targets by finding the two shots that were furthest apart and measuring the distance between them to obtain our group-size. At first glance, this seems reasonable. Those two shots are furthest apart, so that must be the size of the group.
But that's wrong! Below is a super nerdy explanation of why. Feel free to keep reading if you want to know more. Or, just trust that TargetView is measuring your targets correctly and accurately, and never worry about it again!
### Explanation
Figure 1 shows two shots on a target that are exactly 3 inches apart. The shots are labeled 1 and 2.
Figure 1
In figure 2, let's calculate the size of this two-shot group by finding the two shots that are furthest apart. This is pretty easy because we only have two shots! The size of this group is 3 inches. We can draw the group-size circle (in orange), just like TargetView does in the app.
Figure 2
Let's draw a circle with a 3-inch radius around shot 1. You can see this circle (in magenta) in figure 3. Any shot that lands within this circle will be less than 3 inches away from shot 1. It is also true that any shot within the magenta circle is at least as close (or closer) to shot 1 as shot 2 is.
Figure 3
Let's draw the same 3-inch radius circle around shot 2. You can see both magenta circles in figure 4. Any shot that lands within these two large circles is no more than 3 inches away from either shot 1 or shot 2.
Figure 4
Just like on a Venn Diagram, the intersection of the two large magenta circles represents a region where shots are no more than 3 inches away from both shot 1 and shot 2. This special region is colored gray in figure 5. Any shot that lands in the gray region will be less than 3 inches from shots 1 and 2.
Do you see the problem yet?
Figure 5
Let's clean up the drawing a little bit by eliminating extra lines. Figure 6 shows the special intersection region of the two large magenta circles. You can see there's a new shot (shot 3) on the target. Notice that shot 3 is not within the orange group-size circle!
Well, fine, that just means we have to redraw the orange circle. So let's start over by finding the two shots that are furthest apart, just like we did in the first step.
Figure 6
In figure 7, let's measure the distance between all the shots. Shot 1 and shot 2 are still 3 inches apart. Shot 1 and shot 3 are 2.73 inches apart. Shot 2 and shot 3 are 2.73 inches apart. So shots 1 and 2 are still the furthest from each other. So if we measure this 3-shot target, we would still get a group-size of 3 inches.
But this is wrong! If we draw a 3 inch circle on the target, shot 3 will be outside of this circle! You can see this clearly in figure 7. There is a 3 inch circle that connects shots 1 and 2. Shot 3 is outside of this circle.
In fact, any shot that lands within the gray intersection area will throw off the group-size calculation. Because: any shot that lands in the gray area is closer to both shot 1 and shot 2 than shots 1 and 2 are from each other.
Let's reword that last sentence: Any shot that lands in the gray area will be less than 3 inches away from shot 1 and less than 3 inches away from shot 2 and shots 1 and 2 will always be the two shots that are furthest from each other.
Figure 7
Let's draw the correct group-size circle in figure 8. The true/correct size of this three shot group is actually 3.27 inches, not 3 inches. TargetView knows how to do this measurement the right way, every time!
Figure 8
The worst-case scenario for error is shown in figure 9. If we place shot 4 all the way at the edge of the gray area, the true group-size is 3.46 inches. Technically, shots 1, 2, and 4 are all equidistant from each other. If you were to measure this target using the old (incorrect) method, you would still get an (incorrect) group-size of 3 inches. This is almost a half-inch of error, or an error of 13.3%! That's huge!
TargetView knows how to calculate the size of your groups *correctly* so you never have to worry. Your targets will always be measured correctly and reflect your true accuracy.
It's important to note that this explanation is not a recommendation to use 3-shot or 4-shot groups to measure the accuracy of your firearm or of yourself. 3-shot groups are not a statistically significant way to to access accuracy.
10-shot groups are for more representative of the true accuracy of a particular firearm, and 30 (or more) shot groups are even better.
Figure 9
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# What is the domain and range of f(x)=4?
Domain: $\left(- \infty , \infty\right)$ Range: $\left\{4\right\}$
This is really just a horizontal line. It exists for all values of $x ,$ in fact, what value of $x$ we choose will not change its value.
The domain is then $\left(- \infty , \infty\right)$
The line, however, only exists at $y = 4$. There is no interval for the range, range must be $\left\{4\right\}$
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# Logistic regression - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Logistic regression
Logistic regression
Independent variablesIndependent variables
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Dependent variableDependent variable
One categorical with 2 independent groupsOne categorical with 2 independent groups
Null hypothesisNull hypothesis
Model chi-squared test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
in the regression equation $\ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $y = 1$ (or equivalently, the proportion of $y = 1$ in the population) given the scores on the independent variables.
Model chi-squared test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
or in terms of odds ratio:
• H0: $e^{\beta_k} = 1$
in the regression equation $\ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $y = 1$ (or equivalently, the proportion of $y = 1$ in the population) given the scores on the independent variables.
Alternative hypothesisAlternative hypothesis
Model chi-squared test for the complete regression model:
• H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
Model chi-squared test for the complete regression model:
• H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
• H1: $\beta_k \neq 0$
or in terms of odds ratio:
• H1: $e^{\beta_k} \neq 1$
AssumptionsAssumptions
• In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
• In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
Test statisticTest statistic
Model chi-squared test for the complete regression model:
• $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance}$
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
• Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$
• Wald $= \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.
Likelihood ratio chi-squared test for individual $\beta_k$:
• $X^2 = D_{K-1} - D_K$
$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
Model chi-squared test for the complete regression model:
• $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance}$
$D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
• Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$
• Wald $= \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.
Likelihood ratio chi-squared test for individual $\beta_k$:
• $X^2 = D_{K-1} - D_K$
$D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
Sampling distribution of $X^2$ and of the Wald statistic if H0 were trueSampling distribution of $X^2$ and of the Wald statistic if H0 were true
Sampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
• chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
• chi-squared distribution with 1 degree of freedom
Sampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
• chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
• chi-squared distribution with 1 degree of freedom
Significant?Significant?
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
• If defined as Wald $= \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
• If defined as Wald $= \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
Wald-type approximate $C\%$ confidence interval for $\beta_k$Wald-type approximate $C\%$ confidence interval for $\beta_k$
$b_k \pm z^* \times SE_{b_k}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
$b_k \pm z^* \times SE_{b_k}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
Goodness of fit measure $R^2_L$Goodness of fit measure $R^2_L$
$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
Example contextExample context
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
SPSSSPSS
Analyze > Regression > Binary Logistic...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Analyze > Regression > Binary Logistic...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
JamoviJamovi
Regression > 2 Outcomes - Binomial
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Regression > 2 Outcomes - Binomial
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Practice questionsPractice questions
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# Thursday October 17, 2024
## Calculating 2 days before Saturday October 19, 2024 by hand
This page helps you figure out the date that is 2 days before Saturday October 19, 2024. We've made a calculator to find the date before a certain number of days before a specific date. If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculatorto type in a new question or days from a specific date if you want to add 2 days.
But for all you time sickos out there who want to calculate 2 days before Saturday October 19, 2024 - here's how you do it:
1. Start with the Input Date (Saturday October 19, 2024): Write it down! I can't stress this enough
2. Count in Weeks: Recognize that 2 days is approximately 0.2857142857142857 weeks. Count forward 0.2857142857142857 weeks (0.4 work weeks) from the input date. This takes you to .
3. Add Remaining Days: Since you've counted 0.2857142857142857 weeks, you only need to add the remaining days to reach Thursday October 17, 2024
4. Use Mental Math: If Saturday October 19, 2024 is a Thursday, then compare that to if 2 is divisible by 7. That way, you can double-check if October 17 matches that Thursday.
## Thursday October 17, 2024 Stats
• Day of the week: Thursday
• Month: October
• Day of the year: 291
## Counting 2 days backward from Saturday October 19, 2024
Counting backward from today, Thursday October 17, 2024 is 2 before now using our current calendar. 2 days is equivalent to:
2 days is also 48 hours. Thursday October 17, 2024 is 79% of the year completed.
## Within 2 days there are 48 hours, 2880 minutes, or 172800 seconds
Thursday Thursday October 17, 2024 is day number 291 of the year. At that time, we will be 79% through 2024.
## In 2 days, the Average Person Spent...
• 429.6 hours Sleeping
• 57.12 hours Eating and drinking
• 93.6 hours Household activities
• 27.84 hours Housework
• 30.72 hours Food preparation and cleanup
• 9.6 hours Lawn and garden care
• 168.0 hours Working and work-related activities
• 154.56 hours Working
• 252.96 hours Leisure and sports
• 137.28 hours Watching television
## Famous Sporting and Music Events on October 17
• 1939 "Mr. Smith Goes to Washington", directed by Frank Capra and starring James Stewart and Jean Arthur, is released
• 1860 1st British Open Men's Golf, Prestwick GC: Willie Park Sr. wins inaugural event by 2 strokes from fellow Scot Tom Morris Sr
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https://socratic.org/questions/how-do-you-use-transformation-to-graph-the-cosine-function-and-determine-the-amp-8
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# How do you use transformation to graph the cosine function and determine the amplitude and period of y=-4cos(-3x)?
May 29, 2017
As cosine is an even function we have that $\cos \left(- x\right) = \cos x$
so, $\cos \left(- 3 x\right) = \cos \left(3 x\right)$
Hence, $y = - 4 \cos \left(- 3 x\right) = - 4 \cos \left(3 x\right)$
If we start with the graph of $y = \cos x$, we need to apply a horizontal squash factor of $\frac{1}{3}$ to get $y = \cos \left(3 x\right)$
i.e. the period changes from 360 degrees to 120 degrees
From $y = \cos \left(3 x\right) \to y = - 4 \cos \left(3 x\right)$ we apply a stretch factor of 4 parallel to the y-axis and reflect about the x-axis due to the -4
The amplitude is 4.
See the graph below:
:)>
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https://snipplr.com/view/35606/project-euler--problem-17
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# Posted By
chroto on 06/08/10
# Statistics
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Favorited by 0 user(s)
# Project Euler - Problem 17
/ Published in: Python
Save to your folder(s)
Copy this code and paste it in your HTML
`#!/usr/bin/python # Project Euler - Problem 17# In the numbers 1 - 1000 how many letters are needed# to spell out each word. sum = 11 # one thousandfor n in range(1,1000): if n % 900 != n: sum += 14 # 'nine hundred and' x = n % 900 elif n % 800 != n: sum += 15 # 'eight hundred and' x = n % 800 elif n % 700 != n: sum += 15 # 'seven hundred and' x = n % 700 elif n % 600 != n: sum += 13 # 'six hundred and' x = n % 600 elif n % 500 != n: sum += 14 # 'five hundred and' x = n % 500 elif n % 400 != n: sum += 14 # 'four hundred and' x = n % 400 elif n % 300 != n: sum += 15 # 'three hundred and' x = n % 300 elif n % 200 != n: sum += 13 # 'two hundred and' x = n % 200 elif n % 100 != n: sum += 13 # 'one hundred and' x = n % 100 else: x = n # n < 100 if x == 0: sum -= 3 # 'remove "and"' elif x % 90 != x: sum += 6 # 'ninety' x = x % 90 elif x % 80 != x: sum += 6 # 'eighty' x = x % 80 elif x % 70 != x: sum += 7 # 'seventy' x = x % 70 elif x % 60 != x: sum += 5 # 'sixty' x = x % 60 elif x % 50 != x: sum += 5 # 'fifty' x = x % 50 elif x % 40 != x: sum += 5 # 'forty' x = x % 40 elif x % 30 != x: sum += 6 # 'thirty' x = x % 30 elif x % 20 != x: sum += 6 # 'twenty' x = x % 20 elif x % 10 != x: if x == 17: sum += 9 elif x == 11 or x == 12: sum += 6 elif x == 15 or x == 16: sum += 7 elif x == 10: sum += 3 else: sum += 8 x = 0 if x == 1 or x == 2 or x == 6: sum += 3 elif x == 3 or x == 7 or x== 8: sum += 5 elif x == 4 or x == 5 or x == 9: sum += 4 print sum`
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https://socratic.org/questions/how-to-prove-that-2sin-4pi-5-cos-pi-5-sin-2pi-5#621286
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# How to prove that 2sin((4pi)/5)cos(pi/5)=sin((2pi)/5)?
May 28, 2018
$\sin \left(\frac{4 \pi}{5}\right) = \sin \left(\pi - \frac{\pi}{5}\right) = \sin \left(\frac{\pi}{5}\right)$
Using $2 \sin x \cos x = \sin \left(2 x\right)$
we obtain
$2 \sin \left(\frac{\pi}{5}\right) \cos \left(\frac{\pi}{5}\right) = \sin \left(\frac{2 \pi}{5}\right)$
and finally
$2 \sin \left(\frac{4 \pi}{5}\right) \cos \left(\frac{\pi}{5}\right) = \sin \left(\frac{2 \pi}{5}\right)$
May 28, 2018
$\text{see explanation}$
#### Explanation:
$\text{using the "color(blue)"product to sum formula}$
•color(white)(x)2sinxcosy=sin(x+y)+sin(x-y)
$\text{here "x=(4pi)/5" and } y = \frac{\pi}{5}$
$\text{consider left side}$
$\sin \left(\frac{4 \pi}{5} + \frac{\pi}{5}\right) + \sin \left(\frac{4 \pi}{5} - \frac{\pi}{5}\right)$
$= \sin \pi + \sin \left(\frac{3 \pi}{5}\right)$
$= 0 + \sin \left(\pi - \frac{3 \pi}{5}\right)$
$= \sin \left(\frac{2 \pi}{5}\right) = \text{ right side"rArr" verified}$
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https://math.answers.com/other-math/What_is_74_percent_as_a_fraction_in_lowest_terms
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0
# What is 74 percent as a fraction in lowest terms?
Updated: 4/28/2022
Doraexplorer
Lvl 1
10y ago
74% = .74 = 74/100 = 37/50
Divide 74/100 by 2. This is in lowest terms because 37 is a Prime number. Only one number in a fraction must be prime in order for a fraction to be in lowest terms.
~Doraexplorer~
Wiki User
10y ago
Earn +20 pts
Q: What is 74 percent as a fraction in lowest terms?
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https://www.physicsforums.com/threads/laplace-transforms-to-solve-initial-value-de-partial-fractions.711646/
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# Laplace transforms to solve initial value DE / partial fractions
1. Sep 20, 2013
### CrazyCamo
Hey guys, i have read many posts on physics forums but this would be my first post. I am quite stuck so any help would be much appreciated.
1. The problem statement, all variables and given/known data
Use Laplace transforms to solve the initial value problem:
f''(y) + 4f'(y) +8y = u(t-1) where y(0) = 1 and y'(0) = -1
Solve this problem using laplace transforms, showing all steps in your reasoning. State the solution y(t) for each of 0<t<1 and t>1, then sketch it over the range 0<= t <= 10, noting its main features.
2. Relevant equations
3. The attempt at a solution
I have gotten up to F(Y) = (e^-s + s^2 + 3s)/(s(s^2 + 4s+8))
However, from here i am not sure what to do. I tried taking the partial fractions of:
1/(s(s^2 + 4s+8))
but am getting very confused. Again any help would be much appreciated. Cheers
2. Sep 21, 2013
### SteamKing
Staff Emeritus
Why don't you break F(Y) down into the sum of its various components?
E.g., D = (s(s^2+4s+8))
F(Y) = (e^-s)/D + s^2/D + 3s/D
You can tackle each term individually.
PS: finding the PFE of 1/D doesn't help.
3. Sep 21, 2013
### HallsofIvy
Staff Emeritus
The denominator is $$s(s^2+4s+ 8)= s(s^2+ 4s+ 4+ 4)= s((s+ 2)^2+ 4)$$ so you can use "partial fractions to write that as $$\frac{A}s+ \frac{Bs+ C}{(s+2)^2+ 4}$$.
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# A metallic sphere of 1 kg mass, with a surface area of 0.0314 m2, is maintained at an initial temperature of 50°C. The fluid circulating the sphere is maintained at a temperature of 10°C. Specific heat of metallic sphere is 314 J/kg-K and the heat transfer coefficient between the fluid and the sphere is 10 W/m2 -K. The time is taken (in seconds) for the sphere to cool down to 20°C is___
This question was previously asked in
GATE PI 2014 Official Paper
View all GATE PI Papers >
## Detailed Solution
Concept:
Lumped parameter analysis:
Internal/conductive resistance is very little as compared to surface convective resistance and the temperature distribution is given by
$$\frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~}=~{{e}^{\left( \frac{hA}{ρ V{{C}_{p}}} \right)t}}$$
$$\frac{hA}{ρ V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)$$
where
• Ti = Initial temperature of beat at t = 0,
• T = Temperature of body at any instant ‘t’ sec
• T = Ambient fluid temperature
• h = heat transfer coefficient, ρ is the density of the metal sphere
For sphere: $$\frac{V}{A} = \frac{{Volume\;of\;body}}{{surface\;area}} = \frac{{\frac{4}{3}π {r^3}}}{{4π {r^2}}} = \frac{r}{3}$$
V/A = r/3
Calculation:
Given:
Cp = 314 J/kgK, h = 10 W/m2K, m = 1 kg, A = 0.0314 m2
Ti = 50°C, T∞ = 10°C, T = 20°C
surface area of sphere, A = 4 π r2 = 0.0314 m2
⇒ r = 0.05 m
density of the metal sphere,
$$\rho = {mass\over volume \;of \;sphere} = {m\over{\frac{4}{3}\pi r^3}}$$
$$={ 1\over{\frac{4}{3}\pi\;0.05^3 }}$$ =1909.86 kg/m3
Using the equation below and putting the given values:
$$\frac{hA}{ρ V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)$$
$$\ln\left( {\frac{{50 - 10}}{{20 - 10}}} \right) = \left( {\frac{{10 × 3 }}{{0.05× 1909.86 × 314}}} \right) × t$$
$$⇒ t= \frac{{1.386 × \;0.05 × 1909.86 × 314}}{{10 × 3}}$$
⇒ t =1385.29 seconds
Time is taken (in seconds) for the sphere to cool down to 20°C is = 1385.29 sec
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# 6.8 – Areas and Sectors
## Key Terms
• Area – The space taken up by a two-dimensional figure or surface.
• Area is measured in square units, such as square inches, square centimeters, or square feet.
• Sector – A part of the interior of a circle bounded by an arc and the two radii that share the arc’s endpoints.
## Review
Units of Area
• Area is always listed in units “squared”
• Units may be units (units), centimeters (cm), meters (m), feet (ft), inches (in), kilometers (km), etc.
• Ex. $18\:m^2, 9\:cm^2, 3\:units^2, 6ft^2, 102km^2$
Arc Length and Area of a Sector Comparison
## Notes
Area of a Circle
• When we talk about the area of a circle, we mean the area enclosed by the circle (how much space is inside the circle).
• Steps to find the area of a circle
1. Find the radius of the circle.
2. Square the radius.
3. Multiply the result by using the calculator button $\pi$ or 3.14.
4. Round your answer to the nearest hundredth.
• Example: What is the approximate area of the circle shown below?
• Formula: $A=\pi\:r^2$
• Substitute: $A=\pi\:23^2$
• Simplify: $A=\pi\:529$
• Answer: $A=1661.9\:cm^2$
Sectors
• A sector of a circle is a region bounded by an arc of a circle and radii to the endpoints of the arc.
• The area of a sector is the area of the circle multiplied by the fraction of the circle covered by that sector.
## Examples
Area of a Circle
• Ex 1. If a small 12-inch pizza costs $8 and a large 16-inch pizza costs$12, which is the better deal?
• Steps to Find Out
1. Find the area of each pizza.
2. Use the area to find each pizza’s cost per square inch.
3. Compare. The pizza that costs the least per square inch is the better deal.
• Large: diameter = 16 in; so, radius = 8 in
• Formula: $Area\:=\pi r^2$
• Substitution: $A\:=\pi 8^2$
• Simplify: $A\:=\pi\bullet64$
• Solve: $A=201\:in^2$
• Cost per Inch: $cost\:\div size$; so, $12\div 201=0.06\:cents\:per\:inch$
• Small: diameter = 12 in; so, 4adius = 6 in
• Formula: $Area\:=\pi r^2$
• Substitution: $A\:=\pi 6^2$
• Simplify: $A\:=\pi\bullet36$
• Solve: $A=113\:in^2$
• Cost per Inch: $cost\:\div size$; so, $8\div 113=0.07\:cents\:per\:inch$
• Result: The large pizza is about 1 cent cheaper per square inch. It doesn’t seem like much, but it will save you money in the long run if you shop “price per unit” for the lowest amount!
• For the pizza problem, you save about 88 cents per pizza if you buy the large one (201 sq in minus 113 sq in = 88 sq in).
• If each square inch is about 1 cent, multiply 1 cent by 88 sq in to get 88 cents!
• If you buy 10 pizzas, you save \$8.80!
Area of Sectors
• Ex 1. If the area of the circle below is $12\:m^2$, what is the area of the shaded sector?
• Answer: $3\:m^2$
• Ex 2. A circle has an area of $36\:m^2$. What is the area of a 40° sector of this circle?
• Answer: $4\:m^2$
• Ex 3. What is the approximate area of the shaded sector in the circle shown below?
• Answer: $18.8\:cm^2$
• Ex 4. What is the approximate area of the shaded sector in the circle shown below?
• Answer: $297\:cm^2$
• Ex 5. What is the approximate area of the shaded sector in the circle shown below?
• Notice that the marker is NOT a chord. It is the measure of the diameter (5.4cm).
• Answer: $11.45\:cm^2$
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```Question 570724
The town of Foxton lies 10 mi north of an abandoned east-west road that runs through Grimley, as shown in the figure.
The point on the abandoned road closest to Foxton is 60 mi from Grimley.
County officials are about to build a new road connecting the two towns.
They have determined that restoring the old road would cost \$100,000 per mile, while building a new road would cost \$200,000 per mile.
How much of the abandoned road should be used (as indicated in the figure) if the officials intend to spend exactly \$8.8 million?
:
This is a triangle problem has a right angle at a point on the old road directly south of Fox
:
Leg 1 = 10 mi
leg 2 = x, the distance from a point due south of F, to where the new road joins the old road
then
Road dist from F to G = hypotenuse + (60-x)
:
Cost: hypotenuse 200,000 per mile; (60-x) \$100,000 per mile
:
Do this in 100 thousands of dollars, to avoid writing all these zeros
:
{{{2*sqrt(10^2 + x^2)}}} + 1(60-x) = 88
{{{2*sqrt(100 + x^2)}}} = 88 - (60-x)
{{{2*sqrt(100 + x^2)}}} = 88 - 60 + x
{{{2*sqrt(100 + x^2)}}} = x + 28
Square both sides
4(100+x^2) = (x+28)^2
400 + 4x^2 = x^2 + 56x + 784
Combine like terms on the left
4x^2 - x^2 - 56x + 400 - 784 = 0
3x^2 - 56x - 384 = 0
we can use the quadratic formula here, but this will factor to:
(x-24)(3x+16) = 0
The positive solution
x = 24
then
{{{sqrt(10^2+24^2)}}} = 26 mi
:
60 - 24 = 36 mi
:
If we did this right, the cost should be 8.8 million
26(200000) = \$5,200,000
36(100000) = \$3,600,000
------------------------
total road: \$8,800,000 which is 8.8 million
:
:
How much would it cost to build a new road connecting the towns directly?
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When introducing or reinforcing adding decimals, this 2 page worksheet includes word problems that require students to add decimals by coloring parts in hundreds grids. Answer Key included!
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Students don't always have manipulatives available. This sheet of hundreds grids will allow students to have the support that they need! This can be very useful at home for homework, or on a test if it is allowed.
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Fraction and Decimal Folding Hundred Grid This is an empty Hundred grid that you can use to help students make the link between fractions and decimals. Students can fold the grid to the desired fraction and then count the squares to work out what the decimal number is. For example if they fold i
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Match up pictures of decimals in a hundreds grid to it's decimal and written format
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This is a set of 24 hundreds block cards with various colored designs that can be used for a number of activities or centers. Be creative- they lend themselves well as a multipurpose resource! A few examples: Fractions and decimals to the hundredths Finding area of a color/shape Finding perimeter
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Help your students become experts with tenths and hundredths, developing their understanding of how these fractions relate with this set of games, activity cards, reference sheets, and assessment activities. ************************************************************************ Save \$\$\$ by purc
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This is an engaging performance task that I created as a homework project for my 4th graders. This could also easily be done in class! The project requires students to design a garden on a hundreds grid following specific planting rules (i.e. 2 tenths of the garden are carrots, 0.18 are sweet peas).
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Students use decimal cards and place value fraction cards to build decimal numbers and learn about how place value affects the way that decimals are written, compared, modeled, and multiplied. Use the tenth, hundredths, and thousandths decimal place cards to build numbers and explore the relationshi
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Great Product to use for helping students write, read, and convert fractions with denominators of 10 and 100 to decimals. Models including number lines, tens and hundred grids are utilized!!! Less Procedures/More Critical Thinking!!! This Product contains a PowerPoint Math Lesson (Chalkboard Theme)
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This PowerPoint presentation takes students through the process of using a ten frame or hundred grid to make models of division problems involving decimals. The topics include division of decimal by a whole number, whole number by a decimal, and decimal by a decimal. Practice problems are included
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Have students practice writing equivalent forms of numbers as fractions, decimals, or percents using these math center activities. A set of 96 cards in 24 matching sets can be used to create math centers games or used with two extension activities. Money models are used with the half, fourths, ten
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This is a fun activity that reinforces fraction, decimal, and percent conversions. Students use a hundreds grid to design a blueprint for Santa's Village. Students must calculate the fractional amount of each area of the village. Then, depending on which version of the activity (there are three i
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#testprepsale **SAVE 20% WHEN YOU PURCHASE THIS BUNDLE (includes both our Basic & Advanced MATH READY Representing the Value of the Digit in Decimals to the Hundredths Task Cards sets)!** *************************************************************************** Customers who used our 4th grad
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This is a resource used to help students identify decimal place value and represent decimals in several ways. Students can use a hundreds grid, a number line, and represent it in word form.
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$\newcommand{\ds}{\displaystyle}$ $\newcommand{\ve}[1]{\overrightarrow{\mathbf{#1}}}$ $\newcommand{\ora}{\overrightarrow}$ $\newcommand{\com}[1]{\left\langle #1 \right\rangle}$ $\newcommand{\mb}{\mathbf}$ $\newcommand{\ivec}{\hat{\mbox{\bf{\dotlessi}}}}$ $\newcommand{\jvec}{\hat{\mbox{\bf{\dotlessj}}}}$ $\newcommand{\kvec}{\hat{\mbox{\bf{k}}}}$ $\newcommand{\magn}[1]{\left|\left| #1 \right|\right|}$
Series Convergence/Divergence
Click on each problem to expand and view the problem. You can then download a solution to the problem in PDF format. You can collapse the problem by clicking on the problem again.
• Exam 2, Spring 2006, Problem 6
• Exam 2, Fall 2009, Problem 6
• Exam 2, Spring 2010, Problem 5a
• Exam 2, Spring 2010, Problem 5b
• Exam 2, Spring 2012, Problem 5
Determine whether each of the following series converges or diverges. Indicate the method you are using.
(a) $\ds\sum_{k=1}^\infty\frac{k^2}{k^4+1}$
(b) $\ds\sum_{k=1}^\infty\frac{k!}{k^k}$
• Exam 2, Fall 2012, Problem 5
Determine whether or not the following infinite series converge. Justify your answers.
(a) $\ds\sum_{k=1}^\infty\frac{k-1}{k^3+5}$
(b) $\ds\sum_{k=3}^\infty\frac{1}{(\ln k)^{10}}$
• Exam 2, Spring 2013, Problem 3
Determine whether the following series converge or not.
(a) $\ds\sum_{n=1}^\infty\cos\left(\frac{1}{n}\right)$
(b) $\ds\sum_{n=1}^\infty\left(\frac{n}{5n+3}\right)^n$
(c) $\ds\sum_{n=1}^\infty\frac{\sin^2(n)}{n^2}$
• Exam 2, Study Guide 3, Problem 16
• Exam 2, Study Guide 3, Problem 17
• Exam 2, Study Guide 3, Problem 18
• Exam 2, Study Guide 3, Problem 19
• Exam 2, Study Guide 3, Problem 20
• Exam 2, Study Guide 3, Problem 21
• Exam 2, Study Guide 3, Problem 22
• Exam 2, Study Guide 3, Problem 23
• Exam 2, Study Guide 3, Problem 24
• Final Exam, Fall 2008, Problem 7
• Final Exam, Spring 2009, Problem 8
• Final Exam, Fall 2009, Problem 8
• Final Exam, Spring 2010, Problem 3
• Final Exam, Fall 2011, Problem 2
(a) $\ds \sum_{k=2}^\infty\frac{1}{k\ln(k)}$
(b) $\ds \sum_{k=1}^\infty\frac{(-1)^kk^2}{2^k}$
• Final Exam, Spring 2012, Problem 4
Determine whether each of the following series converges or diverges. Moreover, for those that converge, compute their sum. Indicate the method you are using.
(a) $\ds\sum_{n=1}^\infty\frac{-2}{n(n+1)}$
(b) $\ds\sum_{n=3}^\infty\frac{6}{\pi^n}$
(c) $\ds\sum_{k=2}^\infty\left(\frac{3k^3}{2k^3+9k^2+2k}\right)^{k/2}$
• Final Exam, Spring 2012, Problem 5
Determine whether the following series (a) converges absolutely, (b) converges conditionally, or (c) diverges. Justify your answer. $$\sum_{n=20}^\infty\frac{(-1)^n}{\ln(\ln n)}$$
• Final Exam, Study Guide, Problem 9
• Final Exam, Study Guide, Problem 10
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1. Derivatives and Perpendicular Lines
Determine the coordinates of the point of intersection of the two perpendicular lines that intersect on the y-axis and are both tangent to the parabola given below:
$y = 9x^2$
When I sketch this out, I can see visually that this point of intersection cannot occur at a positive y-value. Other things that are quickly apparent:
$f'(x) = 18x$
The point of intersection must have an x-coordinate of 0.
So, if I use the formula of a line, and plug in an x-coordinate of 0, I get:
$y = f(0) + f'(0)(x - 0)$
$y = 0 + 0(x - 0)$
$y = 0$
And I conclude that my point of intersection is (0,0). All fine and dandy, except that this isn't the right answer. Can anybody help?
Thanks.
2. The two tangents should be at right angles to each other, and at the same time tangent to the curve.
Let the gradient of the first line be $m_1$
That of the second will be $m_2$
And $m_1 \times m_2 = -1$
3. Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:
Let $L_1$ = the positively sloping line and $L_2$ be the negatively sloping line.
Let $L_1$ pass through the points, $(x_1, y_1)$ and $(0, y_2)$ where $y_2 < 0$.
Since $y = 9x^2$ is even, this means that $L_2$ passes through $(-x_1, y_1)$ and $(0, y_2)$.
As you point out, the slopes are negative reciprocals of each other, by definition.
Then, there are a couple of things that I think are true, but I don't know just how to prove them. First $m_1 = 1$ and $m_2 = -1$. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).
If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't $y_2$ have to equal $-(y_1)$?
Am I on the right track? Making this to complicated? Any help is appreciated.
Thanks again.
4. That's right! m1 and m2 equal to -1 and 1 (order unimportant).
So that, the gradient at anypoint being 18x, you get:
$18x = 1$
$x = \dfrac{1}{18}$
And:
$18x = -1$
$x = -\dfrac{1}{18}$
Those are your two x coordinates. In your picture, the points where the tangents meet the line have the same y coordinate.
At x = 1/18, find y and from the points and slope, get the equation of the lines (in fact, one only is necessary)
Then, find where the line meets the y axis.
5. Originally Posted by joatmon
Yeah, I thought of that (and probably should have listed it in my original post), but I still don't know how to integrate it into this solution. Here's some further thoughts:
Let $L_1$ = the positively sloping line and $L_2$ be the negatively sloping line.
Let $L_1$ pass through the points, $(x_1, y_1)$ and $(0, y_2)$ where $y_2 < 0$.
Since $y = 9x^2$ is even, this means that $L_2$ passes through $(-x_1, y_1)$ and $(0, y_2)$.
As you point out, the slopes are negative reciprocals of each other, by definition.
Then, there are a couple of things that I think are true, but I don't know just how to prove them. First $m_1 = 1$ and $m_2 = -1$. I say this because, in order to be tangent to an even function, the only way that I can see two perpendicular lines meeting at the y-axis would be if they both had symmetrical x-coordinates (e.g. (5, 0) and (-5, 0)).
If that is true, this final one might be the key to the whole thing. If the slope is 1 or -1, and the lines intersect at the y-axis, then doesn't $y_2$ have to equal $-(y_1)$?
Am I on the right track? Making this to complicated? Any help is appreciated.
Thanks again.
you are on the right track ... the use of symmetry is key to this problem.
6. Got it. Thanks to both of you for your help.
$f(\frac{1}{18}) = 9(\frac{1}{18})^2 = \frac{1}{36}$
So the tangental points are $(\frac{1}{18}, \frac{1}{36})$ and $(\frac{-1}{18}, \frac{1}{36})$
And that means that the lines intersect at $(0, \frac{-1}{36})$.
Can't thank you enough for your help.
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Anda di halaman 1dari 3
# Assignment
Problem 1
The names and locations of the rain gauge stations with their mean annual precipitation are as
given in the table below. Calculate average precipitation of the catchment area by Isohyetal
method.
Name of
Station
Latitude
Degrees
Minutes
Longitude
Degrees
Minutes
Mean Annual
Precipitation
(in)
A
B
C
D
E
F
G
H
I
J
K
L
16
16
14
17
17
17
14
16
15
17
15
16
03
35
58
25
27
10
53
25
15
26
45
26
103
104
102
104
101
104
103
101
104
102
102
102
04
44
07
50
44
09
29
08
53
46
02
51
39.85
56.58
45.73
85.12
47.51
70.59
52.67
50.83
60.57
56.22
41.50
46.81
Latitude and Longitude of the Catchment Area are 140 30 to 180 30 and 1010 00 to 1040
00, respectively.
Problem 2
Assuming the rain falling vertically, express the catch of a gauge inclined 15 from the vertical
as a percentage of the catch of a gauge installed vertically.
Problem 3
A 3-hour storm occurred at a place and the precipitation in the neighboring rain-gage stations P,
Q and R were measured as 3.8, 4.1 and 4.5 cm respectively. The precipitation in the neighboring
station S could not be measured since the rain gauge was damaged. The normal annual
precipitation in the four stations P, Q, R and S were 45, 48, 53 and 50 cm, respectively. Estimate
the storm precipitation at station S.
Problem 4
In a catchment one precipitation station A, was inoperative for part of a month during which
storm occurred. The respective storm totals at three surrounding stations 1, 2 and 3 are 35, 40
and 30 mm, respectively. The normal annual precipitations at A, 1, 2 & 3 are respectively 985,
1125, 940 and 1210 mm. Estimate the storm precipitation for Station A.
Problem 5
The average annual precipitation for the four sub-basins constituting a large river basin is 58, 67,
85 and 80 cm. The sub-basin areas are 900, 690, 1050 and 1650 km, respectively. What is the
average annual precipitation for the basin as a whole?
Problem 6
The annual precipitation at station A and the
15 surrounding stations are given in the Table 3.19 given below:
a. Determine the consistency of the record at station A.
average
precipitation
at
## b. In which year, there is a regime change indicated?
Table 3.19 Data Table
Year
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
Annual Precipitation
Station A
15 Station
(cm)
Average (cm)
50.5
90.0
16.0
21.5
50.5
62.5
69.5
36.0
42.0
42.0
71.5
57.0
27.5
25.0
60.0
22.0
55.0
57.0
36.5
19.5
Year
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
Annual Precipitation
Station A
15 Station
(cm)
Average (cm)
36.0
42.0
18.0
30.0
54.0
48.0
12.0
36.0
42.0
36.0
27.5
60.5
55.0
38.5
38.5
47.5
49.5
24.0
44.0
60.5
Problem 7
The Table 3.20 given below gives the annual rainfall at a station A and the average annual
rainfall of 10 stations in the vicinity for a period of 17 years. It is suspected that there has been a
change in the location of the rain gauge at Station A during the period of this record. Determine
the year when this change occurred and the corrected rain gauge readings prior to this year.
## Table 3.20 Rainfall Data
Rainfall
Year
Station A
(mm)
Average
Annual
Rainfall of 10
Stations (mm)
1931
1932
1933
1934
1935
1936
1937
1938
1939
1940
505
900
60
215
505
625
695
360
420
360
615
570
275
250
605
220
550
570
365
275
Year
Rainfall
Average
Station A
Annual
(mm)
Rainfall of 10
Stations (mm)
1941
1942
1943
1944
1945
1946
1947
420
180
300
540
480
120
360
605
550
385
385
475
495
240
Problem 8
Point rainfalls due to a storm at several rain-gauge stations in a basin are shown in Fig. 3.16.
Determine the mean areal depth of rainfall over the basin.
8 cm
6 cm
zone
Area(km)
I
410
II
900
III
2850
IV
1750
V
720
VI
550
Area of the Basin = 7180 km
8 cm
zone I
N (9.2)
zone VI
O (7.4 cm)
D (9.2)
G (8.5)
A (8.6)
L (5.2)
C (10.8)
M (5.6)
H (10.5)
E (13.8)
K (7.8)
B (7.6)
F (10.4)
I (11.2)
J (9.5)
zone II
10 cm
zone V
zone IV
zone III
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# Differences between MI and MIGRAM
There is a minor difference between the results of the mutual information derived with mi and migram. The reason is that mi uses an algorithm as suggested by Roulston, M. S.: Estimating the errors on measured entropy and mutual information, Physica D, 125, 1999, which considers estimation errors (Px and Py are the probabilities of x and y, Pxy is the joint probability, and Nx, Ny and Nxy are the numbers of non-zero elements in Px, Py and Pxy):
• Ix = -sum(Px * log(Px))
• Iy = -sum(Py * log(Py))
• Ixy = -sum(Pxy * log(Pxy))
• Ixy_syserr = (Nx + Ny - Nxy - 1) / (2*length(x))
• MI = Ix + Iy - Ixy + Ixy_syserr
In contrast, migram only uses the simple definition of mutual information:
• Ix = -sum(Px * log(Px))
• Iy = -sum(Py * log(Py))
• Ixy = -sum(Pxy * log(Pxy))
• MI = Ix + Iy - Ixy
However, the trend in the results are quite similar.
Let us consider two oscillations
```x = cos(0:.01:10*pi)';
y = sin(0:.01:10*pi)' + .1 * randn(length(x),1);
```
between which we will estimate the mutual information up to a lag of 10 steps.
Using migram we need to call this command with the arguments specifying the window length, which should be equal to the data length. In order to ensure that we use 10 bins we can also specify this argument.
```m1 = migram(x,y,10,length(x),[],10);
```
The output m is a vector containing the mutual information between x and y from lag -10 up to +10 (hence, the length is 21)
```m1'
```
```ans =
Columns 1 through 9
0.6564 0.6608 0.6602 0.6570 0.6550 0.6528 0.6512 0.6519 0.6484
Columns 10 through 18
0.6461 0.6459 0.6492 0.6499 0.6522 0.6569 0.6534 0.6504 0.6501
Columns 19 through 21
0.6522 0.6510 0.6475
```
Using mi we again specify the number of bins and the maximal length (both 10).
```m2 = mi(x,y,10,10);
```
The output is a 3-dimensional matrix, where the third dimension corresponds to the lag, and the first two dimensions to the first and second time series. E.g.,
```squeeze(m2(1,1,:))'
```
```ans =
Columns 1 through 9
2.1842 2.0405 1.9405 1.8567 1.7839 1.7198 1.6628 1.6122 1.5673
Columns 10 through 11
1.5278 1.4933
```
gives the auto mutual information of the first time series x for the lags 0:10, wheras
```squeeze(m2(2,1,:))'
```
```ans =
Columns 1 through 9
0.6378 0.6412 0.6419 0.6441 0.6489 0.6453 0.6423 0.6419 0.6440
Columns 10 through 11
0.6427 0.6388
```
gives the mutual information between x and y for lags 0:10. If we need the mutual information between x and y for lags -10:0, i.e. between y and x for lags 0:10, then we call
```squeeze(m2(1,2,:))'
```
```ans =
Columns 1 through 9
0.6378 0.6394 0.6427 0.6476 0.6478 0.6501 0.6531 0.6563 0.6604
Columns 10 through 11
0.6619 0.6582
```
Finally we compare the results derived with migram and mi in a plot. Note that we need to concatenate the components (1,2) and (2,1) of the results matrix derived with mi in order to get the mutual information of lags -10:10, and that the component (1,2) has to be flipped because it runs from 0:-10 but we need it to run from -10:0.
```subplot(2,1,1)
plot(-10:10, m1)
grid on, xlabel('Lag'), ylabel('MIGRAM')
subplot(2,1,2)
plot(-10:10, [flipud(squeeze(m2(1,2,:))); squeeze(m2(2,1,2:end))])
grid on, xlabel('Lag'), ylabel('MI')
```
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# Homework Help: Finding Speed and Coefficient of Friction
1. Oct 2, 2004
Finding Speed with Coefficient of Friction
You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.5 m/s.
With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
I've figured out this equation since there is constant acceleration:
V^2=Vo^2+2a(x-xo) where a=-(Mk)g. I haven't taken into account the incline and I'm not sure how. How do I incorporate the incline into this equation, if my equation is even right.
#### Attached Files:
• ###### W0106-N.jpg
File size:
4.2 KB
Views:
87
Last edited: Oct 2, 2004
2. Oct 2, 2004
### Pyrrhus
Let me point out change of Mechanical Energy will be equal to the work done by friction.
$$\Delta E = W_{f}$$
Last edited: Oct 2, 2004
3. Oct 2, 2004
I don't see the relation....
4. Oct 2, 2004
### Pyrrhus
$$E - E_{o} = W_{f}$$
What's not to see?
Tell me exactly what you don't understand.
5. Oct 2, 2004
Well in my class we haven't dealt with that equation yet, and this problem should have nothing to do with Mechanical Energy, just Newton's Laws and Acceleration and Friction.
6. Oct 2, 2004
### Pyrrhus
Oh, should had said so , then use newton's 2nd law (to find acceleration) and kinematics for final speed.
7. Oct 2, 2004
I can't do that lol I wrote out my equation in my original post I just can't incorporate the 12 degree angle.
8. Oct 2, 2004
### Pyrrhus
Put your X-axis along the 12 degree straight line, so the normal force will have only a non-zero component, and the weight of the car will have 2 non-zero components one pointing left, and the other pointing down., and the friction force will be on the right, against movement with 1 non-zero component.
Last edited: Oct 2, 2004
9. Oct 2, 2004
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# cross correlation sound
Started by August 24, 2008
```Yes, you are definitely using too many points, so cut that number down
by a significant amount. Given the physical parameters of your set-
up, you really don�t need nearly a million points to compute
something.
I know you�re under deadline, and you�re probably feeling a bit
stressed at this point, so let me suggest something that could make
things easier for you. It has to do with how you�re computing your
solution once you get the time delays.
You don�t really need to solve simultaneous equations or anything like
that. You can simply use brute force computer power to get an answer
to �what is the x,y location of the sound source, given the following
delays between microphones 1, 2 and 3?�
Let�s presume we have the time delays between the microphones. Now
the problem becomes: �given the physical arrangement of the set-up,
might we use a look-up table to determine the x,y location of the
sound source, given that we have the following delays between
microphones?�
know exactly how long it takes for a sound to travel from one point to
another. Let's say that your microphones are placed at positions M0
(x=0, y=0), MX (x=1, y= 0), and MY (x=0, y=1). OK. I know that your
somewhere in the upper right quadrant bounded by M0 on the lower left,
MY on the upper left, and MX on the lower right. The sound source is
somewhere in the upper right quadrant of a Cartesian coordinate
system, and you're trying to find its x,y location. Now place an x,y
grid on top of the Cartesian quadrant. The spacing of the grid points
is up to you, but your precision is limited by the physical parameters
of your problem. Then compute the time delay from every grid point to
the known microphone locations.
For instance, presuming that the sound source is exactly at the M0
location, then, based on your spacing, you know precisely how long it
will take the sound to reach M0, MX and MY (the time to M0 is 0, and
the travel time to the others can be precisely computed).
Now sequence through every single one of the grid points and calculate
the delay from that point to M0, MX and MY. This can be used to
generate a look-up table for any x,y grid point. Each x,y point thus
contains the delays to each microphone. All of this can be pre-
computed.
Now obtain the delays by using max( ) or xcorr( ) on your recorded
data. They are your measured results. Compare the measured delays
with your previously generated grid data by going through the entire
list of grid points and compare them with your pre-computed delays.
Start with grid point 0,0, and run through all the rest. Find the
best match between the measured delays and your pre_computed ones.
You�ll have a problem in that you need to account for positive or
negative delays (e.g.: m1 might be ahead of m2, or m2 might be ahead
of m1). You'll also actually need multiple tables, because you have
multiple microphones.
Given the physical parameters of your problem, it shouldn't be too
difficult to compute a look-up table(s) with pre-computed time delays
from any x,y point in the grid to the known locations of M0, MY and
MX. Then, using your measured delays ( from max( ) or xcorr( ) ),
you�d search through the look-up table(s) to find the closest match.
The physical parameters of your problem indicate that the maximum
travel time will be 155 sample time delays, so you don�t need very big
table(s) to cover all the possible time delays.
```
```>
>Yes, you are definitely using too many points, so cut that number down
>by a significant amount. Given the physical parameters of your set-
>up, you really don=92t need nearly a million points to compute
>something.
>
>I know you=92re under deadline, and you=92re probably feeling a bit
>stressed at this point, so let me suggest something that could make
>things easier for you. It has to do with how you=92re computing your
>solution once you get the time delays.
>
>You don=92t really need to solve simultaneous equations or anything like
>that. You can simply use brute force computer power to get an answer
>to =93what is the x,y location of the sound source, given the following
>delays between microphones 1, 2 and 3?=94
>
>Let=92s presume we have the time delays between the microphones. Now
>the problem becomes: =93given the physical arrangement of the set-up,
>might we use a look-up table to determine the x,y location of the
>sound source, given that we have the following delays between
>microphones?=94
>
>know exactly how long it takes for a sound to travel from one point to
>another. Let's say that your microphones are placed at positions M0
>(x=3D0, y=3D0), MX (x=3D1, y=3D 0), and MY (x=3D0, y=3D1). OK. I know
tha=
>t your
>somewhere in the upper right quadrant bounded by M0 on the lower left,
>MY on the upper left, and MX on the lower right. The sound source is
>somewhere in the upper right quadrant of a Cartesian coordinate
>system, and you're trying to find its x,y location. Now place an x,y
>grid on top of the Cartesian quadrant. The spacing of the grid points
>is up to you, but your precision is limited by the physical parameters
>of your problem. Then compute the time delay from every grid point to
>the known microphone locations.
>
>For instance, presuming that the sound source is exactly at the M0
>location, then, based on your spacing, you know precisely how long it
>will take the sound to reach M0, MX and MY (the time to M0 is 0, and
>the travel time to the others can be precisely computed).
>
>Now sequence through every single one of the grid points and calculate
>the delay from that point to M0, MX and MY. This can be used to
>generate a look-up table for any x,y grid point. Each x,y point thus
>contains the delays to each microphone. All of this can be pre-
>computed.
>
>Now obtain the delays by using max( ) or xcorr( ) on your recorded
>data. They are your measured results. Compare the measured delays
>with your previously generated grid data by going through the entire
>list of grid points and compare them with your pre-computed delays.
>Start with grid point 0,0, and run through all the rest. Find the
>best match between the measured delays and your pre_computed ones.
>
>You=92ll have a problem in that you need to account for positive or
>negative delays (e.g.: m1 might be ahead of m2, or m2 might be ahead
>of m1). You'll also actually need multiple tables, because you have
>multiple microphones.
>
>Given the physical parameters of your problem, it shouldn't be too
>difficult to compute a look-up table(s) with pre-computed time delays
>from any x,y point in the grid to the known locations of M0, MY and
>MX. Then, using your measured delays ( from max( ) or xcorr( ) ),
>you=92d search through the look-up table(s) to find the closest match.
>The physical parameters of your problem indicate that the maximum
>travel time will be 155 sample time delays, so you don=91t need very big
>table(s) to cover all the possible time delays.
>
Hi,
Thank you. This seems ok because it will also give me an idea whether the
way we were calculating the delay was right or wrong.
I will try it out and get back to you.
But yea, one more thing. I have the formula and the functions to calculate
the co-ordinates of the sound source using non-linear equations. I get
perfect results as far as the center location is concerned (i.e. when i
generate a sound in the center of the table). BUT for any other location
other than the center, I still get the result as the center which is
weird.
I figured out where the problem is, and the problem is right here as what
we are trying to solve i.e. the delay.
The delay is converted into relative distance as (delay * 340 i.e. v of
sound)/44100 Hz i.e. fz . This ans is very very small and therefore, when
applied to the function; there is no big enough change in the value, thus
showing the location as center always.
```
```>>
>>Yes, you are definitely using too many points, so cut that number down
>>by a significant amount. Given the physical parameters of your set-
>>up, you really don=92t need nearly a million points to compute
>>something.
>>
>>I know you=92re under deadline, and you=92re probably feeling a bit
>>stressed at this point, so let me suggest something that could make
>>things easier for you. It has to do with how you=92re computing your
>>solution once you get the time delays.
>>
>>You don=92t really need to solve simultaneous equations or anything
like
>>that. You can simply use brute force computer power to get an answer
>>to =93what is the x,y location of the sound source, given the following
>>delays between microphones 1, 2 and 3?=94
>>
>>Let=92s presume we have the time delays between the microphones. Now
>>the problem becomes: =93given the physical arrangement of the set-up,
>>might we use a look-up table to determine the x,y location of the
>>sound source, given that we have the following delays between
>>microphones?=94
>>
>>know exactly how long it takes for a sound to travel from one point to
>>another. Let's say that your microphones are placed at positions M0
>>(x=3D0, y=3D0), MX (x=3D1, y=3D 0), and MY (x=3D0, y=3D1). OK. I know
>tha=
>>t your
>>somewhere in the upper right quadrant bounded by M0 on the lower left,
>>MY on the upper left, and MX on the lower right. The sound source is
>>somewhere in the upper right quadrant of a Cartesian coordinate
>>system, and you're trying to find its x,y location. Now place an x,y
>>grid on top of the Cartesian quadrant. The spacing of the grid points
>>is up to you, but your precision is limited by the physical parameters
>>of your problem. Then compute the time delay from every grid point to
>>the known microphone locations.
>>
>>For instance, presuming that the sound source is exactly at the M0
>>location, then, based on your spacing, you know precisely how long it
>>will take the sound to reach M0, MX and MY (the time to M0 is 0, and
>>the travel time to the others can be precisely computed).
>>
>>Now sequence through every single one of the grid points and calculate
>>the delay from that point to M0, MX and MY. This can be used to
>>generate a look-up table for any x,y grid point. Each x,y point thus
>>contains the delays to each microphone. All of this can be pre-
>>computed.
>>
>>Now obtain the delays by using max( ) or xcorr( ) on your recorded
>>data. They are your measured results. Compare the measured delays
>>with your previously generated grid data by going through the entire
>>list of grid points and compare them with your pre-computed delays.
>>Start with grid point 0,0, and run through all the rest. Find the
>>best match between the measured delays and your pre_computed ones.
>>
>>You=92ll have a problem in that you need to account for positive or
>>negative delays (e.g.: m1 might be ahead of m2, or m2 might be ahead
>>of m1). You'll also actually need multiple tables, because you have
>>multiple microphones.
>>
>>Given the physical parameters of your problem, it shouldn't be too
>>difficult to compute a look-up table(s) with pre-computed time delays
>>from any x,y point in the grid to the known locations of M0, MY and
>>MX. Then, using your measured delays ( from max( ) or xcorr( ) ),
>>you=92d search through the look-up table(s) to find the closest match.
>>The physical parameters of your problem indicate that the maximum
>>travel time will be 155 sample time delays, so you don=91t need very
big
>>table(s) to cover all the possible time delays.
>>
>Hi,
>
>Thank you. This seems ok because it will also give me an idea whether
the
>way we were calculating the delay was right or wrong.
>
>I will try it out and get back to you.
>
>But yea, one more thing. I have the formula and the functions to
calculate
>the co-ordinates of the sound source using non-linear equations. I get
>perfect results as far as the center location is concerned (i.e. when i
>generate a sound in the center of the table). BUT for any other location
>other than the center, I still get the result as the center which is
>weird.
>I figured out where the problem is, and the problem is right here as
what
>we are trying to solve i.e. the delay.
>The delay is converted into relative distance as (delay * 340 i.e. v of
>sound)/44100 Hz i.e. fz . This ans is very very small and therefore,
when
>applied to the function; there is no big enough change in the value,
thus
>showing the location as center always.
>
>
Hi,
Thank you. This seems ok because it will also give me an idea whether the
way we were calculating the delay was right or wrong.
I have captured the sound from different co-ordinates on the board in sets
of 4 (i.e. from 4 mics).
But yea, one more thing. I have the formula and the functions to
calculate
the co-ordinates of the sound source using non-linear equations. I get
perfect results as far as the center location is concerned (i.e. when i
generate a sound in the center of the table). BUT for any other location
other than the center, I still get the result as the center which is
weird.
I figured out where the problem is, and the problem is right here as what
we are trying to solve i.e. the delay.
The delay is converted into relative distance as (delay * 340 i.e. v of
sound)/44100 Hz i.e. fz . This ans is very very small and therefore, when
applied to the function; there is no big enough change in the value, thus
showing the location as center always.
```
```>>>
>>>Yes, you are definitely using too many points, so cut that number down
>>>by a significant amount. Given the physical parameters of your set-
>>>up, you really don=92t need nearly a million points to compute
>>>something.
>>>
>>>I know you=92re under deadline, and you=92re probably feeling a bit
>>>stressed at this point, so let me suggest something that could make
>>>things easier for you. It has to do with how you=92re computing your
>>>solution once you get the time delays.
>>>
>>>You don=92t really need to solve simultaneous equations or anything
>like
>>>that. You can simply use brute force computer power to get an answer
>>>to =93what is the x,y location of the sound source, given the
following
>>>delays between microphones 1, 2 and 3?=94
>>>
>>>Let=92s presume we have the time delays between the microphones. Now
>>>the problem becomes: =93given the physical arrangement of the set-up,
>>>might we use a look-up table to determine the x,y location of the
>>>sound source, given that we have the following delays between
>>>microphones?=94
>>>
>>>know exactly how long it takes for a sound to travel from one point to
>>>another. Let's say that your microphones are placed at positions M0
>>>(x=3D0, y=3D0), MX (x=3D1, y=3D 0), and MY (x=3D0, y=3D1). OK. I
know
>>tha=
>>>t your
>>>somewhere in the upper right quadrant bounded by M0 on the lower left,
>>>MY on the upper left, and MX on the lower right. The sound source is
>>>somewhere in the upper right quadrant of a Cartesian coordinate
>>>system, and you're trying to find its x,y location. Now place an x,y
>>>grid on top of the Cartesian quadrant. The spacing of the grid points
>>>is up to you, but your precision is limited by the physical parameters
>>>of your problem. Then compute the time delay from every grid point to
>>>the known microphone locations.
>>>
>>>For instance, presuming that the sound source is exactly at the M0
>>>location, then, based on your spacing, you know precisely how long it
>>>will take the sound to reach M0, MX and MY (the time to M0 is 0, and
>>>the travel time to the others can be precisely computed).
>>>
>>>Now sequence through every single one of the grid points and calculate
>>>the delay from that point to M0, MX and MY. This can be used to
>>>generate a look-up table for any x,y grid point. Each x,y point thus
>>>contains the delays to each microphone. All of this can be pre-
>>>computed.
>>>
>>>Now obtain the delays by using max( ) or xcorr( ) on your recorded
>>>data. They are your measured results. Compare the measured delays
>>>with your previously generated grid data by going through the entire
>>>list of grid points and compare them with your pre-computed delays.
>>>Start with grid point 0,0, and run through all the rest. Find the
>>>best match between the measured delays and your pre_computed ones.
>>>
>>>You=92ll have a problem in that you need to account for positive or
>>>negative delays (e.g.: m1 might be ahead of m2, or m2 might be ahead
>>>of m1). You'll also actually need multiple tables, because you have
>>>multiple microphones.
>>>
>>>Given the physical parameters of your problem, it shouldn't be too
>>>difficult to compute a look-up table(s) with pre-computed time delays
>>>from any x,y point in the grid to the known locations of M0, MY and
>>>MX. Then, using your measured delays ( from max( ) or xcorr( ) ),
>>>you=92d search through the look-up table(s) to find the closest match.
>>>The physical parameters of your problem indicate that the maximum
>>>travel time will be 155 sample time delays, so you don=91t need very
>big
>>>table(s) to cover all the possible time delays.
>>>
>>Hi,
>>
>>Thank you. This seems ok because it will also give me an idea whether
>the
>>way we were calculating the delay was right or wrong.
>>
>>I will try it out and get back to you.
>>
>>But yea, one more thing. I have the formula and the functions to
>calculate
>>the co-ordinates of the sound source using non-linear equations. I get
>>perfect results as far as the center location is concerned (i.e. when i
>>generate a sound in the center of the table). BUT for any other
location
>>other than the center, I still get the result as the center which is
>>weird.
>>I figured out where the problem is, and the problem is right here as
>what
>>we are trying to solve i.e. the delay.
>>The delay is converted into relative distance as (delay * 340 i.e. v of
>>sound)/44100 Hz i.e. fz . This ans is very very small and therefore,
>when
>>applied to the function; there is no big enough change in the value,
>thus
>>showing the location as center always.
Also I have read about ways such as frequency cross-correlation using FFT
and generalized cross correlation. Will this help? How do I do this?
```
```In my post of Aug. 24, 10:59 PM, I described how a simple cross
correlation is done with FFT, and mentioned 3 references for
generalized cross correlation. Basically, a generalized cross
correlator requires adding a filter in the frequency domain. For
instance, one particular filter function is known as the 'smoothed
coherence transform' or SCOT:
G. C. Carter, A. H. Nuttal, P. G. Cable, " The smoothed coherence
transform," Proc. IEEE (Lett), vol. 61, pp. 1497-1498, Oct., 1973
There are many other filters (see previous 3 references). But those
filter functions are not easy for a beginner to figure out, and it may
be very difficult to do it on your own. Many years ago, I took a
graduate course on time delay estimation, and we spent a great deal of
time computing and comparing various filters and running computer
simulations. I just don't think that it's something that anyone can
learn on a weekend.
You might want to try the grid method first. Just superimpose a grid
on you data space and compute the time delays from any given point x,y
to the known microphone locations. Then, after getting the measured
time delays from your data (from xcorr( ) or max( ) ), sift through
all x,y points of the grid until you find the best match between
measured and pre-computed delays. It'd be a lot easier.
```
```On Sep 11, 12:07 am, [email protected] wrote:
> In my post of Aug. 24, 10:59 PM, I described how a simple cross
> correlation is done with FFT, and mentioned 3 references for
> generalized cross correlation. Basically, a generalized cross
> correlator requires adding a filter in the frequency domain.
so, is this in lieu of windowing in the time domain? or to undo some
windowing in the time domain?
just curious.
r b-j
```
```>On Sep 11, 12:07 am, [email protected] wrote:
>> In my post of Aug. 24, 10:59 PM, I described how a simple cross
>> correlation is done with FFT, and mentioned 3 references for
>> generalized cross correlation. Basically, a generalized cross
>> correlator requires adding a filter in the frequency domain.
>
>so, is this in lieu of windowing in the time domain? or to undo some
>windowing in the time domain?
>
>just curious.
>
>r b-j
>
Hi,
Ok, lets leave the generalized cross-correlation for later (if i have
time). You told me you can help me with the localization as well once i get
the delays. I think the delays we calculate using xcorr are right.
I am working on the grid technique, but its obviously not the best if the
area is large. Can you suggest any other method? I have papers which say
about over-determined systems using least square or least mean square
technique.
```
```>>On Sep 11, 12:07 am, [email protected] wrote:
>>> In my post of Aug. 24, 10:59 PM, I described how a simple cross
>>> correlation is done with FFT, and mentioned 3 references for
>>> generalized cross correlation. Basically, a generalized cross
>>> correlator requires adding a filter in the frequency domain.
>>
>>so, is this in lieu of windowing in the time domain? or to undo some
>>windowing in the time domain?
>>
>>just curious.
>>
>>r b-j
>>
>
>Hi,
>
>Ok, lets leave the generalized cross-correlation for later (if i have
>time). You told me you can help me with the localization as well once i
get
>the delays. I think the delays we calculate using xcorr are right.
>
>I am working on the grid technique, but its obviously not the best if
the
>area is large. Can you suggest any other method? I have papers which say
>about over-determined systems using least square or least mean square
>technique.
>
Hi,
I did try for the 1st location i.e. (30,10).
the co-ordinates of the mics are:
m1(0,0), m2(0,60), m3(120,60), m4(120,0)
the location at which the sound is made is: (30,10)
(all in centimeter)
Now,
distance bet mic 1 and source is: 31.62
therefore, time1=0.00093s (t=d/s i.e. t=31.62/34000)
similarly,
for mic 2, time2 = 0.001715
for mic 3, time3 = 0.00303
for mic 4, time4 = 0.00266
therefore, the time diff bet mic 1 and 2 = time2-time1 = 0.000785 (do we
have to divide by the fz?)
therefore, the distance between mic 1 to the source and and mic 2 to the
source is: d = 26.69 (d=t*s i.e. 0.000785 * 34000)
Now, using xcorr, the delay between mic 1 and 2 is 30 sample units.
converting it to relative distance (d1)= (delay *34000)/fz = 23.1293
Thus, d and d1 are not same.
Is this what you are trying to say? Is the method and values correct?
```
```To robert bristow-johnson: Actually, it's not really in lieu of
windowing or undoing some windowing in the time domain. The two
signals are appropriately FFT'd, then one result is conjugated. This
is equivalent to time reversing one the waveforms in the time domain.
Then the two results are multiplied in the frequency domain, the
filtering function multiplication is applied, and the result is
inverse transformed. The output is the cross correlation. If there
were no conjugation, then you'd get the convolution of the two inputs.
When doing a generalized cross correlation, the filter is applied
AFTER the conjugation/multiplication in the frequency domain, and just
before the inverse. I suppose you could move the filter past the
inverse transform and then convolve it with your unfiltered result
(i.e.: do a simple cross correlator and then convolve the output with
the inverse transform of your frequency domain filter), but that seems
more difficult. I suspect it would be even more difficult if you were
to try to move the filter function to the front end of the problem,
because you'd have to move it past the multiplication, conjugation and
FFT parts. I suppose it could be done, and in that case, you'd window
both inputs in the time domain before doing the simple cross
correlation. But those windows will be different for each time domain
input, and perhaps very difficult to compute (they're difficult enough
as it is). It just seems easier to use the filter in the frequency
domain.
The main purpose of the frequency domain filter is to overcome the
problems with a simple cross correlator by taking into account the
characteristics of the signals and noise. With a simple cross
correlator, there are two major problems: 1) any noise added to the
signals may cause the output to indicate a false time delay peak, and
2) sinusoidal inputs can give you time delay outputs that oscillate.
In a generalized cross correlator, the frequency domain filter is used
to get a good time delay estimate based on the characteristics of the
signal and noise.
To Maz: Your problem has two distinct parts - 1) estimate the time
delay based on the measured data, and 2) using the estimates.
determine the x,y location of the source. Suppose you could send your
measured data off to a lab somewhere and tell them: "determine the
exact time delays." Now suppose they handed an answer back to you:
"We've used some of the latest techniques in physics, math and
engineering to determine that the delays are precisely: (insert some
numbers here up to 10 to the million decimal points)." Now the
question is: "What do you do with those estimates?"
I was hoping that you'd try the grid estimation method first because
that gives you a good idea of how your time delays physically relate
And I've just noticed that you've posted again (2:59 AM) - I'll look
at the new post later today.
```
```For one half of your problem, all you need to do is fill in a look-up
table. You pick a grid point x,y and determine the distance between
that point and your microphones (simple high school geometry). Then
you calculate the delay between that point and the microphones. As
per your example, for location (30,10), the time delay between that
location and microphone 1 is .00093 sec. You compute the delays to
all the other microphones. Then you should multiply the absolute
delays by 44100 (the sample rate) to get the delay expressed in terms
of the 'number of sample times.' Those are the values you put into
the look-up table (e.g.: for grid location 30,10 - the time delays to
the microphones are: t1, t2, t3, t4). I have no idea why you're
computing distances after computing the time delays.
Once you've populated the look-up table, you turn to the other half of
your problem - computing the actual time delays using your
measurements (do so with xcorr( ), max( ), or some other method to be
determined later). Your measured results will be a delay expressed as
'number of sample times.' Now, using your measured results, search
through the look-up table to find the best match between the measured
and pre-computed time delays.
Aa an aside, I also can't help but notice that you frequently start
other threads under a different title for the same problem, and you
ask for MATLAB code examples. I strongly suspect that you are not
very experienced with programming. Most people posting here, while
they can be very generous with their time and knowledge, won't do
something like "Here's how to solve your problem analytically, and,
oh, by the way, here's some MATLAB code to do it." They rightfully
expect that the person who poses the problem has some basic analytical
and programming skills. So I think you might have to reconsider your
deadline and include some unscheduled 'learn how to program in MATLAB'
time (or C, C++, FORTRAN, etc.).
It takes time to learn DSP techniques, and it can be very
frustrating. But you should at least have some kind of programming
experience.
```
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# Lesson 3
Are the Expressions Equal?
## Warm-up: How Many Do You See: Sums within 10 (10 minutes)
### Narrative
The purpose of this How Many Do You See is for students to subitize or use grouping strategies to describe the images they see. Students see two-color counters on the 10-frame and may know that when the 10-frame is filled, it is 10. Then they may see how many are not filled and subtract that many from 10 or may see how many are filled in each row and add those together. This deepens their understanding of the structure of 10 (MP7).
### Launch
• Groups of 2
• “How many do you see? How do you see them?”
• Flash the image.
• 30 seconds: quiet think time
### Activity
• Display the image.
• 1 minute: partner discussion
• Record responses.
• Repeat for each image.
### Student Facing
How many do you see?
How do you see them?
### Activity Synthesis
• “How does the structure of the 10-frame help you ‘see’ the total?” (I know that when the 10-frame is filled it is 10. I can see how many are not filled and subtract that many from 10, or I can see how many are filled in each row and add those together.)
## Activity 1: Sort Addition Expressions (20 minutes)
### Narrative
The purpose of this activity is for students to sort addition expressions by their value. Students find the value of each sum on their own and share their method with a partner, moving students towards fluency.
During the synthesis the teacher introduces an equation with addition expressions on both sides of the equal sign.
MLR8 Discussion Supports. Synthesis: Before students share, remind students to use “sum” and “expression.”
### Required Materials
Materials to Gather
### Required Preparation
• Each student needs their addition expression cards from a previous lesson.
### Launch
• Groups of 2
• Give students their addition expression cards.
• “Sort the cards into groups with the same value.”
• Display an addition expression card, such as $$2 + 5$$.
• “I know the value of this sum is seven. It is a sum that I just know. I will start a pile for sums of seven.”
### Activity
• “Work with your partner. Make sure that each partner has a chance to find the value before you place the card in a group. If you and your partner disagree, work together to find the value of the sum.”
• 12 minutes: partner work time
### Activity Synthesis
• “What sums have a value of seven?” (1 + 6, 6 + 1, 2 + 5, 5 + 2, 3 + 4, 4 + 3)
• Display $$4 + 3 = 3 + 4$$.
• “What do you notice about this equation?” (Each side has a 3 and a 4, but in a different order. Each side equals 7.)
## Activity 2: Are Both Sides Equal? (15 minutes)
### Narrative
The purpose of this activity is for students to determine whether equations are true or false. Students may use a combination of computation and reasoning about the commutative property to determine whether each equation is true or false. The synthesis focuses on how students can use the structure of the expressions to determine if they are equal without finding their values (MP7).
Representation: Internalize Comprehension. Provide students with a graphic organizer, such as a two-column table or sorting mat, to visually represent the expressions on each side of the equations.
Supports accessibility for: Visual Spatial Processing, Conceptual Processing
### Launch
• Groups of 4
• “We just found expressions that were equal to each other. Look at this equation.”
• Display $$4 + 2 = 6 + 1$$.
• “Is this equation true or false? How do you know?” (False. $$4 + 2 = 6$$, but the other side of the equal sign is 1 more than 6.)
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Share responses.
### Activity
• “You will work on these problems independently. I will let you know when it is time to share with a partner.”
• 4 minutes: independent work time
• “Share your thinking with a partner. Find a different partner for each problem. If you and your partner do not agree, work together to agree on the answer.”
• 3 minutes: partner discussion
### Student Facing
Determine whether each equation is true or false.
Be ready to explain your reasoning in a way that others will understand.
1. $$4 + 2 = 2 + 4$$
2. $$3 + 6 = 6 + 4$$
3. $$5 + 3 = 1 + 7$$
4. $$6 + 4 = 5 + 3$$
5. $$6 + 3 = 9 + 2$$
If you have time: Change the false equations to make them true.
### Student Response
If students circle true for an equation where the value to the left of the equal sign is the same as the first number on the right of the equal sign, consider asking:
• “How did you decide this equation is true?”
• “How can you use two-color counters to represent both sides of the equation? Can you use these counters to decide if the equation is true?”
### Activity Synthesis
• “Which equations could you tell were true or false without finding the value of both sums?” (Problem 1. That’s the add in any order property. Problem 2. You can see that the number you are adding to 6 is different on each side of the equal sign. Problem 5. $$6 + 3$$ is 9. The other side of the expression is 9 and some more.)
## Lesson Synthesis
### Lesson Synthesis
Display $$6 + 3 = 9 + 2$$
“Today we worked with equations that have expressions on both sides of the equal sign. What would you tell someone who said this equation was true because $$6 + 3 = 9$$?” (This side of the equal sign is 9 and the other side is 11. 9 does not equal 11.
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You are on page 1of 9
# 1
Q1: [ 3 Marks]
CLO 1: Perform algebraic manipulations including working with exponents,
radicals, polynomial operations, factoring and algebraic fractions.
Simplify the following expression:
Solution:
( )
( )( )
( )
( ) ( )
1
1
1
1
1 ) 1 (
1 1
1
2 2
2 3
=
+
+ +
+ +
+
+
x x x
x x x
x x
x
( )
( ) ( ) ( ) 1 2
1
1
) 1 (
1
1
2 2
3
2
3
+ +
+ +
+
x x x x
x
x
x
2
Q2: [ 4 Marks]
Present methods for solving first and second degree equations and inequalities
and related topics.
Find the values of (x) that satisfy the following equation:
0 1
2
= + + x x
Are these values (roots) real, imaginary or complex? What is the relation between
these roots?
1 = a , 1 = b , 1 = c
a
ac b b
x
2
4
2
=
1 2
1 1 4 1 1
2
= x
2
3 1
= x
i x .
2
3
2
1
1
|
|
.
|
\
|
+ =
i x .
2
3
2
1
2
|
|
.
|
\
|
=
these values (roots) are complex.
These two roots are complex conjugates.
3
Q3: [ 6 Marks]
Recognize and graph linear functions, rational functions, absolute value
functions and graph inequalities in two variables.
1. Graph the inequality : [ 3 Marks]
x y x 2 3 >
x x y + s 2 3
x y 3 3 s
x y s
x y =
Plot y against x and the region below the line y=x (including line points) is
the solution.
4
2. [ 3 Marks]
The manager of a weekend flea market knows from past experience that if
he charges (x) dollars for a rental space at the market, the number (y) of
spaces he can rent is given by the equation:
x y 2 100 =
a) Sketch a graph of this linear function.
b) What do the slope, the y-intercept and the x-intercept of the graph
represent?
Slope = -2 and it represents the number of spaces per dollar.
y-intercept = 100 and it represents the number of spaces if he charges
(zero dollars.)
x-intercept = 50 dollars for a rental space at the market when the number
of spaces is zero.
5
Q4: [ 5 Marks]
CLO 4: form and solve matrices and operations involving matrices.
Find the value of (x) that satisfies the solution for the following system of linear
equations using Cramers Rule:
3 2 4 = + z y x
z y x 7 1 3 = +
z y x = 9 8 2
|
|
|
.
|
\
|
=
1 9 2
7 3 1
2 4 1
A
, det (A) =1(3-63)-1(4+18)+2(28+6) = -14
|
|
|
.
|
\
|
=
8
1
3
b
|
|
|
.
|
\
|
=
1 9 8
7 3 1
2 4 3
x
A
, det (Ax) = 3(3-63)-1(4+18)+8(28+6) = -180 22 + 272 = 70
5
14
70
=
= =
A
A
x
x
6
Q5: [5 Marks]
CLO 5: Identify limits of algebraic, trigonometric, and composite functions.
1. On what intervals is the following function continuous? [3 Marks]
x
x
x g
2 cos 2
2 sin
) (
+
=
Then find:
x
x
x
2 cos 2
2 sin
lim
4
+
t
Solution:
The function is continuous on the open interval ( + , )
2
1
0 2
1
2
cos 2
2
sin
4
2 cos 2
4
2 sin
2 cos 2
2 sin
lim
4
=
+
=
+
=
|
.
|
\
|
+
|
.
|
\
|
=
+
t
t
t
t
t x
x
x
2. Find the value of the following limit:
[2 Marks]
2
4
2
2
lim
x
x
x
Solution:
( )
) 2 (
2 ) 2 (
lim
2
+
=
x
x x
x
( ) 4 2 2 2
lim
2
= + = +
=
x
x
7
Q6: [6 Marks]
CLO 6: solve for the derivatives of algebraic, trigonometric and composite
functions.
Find the equation of the tangent line to the curve:
2
3 6 x x y =
at the point where x=2.
Also find the point at which the rate of change of (y) with respect to (x) is equal to
zero. Is this point a maximum, a minimum or a turning point?
Solution:
x
dx
dy
6 6 =
= = = |
.
|
\
|
=
6 2 6 6
2 x
dx
dy
slope of tangent to curve at x = 2.
0 12 12 2 3 2 6
2
= = = y
Equation of straight line is:
c mx y + =
( ) 12 2 6 0 = + = c c
12 6 + = x y
1 0 6 6 = = = x x
dx
dy
3 1 3 1 6
2
= = y y so the point is (1,3)
6
2
2
=
dx
y d
Since second derivative is zero, therefore, the point (1,3) is a maximum.
8
Q7: [6 Marks]
CLO 7: find indefinite and definite integrals of algebraic, trigonometric, and
composite functions.
1. Find the area between the curve [3 Marks]
1
2
+ = x y
And the x-axis from x=0 to x=3.
( ) 12 0 ) 3
3
3
(
3
1
3
3
0
3 3
0
2
=
(
+ =
(
+ = + =
}
x
x
dx x A
2. Find the value of the following integral: [3 Marks]
( )dx x x
}
t
0
sin 5
( ) ( ) ( )
(
+ =
(
+ + =
(
+ =
}
) 1 ( 5 1 5
2
0 cos 5 0 ) cos 5
2
( cos 5
2
sin 5
2 2
0
2
0
t
t
t
t
t
x
x
dx x x
10
2
2
=
t
9
Q8: [5 Marks]
CLO 8: Graph and solve the equations of lines, circles, parabolas, ellipses and
Hyperbolas.
Decide which of the following equations represents a straight line, a parabola, a
circle, an ellipse or a hyperbola. Then sketch the graph of the equation that
represents an ellipse.
Circle x y = + 1
2 2
Line Straight x y . 1 + =
Hyperbola x y = 1
2 2
Parabola x y =
2
Ellipse
y x
= + 1
9 16
2 2
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You are on page 1of 4
# BML Study Guide: 11th Grade, Logarithmic/Exponential
Equations
Joshua Kim
Meet 2, December 3, 2014
>Took minimal effort
>Enjoy and blame dohyun cheon for any mistakes
Let us begin with a definition: For any real a, b, and c, if ab = c, then loga c = b.
This is the definition of a logarithm. The latter expression is read as the log base a of c equals b.
Example 1: Evaluate log27 243.
Solution 1: Turning this expression into exponential form results in:
27x = 243
A careful reader will notice that 27 is 33 and 243 is 35 ; therefore, this expression simplifies out into:
33x = 35
The exponents have to equal; therefore, x =
5
3
Here are the exponential properties, most of them from alg 2 or precalc:
ab ac = ab+c
ab
= abc
ac
(ab )c = abc
c
b
ac = a b
Using those properties, you can prove most, if not all, of the logarithmic properties presented below.
(loga b)(logb c) = loga c
logb c
= loga c
logb a
loga b + loga c = loga bc
b
loga b loga c = loga
c
loga bc = c loga b
logc b
loga b =
logc a
1
loga b =
logb a
1
The last one on the list is not common and most likely not necessary for competition, but its better to
know than not know. It may help, I dont know.
Example 2: Prove that xlogx a = a.
Solution 2: This is a straightforward application of the definition of a logarithm. Plug in the values and
see if it works yourself.
Also an interesting fact: logx xa = a. You can use this fact to prove that the logarithmic function and the
exponential function are inverses of each other.
There are two main types of logarithms: a common logarithm is a logarithm with base 10 (signified
simply by log b), and a natural logarithm is a logarithm with base e (signified by ln b). e, also known as
Eulers number, is approximately 2.71828182846... Derivations of this number should be easily found in any
calculus book.
The change of base formula is basically equivalent to one of the logarithm properties. It allows you to
take any logarithm and turn that into an expression you can plug into your calculator (most four-function
and scientific calculators have buttons for common and natural logs).
loga b =
ln b
log b
=
log a
ln a
Practice time! (There will be more exercises to practice from at the end of this worksheet)
3
Example 3: Solve the following equation (for all real x) = log4 x + log4 (x + 2) =
2
Solution 3: Its all log properties and changing the expression to exponential form.
log4 x + log4 (x + 2) =
3
3
= log4 x(x + 2) =
2
2
= x(x + 2) = 8
= x = 4, 2
However, this is not our final answer. log4 (4) doesnt make sense, because there is no value which 4 can
be raised, to result in negative 4. Our only answer is 2 .
This example problem raises up a good point: loga b = c has no solution if a, b, and c are real and b is
negative. LOOK OUT FOR THESE EXTRANEOUS SOLUTIONS! (BML can shaft you quite easily for
these)
2
## Example 4: Solve for x = 9(x 5x5) = 27(x +5)
Solution 4: This doesnt even have to be a logarithm problem. Keep your eyes peeled for the crucial step.
2
9(x
5x5)
= 27(x
+5)
= 32(x
5x5)
= 3(2x
10x10)
= 33(x
+5)
= 3(3x
+15)
## = 2x2 10x 10 = 3x2 + 15
= 0 = x2 + 10x + 25
= x = 5
No extraneous solutions here, life is good :)
2
Example 5: During a Science Bowl meeting, Daw tells Cowbo to find a number a such that three to the
power of that number will equal to 41. Ricardo, being the master ruseman he is, calculates the numerical
value of three to the power of three taken from twice that number and begins to yell out, The answer is
... What number did Cowbo hear Ricardo say? Express your answer as a common fraction.
Solution 5: I will do this problem two ways, especially since this type of problem comes up often in many
competitions (of course, not in this form lol). First, exponents.
3a = 41
(3a )2
32a
=
3
3
27
(41)2
=
27
1681
=
27
32a3 =
Second, logarithms.
3a = 41 = log3 41 = a
2a 3 = 2 log3 41 3
= log3 412 log3 33
= log3
32a3 = 3log
412
27
412
27
412
27
1681
=
27
=
And that should be it. Have fun with the exercises! Starred ones are notably more difficult than the others.
Exercises
1. Solve for x:
3x
9(x+1)
= 81(x+2)
## 2. Simplify to lowest terms:
1
1
+
log3 2 log9 2
3. If log 2 = 0.301 and log 3 = 0.477, find, to three decimal places, log 288.
4. Simplify to lowest terms:
ln 64
2 ln 2
## 5. log1 1 + log2 1 + log3 1 + log8192 1 = ?
6. In how many distinct points do the lines y = 3 log x and y = log 3x intersect? (Based off of a problem
on AHSME 1962)
7. Find all of the solutions of x2 log x =
x2
. (Based off of a problem on AHSME 1978)
25
## 8. ** If 60a = 3 and 60b = 5, then find 12[(1ab)/2(1b)] . (AHSME 1983)
9. If log
b
a
+ log = log(a + b), find a2 + b2 + 2ab.
b
a
10. ** Find the sum of all values of x that satisfy the equation log2 x +
## 11. Find the value of
3
log2 x +
3
log2 x+ log
=4
3
2 x+
x
1
if 2 log5 (x 3y) = log5 (2x) +
.
y
log2y 5
12. The function y = axn passes through the points (2, 1) and (32, 4). Find
13. ** If logy 8 + log9 x = 2 and
2
log2 y
8
+
= , find the product of the possible values of x.
log3 x
3
3
1.
## x equals negative two.
2.
log base two of twenty seven or three times the log base two of three
3.
4.
three
5.
zero
6.
one
7.
8.
t wo
9.
one
10.
11.
nine
12.
13.
a
.
n
eighty one
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# Basic Relationships, Concepts and Laws of Electric Circuits
March 22, 2020 by Robert Keim
This video tutorial explains fundamental laws and relationships that help us to understand, analyze, and design electrical systems and electronic devices.
The primary task of hardware-focused electrical engineers is circuit design. Notice how we repeatedly use this word “circuit” when referring to an interconnected group of sources, resistors, capacitors, and so forth. “Circuit” is not only an engineering word; it is a normal English word that refers to (among other things) some sort of route or movement that starts and ends at the same place.
Electronic design is fundamentally about electric circuits, that is, networks that allow electric charge to start at a source, move through a continuous pathway of conductors and components, and return to the source. This video tutorial will provide an overview of important scientific laws and mathematical relationships that describe the behavior of voltages and currents in the context of an electric circuit. If you want to explore these topics in greater detail, please refer to the AAC textbook.
## Electron Flow vs. Current Flow
In the world of electrical engineering, current flows from higher voltage to lower voltage. For example:
If you are accustomed to thinking about electricity as the movement of electrons, which flow from lower voltage to higher voltage, you need to remember that in the context of circuit design and analysis, we assume that current is the movement of electric charge from higher voltage to lower voltage.
## Kirchhoff’s Current Law
This law, abbreviated KCL, states that the total current entering a node is equal to the total current exiting a node. This follows naturally from the idea that electric charge cannot disappear. KCL allows us to find unknown branch currents by way of known branch currents. In the example below, if we measure I1 and I2, we can calculate I3, because I1 = I2 + I3.
## Kirchhoff’s Voltage Law
A typical circuit consists of a voltage source and various components that “claim” part of this supplied voltage as a result of their electrical behavior. The change in voltage between one terminal of a component and another terminal of the same component is called a voltage drop
Kirchhoff’s Voltage Law, abbreviated KVL, tells us that the sum of the voltage drops in a closed circuit is equal to the voltage supplied by the source. In other words, all of the voltage generated by the source must be accounted for somewhere in the circuit. In the example below, we have used a voltmeter to find the voltage across R1. Since VR1 = 2 V and VSUPPLY = 5 V, we know from KVL that VR2 = 3 V.
## Ohm’s Law
We have already learned about voltage sources and resistance, and we know that a voltage source will generate current flow when it is connected to a circuit consisting of one or more resistive components. In order to determine the exact amount of current, we need Ohm’s Law, which is an indispensable piece of information for anyone who works with electronic circuits. It states that the current through a resistive component is equal to the voltage across the component divided by the resistance.
$I=\frac {V}{R}$
We can rearrange this equation as follows:
$V =IR$
In this form, Ohm’s Law tells us that the voltage across a resistive component is equal to the current flowing through the component multiplied by the resistance. The diagram below shows how Ohm’s Law can be used to determine, first, the total current in the circuit, and second, the voltage dropped across each resistor.
## Power Dissipation
Neither voltage nor current is a direct indication of the way in which a circuit is using energy. However, energetic calculations are very important because energy, not voltage or current, is the quantity that corresponds to a system’s ability to perform useful work.
The energetic characteristics of an electric circuit or component are analyzed by means of power, which tells us the rate at which energy is consumed or transferred. Electric power (in watts) is the product of voltage (in volts) and current (in amperes):
$P=IV$
By combining this equation with Ohm’s Law, we can create alternative expressions that emphasize the relationship between power and voltage or between power and current:
$P=\frac {V^2}{R}$
$P=I^2R$
Resistive materials convert electrical energy into heat. In most cases this heat is not desired. Consequently, the power associated with a resistive material is “dissipated” into the environment, and we often refer to electrical power as power dissipation.
## Recap
• Electric circuits are energetic systems in which we analyze currents, voltages, and power dissipation.
• For the purposes of circuit analysis, electric current flows from higher voltage to lower voltage.
• The behavior of electric current at a node and the behavior of voltages in a closed circuit are described, respectively, by Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.
• Ohm’s Law states that the current flowing through a resistive component is equal to the voltage across the component divided by its resistance.
• Power dissipation is equal to voltage multiplied by current.
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Next: Properties Up: Functions, continuity and differentiability Previous: Functions, continuity and differentiability
## Definitions
The study of differentiable functions is the study of functions that mimic the behaviour of polynomials approximately''. To begin with we must formally define the notion of approximation.
Exercise 34 For any real number 0 < x < 1 show that xn is a decreasing sequence with limit 0.
In particular, we see that a polynomial that vanishes to order (n + 1) at 0 satisfies the following condition on functions of one variable.
Definition 1 A function g(x) of one variable is said to be in o(xn) if for any > 0 there is a > 0 so that
| g(x)| < | xn| for all x so that | x| < .
An alternative notion is
Definition 2 A function g(x) one variable is said to be in O(xn) is there is a > 0 and a constant C so that
| g(x)| < C| xn| for all x so that | x| <
Clearly, any polynomial that vanishes to order n is O(xn). Further, it is clear that an function g(x) that is O(xn) is o(xn - 1) and any function that is o(xn) is O(xn).
We can extend these notions to many variables as well. A function g(x1,..., xn) of n variables is said to be in o(xn) (respectively O(xn)) if for all lines (x1,..., xn) = (xc1,..., xcn) through the origin the restricted function f (x) = g(xc1,..., xcn) is in o(xn) (respectively O(xn)). We can further extend this to define o((x - b)n) and O((x - b)n) where b = (b1,..., bn) is some point, as a way of approximating functions near this point.
We say that g and f agree upto o((x - b)n) (or f approximates g upto o((x - b)n)) if f - g is in o((x - b)n). Note in particular, that f and g take the same value at b.
A function is differentiable n times at the point c if it is approximated upto o((x - b)n) by a polynomial (of degree n). Clearly, a polynomial of any degree is differentiable by the results of the previous section. In the one variable case we write this as follows
f (x) = a0 + a1x + ... + anxn + o((x - c)n)
Exercise 35 Show that for any two functions f and g in o(xn) and a function h which is differentiable n times at the origin, the function h . f + g is in o(xn).
Exercise 36 Show that the numbers ak are uniquely determined by the function f.
Now the number a1 depends on f and the point c. Now suppose that f is differentiable (1 times) at all points c so that it can be written as above near every point c. Then we can define the derived function f' by letting f'(c) = a1 for each point c; the function f' is also called the derivative of f. Now it clear that if f is the function given by a polynomial P then f' is dP/dx. Thus we also use the notation df /dx for f'. We have the derivation property as well.
Exercise 37 If f, g and h are differentiable then so is hf + g and
(hf + g) = f + h +
However, unlike the condition of vanishing to order n at c, the condition o((x - c)n) is not very well behaved.
Exercise 38 Show that f (x) = x2sin(1/x) is o(x) but the derivative of f' is not o(x0).
A function f (x) is called continuous at a point c if f (x) - f (c) is o(x - c) (i. e. it is differentiable 0 times!). It is called continuous it it has this property at all points. Thus we would like to study functions f which are differentiable and in addition the derivative f' is continuous. Such functions are provided by the fundamental theorem of calculus.
Next: Properties Up: Functions, continuity and differentiability Previous: Functions, continuity and differentiability
Kapil H. Paranjape 2001-01-20
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# 180 Days of Math for First Grade Day 3 Answers Key
By accessing our 180 Days of Math for First Grade Answers Key Day 3 regularly, students can get better problem-solving skills.
## 180 Days of Math for First Grade Answers Key Day 3
Directions Solve each problems.
Question 1.
Write the numeral for six.
__________
A numeral is nothing but expressing the numbers.
The numeral for six is 6.
Question 2.
How many in all?
In the first figure, there are 4 fishes.
In the second figure, there are 3 fishes.
By adding both the numbers we can find the total number of fish.
4 + 3 = 7
The total number of fish is 7.
Question 3.
One is crossed off. How many are left?
__________
Total number of apple juice tins = 8
One tin is crossed off from 8.
8 – 1 = 7
Thus 7 tins are left out of 8.
Question 4.
4 – = 2
Explanation:
Let the missing number be x.
4 – x = 2
4 – 2 = x
x = 2
So, the unknown number is 2.
Question 5.
Will the object stack?
Circle: yes no
The name of the object is a book.
The book is in the shape of a rectangular prism.
So, the answer is yes, the object can stack.
Question 6.
What time is it?
____________ o’clock
As per the given clock the time is 2 o’clock.
The shorthand is on 2 and the minute hand is on 12. So, the time is 2 o’clock.
Question 7.
How many are blue?
____________
Number of red color pencils = 5
Number of green color pencils = 8
Number of blue color pencils = 3
The number of pencil colors that are blue is 3.
Question 8.
Mom has 3 flowers. I give her 1 more. How many flowers does she have in all?
________________
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http://www.qacollections.com/How-to-Graph-a-Slope-Intercept-Form-for-Linear-Inequalities
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# How to Graph a Slope Intercept Form for Linear Inequalities?
Linear inequalities have an infinite number of solutions, so it's impossible to list all solutions of an inequality. Drawing a graph is one way to illustrate a linear inequality. If you have a line... Read More »
http://www.ehow.com/how_8270815_graph-intercept-form-linear-inequalities.html
Top Q&A For: How to Graph a Slope Intercept Form for Linear ...
## How to Graph Linear Equations Using the Slope Intercept Form for the Seventh Grade?
Linear equations are straight lines when graphed, and have simple variables, with only x and y. If the values are squared or cubed, they are no longer lines. The slope-intercept form of a linear eq... Read More »
## How to Rewrite Linear Equations Into Slope-Intercept Form?
A linear equation is an equation that produces a line when graphed. The form y = mx + b is called the slope-intercept form of the equation because it tells you exactly what the slope (m) is and whe... Read More »
http://www.ehow.com/how_6393750_rewrite-linear-equations-slope_intercept-form.html
## How to Convert Linear Equations to Slope Intercept Form?
Solving equations for lines and graphing lines are common tasks in algebra and calculus courses. The standard form for a linear equation is ax + by = c. The slope-intercept form of a line is y = mx... Read More »
http://www.ehow.com/how_8429968_convert-equations-slope-intercept-form.html
## How to Convert a Slope-Intercept Form Equation Into a General Linear Equation?
When writing linear equations, you can use either the slope-intercept form or general form. The advantage of slope-intercept form, or y = mx + b, is that you can immediately see the slope, m, and t... Read More »
http://www.ehow.com/how_8444467_convert-equation-general-linear-equation.html
Related Questions
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Tooltip
These resources have been reviewed and selected by STEM Learning’s team of education specialists for factual accuracy and relevance to teaching STEM subjects in UK schools.
In this Year Seven textbook from Task Maths, there are 28 mathematics tasks. There are also useful sets of review exercises after every three tasks and revision exercises at the end. Each task consists of a number of differentiated activities. Many activities are investigative or exploratory and encourage discussion of ideas. Task Maths contains some innovative content in most tasks with some containing material which is very different, and possibly more engaging, than that found in many texts.
1. Monty - Number patterns on a 100 Square using a computer program
2. How big is your classroom? - Area and volume
3. Birthdays - Data handling using birth dates and ages
4. Halving - Particularly based on symmetry and halving of various grids
5. Bags of cubes - Introductory probability
6. Slot machines - Coin problems and paying for various tickets
7. Leaves - Area of a leaf
8. What is a square? - Exploring properties and making other shapes starting with squares
9. A slice of pie - Angles and pie charts
10. Words - Handling data – includes word length, starting letters , house numbers
11. Cube models - Problems based on shapes made from cubes including isometric drawings
12.Twelve days of Christmas - Puzzles based on the song including changing the rules
13. The answer is 42 - Explorations – what is the question
14. Models made from squares and triangles - Making 3D models
15. Bingo - Probability –includes using dice and random numbers
16. Number chains - Variety of number chains generated by flow charts
17. Polygon arrangements - Symmetry mainly based on ATM Polygonal MATS
18. Palindromes - Variety of tasks and includes the ‘Palindromic Number” investigation
19. Food - Includes a range of ancient recipes, changing units, favourites, network diagrams
20. Number shapes - Number patterns
21. Optical illusions - Measuring and angles in an interesting context
22. What is a third? - As a vulgar fraction, decimal and percentage
23. It's magic! - A resource for magic squares and crosses includes construction methods
24. People and calculators - Methods of calculation and some ways of using a calculator
25. What is a circle? - Circles in everyday contexts and circle patterns
26. Folding Paper and Card - Folding in half and folding strips to make shapes
27. Number ladders - A number-based strategy game
28. What a load of rubbish! - Handling data: re-cycling and proportions of different types of rubbish
The Teachers' Resource Book Contents *Introduction The first part of the book covers the approach taken, resources and materials needed and the teacher’s role. *Guide to tasks There are typically two or three pages of information on each of the 28 tasks which give answers and brief suggestions for approaches to the activities as well as listing any resources needed.
*Answers to the review exercises and to the revision exercise
*Copymasters.
#### Show health and safety information
Please be aware that resources have been published on the website in the form that they were originally supplied. This means that procedures reflect general practice and standards applicable at the time resources were produced and cannot be assumed to be acceptable today. Website users are fully responsible for ensuring that any activity, including practical work, which they carry out is in accordance with current regulations related to health and safety and that an appropriate risk assessment has been carried out.
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# yuur 2.0
More Options: Make a Folding Card
#### Storyboard Description
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#### Storyboard Text
• Now if we are subtracting we will do this, (4x-7)-(8x-5), first instead of going right to the like terms we have to make the 8x negative and since -5 would be double negative it would just be 5, so then we just combine the like terms to get -4x-2.
• Now we are going to multiply, the problem is (4x-7) (8x-5) and we are doing the Foil method. FIrst, we have to distribute from the terms from the first parenthesis to the term in the second parenthesis and then once you get those numbers then you have to add and subtract. You will get 32^2x-76x+35.
• Now we are going to factor 4x2 + 12x + 5. You would multiply 4 and 5 to get 20 by the AC method. You then get the factors of 20 and use the one that adds up to 12 which would be 10 and 2. Then you would factor to get the answer of (2x + 5)(2x + 1)
• Now were are going to solve the problem 12x5 – 18x3 – 3x2 using greatest common factor. First off we will, Determine the greatest common factor of the given terms. The greatest common factor or GCF is the largest factor that all terms have in common. Then we will Factor out (or divide out) the greatest common factor from each term. This gives us the gcf of 3x2. Then the final answer is 3x2(4x3-6x-1).
• For our next problem we will try to solve the problem x2-36. Our first step to finding the solution of this problem is to decide if the four terms have anything in common, called the greatest common factor or GCF. If so, factor out the GCF. Do not forget to include the GCF as part of your final answer. In this case, the two terms only have a 1 in common which is of no help. But to factor this problem into the form (a + b)(a – b), you need to determine what squared will equal x2 and what squared will equal 36. In this case the choices are x and 6 because (x)(x) = x2 and (6)(6) = 36. This gives us (x+6)(x-6) or (x-6)(x+6). Our last step is to determine if any of the remaining factors can be factored further. In this case they can not so the final answer is (x+6)(x-6) or (x-6)(x+6)
• Now we are going to factor 6x² -11x +4. Here we use the AC method to get 24. Now you find the factors of 24 that would add up to -11. This answer would be -8 and -3. The equation then becomes (6x^2 -8x) (-3x+4). Now you factor the equation to get the answer of (2x-1) (3x-4). And that is it for me everyone, keep learning Polyman Out!!
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# Genetic Algorithms, Why does random crossover work?
Hi all, I understand Genetic algorithms aside form why crossover helps things, it has no guarantee of getting the best characteristics of each chromosome.
Ie say you had a genetic algorithm to calculate square roots, fitness = 1(Abs(NumToFindRoot of-(Guess*Guess))
and your guess was a binary string 00100111001 where the bottom half represents the decimals and the top half the integer part. say our initial population was
110011= 51
001010= 20
111000 =7
001011= 52
now select using tournament or roulette method:
111000 =7
001010 =20
and cross them:
111010= 23
then mutate
110010 =19
Now say we keep repeating this:
111000=7
cross->
110011=51
equals->
111111=63
mutate->
111110 31
---------
001010 20
cross
110011 51
equals
001011 52
mutate
000011 = 48
Continue repeating with population of 3, assuming we trying to find square root of 16
000011 = 48
111110 = 31
110010 = 19
----------
010010 = 18
111000 = 7
001010 = 20
---------
010000 = 2
101010 = 21
110010 =19
-------
010110 = 26
100010 = 17
010000 =2
Seems to be getting circular. Am I going about this wrong? can someone explain why random crossover can actually work?
My code just infinite loops, so i found a GA online and modified it to do this job and it still infinite loops, unless the fitness function needs to be a lot more in depth perhaps? I'm thinking my issue is that there isn't a "close to square root" because of the way im representing my number?
Code:
for (int i = 0; i <30; i++)
{
Individual indiv1 = tournamentSelection(pop);
Individual indiv2 = tournamentSelection(pop);
Individual newIndiv = crossover(indiv1, indiv2);
newPopulation.saveIndividual(i, newIndiv);
}
for (int i = elitismOffset; i < newPopulation.size(); i++)
{
mutate(newPopulation.getIndividual(i));
}
Last edited:
mfb
Mentor
I don't understand your example, but in general most crossovers won't give good results - but there is a small chance to get an interesting result that would be hard to reach with smaller mutations.
but even with those, the premise seems flawed. It seems to just randomly reach instead of converge since the best traits arnt carried over, how does it converge? Can someone explain?
Stephen Tashi
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Author Topic: FE Problem 3 (Read 3853 times)
Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
FE Problem 3
« on: April 14, 2015, 07:41:13 PM »
Solve by the method of separation of variables
\begin{equation*}
\left\{\begin{aligned}
& u_x (0,t)= u _x (1,t)=0,\\[2pt]
& u(x,0)=f(x),\\[2pt]
& u_t(x,0)=g(x)\end{aligned}\right.
\end{equation*}
with $f(x)=x(1 -x)$, $g(x)=0$. Write the answer in terms of Fourier series.
Zacharie Leger
• Jr. Member
• Posts: 6
• Karma: 0
Re: FE Problem 3
« Reply #1 on: April 16, 2015, 02:16:09 AM »
Hopeful solution
By letting $u(x,t)=X(x)T(t)$ and plug into the wave equation we can get
$$\frac{X''(x)}{X(x)}-\frac{T''(t)}{4T(t)}=0 \Rightarrow \frac{X''(x)}{X(x)}=\frac{T''(t)}{4T(T)}=-\lambda$$
We know that $\lambda\geq 0$, so let of first develop the the case where $\lambda =0$. Hence,
$$X''(x)=0 \Rightarrow X(x)=Ax+B \Rightarrow X'(0)=A=0 \Rightarrow X(x)=B$$
Now if $\lambda =\omega ^2,\,\omega >0$ we get the following solution
$$X(x)=C\sin(\omega X)+D\cos(\omega X)$$
Applying the boundry conditions on the solution we see that
$$X'(0)=C\omega\cos(\omega (0))+D\omega\sin(\omega (0))=C=0\,\mathrm{and}\, X(1)=D\omega\sin(\omega (1))=0\Rightarrow \omega=n\pi$$
$$\Rightarrow X_n (x)=D_n \cos(n\pi x)$$
As for the solution for $T(t)$, we again consider first when $\lambda=0$
$$T''(t)=0 \Rightarrow T(t)=Et+F \Rightarrow T'(0)=E=0 \Rightarrow T(t)=F$$
Now when $\lambda =(n\pi)^2$
$$T(t)=G\cos(2n\pi x)+H\sin(2n\pi x)$$
Applying the boundary conditions we see that
$$T'(0)=2n\pi G\sin(n(0))+2n\pi H\cos(n(0))=2n\pi F=0$$
Hence,
$$T_n (t)=G_n \cos(2n\pi t)$$
So the full solution can be expressed as
$$u(x,y)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)\cos(2n\pi t)$$
where we let $\alpha _0 =B+F$ and $\alpha _n = D_n G_n$. We have one last boundary conditions to apply to get our final answer:
$$u(x,0)=x(1-x)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)$$
Fourier tells us that the coefficiants will be given by
$$\alpha _0 =\int_0^1 x-x^2\mathrm{d}x=[\frac{x^2}{2}-\frac{x^3}{3}]_0^1=\frac{1}{6}$$
$$\alpha _n =2\int_0^1 (x-x^2)\cos(n\pi x)\mathrm{d}x=2[(-n\pi(-1+2 x)\cos(n\pi x)+(2-n^2\pi^2 (-1+x)x)\sin(n\pi x))/(n^3\pi^3)]_0^1=\frac{(1+(-1)^n)}{\pi^2 n^2}$$
Therefore, our final solution is:
$$u(x,y)=\frac{1}{6} +\sum\limits_{n=1}^\infty \frac{2(1+(-1)^n)}{\pi^2 n^2} \cos(n\pi x)\cos(2n\pi t)$$
Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
Re: FE Problem 3
« Reply #2 on: April 17, 2015, 08:17:08 AM »
Separation of variables results in $X''+\lambda X=0$, $X'(0)=X'(1)=0$ and thus $\lambda_0=0$, $X_0= \frac{1}{2}$ and $\lambda_n=\pi^2n^2$, $X_n=\cos (\pi n x)$ with $n=1,2,\ldots$; also $T''+4\pi^2 T=0$ and thus $T_0=A_0+B_0t$, $T_n= A_n \cos (2\pi n t)+B_n \sin (2\pi n t)$, and
u=\frac{1}{2}(A_0+B_0t) + \sum_{n=1}^\infty \Bigl(A_n \cos (2\pi n t)+B_n \sin (2\pi n t)\Bigr)\cos (\pi n x).
The initial conditions result in
\begin{equation*}
\frac{1}{2}A_0+ \sum_{n=1}^\infty A_n \cos (\pi n x)=x(1-x),\qquad
\frac{1}{2}B_0 + \sum_{n=1}^\infty 2\pi n B_n \cos (\pi n x)=0
\end{equation*}
and $B_n=0$ ($n=0,1,2,\ldots$) and
\begin{multline*}
A_n =2 \int_0^1 x(1-x)\cos (\pi n x)\,dx=- \frac{2}{\pi n} \int_0^1 (1-2x) \sin (\pi nx)=\\
-\frac{2}{\pi ^2n^2 } (1-2x) \cos (\pi nx)\Bigr|_{x=0}^{x=1} +
\frac{4}{\pi ^2n^2 } \int_0^1 \cos (\pi nx)\,dx =\left\{\begin{aligned} -\frac{1}{\pi ^2m^2 } & && n=2m,\\
0& &&n=2m+1\end{aligned}\right.
\end{multline*}
$m=1,2, \ldots$. Meanwhile $A_0=\frac{1}{3}$. Then
\begin{equation*}
u=\frac{1}{12}- \sum_{m=1}^\infty \frac{1}{\pi^2m^2} A_n \cos (4\pi m t)\cos (2\pi m x).
\end{equation*}
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http://oeis.org/A030588
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A030588 Write odd numbers in base 6 and juxtapose. 16
1, 3, 5, 1, 1, 1, 3, 1, 5, 2, 1, 2, 3, 2, 5, 3, 1, 3, 3, 3, 5, 4, 1, 4, 3, 4, 5, 5, 1, 5, 3, 5, 5, 1, 0, 1, 1, 0, 3, 1, 0, 5, 1, 1, 1, 1, 1, 3, 1, 1, 5, 1, 2, 1, 1, 2, 3, 1, 2, 5, 1, 3, 1, 1, 3, 3, 1, 3, 5, 1, 4, 1, 1, 4, 3, 1, 4, 5, 1, 5, 1, 1, 5, 3, 1, 5, 5, 2, 0, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Antti Karttunen, Table of n, a(n) for n = 1..100000 EXAMPLE From Antti Karttunen, Dec 26 2018: (Start) The first twelve odd numbers and their base-6 representations (A007092) are: 1 1 3 3 5 5 7 11 9 13 11 15 13 21 15 23 17 25 19 31 21 33 23 35 thus the sequence begins with terms 1, 3, 5, 1, 1, 1, 3, 1, 5, 2, 1, 2, 3, 2, 5, 3, 1, 3, 3, 3, 5. (End) MATHEMATICA Flatten[Table[IntegerDigits[2n - 1, 6], {n, 50}]] (* Harvey P. Dale, Jul 22 2012 *) PROG (PARI) up_to = 100000; A030588list(up_to) = { my(v=vector(up_to), ds=List([]), o=1, j=1); for(i=1, up_to, if(j>#ds, ds=digits(o, 6); j=1; o+=2); v[i] = ds[j]; j++); (v); }; v030588 = A030588list(up_to); A030588(n) = v030588[n]; \\ Antti Karttunen, Dec 24 2018 (Scala) (1 to 99 by 2).map(Integer.toString(_, 6).toCharArray).flatten // Alonso del Arte, Apr 02 2020 CROSSREFS Cf. A007092. Cf. A030589, A030590, A030591, A030592, A030593, A030594. Cf. A030595, A030601, A030602, A030603. Cf. A030596, A030597, A030598, A030599, A030600. Sequence in context: A016452 A091717 A154512 * A307860 A123701 A143303 Adjacent sequences: A030585 A030586 A030587 * A030589 A030590 A030591 KEYWORD nonn,base,easy AUTHOR STATUS approved
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