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https://ccssmathanswers.com/eureka-math-grade-8-module-1-lesson-10/ | 1,701,992,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00601.warc.gz | 182,269,923 | 56,216 | ## Engage NY Eureka Math 8th Grade Module 1 Lesson 10 Answer Key
### Eureka Math Grade 8 Module 1 Lesson 10 Exercise Answer Key
Exercise 1.
The speed of light is 300,000,000 meters per second. The sun is approximately 1.5×1011 meters from Earth. How many seconds does it take for sunlight to reach Earth?
300 000 000=3×108
$$\frac{1.5 \times 10^{11}}{3 \times 10^{8}}$$ = $$\frac{1.5}{3}$$×$$\frac{10^{11}}{10^{8}}$$
=0.5×103
=0.5×10×102
=5×102
It takes 500 seconds for sunlight to reach Earth.
Exercise 2.
The mass of the moon is about 7.3×1022 kg. It would take approximately 26,000,000 moons to equal the mass of the sun. Determine the mass of the sun.
26 000 000=2.6×107
(2.6× 107)(7.3×1022)=(2.6×7.3)(107×1022)
=18.98×1029
=1.898×10×1029
=1.898×1030
The mass of the sun is 1.898×1030 kg.
Exercise 3.
The mass of Earth is 5.9×1024 kg. The mass of Pluto is 13,000,000,000,000,000,000,000 kg. Compared to Pluto, how much greater is Earth’s mass than Pluto’s mass?
13 000 000 000 000 000 000 000=1.3×1022
5.9×1024-1.3×1022=(5.9×102)×1022-1.3×1022
=(590-1.3)×1022
=588.7×1022
=5.887×102×1022
=5.887×1024
The mass of Earth is 5.887×1024 kg greater than the mass of Pluto.
Exercise 4.
Using the information in Exercises 2 and 3, find the combined mass of the moon, Earth, and Pluto.
7.3×1022+1.3×1022+5.9×1024=(7.3×1022+1.3×1022)+5.9×1024
=8.6×1022+5.9×1024
=(8.6+590)×1022
=598.6×1022
=5.986×102×1022
=5.986×1024
The combined mass of the moon, Earth, and Pluto is 5.986×1024 kg.
Exercise 5.
How many combined moon, Earth, and Pluto masses (i.e., the answer to Exercise 4) are needed to equal the mass of the sun (i.e., the answer to Exercise 2)?
$$\frac{1.898 \times 10^{30}}{5.986 \times 10^{24}}$$ = $$\frac{1.898}{5.986}$$ × $$\frac{10^{30}}{10^{24}}$$
=0.3170…×106
≈0.32×106
=0.32×10×105
=3.2×105
It would take 3.2×105 combined masses of the moon, Earth, and Pluto to equal the mass of the sun.
### Eureka Math Grade 8 Module 1 Lesson 10 Problem Set Answer Key
Have students practice operations with numbers written in scientific notation and standard notation.
Question 1.
The sun produces 3.8×1027 joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31,000,000 seconds).
31 000 000=3.1×107
(3.8×1027)(3.1×107)=(3.8×3.1)(1027×107)
=11.78×1034
=1.178×10×1034
=1.178×1035
The sun produces 1.178×1035 joules of energy in a year.
Question 2.
On average, Mercury is about 57,000,000 km from the sun, whereas Neptune is about 4.5×109 km from the sun. What is the difference between Mercury’s and Neptune’s distances from the sun?
57 000 000=5.7×107
4.5×109-5.7×107=(4.5×102)×107-5.7×107
=450×107-5.7×107
=(450-5.7)×107
=444.3×107
=4.443×102×107
=4.443×109
The difference in the distance of Mercury and Neptune from the sun is 4.443×109 km.
Question 3.
The mass of Earth is approximately 5.9×1024 kg, and the mass of Venus is approximately 4.9×1024 kg.
a. Find their combined mass.
5.9×1024+4.9×1024=(5.9+4.9)×1024
=10.8×1024
=1.08×10×1024
=1.08×1025
The combined mass of Earth and Venus is 1.08×1025 kg.
b. Given that the mass of the sun is approximately 1.9×1030 kg, how many Venuses and Earths would it take to equal the mass of the sun?
$$\frac{1.9 \times 10^{30}}{1.08 \times 10^{25}}$$=$$\frac{1.9}{1.08} \times \frac{10^{30}}{10^{25}}$$
=1.75925…×105
≈1.8×105
It would take approximately 1.8×105 Venuses and Earths to equal the mass of the sun.
### Eureka Math Grade 8 Module 1 Lesson 10 Exit Ticket Answer Key
Question 1.
The speed of light is 3×108 meters per second. The sun is approximately 230,000,000,000 meters from Mars. How many seconds does it take for sunlight to reach Mars?
230 000 000 000=2.3×1011
$$\frac{2.3 \times 10^{11}}{3 \times 10^{8}}$$=$$\frac{2.3}{3}$$×$$\frac{10^{11}}{10^{8}}$$
=0.7666…×103
≈0.77×10×102
≈7.7×102
It takes approximately 770 seconds for sunlight to reach Mars.
Question 2.
If the sun is approximately 1.5×1011 meters from Earth, what is the approximate distance from Earth to Mars? | 1,651 | 4,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-50 | latest | en | 0.649763 |
https://dev.to/gabrieleboccarusso/how-i-am-learning-machine-learning-week-6-python-and-matplotlib-part-two-3fdf | 1,623,530,349,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586390.4/warc/CC-MAIN-20210612193058-20210612223058-00427.warc.gz | 212,703,470 | 34,887 | DEV Community is a community of 636,345 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
How I am learning machine learning - week 6: python and matplotlib (part two)
Gabriele Boccarusso Updated on ・5 min read
Last week we saw the fundamentals of the matplotlib package and what we can do with it, now we'll see how to use it with pandas and numpy to use it effectively.
Pandas dataframes
Pandas dataframes are the most common data structure that we'll use and we can obviously plot them with matplolib. When we introduced pandas we used a sample about baseball players with just ten rows because I have deleted all of them to simplify the learning. Now we can use the complete one that you can find here
But we have a problem: the names of the columns are badly formatted. We can resolve this in few steps: first, we see the name of the columns with the keys function, and then we rename them with the rename function:
Plotting a dataframe
We can quickly plot through our dataframe with the plot function. This is not the recommended way, but for quick plots works exactly fine:
This way we can have a quick overview of our data, another example is the hist function:
The object-oriented method
Now we know how to quickly see the data we need, but in the majority of cases we'll need well structured and customized plots, and to do this is more efficient the object-oriented method or the OO method. Let's see an example:
Now that we have our plot let's take a moment to understand what's going on here: We create a fig and an ax and gave to them a size, then we call the plot function on our dataframe, and into the function we declare
• the type of plot that we want (scatter)
• the x-axis (the age of every player)
• the y-axis (the weight of our players)
• the (c)olor of the plot, the range of every element based on a third value (their height in inches)
• the ax that we are plotting. In this case there was only one axis, but if there was more we could have typed ax = ax[0, 0] or ax = axis_name.
But we are not still completely using the OO method. To do it, we should use the subplots function with using all that derives from it:
``````figure, axis = plt.subplots(figsize=(10, 7))
#creating the plot, same as before
first_plot = axis.scatter(x = bb_players['Age'],
y = bb_players['Weight (lbs)'],
c = bb_players['Height (inches)'])
axis.set(title="correlations between baseball players weight and age",
xlabel="players age",
ylabel="players weight")
axis.legend(*first_plot.legend_elements(), title = 'Weight in lbs');
# the "*" query all the elements in first_plot
``````
What we are doing here is simply creating a figure and customizing its plot.
The result will be:
here there are some typing error here, the most obvious is the legend referring to weight instead of height
Now should be clear how to do a figure with the OO method but with more plots in it. An example of code may be:
``````# creating a figure with more than 1 subplots
figure, (first_plot, second_plot) = plt.subplots(nrows = 2,
ncols = 1,
figsize = (12, 20))
scatter_plot = first_plot.scatter(x = bb_players['Age'],
y = bb_players['Weight (lbs)'],
c = bb_players['Height (inches)'])
first_plot.set(title="correlations between baseball players weight and age",
xlabel="age of players",
ylabel="weight of players")
first_plot.legend(*scatter_plot.legend_elements(),
title = 'Height in inches');
# putting a line on the average data
first_plot.axhline(y = bb_players['Weight (lbs)'].mean(),
color = 'r',
linestyle = '-')
# second plot
second_plot.hist(x = bb_players['Age'],
bins = 20)
second_plot.set(title = "Numbers of players with a certain age",
xlabel="age of players",
ylabel="number of players");
``````
As you can see from the code we have set the rows and the columns of the figure and set the bins of the histogram. The bins are simply the unit that spaces through the x-axis. A histogram with seven bins will be similar to:
source
The axhline is a very simple function explained here. In this case, we set the mean value of the weight columns as the "y".
Now that the code should be clear we can see the result:
click to enlarge
Customizing and saving the plot
There various ways to customize our plot in matplotlib with general styles which we can modify details if we want.
the basic command is:
``````plt.style.available
``````
Using matplotlib style are important to add meaning to our graph, here the difference between the default style and one that you can choose:
But despite being now slightly different is may still be confusing. Matplotlib offers us the cmap function, among others, to change the details of a style:
The full reference is available at the official documentation page
Now the last step would be creating an image of your figure with the savefig function that we saw in the last post and everything would be ready to show our data to others.
Final thoughts
Today we saw the last part of the matplotlib python package. Not that we know well enough pandas, numpy, and matplotlib, we can see how scikit works and begin working on some model. | 1,177 | 5,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | longest | en | 0.947791 |
https://mathematica.stackexchange.com/questions/152837/how-to-count-frequency-of-numbers-while-accounting-for-the-time-of-the-data | 1,725,729,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00383.warc.gz | 374,064,247 | 41,330 | # How to count frequency of numbers while accounting for the time of the data?
I am hoping to get some guidance as to how to tackle this problem. I have the following sample dataset with the dates in YYYYMMDD format:
data={{{20120802}, 193.9}, {{20120912}, 493.9}, {{20130412}, 33.9},{{20130502}, 193.9},
{{20130802}, 193.9}, {{20130822}, 193.9}, {{20131002}, 193.9}, {{20131022}, 193.9}};
The "Y" values themselves don't matter for this question since I am looking at frequency. I am wondering how I would be able to calculate the frequency that such data points occur relative to the time period. I want to divide it so when I graph it in a line plot, it is divided into 4 quarters (3 months each) over the span of several years (in the example above it is 2 years) and for each quarter it includes all the points that occurred within that date range.
For visualization sake with the data above, the line plot that I want to make would have an x-axis with the tickers: 2012Q1, 2012Q2, 2012Q3, 2012Q4, 2013Q1, 2013Q2, 2013Q3, 2013Q4 and the data points would only be located on those tickers. So for points {{20130412},33.9},{{20130502},193.9}, this would be a "2" on the y-axis and have the x-coordinate of 2013Q2.
If any has any pointers on how to tackle this I would really appreciate it.
Update
Show zero values and some other improvements
data =
{{{20120802}, 193.9}, {{20120912}, 493.9}, {{20130412}, 33.9},
{{20130502}, 193.9}, {{20130802}, 193.9}, {{20130822}, 193.9},
{{20131002}, 193.9}, {{20131022}, 193.9}};
Convert dates
dat = MapAt[DateList@*ToString@*First, data, {All, 1}]
{{{2012, 8, 2, 0, 0, 0.}, 193.9}, {{2012, 9, 12, 0, 0, 0.}, 493.9}...}
All months with default value zero - must start with year or quarter
amo = Transpose[{#, Array[0 &, {Length@#}]}] &[DateRange[{2012}, {2014}, "Month"]]
{{{2012, 1, 1}, 0}, {{2012, 2, 1}, 0} ... {{2014, 1, 1}, 0}}
DateListPlot[
TimeSeriesAggregate[Join[dat, amo], {"Quarter", Left}, Count[#, _?Positive] &],
DateTicksFormat -> {"Year", "/", "QuarterNameShort"},
Filling -> Bottom,
GridLines -> Automatic,
InterpolationOrder -> 0,
PlotRange -> {{{2012, 4, 1}, {2013, 12, 1}}, Automatic},
Mesh -> Full]
To get an accurate aggregation we must start at {20120101}:
data = {{{20120101}, Missing[]}, {{20120802}, 193.9}, {{20120912},
493.9}, {{20130412}, 33.9}, {{20130502}, 193.9}, {{20130802},
193.9}, {{20130822}, 193.9}, {{20131002}, 193.9}, {{20131022}, 193.9}};
Then we must transform the dates in a Mathematica-like form:
dates =
ToExpression[{StringTake[#, 4], StringTake[#, {5, 6}],
StringTake[#, -2]}] & /@ (ToString /@ Flatten@data[[All, 1]]);
Now we use TimeseriesAggregate to count the frequency per quarter
DateListPlot[
TimeSeriesAggregate[Transpose[{dates, data[[All, 2]]}], "Quarter", Length],
DateTicksFormat -> {"Year", "/", "QuarterNameShort"},
GridLines -> Automatic,
PlotRange -> {{{2012, 8, 1}, {2013, 12, 31}}, Automatic},
Mesh -> Full]
And here are the monthly figures
DateListPlot[
TimeSeriesAggregate[Transpose[{dates, data[[All, 2]]}], "Month", Length],
DateTicksFormat -> {"Year", "/", "Month"},
GridLines -> Automatic,
PlotRange -> {{{2012, 8, 1}, {2013, 10, 31}}, Automatic},
Mesh -> Full]
• Thanks eldo! Really helpful. Thanks for taking the time to write this out.
– jeff
Commented Aug 2, 2017 at 14:22
• Hi eldo, just one last quick question. Do you have a recommendation of how to format the graph so when there isn't a data input for a period e,g, 2013Q2, it goes straight to 0 instead of connecting with the next non-zero datapoint? Thanks.
– jeff
Commented Aug 2, 2017 at 16:40
• Please wait, will respond in some hours - and thanks for acceptance
– eldo
Commented Aug 2, 2017 at 17:14
• Sounds good, thanks!
– jeff
Commented Aug 2, 2017 at 17:49 | 1,223 | 3,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.816244 |
https://stonespounds.com/2899-pounds-in-stones-and-pounds | 1,611,544,557,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00522.warc.gz | 552,133,742 | 4,660 | # 2899 pounds in stones and pounds
## Result
2899 pounds equals 207 stones and 1 pounds
You can also convert 2899 pounds to stones.
## Converter
Two thousand eight hundred ninety-nine pounds is equal to two hundred seven stones and one pounds (2899lbs = 207st 1lb). | 69 | 270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-04 | latest | en | 0.788118 |
http://www.netlib.org/lapack/lapack-3.1.1/manpages/man/manl/zrot.l | 1,555,930,575,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-18/segments/1555578551739.43/warc/CC-MAIN-20190422095521-20190422121111-00043.warc.gz | 275,664,380 | 1,099 | .TH ZROT 1 "November 2006" " LAPACK auxiliary routine (version 3.1) " " LAPACK auxiliary routine (version 3.1) " .SH NAME ZROT - a plane rotation, where the cos (C) is real and the sin (S) is complex, and the vectors CX and CY are complex .SH SYNOPSIS .TP 17 SUBROUTINE ZROT( N, CX, INCX, CY, INCY, C, S ) .TP 17 .ti +4 INTEGER INCX, INCY, N .TP 17 .ti +4 DOUBLE PRECISION C .TP 17 .ti +4 COMPLEX*16 S .TP 17 .ti +4 COMPLEX*16 CX( * ), CY( * ) .SH PURPOSE ZROT applies a plane rotation, where the cos (C) is real and the sin (S) is complex, and the vectors CX and CY are complex. .SH ARGUMENTS .TP 8 N (input) INTEGER The number of elements in the vectors CX and CY. .TP 8 CX (input/output) COMPLEX*16 array, dimension (N) On input, the vector X. On output, CX is overwritten with C*X + S*Y. .TP 8 INCX (input) INTEGER The increment between successive values of CY. INCX <> 0. .TP 8 CY (input/output) COMPLEX*16 array, dimension (N) On input, the vector Y. On output, CY is overwritten with -CONJG(S)*X + C*Y. .TP 8 INCY (input) INTEGER The increment between successive values of CY. INCX <> 0. .TP 8 C (input) DOUBLE PRECISION S (input) COMPLEX*16 C and S define a rotation [ C S ] [ -conjg(S) C ] where C*C + S*CONJG(S) = 1.0. | 412 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-18 | latest | en | 0.57583 |
https://albertteen.com/uk/gcse/physics/atoms-and-radioactivity/electrons-and-their-orbits | 1,726,889,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00638.warc.gz | 65,864,621 | 46,489 | Albert Teen
YOU ARE LEARNING:
Electrons and Their Orbits
# Electrons and Their Orbits
### Electrons and Their Orbits
Electrons orbit atoms in shells, and can be removed or added to created ions.
1
The table shows the charge and relative mass of subatomic particles, but the values for an electron are missing! An electron has a negative charge, so what value do you think is missing for the charge of an electron?
2
The mass of an electron is so small that it is insignificant to the overall mass of the atom, so what value do you think is missing from the table for the relative mass of an electron?
3
Now look at the image of an atom. Which arrow is pointing to the nucleus? Answer A, B, C or D.
4
Which arrow is pointing towards and electron? Answer A, B, C, or D.
What number in the periodic table do you use to find out how many electrons an atom has?
Electrons are the subatomic particles which orbit the nucleus of an atom. They are negatively charged, and they are much smaller and have much less mass than the nucleus. A neutral atom has the same number of protons and neutrons, so the atomic number can tell you how many electrons an atom has.
How electrons are arranged in an atom
1
Electrons are arranged in shells around the nucleus.
Each shell refers to an orbital for electrons which have a certain energy. This is why shells are also called energy levels.
2
Where do you think the shell with the lowest energy level lies?
3
The nucleus of an atom is surrounded by electrons. These are arranged in energy levels at different distances from the nucleus, where the lowest energy level is closest to the nucleus.
Electrons usually occupy the lowest energy levels available to them.
4
The arrangement of electrons differs from element to element, and is responsible for the chemical properties of each element. Depending on how many electrons an atom has, they will have a different number of shells. What do you think we call the highest energy level an atom has?
5
A sodium atom has 2 electrons in first shell, 8 in second shell and 1 in third shell. We call the arrangement of electrons the electronic configuration of the atom. So in this case it is 2.8.1 How many electrons does the atom have in total?
6
A sodium atom has electron configuration 2.8.1. A potassium atom has configuration 2.8.8.1. How many electrons can the first shell of an atom hold?
7
The maximum number of electrons in each shell, going from the middle to the outside, is 2, 8, 8, 18. So if a sodium atom has a configuration of 2.8.1, is its third and outer shell full?
8
When the outer shell has the maximum number of electrons, the electron shells are said to be full. The inner shells of an atom are always full.
An atom that has the maximum number of electrons in its outer shell will be stable. This means that it will not react with other atoms.
1
Atoms tend to want to have a full outer shell of electrons.
That way they are more stable.
2
Look at the electron configurations of sodium and chlorine. If sodium were to achieve a full outer shell, how many electrons would it have to gain?
3
Gaining 7 electrons is really difficult!
Instead, sodium could lose its 1 electron in the third shell. That way its outer shell would become its full second shell.
4
How many electrons would chlorine need to gain to have a full outer shell?
5
So sodium wants to lose an electron, and chlorine wants to gain one. Handy!
A) Sodium and chlorine are not reactive together. B) Sodium and chlorine are reactive together. C) Sodium and chlorine have similar properties.
6
This explains why sodium and chlorine easily bond to form sodium chloride, NaCl.
They just have to transfer just one electron. We call this ionic bonding.
Atoms are most stable when their outer shell is full, and are most reactive when it is not. This explains why the noble gases in group 8 of the periodic table are very unreactive. It is also why group 1 and group 7 elements bond well with each other.
If an atom of sodium were to transfer an electron to an atom of chlorine, what charge would each atom have? Pick the correct one for sodium and the correct one for chlorine.
When atoms lose or gain electrons in the process of creating a full outer shell, the overall charge of the atom is affected. If electrons are lost, the atom will become a positively charged ion, whereas if electrons are gained the atom becomes a negatively charged ion. This is what is called ionisation.
Exciting electrons
1
An atom can become ionised when an electron leaves its outer shell. In order to do this, the electron must...
2
When an electron absorbs energy, it jumps up to a higher energy level. We say that it has become excited.
If the electron gains enough energy, it can leave the atom altogether and the atom becomes a positively charged ion.
3
Most often however, electrons don't gain enough energy to leave the atom. Instead they jump to a higher energy level. After some time however, they will want to go back to the lower energy level to be more stable. In this case they will...
4
The energy the electrons absorb and emit are in the form of electromagnetic radiation.
The difference in energy levels that an electron jumps between corresponds to a particular frequency of the electromagnetic spectrum.
1
The absorption of radiation can be analysed by an ‘absorption spectrum’. The different black lines show the frequencies __________ by electrons in atoms, allowing them to jump to higher energy levels.
2
The emission of radiation can be analysed by an ‘emission spectrum’. The different coloured lines show the frequencies __________ by excited electrons falling back down to lower and more stable orbits. | 1,249 | 5,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.913599 |
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Discussion Board for collaboration related to Creating Analytics for QlikView.
New Contributor III
Count product Help
Hi
I need to count how many transactions have been made for each of the product from the every solddate + 7 days. This is my table:
Product SoldDate 1001 01-Jul-13 1001 03-Jul-13 1001 03-Jul-13 1001 04-Jul-13 1001 10-Jul-13 1002 04-Jul-13 1002 05-Jul-13 1002 06-Jul-13 1002 26-Jul-13
I wish to achieve the following result:
Product SoldDate Count 1001 01-Jul-13 4 1001 03-Jul-13 4 1001 03-Jul-13 3 1001 04-Jul-13 2 1001 10-Jul-13 1 1002 04-Jul-13 3 1002 05-Jul-13 2 1002 06-Jul-13 1 1002 26-Jul-13 1
For example, for product 1001, count for 01-Jul-13 is 4 because there are 4 transactions before 08-Jul-13 where solddate>=solddate but <=solddate+7.
Count for 04-Jul-13 is 2 because there are 2 transactions before 11-Jul-13.
It would be great if anyone can help me on this. Thanks in advance.
1 Solution
Accepted Solutions
Highlighted
Re: Count product Help
See attached example. I don't see a way to get two records for 3/7/2013 for product 1001 with different counts. Other than that it's what you want afaict.
talk is cheap, supply exceeds demand
5 Replies
New Contributor III
Re: Count product Help
anyone can help me on this?
Highlighted
Re: Count product Help
See attached example. I don't see a way to get two records for 3/7/2013 for product 1001 with different counts. Other than that it's what you want afaict.
talk is cheap, supply exceeds demand
New Contributor III
Re: Count product Help
Thanks Gysbert this is exactly what i want!!!
Could you explain to me the script? how does this "IterNo" function works?
And why i need to count(Amount)? instead of count(Product)?
Re: Count product Help
The iterno() is used to get the iteration number of the while loop. Your example didn't have a field named CIF. If you rather use another field then Amount to count that should work too as long as it isn't Product or SoldDate.
talk is cheap, supply exceeds demand
New Contributor III
Re: Count product Help
my mistake... what i mean is why we can't use Product? Count(Product)? | 617 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.901428 |
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Topic: binary search with arbitrary or random split
Replies: 1 Last Post: Aug 7, 2012 8:22 PM
Ben Bacarisse Posts: 1,972 Registered: 7/4/07
Re: binary search with arbitrary or random split
Posted: Aug 7, 2012 8:22 PM
[Followups-to set since I don't think sci.math.num-analysis is topical
anymore.]
Ralph Hartley <hartley@aic.nrl.navy.mil> writes:
> On 07/24/2012 06:02 PM, Ben Bacarisse wrote:
>> What's going on here? If there were any down side at to using n/2 I can
>> see why one might consider alternatives -- even random ones (some people
>> do for quicksort, partitioning for example) --- but since n/2 is simple,
>> easy to calculate and optimal you can't be proposing an alternative.
>
> For searching a sorted list, I agree it doesn't make much sense, but
> consider sorting.
>
> Quicksort is a simple algorithm, and is close to O(n log(n)) for most
> inputs, but you can't use it when the input is chosen by an adversary
> because in worst case it is O(n^2).
>
> It is reasonable to use quicksort with each pivot point chosen
> randomly from a uniform distribution over the list. Unlike search,
> there is no consistently good choice, and any deterministic rule could
> be exploited to give O(n^2) cost *every* time.
That's not true. You can pick the median in O(n) time and that means
(if you're careful about a few other details) that you can have a
deterministic quicksort that is O(n log(n)) in the worst case. It's not
really practical, but that's not the point.
> Worst case is *still* O(n^2), but that's ok. What you really care
> about is that the average over your random choices is O(n log(n)),
> even for the worst possible input.
Depends on who "you" refers to! People interested in complexity theory
(this is comp.theory after all) are more likely to be interested in
worst-case analysis. This is not to deny the large body of theory on
probabilistic analysis -- it's just not the majority concern for
complexity theorists.
> I think you can actually say something slightly stronger than that,
> and show that the probability of very bad cases is small.
Yes, I think so too, but I don't know if there is an agreed measure for
"bad cases are sufficiently rare". The one paper cited in this thread,
> It's OK to assume that your random choices follow a known
> distribution, you can make them so, but it can be a disaster to assume
> that the input array is anything but worse case.
>
> There are sorting algorithms with better worst case, but they are
> often not worth the trouble.
>
> None of this is *remotely* new.
No, but probabilistic measures of "almost always" O(f) are newer (if not
actually new). I'd be interested to hear from anyone who know what
recent thinking is about such things.
--
Ben. | 748 | 2,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-13 | latest | en | 0.93768 |
http://www.tutorialtext.com/%09atoi-Function/ | 1,566,272,014,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.14/warc/CC-MAIN-20190820024110-20190820050110-00124.warc.gz | 331,766,613 | 7,698 | # Education is not limited to just classrooms. It can be gained anytime, anywhere... - Ravi Ranjan (M.Tech-NIT)
Atoi Function :
• Atoi = A to I = Alphabet to Integer
• Convert String of number into Integer
Syntax:
`num = atoi(String);`
```num - Integer Variable
String- String of Numbers```
Example :
```num = atoi("1947");
printf("%d",num);```
Output :
`1947`
Significance :
1. Can Convert any String of Number into Integer Value that can Perform the arithmetic Operations like integer
Ways of Using Atoi Function :
Way 1 : Passing Variable in Atoi Function
```// Variable marks is of Char Type
int num;
char marks[3] = "98";
num = atoi(marks);
printf("
Marks : %d",num);```
Way 2 : Passing Direct String in Atoi Function
```int num;
num = atoi("98");
printf("
Marks : %d",num);
``` | 213 | 801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-35 | latest | en | 0.690554 |
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» Continue «
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L O A D I N G
### KS 3
Topic & lesson View
lesson
Work
sheet
sheet
### How it works
Click on this icon
below, to view the information about each lesson.
### Area
1
Calculating the Area of a Rectangle
2
Calculating the Area of a Rectangle Common Mistakes
3
The formula of the Area of a Triangle
4
Calculating the Area of a Triangle
5
The formula of the Area of a Parallelogram
6
Calculating the Area of a Parallelogram
7
Calculating the Area of a Trapezium
### Compound Areas
8
Calculating Compound Areas
9
Compound Areas Involving Different Shapes Part 1
10
Compound Areas Involving Different Shapes Part 2
11
Compound Area where you're not given all the lengths
### Surface Area
12
Surface area of a cuboid
13
Surface area of a triangular prism
14
Surface area of a cylinder
15
Surface area of a sphere (Higher Only)
16
Surface area of a hemisphere exam question (Higher Only)
17
Curved surface area of a cone (Higher Only)
18
Calculating the surface area of a frustum (Higher Only)
### Volume
19
Volume of a cuboid
20
Calculating the volume of any prism
21
Volume of triangular prism
22
Volume of a cylinder
23
Finding the height of a cylinder using its volume (Higher Only)
24
Volume of a sphere (Higher Only)
25
Volume of a cone (Higher Only)
26
Volume of a cone exam question (Higher Only)
27
Volume of a frustum (Higher Only)
28
Volume of a square based pyramid
### Circles
29
Circumference of a circle
30
Circumference of a circle leaving your answer in terms of Pi (Higher Only)
31
Perimeter of a semi-circle
32
Length of an arc (Higher Only)
33
Area of a circle
34
Area of a semi-circle
35
Area of a circle, leaving your answer in terms of Pi (Higher Only)
36
Area of a sector (Higher Only)
### Circle Geometry (Higher Only)
37
Angle in a semi-circle is 90 degrees
38
Sneaky isosceles
39
Angle at centre is twice angle at circumference part 1
40
Angle at centre is twice angle at circumference part 2
41
Angle at centre is twice angle at circumference part 3
42
43
Angles in the same segment are equal
44
Tangent meets radius at 90 degrees
45
Tangents that meet at a point are equal in length
46
Alternate segment theorem part 1
47
Alternate segment theorem part 2
48
3 letter angle notation in circle geometry
### Angles
49
Naming angles
50
Proving angles on a straight line add to 180
51
Angles on a straight line
52
Angles on a straight line - Common Mistake
53
Angles on a straight line - Exam style question
54
Proving the angles around a point add up to 360
55
Angles around a point
56
Angles around a point - Exam style question
57
58
Finding angles in a triangle
59
Introduction to Isosceles triangles & finding the base angles
60
Finding angles in Isosceles triangles
61
Finding angles in Isosceles triangles - Exam style question
62
Introduction to Equilateral triangles
63
The 4 types of Triangles
64
Vertically opposite angles
65
Vertically opposite angles exam question
66
3 letter angle notation
### Angles on Parallel Lines
67
Alternate angles on parallel lines
68
Corresponding angles
69
Co-interior angles
70
Finding angles on parallel lines exam question
### Measuring Angles
71
Introduction to the parts of a protractor
72
How to use a protractor
73
Exam tip for measuring difficult angles
### Polygons
74
Introduction to polygons
75
Introduction to interior angles and notation
76
Finding interior angles
77
Finding external angles
78
Polygons exam question
### Properties of Shapes
79
Kites
80
Parallelograms and Rhombuses
81
Trapeziums
### Midpoints of Lines
82
Finding the Midpoint of a Line Joining 2 Points
83
Finding the Midpoint of a Line Joining 2 Points Exam Question
### 3D Coordinates (Higher Only)
84
x, y, z coordinates
85
Finding the Midpoint of a Line Joining 2 3D coordinates
### Converting Units of Length, Area and Volume
86
Converting units of area
87
Converting units of volume part 1
88
Converting units of volume part 2
### Similar Triangles and Similar Shapes (Higher Only)
89
Similar triangles introduction
90
Similar triangles part 1
91
Similar triangles part 2
92
Similar triangles part 3
93
Proving that triangles are similar using lengths
94
Proving that triangles are similar using angles
95
Scale factors for area's of similar shapes
96
Scale factors for volume's of similar shapes
97
Similar Shapes exam question
### Loci and Constructions
98
Perpendicular Bisector
99
Perpendicular line from a point on a line
100
Perpendicular line from a point to a line
101
Angle bisector
102
Drawing an equilateral triangle
103
Drawing an isosceles triangle
104
Drawing a scalene triangle
105
Drawing a 60$^\circ$ angle
106
Drawing a 45$^\circ$ and 30$^\circ$ angle
107
The rules of Loci
108
Drawing boundaries around shapes
109
### Bounds (Higher Only)
110
Upper and lower bounds of discrete numbers
111
Upper and lower bounds of continuous numbers
112
113
Calculations with bounds subtracting bounds
114
Calculations with bounds multiplying bounds
115
Calculations with bounds dividing bounds
### Converting Metric Units
116
Converting metric units of length part 1
117
Converting metric units of length part 2
118
Converting units exam question (length)
119
Converting metric units of mass part 1
120
Converting metric units of mass part 2
121
Converting metric units of volume and capacity part 1
122
Converting metric units of volume and capacity part 2
123
Converting units exam question (capacity)
124
Converting speeds 1 step problems
125
Converting speeds 2 step problems
### Converting Imperial Units
126
Converting imperial units of length
127
Converting lengths 1 step problems
128
Converting lengths 2 step problems
129
Converting imperial units of mass
130
Converting mass 1 step problems
131
Converting mass 2 step problems
132
Converting imperial units of capacity
133
Converting capacity 1 step problems
134
Converting volume and capacity 2 step problems
### Pythagoras Theorem
135
Pythagoras theorem finding the longest side
136
Pythagoras theorem finding the shorter side
137
Pythagoras theorem problems (exam question)
### Right-angled Trigonometry
138
How to label a triangle
139
Finding a length
140
Finding an angle
### Non-right Angled Trigonometry (Higher Only)
141
Sine rule finding a length
142
Choosing which rule to use and labelling the triangle
143
Cosine rule finding a length
144
Cosine rule finding an angle
145
Area of a non-right angled triangle
146
Sine Rule Finding an Angle Part 1
147
Sine Rule Finding an Angle Part 2
### 3D Pythagoras (Higher Only)
148
3D Pythagoras in a cuboid
149
3D Pythagoras in a pyramid
### 3D Right-angled Trigonometry (Higher Only)
150
3D Trigonometry in a cuboid
151
3D Trigonometry in a Pyramid
### Bearings
152
Rules of bearings
153
Introduction to bearings Part 1
154
Introduction to bearings Part 2
155
Simple bearings questions 1
156
Simple bearings question 2
157
Sketching simple Bearings questions
158
Bearings questions involving Trigonometry part 1
159
Bearings questions involving trigonometry part 2
160
Bearings questions involving Trigonometry part 3
### Density
161
The different units of density
162
Simple Density Questions Part 1
163
Simple Density Questions part 2
164
Questions involving mixing 2 different liquids
### Congruence
165
Congruent Shapes
166
Proving congruence SSS
167
Proving congruence RHS
168
Proving congruence SAS
169
Proving congruence AAS
170
Proving congruence exam style question 1
171
Proving congruence exam style question 2
### Translations
172
Translations
173
Translations exam question
### Reflection
174
Reflecting shapes in the $x$ and $y$ axis
175
Reflecting shapes in the line $y=5$ ${ }$ $x=6$
176
Reflecting shapes in the line $y=x$
### Rotation
177
Rotation about a centre part 1
178
Rotation about a centre part 2
179
Finding the direction and angle of rotation
180
Finding the COR (centre of rotation) (Higher Only)
### Enlargements
181
Enlarging shapes
182
Enlarging shapes exam tip
183
Enlargement with a centre of enlargement
184
Enlargement with a fractional scale factor (Higher Only)
185
Enlargements with a negative scale factor (Higher Only)
186
Finding the centre of enlargement
187
Finding the scale factor of enlargement
188
Finding the scale factor of an inverted enlargement (Higher Only)
### Describing Single Transformations
189
Describing single transformations - How to score full marks
190
Describing single transformations - Exam Tip
### Vectors
191
Introduction to vectors
192
Writing and drawing column vectors
193
Subtracting vectors
194
195
Reversing the direction of vectors
196
Multiplying vectors
197
Vector problems
198
Vector problems with midpoints
199
Vector problems with ratios
200
Proving vectors are parallel
Move to class | 2,311 | 9,124 | {"found_math": true, "script_math_tex": 9, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.844109 |
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11 Plus Probability Practice Tests
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Many 11 Plus Probability Practice Tests available @ 11PluseHelp.co.uk
11 Plus Probability
Probability:
Probability is a ratio of the number of favourable outcomes to the number of possible outcomes.
Probability formula:
Number of favourable outcomes=f
Number of possible outcomes=n
Probability= f/n
Key points to remember in Probability:
1. The minimum value of probability=0
2. The maximum value of probability=1
3. The set of all possible outcomes is called sample space.
4. The Probability of sample space is 1
5. The sum of probabilities of all possible outcomes in a sample space=1
Probability is one of the 11 plus exam maths topics, and probability topics are regularly asked, so practising these questions is necessary.
Examples papers from our websites:
EXample:1
Example:2
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Thanks,
11Plusehelp.co.uk | 1,278 | 5,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-27 | latest | en | 0.884299 |
https://everything2.com/title/additive+order | 1,620,339,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988763.83/warc/CC-MAIN-20210506205251-20210506235251-00179.warc.gz | 274,632,392 | 6,095 | A ring whose operations are "multiplication" and "addition" can have additive orders and multiplicative orders for each of its elements. The additive order of a given element r in a ring R is the number of times n which r must be 'added' to itself (under the ring's "additive" operation) to get the zero of R. In basic mathematical notation: n ∈ N such that n*r = 0R. (Note: even though integers and ring elements are being mixed/multiplied together, we define this kind of 'multiplication' as 'multiple additions', so something like 3*r is understood to mean r+r+r under the ring's addition operation.) If no finite n exists such that n*r = 0R, we let n = ∞.
### In the context of the integers
The set of integers form a ring (in fact, an integral domain) over the multiplication and addition operations. The additive order of any given integer within the context of the integers is ∞, so we usually talk about the additive order of integers modulo a specifc integer.
Given integers a and n (n positive), the additive order of a mod n is the smallest positive integer m such that m*a=0(mod n). In fact, the additive order m of a mod n is always m=n/gcd(a,n).
Proof (m=n/gcd(a,n) is the additive order of a mod n):
Let a and n be integers, n positive. Let m=n/gcd(a,n). Then m*a=a*n/gcd(a,n)=n*(a/gcd(a,n))=0(mod n) (since gcd(a,n) divides a).
Let a and n be integers, n positive. Notice that n*a=0(mod n), since n|n*a. But n|n*(a/gcd(a,n)) as well, which is the same as saying n|(n/gcd(a,n))*a. Then d=gcd(a,n) is the greatest integer such that d|a and d|n, so n/gcd(a,n) is the smallest number with the property n|n*a/gcd(a,n). Thus m=n/gcd(a,n) is the additive order of a mod n.
Log in or register to write something here or to contact authors. | 475 | 1,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-21 | latest | en | 0.871342 |
https://wellsr.com/vba/2018/excel/reverse-order-of-array-with-vba/ | 1,542,742,974,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746639.67/warc/CC-MAIN-20181120191321-20181120213321-00442.warc.gz | 766,327,396 | 10,249 | Introduction | Example | Tutorial | Applications | Comments
## Introduction - Reverse Array
This macro will reverse the order of an array using Excel VBA. This is a useful macro if you have an array sorted smallest to largest and you want to sort it largest to smallest. Instead of rewriting your sorting algorithm, you can reverse the order of your existing array.
This tutorial was inspired by a comment I received on my super fast VBA Quicksort macro, which sorts your VBA arrays from smallest to largest.
## Example - Reverse Array
### Reverse the order of your array
``````Sub ReverseArray(vArray As Variant)
'Reverse the order of an array, so if it's already sorted
'from smallest to largest, it will now be sorted from
'largest to smallest.
Dim vTemp As Variant
Dim i As Long
Dim iUpper As Long
Dim iMidPt As Long
iUpper = UBound(vArray)
iMidPt = (UBound(vArray) - LBound(vArray)) \ 2 + LBound(vArray)
For i = LBound(vArray) To iMidPt
vTemp = vArray(iUpper)
vArray(iUpper) = vArray(i)
vArray(i) = vTemp
iUpper = iUpper - 1
Next i
End Sub``````
Write better macros in half the time
I see people struggling with Excel every day and I want to help. That's why I'm giving away my personal macro library for free. This powerful gift lets you automatically import all my macros directly into your spreadsheets with just one click.
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## Tutorial - Reverse Array
### How the macro reverses the order of your array
This macro loops through half of your array using a combination of the LBound and UBound functions to find the midpoint. It only needs to go through half your array because it’s essentially swapping the first half (counting up) with the last half (counting donw).
It takes the first element if your array and moves it to the last element, then it takes your last element and moves it to your first element. It keeps going until the entire array is in the opposite order.
I’ve come across a number of functions online that claim to reverse your VBA arrays, but the way they define the midpoint is incorrect. Many of them either add a `+ 1` or nothing to the end, where this macro properly adds the lower bound of your array.
If you were to hardwire the “adder” on the end if your midpoint, you’re inherently forcing your macro to only work with Base 0 or Base 1 arrays, depending on how you do it.
This macro is different. It will work with Base 0, Base 1 and any other base array you pass it. For example, even if your array was defined with elements from -1 to 5, it will still be properly reversed.
Other macros online will mess up the central elements in your array and you’d be left with a hodgepodge of orders in the middle. That’s not what you want!
### Calling the ReverseArray Macro
Here’s a basic example to show you how to call the ReverseArray macro to reverse the order of the elements in your array. All you have to do is pass the macro 1 argument: the name of your array. The array is named v in this example.
``````Sub Reverse_Example()
Dim v() As Variant
v() = Array(1, 2, 3, 4, 5, 6)
Call ReverseArray(v)
'From here on, the array "v" is in reverse order (6,5,4,3,2,1)
End Sub``````
This macro will work whether you use `Option Base 0` (default) or `Option Base 1` to define your arrays.
It will also work if you have a custom lower bound, like in the following example:
``````Sub Reverse_Example2()
Dim v(-5 To 5) As Variant
For i = LBound(v) To UBound(v)
v(i) = i
Next i
Call ReverseArray(v)
'From here on, the array "v" is in reverse order (5,4,3...-4,-5)
End Sub``````
Not very many people know how to reverse the order of an array, but there are even rarer array tricks out there for you to learn. Grab a copy of our comprehensive VBA Arrays Cheat Sheet with over 20 pre-built macros and dozens of tips designed to show you even more tricks for handling arrays.
## Application Ideas
Reversing the order of your array is most useful when you already have a sorted array, but you want it sorted the other way. Sometimes you just don’t want to use a For Loop with the Step option set to -1 to walk through your array backwards.
In these instances, instead of changing the logic of the sorting algorithms you already use, just switch the order of your array so it’s sorted the other way. The first element will now be your last element and your last element will now be your first element.
Doing it this way may take a little longer than sorting it the way you want it up front, but the added time is miniscule compared to the aggravation of rewriting your sorting algorithms!
Whether you use the VBA bubble sort or VBA Quicksort algorithms, this ReverseArray example is a nice compliment to your existing array manipulation tools. Just call the procedure immediately after sorting your array, and your array will be sorted the other way. | 1,170 | 4,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-47 | longest | en | 0.834124 |
http://forums.wolfram.com/mathgroup/archive/2008/May/msg00043.html | 1,591,273,172,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00592.warc.gz | 44,971,058 | 7,615 | Re: Convert funcion
• To: mathgroup at smc.vnet.net
• Subject: [mg88332] Re: Convert funcion
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 2 May 2008 03:42:01 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <fvc620\$9n2\$1@smc.vnet.net>
```Johum wrote:
> Someone could help me with this function?
>
> fConver[cmatrix], this function will find all subsets of rows of the rectangular matrix which have exactly one nonzero element in each column.
>
> For example:
>
> cmatrix=({{0,0,1,0,1,1,0},{1,0,0,1,0,0,1},{0,1,1,0,0,1,0},
> {1,0,0,1,0,0,0},{0,1,0,0,0,0,1},{0,0,0,1,1,0,1}});
>
>
> fConvert[cmatrix]
>
> the out put will be
> {{{1,4},{3,5,6},{2,7}}}
Since you seem to have taken your "example" from the documentation of
the Imtek`ExactCover` function, available at
http://www.imtek.de/simulation//mathematica/IMSweb/imsTOC/Game%20Theory/ExactCoverDocu.html
it is not clear why you would not use it as it is? Dis you try it? Did
you face any difficulty or get any erroneous results? What would you do
with such a function? Perhaps solving the Latin Square Puzzle or the
Height Queens Puzzle?
Regards,
-- Jean-Marc
```
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• Next by thread: Re: Convert funcion | 444 | 1,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-24 | latest | en | 0.818996 |
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also what did you guys get for that rectangle question, where you had to find the missing length?
0
1 month ago
#2
2 packs of BUTTER
4
1 month ago
#3
rectangle question
x=8.5
4
1 month ago
#4
96 green pens
And for the last one it was 12 red counters and 9 green
2
1 month ago
#5
they gave a length 10cm and the area was 45cm^2.
divide 45 by 10 =4.5
then it said that length is the same as the other length
4.5+4.5=9
26cm (perimeter) -9= 17
17/2= 8.5
sorry for bad explanation i forgot the letters of the shape
2
1 month ago
#6
wasnt the function question bae
0
#7
(Original post by Tbn2002)
2 packs of BUTTER
i got that too!
1
#8
yess thanks, i got that too
(Original post by Rigeld)
rectangle question
x=8.5
(Original post by bob200328)
they gave a length 10cm and the area was 45cm^2.
divide 45 by 10 =4.5
then it said that length is the same as the other length
4.5+4.5=9
26cm (perimeter) -9= 17
17/2= 8.5
sorry for bad explanation i forgot the letters of the shape
0
1 month ago
#9
The percentage anyone else get 25%?
0
1 month ago
#10
anyone else get 48 pi for cone + hemisphere volume?
0
1 month ago
#11
got 36 pi
(Original post by ay._bee)
anyone else get 48 pi for cone + hemisphere volume?
1
1 month ago
#12
(Original post by bob200328)
got 36 pi
What was ur method?
0
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# OG 11 Diagnostic Test- Verbal Question 25
Author Message
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OG 11 Diagnostic Test- Verbal Question 25 [#permalink]
### Show Tags
07 Dec 2008, 12:58
Hi All,
I have serious doubt regarding the reasoning provided to support the answer of Question 25 in Diagnostic Verbal Test.
The question:
Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running toward an insect, a tiger bettle will intermittently stop and then, a moment later, resume its attack. Perhaps the beetles cannot maintain their pace and must pause for a moment's rest; but an alternative hypothesis is that while running, tiger bettles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.
Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other?
Options:
A) When a prey insect is moved directly toward a bettle that has been chasing it, the bettle immediately stops and runs away without its usuall intermittent stopping.
B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.
C) In pursuing a moving insect, a bettle usually responds immediately to changes in the insect's direction, and it pauses equally frequently whether the chase is up or down an incline.
D) If, when a bettle pauses, it has not gained on the insect it is pursuing, the bettle generally ends its pursuit.
E) The fasta bettle pursues an insect fleeing directly away from it, the more frequently the bettle stops.
Answer I selected was (A), because it shows that it could make anything from the sudden visual changes (the prey running towards it) but it could keep up its pace ( without its usual intermittent stopping)
Kindly help
If you have any questions
New!
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Re: OG 11 Diagnostic Test- Verbal Question 25 [#permalink]
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07 Dec 2008, 14:43
My answer is B and here's why:
The two hypotheses:
1) Beetles pauses because they get tired
2) Beetles go blind in motion.
We need to find which of the following supports one of the two AND undermines the other.
A) When a prey insect is moved directly toward a bettle that has been chasing it, the bettle immediately stops and runs away without its usuall intermittent stopping. ....> Bettle is not blind, not pausing coz of tiredness. => supports neither hypotheses
B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses......>Bettle is not blind, is tired => supports 1, undermines 2
C) In pursuing a moving insect, a bettle usually responds immediately to changes in the insect's direction, and it pauses equally frequently whether the chase is up or down an incline......>not blind, pauses not caused by tiredness => supports neither
D) If, when a bettle pauses, it has not gained on the insect it is pursuing, the bettle generally ends its pursuit. => could be blind, but no conclusion about pausing coz of tiredness. => supports 2, does not undermine 1
E) The fasta bettle pursues an insect fleeing directly away from it, the more frequently the bettle stops. => supports 1, does not undermine 2
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Re: OG 11 Diagnostic Test- Verbal Question 25 [#permalink]
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07 Dec 2008, 14:54
Hi,
In the explanation of Option (A) you have said, that its not blind.
How can it not be blind?
It stopped either because it got blind or because it got tired.
It cannot get tired because the prey suddenly came towards beetle and the 2nd stmt tells tht it ran away from it without pauses.
So logically it means that it stopped because it got blind.
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Re: OG 11 Diagnostic Test- Verbal Question 25 [#permalink]
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07 Dec 2008, 14:57
Interestingly OG gives another explanation to selecting Option B as answer.
Hypotheses
1) Cannot maintain the rapid pace and must rest.
2) Temporarily go blind.
B. Correct. This statement provides information that strengthens the second hypothesis: the swerving pursuit and the resulting continual course adjustments appear to be forcing the beetle to stop with increasing frequency to sort out the erratic visual information.
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Re: OG 11 Diagnostic Test- Verbal Question 25 [#permalink]
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07 Dec 2008, 15:53
jeril842002 wrote:
Hi,
In the explanation of Option (A) you have said, that its not blind.
How can it not be blind?
It stopped either because it got blind or because it got tired.
It cannot get tired because the prey suddenly came towards beetle and the 2nd stmt tells tht it ran away from it without pauses.
So logically it means that it stopped because it got blind.
the beetle stopped because it saw the prey suddenly approaching it while it(the beetle) was in motion.
hence was not blinded
Re: OG 11 Diagnostic Test- Verbal Question 25 [#permalink] 07 Dec 2008, 15:53
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2.7. Mathematical optimization: finding minima of functions¶. Authors: Gaël Varoquaux. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function.
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However, when the problem sizenis large, evaluating the Jacobian matrix in each iteration of the Newton method is expensive. Thus, approximations of the Jacobian matrix are often used to replace the exact Jacobian matrix in the Newton iteration [3, 30].
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Dec 16, 2009 · I want to implement this on Matlab using Simpson's Rule. The answer should fill an NxN matrix A, with A i,j =Integral(log(r i)*log(r j))ds So far I have calculated the integrand for a chosen (x,y) and have the values stored in an NxN matrix. I have written working code for Simpson's rule but am not sure how to apply it to this problem.
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Jan 31, 2010 · Good evening, dear respective Mr. César de Souza! If is it possible could you please answer on few questions concerning Jacobian calculation? You have wrote that in Jacobian Matrix F(xi,w) is "Where F(xi, w) is the network function evaluated for the i-th input vector of the training set using the weight vector w and wj is the j-th element of the weight vector w of the network.".
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Use the zoom feature of plots in Matlab (the magnifying glass icon with a + sign) to look at region with horizontal extent around [-20,20]. It is clear that most of the iterates appear to be on the -axis. Please send me a copy of this plot. Look at the formula you used for the Jacobian in f8v.m.View MATLAB Command Use the diff function to approximate partial derivatives with the syntax Y = diff (f)/h, where f is a vector of function values evaluated over some domain, X, and h is an appropriate step size. For example, the first derivative of sin (x) with respect to x is cos (x), and the second derivative with respect to x is -sin (x). | 531 | 2,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-17 | latest | en | 0.895536 |
http://encyclopedia2.thefreedictionary.com/wear+pattern | 1,511,401,983,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00506.warc.gz | 93,668,874 | 13,990 | # pattern
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## pattern
Manufacturing a wooden or metal shape or model used in a foundry to make a mould
## Pattern
The juxtaposition of repetitive elements in a design, organized so as to produce an arrangement of parts that are viewed as an unit; may occur at various scales and sizes.
## Pattern
(also, development), in geometry. The pattern of a polyhedron is a set of polygons for which it is shown how the sides and vertices of the polygons must be joined in order to ob-
Figure 1
tain the polyhedron. Several conditions must be met here: each side of a polygon must be joined to no more than one side of another polygon; it must be possible to pass from one polygon to another polygon by traversing pairwise joined polygons; and joined sides must have equal lengths. The pattern of a cube is shown in Figure 1.
Figure 2
In such fields as descriptive geometry and drafting, the concept of pattern is sometimes applied to curved surfaces. Thus, the pattern of the lateral surface of a cone is a sector of a circle (Figure 2).
## Pattern
(Russian, shablom), in foundry production, an element of a gated pattern, consisting of a flat device whose working side has a highly accurate profile. A distinction is made between molding and control patterns. The former, in the form of wooden boards, are used in individual and small-series production; they take the place of a solid casting pattern or corebox in the manufacture of casting molds and mold cores for medium-size and large castings having the outline of a body of revolution (such as cups, vats, covers, and pulleys). The cavity of the mold or the working surface of the core is produced by rotating the molding pattern around its central axis (a core spindle is positioned at this place in the mold). Control patterns are made of plywood or sheet steel and are used for checking the accurate placement of cores in the cavity of a casting mold during assembly and preparation for pouring.
M. N. SOSNENKO
## pattern
(aerospace engineering)
The flight path flown by an aircraft, or prescribed to be flown, as in making an approach to a landing.
(engineering)
A form designed and used as a model for making things.
(graphic arts)
A design or form.
(mathematics)
An equivalence class of colorings of the elements of a finite set, which are indistinguishable with respect to a group of permutations of the colors.
(ordnance)
The distribution of a series of shots fired from one gun or a battery of guns under conditions as nearly identical as possible, the points of impact of the projectiles being dispersed about a point called the center of impact.
## pattern
1. A model made in some easily worked material (such as plaster or wood) which serves as a guide, with respect to form and dimensions, in laying out any piece of work, esp. to preserve and secure uniformity and accuracy.
2. A design, considered as a unit, of which an idea can be given by a fragment, as a diaper pattern.
3. In molding, a form used to provide the interior shape of the mold.
## pattern
i. A flight pattern that an aircraft must follow when approaching for landing and when leaving the airport after takeoff.
ii. Radiation of the transmitting aerial as plotted on a diagram of the field strength for each bearing.
iii. A shape traced out on the ground by the track of the aircraft while following certain procedures, such as making the circuit, making procedure turns, while holding, and while carrying out demonstrations. See circuit and holding pattern.
References in periodicals archive ?
The Lineage XLC's technology eliminates this flare along the back of the cutting edge from occurring and reduces the width and depth of the concave wear pattern by more than 50% for a consistently cleaner, quality cut, FS Tool says.
With advanced capabilities for surface analysis, Lucideon's experts are also able to provide information on what is happening at the interface between materials, such as the effect that coatings have on base materials and whether wear patterns are consistent.
Figures 6 and 7 show the wear pattern for the spiral pointed taps.
It is a cast 3-D set of jawbones showing the exact tooth wear patterns from bucks ranging from 1.
It is not unusual for a different wear pattern to develop and if this other wear pattern goes unnoticed, it propagates to the point where the insert fails completely.
Byram also suggests maintaining a wear pattern and following the maintenance schedule recommended by the machine's manufacturer.
127 cm) depth of cut provided a uniform wear pattern on the flank of the insert without spinning the disc in the chuck of the turning center.
Like Grissom, Holmes loved putting his crime scene evidence under the magnifying glass, checking out footprints, soil samples, and the wear pattern on a pair of boots.
Its ScramblePad is an access control reader that scrambles the digits on the control pad so bystanders cannot learn the code by watching the pattern; scrambling also prevents a wear pattern on the buttons.
You will see a more consistent wear pattern for the machine, and any backlash that occurs can be taken up correctly through the control.
The longevity and wear pattern of the tread depends on how the tire is used (load, speed, road surface, condition of the vehicle, driving style, etc.
Methodologically, Aldenderfer does everything else - extensive spatial analysis at both intrasite and regional levels, wear pattern, source, and flake aggregate analysis within the lithic assemblage, and faunal analysis with an emphasis on both species and packet identification.
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Open / Close | 1,183 | 5,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-47 | latest | en | 0.944889 |
https://laid-back-scientist.com/en/multivariate-statistics | 1,726,639,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00820.warc.gz | 308,775,371 | 41,991 | # 【Multivariate Data】 Scatter Plots and Correlation Coefficients
This article is available in: 日本語
## Introduction
In the previous issue, we discussed how to handle the most basic univariate data.
This article discusses scatter plots and scatter matrix as basic ways to handle multivariate data, and correlation coefficient, rank correlation coefficient, and variance-covariance matrix as summarization methods.
The program used is described in python and it can be tried in Google Colab below.
Google Colab
\begin{align*} \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\f}[2]{\frac{#1}{#2}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\d}[2]{\frac{{\rm d}#1}{{\rm d}#2}} \newcommand{\T}{\mathsf{T}} \newcommand{$}{\left(} \newcommand{$}{\right)} \newcommand{\{}{\left\{} \newcommand{\}}{\right\}} \newcommand{$}{\left[} \newcommand{$}{\right]} \newcommand{\dis}{\displaystyle} \newcommand{\eq}[1]{{\rm Eq}(\ref{#1})} \newcommand{\n}{\notag\\} \newcommand{\t}{\ \ \ \ } \newcommand{\tt}{\t\t\t\t} \newcommand{\argmax}{\mathop{\rm arg\, max}\limits} \newcommand{\argmin}{\mathop{\rm arg\, min}\limits} \def\l<#1>{\left\langle #1 \right\rangle} \def\us#1_#2{\underset{#2}{#1}} \def\os#1^#2{\overset{#2}{#1}} \newcommand{\case}[1]{\{ \begin{array}{ll} #1 \end{array} \right.} \newcommand{\s}[1]{{\scriptstyle #1}} \definecolor{myblack}{rgb}{0.27,0.27,0.27} \definecolor{myred}{rgb}{0.78,0.24,0.18} \definecolor{myblue}{rgb}{0.0,0.443,0.737} \definecolor{myyellow}{rgb}{1.0,0.82,0.165} \definecolor{mygreen}{rgb}{0.24,0.47,0.44} \newcommand{\c}[2]{\textcolor{#1}{#2}} \newcommand{\ub}[2]{\underbrace{#1}_{#2}} \end{align*}
## How to draw a scatter plot
We will again use the iris dataset as the data to be handled. iris dataset consists of petal (petal) and gak (sepal) lengths of three varieties: Versicolour, Virginica, and Setosa.
First, as bivariate data, we will restrict the variety to Setosa and deal only with sepal_length and sepal_width.
Now we will import the iris dataset in python.
import numpy as np
import pandas as pd
import seaborn as sns
import scipy.stats as st
import matplotlib.pyplot as plt
df_iris = sns.load_dataset('iris')
sepal_length = df_iris[df_iris['species']=='setosa']['sepal_length']
sepal_width = df_iris[df_iris['species']=='setosa']['sepal_width']
The most basic way to visually see the relationship between sepal_length and sepal_width is to draw a scatter plot.
To draw a scatterplot, use the scatterplot function in the seaborn library.
sns.scatterplot(x=sepal_length, y=sepal_width)
plt.show()
The scatterplot in the figure above shows that the points are aligned rightward. This means that flowers with larger sepal_length tend to have larger sepal_width.
The seaborn library also has a jointplot function that depicts a histogram along with a scatterplot.
sns.jointplot(x=sepal_length, y=sepal_width)
plt.show()
## correlation coefficient
Correlation coefficients (Pearson’s product-moment correlation coefficient) are often used to identify “rightward” or “downward” trends between data; the correlation coefficient between two variables $x$ and $y$ takes values ranging from $-1$ to $1$, where $x$ tends to increase as $y$ increases. It takes a positive value when $x$ tends to increase and a negative value when $y$ tends to decrease.
There is a positive correlation between $x$ and $y$ when the correlation coefficient is close to $1$, a negative correlation when close to $-1$, and no correlation when close to $0$.
The correlation coefficient $r_{xy}$ is defined as
\begin{align*} r_{xy} = \f{\sum_{i=1}^n (x^{(i)} – \bar{x}) (y^{(i)} – \bar{y}) }{\sqrt{\sum_{i=1}^n (x^{(i)} – \bar{x})^2} \sqrt{\sum_{i=1}^n (y^{(i)} – \bar{y})^2}}. \end{align*}
where as the mean deviation vector
\begin{align*} \bm{x} &= (x^{(1)} – \bar{x}, x^{(2)} – \bar{x}, \dots, x^{(n)} – \bar{x})^\T, \n \bm{y} &= (y^{(1)} – \bar{y}, y^{(2)} – \bar{y}, \dots, y^{(n)} – \bar{y})^\T \end{align*}
The correlation coefficient $R_{xy}$ coincides with the cosine $\cos \theta$ of the angle $\theta$ between the vectors $\bm{x}, \bm{y}$. \begin{align*} r_{xy} = \cos \theta = \f{\bm{x}^\T \bm{y}}{\|\bm{x}\| \|\bm{y}\|}. \end{align*}
From this we see that $-1 \leq r_{xy} \leq 1$.
Also, if there is a positive correlation, $\bm{x}, \bm{y}$ point in the same direction, and if there is no correlation, $\bm{x}, \bm{y}$ can be interpreted as pointing in orthogonal directions.
The numerator of the defining formula for the correlation coefficient divided by the number of samples $n$ is called the covariance of $x, y$.
\begin{align*} \sigma_{xy} = \f{1}{n} \sum_{i=1}^n (x^{(i)} – \bar{x}) (y^{(i)} – \bar{y}) \end{align*}
Using this value and the standard deviation $\sigma_x, \sigma_y$ of $x, y$, the correlation coefficient can be expressed as
\begin{align*} r_{xy} = \f{\sigma_{xy}}{\sigma_x \sigma_y} \end{align*}
When the correlation coefficient is $r_{xy} = \pm 1$, there is a linear relationship between $x, y$. The proof is given below.
Proof
In the following, the variance $\sigma_x^2, \sigma_y^2$ of $x, y$ is not $0$.
If $R_{xy}$ is $1$ or $-1$, then the cosine $\cos \theta$ of the angle $\theta$ between the mean deviation vectors $\bm{x}, \bm{y}$ is $1$ or $-1$. Therefore,
\begin{align*} \bm{y} = \gamma \bm{x} \end{align*}
There exists a constant $\gamma \neq 0$ satisfying $\gamma$ (if $r_{xy}=1$, $\gamma$ is positive, if $r_{xy}=-1$, $\gamma$ is negative). From this, $(y^{(i)} – \bar{y}) = \gamma (x^{(i)} – \bar{x}), \ (i = 1, \dots, n)$. In other words, $x, y$ have a linear relationship.
Also, taking the average of the squares of both sides in the above equation, we obtain
\begin{align*} \f{1}{n}\sum_{i=1}^n (y^{(i)} – \bar{y})^2 &= \gamma^2 \f{1}{n}\sum_{i=1}^n (x^{(i)} – \bar{x})^2 \n \therefore \sigma_y^2 &= \gamma^2 \sigma_x^2. \end{align*}
Therefore, $\gamma = \pm \sqrt{\sigma_y^2 / \sigma_x^2}$ and the following linear relationship holds for $x, y$.
\begin{align*} y = \pm \sqrt{\f{\sigma_y^2}{\sigma_x^2}} (x – \bar{x}) + \bar{y}. \end{align*}
The sign of the slope is the same as the sign of $r_{xy}$.
In python, the correlation coefficient can be calculated as follows
corr = np.corrcoef(sepal_length, sepal_width)[0, 1]
print(f'correlation coefficient: {corr}')
# Output
correlation coefficient: 0.7425466856651597
One caveat to the correlation coefficient is that it is valid when the two variables are in a linear relationship, but not when they are not (in the case of a non-linear relationship). In fact, as shown in the figure below, for data with a relationship other than a linear relationship, the correlation coefficient will determine that there is “no correlation” and will not work effectively.
Thus, the correlation coefficient is a quantitative expression of how close a linear relationship is; “correlated” is different from “there is a relationship between the data.
Another point to note is the phenomenon that when the data is truncated, the correlation coefficient approaches $0$ compared to the original data. As an example, the correlation coefficient between pre-enrollment grades and post-enrollment grades is inherently positive, but since we can observe post-enrollment grades only for accepted students and have no data for those who did not enroll, the correlation coefficient becomes low.
This phenomenon is called the cutting or selection effect.
## rank correlation coefficient
In addition to the correlation coefficient just mentioned (Pearson’s product-moment correlation coefficient), various other correlation coefficients are known. Here we discuss Spearman’s rank correlation coefficient.
The rank correlation coefficient is valid when only the rank of the data is known. For example, if we only know the ranks of the following achievement tests.
The order correlation coefficient is the one that captures the correlation using only the order of such data.
If we denote the order of the observed values $x and y$ as $\tilde{x} and \tilde{y}$, respectively, as shown in the following table,
Spearman’s rank correlation coefficient $\rho_{xy}$ is calculated by:
\begin{align*} \rho_{xy} = 1\ – \f{6}{n(n^2-1)} \sum_{i=1}^n (\tilde{x}^{(i)}\ – \tilde{y}^{(i)})^2. \end{align*}
For more information on why this formula is used, please see the following article.
Derivation of Spearman's rank correlation coefficient and example calculation using python
Correlation coefficients are often used as a method of summarizing relationships between data. There are different types of correlation coefficients, and one is called a rank correlation coefficient, which is used in cases where only the order of the data is known. In this article, I will focus on Spearman's rank correlation coefficient in particular, and describe its derivation and examples of calculations using python.
In python, you can do this:
math = [1, 3, 2, 4, 5, 6]
phys = [1, 4, 2, 5, 3, 6]
corr, pvalue = st.spearmanr(math, phys)
print(corr)
# Outputs
0.8285714285714287
## Description of data over 3 Variables: scatter plot matrix and variance-covariance matrix.
When the data is more than three variables, it becomes difficult to draw a scatter plot using all the variables. So, let’s use a scatter plot matrix that displays two pairs of scatter plots of each variable side by side on a panel.
Let’s draw a scatter plot matrix using the four variables from the iris data set. Python uses the pairplot function of the seaborn library.
df_setosa = df_iris[df_iris['species']=='setosa'] # 品種はSetosaに限定する
sns.pairplot(data=df_setosa)
plt.show()
By looking at the scatter plot matrix in this way, we can capture the relationship between each variable at once.
The correlation coefficients are also grouped together in a matrix format. As a general argument, consider the data of a $m$ variable with a sample size of $n$. In this case, the matrix $\tilde{X}$ is defined below.
\begin{align*} \tilde{X} = \mat{ x_1^{(1)} – \bar{x}_1 & x_2^{(1)} – \bar{x}_2 & \cdots & x_m^{(1)} – \bar{x}_m \\ x_1^{(2)} – \bar{x}_1 & x_2^{(2)} – \bar{x}_2 & \cdots & x_m^{(2)} – \bar{x}_m \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{(n)} – \bar{x}_1 & x_2^{(n)} – \bar{x}_2 & \cdots & x_m^{(n)} – \bar{x}_m }. \end{align*}
Then, the matrix $\Sigma$, called the variance-covariance matrix, is given by.
\begin{align*} \Sigma = \f{1}{n} \tilde{X}^\T \tilde{X}. \end{align*}
Here, the $(i, j)$ component of the variance-covariance matrix, $\sigma_{ij}$, is defined as follows.
\begin{align*} \sigma_{ij} = \f{1}{n} \sum_{k=1}^n (x^{(k)}_i – \bar{x}_i) (x^{(k)}_j – \bar{x}_j) \end{align*}
$\sigma_{ij}$ is the covariance of the $i$ and $j$ variables. In particular, the diagonal component is the variance of the $i$ variable.
Similarly, the target matrix $R$ whose $(i, j)$ component is the correlation coefficient (Pearson’s product-moment correlation coefficient) $r_{ij}$ of the $i$ and $j$ variables is called a correlation matrix.
\begin{align*} R = \mat{ 1 & r_{11} & \cdots & r_{1m} \\ r_{11} & 1 & \cdots & r_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ r_{m1} & r_{m2} & \cdots & 1 }. \end{align*}
In python, the correlation matrix can be computed as follows:
corr_mat = df_setosa.corr()
corr_mat
Correlation matrices are easier to understand using heatmaps.
cmap = sns.diverging_palette(255, 0, as_cmap=True) # カラーパレットの定義
sns.heatmap(corr_mat, annot=True, fmt='1.2f', cmap=cmap, square=True, linewidths=0.5, vmin=-1.0, vmax=1.0)
plt.show()
Next time: ▼Events, probabilities and random variables.
Yukkuri Machine Learning
タイトルとURLをコピーしました | 3,513 | 11,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.586964 |
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# Lecture 12 - Probability and Statistics in the Life...
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Probability and Statistics in the Life Sciences (Spring 2010) AMS 110.02 Lecture 12 (chap7) Donghyung Lee
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Chapter 7 (supplements) Tests About a Population Mean Example 7-3 The board of health of a particular state was called to investigate claims that raw pollutants were being released into the river flowing past a small residential community. By applying financial pressure, the state was able to get the violating company to make major concessions toward the installation of a new water purification system. In the interim, different production systems were to be initiated to help reduce the pollution level of water entering the stream. To monitor the effect of the interim system, a random sample of 50 water specimens was taken throughout the month at a location downstream from the plant. If use the sample data to determine whether the mean dissolved oxygen count of the water (in ppm) is less than 5.2, the average reading at this location over the past year. Conduct the statistical test and state your conclusion using 5.0 .70, x and s = = .05. α =
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Jill Tulane University ‘16, Course Hero Intern | 497 | 2,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-26 | latest | en | 0.944288 |
https://www.it1352.com/2892963.html | 1,652,701,123,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00237.warc.gz | 982,331,974 | 7,024 | # 得到质数的错误答案 [英] Getting wrong answers for prime numbers
### 问题描述
``````for number in range(0, 1000):
for x in range(2, number):
if (number % x == 0):
break
else:
print x
break
``````
### 推荐答案
`for`循环运行后的`else`关键字仅当循环正常退出。因此，只有在没有找到除数的情况下，才会打印"is Prime"部分。
``````for number in range(0,1000):
for x in range(2,number):
if(number % x == 0):
print number,"divisible by",x
break
else:
print number, "is prime"
``````
``````from math import sqrt
for number in range(0,1000):
for x in range(2,int(sqrt(number/2))):
# Rest of code as above.
`````` | 206 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.461507 |
https://oeis.org/A060835 | 1,670,002,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710909.66/warc/CC-MAIN-20221202150823-20221202180823-00344.warc.gz | 461,485,026 | 4,262 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A060835 Least number of the form x^2 + y^3 (x, y positive) in n ways. 3
2, 17, 1737, 1025, 92025, 3375900, 5472225, 35964225, 930860225, 1000837225, 4979585600, 38515961025, 88154795025, 203947076025, 88813460025, 5684061441600, 77806025000000, 64745012358225 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS 10^14 < a(19) <= 4143680790926400. a(20) <= 203947076025000000. a(21) <= 56720592225000000. a(22) <= 6936693633907329600. [From Donovan Johnson, May 30 2010] LINKS FORMULA a(3) = 1737 because 1737 = 3^2 + 12^3 = 35^2 + 8^3 = 39^2 + 6^3. CROSSREFS Sequence in context: A327020 A058233 A062635 * A122054 A284706 A092415 Adjacent sequences: A060832 A060833 A060834 * A060836 A060837 A060838 KEYWORD nonn AUTHOR David W. Wilson, May 02 2001 EXTENSIONS a(14)-a(18) from Donovan Johnson, May 30 2010 STATUS approved
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Last modified December 2 12:32 EST 2022. Contains 358494 sequences. (Running on oeis4.) | 540 | 1,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.70813 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/336/4/a/n/1/1/ | 1,620,285,215,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988741.20/warc/CC-MAIN-20210506053729-20210506083729-00372.warc.gz | 885,793,722 | 81,835 | # Properties
Label 336.4.a.n.1.1 Level $336$ Weight $4$ Character 336.1 Self dual yes Analytic conductor $19.825$ Analytic rank $0$ Dimension $2$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$336 = 2^{4} \cdot 3 \cdot 7$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 336.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$19.8246417619$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{177})$$ Defining polynomial: $$x^{2} - x - 44$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$2$$ Twist minimal: no (minimal twist has level 168) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Root $$7.15207$$ of defining polynomial Character $$\chi$$ $$=$$ 336.1
## $q$-expansion
$$f(q)$$ $$=$$ $$q-3.00000 q^{3} -6.30413 q^{5} -7.00000 q^{7} +9.00000 q^{9} +O(q^{10})$$ $$q-3.00000 q^{3} -6.30413 q^{5} -7.00000 q^{7} +9.00000 q^{9} -48.9124 q^{11} -2.60827 q^{13} +18.9124 q^{15} +136.737 q^{17} -45.2165 q^{19} +21.0000 q^{21} +38.1289 q^{23} -85.2579 q^{25} -27.0000 q^{27} +52.7835 q^{29} +14.7835 q^{31} +146.737 q^{33} +44.1289 q^{35} +333.908 q^{37} +7.82481 q^{39} +227.263 q^{41} +398.433 q^{43} -56.7372 q^{45} +184.608 q^{47} +49.0000 q^{49} -410.212 q^{51} +359.825 q^{53} +308.350 q^{55} +135.650 q^{57} -99.9075 q^{59} -674.516 q^{61} -63.0000 q^{63} +16.4429 q^{65} +376.959 q^{67} -114.387 q^{69} +1187.60 q^{71} -735.825 q^{73} +255.774 q^{75} +342.387 q^{77} +836.774 q^{79} +81.0000 q^{81} -293.732 q^{83} -862.010 q^{85} -158.350 q^{87} +1298.89 q^{89} +18.2579 q^{91} -44.3504 q^{93} +285.051 q^{95} -201.041 q^{97} -440.212 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 6q^{3} + 14q^{5} - 14q^{7} + 18q^{9} + O(q^{10})$$ $$2q - 6q^{3} + 14q^{5} - 14q^{7} + 18q^{9} - 18q^{11} + 48q^{13} - 42q^{15} + 34q^{17} + 16q^{19} + 42q^{21} - 110q^{23} + 202q^{25} - 54q^{27} + 212q^{29} + 136q^{31} + 54q^{33} - 98q^{35} - 24q^{37} - 144q^{39} + 694q^{41} + 584q^{43} + 126q^{45} + 316q^{47} + 98q^{49} - 102q^{51} + 560q^{53} + 936q^{55} - 48q^{57} + 492q^{59} - 604q^{61} - 126q^{63} + 1044q^{65} + 1020q^{67} + 330q^{69} + 1710q^{71} - 1312q^{73} - 606q^{75} + 126q^{77} + 556q^{79} + 162q^{81} + 264q^{83} - 2948q^{85} - 636q^{87} + 70q^{89} - 336q^{91} - 408q^{93} + 1528q^{95} - 136q^{97} - 162q^{99} + O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ −3.00000 −0.577350
$$4$$ 0 0
$$5$$ −6.30413 −0.563859 −0.281929 0.959435i $$-0.590974\pi$$
−0.281929 + 0.959435i $$0.590974\pi$$
$$6$$ 0 0
$$7$$ −7.00000 −0.377964
$$8$$ 0 0
$$9$$ 9.00000 0.333333
$$10$$ 0 0
$$11$$ −48.9124 −1.34069 −0.670347 0.742047i $$-0.733855\pi$$
−0.670347 + 0.742047i $$0.733855\pi$$
$$12$$ 0 0
$$13$$ −2.60827 −0.0556464 −0.0278232 0.999613i $$-0.508858\pi$$
−0.0278232 + 0.999613i $$0.508858\pi$$
$$14$$ 0 0
$$15$$ 18.9124 0.325544
$$16$$ 0 0
$$17$$ 136.737 1.95080 0.975401 0.220436i $$-0.0707482\pi$$
0.975401 + 0.220436i $$0.0707482\pi$$
$$18$$ 0 0
$$19$$ −45.2165 −0.545968 −0.272984 0.962019i $$-0.588011\pi$$
−0.272984 + 0.962019i $$0.588011\pi$$
$$20$$ 0 0
$$21$$ 21.0000 0.218218
$$22$$ 0 0
$$23$$ 38.1289 0.345671 0.172836 0.984951i $$-0.444707\pi$$
0.172836 + 0.984951i $$0.444707\pi$$
$$24$$ 0 0
$$25$$ −85.2579 −0.682063
$$26$$ 0 0
$$27$$ −27.0000 −0.192450
$$28$$ 0 0
$$29$$ 52.7835 0.337988 0.168994 0.985617i $$-0.445948\pi$$
0.168994 + 0.985617i $$0.445948\pi$$
$$30$$ 0 0
$$31$$ 14.7835 0.0856512 0.0428256 0.999083i $$-0.486364\pi$$
0.0428256 + 0.999083i $$0.486364\pi$$
$$32$$ 0 0
$$33$$ 146.737 0.774051
$$34$$ 0 0
$$35$$ 44.1289 0.213119
$$36$$ 0 0
$$37$$ 333.908 1.48362 0.741812 0.670608i $$-0.233967\pi$$
0.741812 + 0.670608i $$0.233967\pi$$
$$38$$ 0 0
$$39$$ 7.82481 0.0321275
$$40$$ 0 0
$$41$$ 227.263 0.865670 0.432835 0.901473i $$-0.357513\pi$$
0.432835 + 0.901473i $$0.357513\pi$$
$$42$$ 0 0
$$43$$ 398.433 1.41303 0.706517 0.707696i $$-0.250265\pi$$
0.706517 + 0.707696i $$0.250265\pi$$
$$44$$ 0 0
$$45$$ −56.7372 −0.187953
$$46$$ 0 0
$$47$$ 184.608 0.572934 0.286467 0.958090i $$-0.407519\pi$$
0.286467 + 0.958090i $$0.407519\pi$$
$$48$$ 0 0
$$49$$ 49.0000 0.142857
$$50$$ 0 0
$$51$$ −410.212 −1.12630
$$52$$ 0 0
$$53$$ 359.825 0.932561 0.466281 0.884637i $$-0.345594\pi$$
0.466281 + 0.884637i $$0.345594\pi$$
$$54$$ 0 0
$$55$$ 308.350 0.755963
$$56$$ 0 0
$$57$$ 135.650 0.315215
$$58$$ 0 0
$$59$$ −99.9075 −0.220455 −0.110228 0.993906i $$-0.535158\pi$$
−0.110228 + 0.993906i $$0.535158\pi$$
$$60$$ 0 0
$$61$$ −674.516 −1.41579 −0.707893 0.706320i $$-0.750354\pi$$
−0.707893 + 0.706320i $$0.750354\pi$$
$$62$$ 0 0
$$63$$ −63.0000 −0.125988
$$64$$ 0 0
$$65$$ 16.4429 0.0313767
$$66$$ 0 0
$$67$$ 376.959 0.687356 0.343678 0.939088i $$-0.388327\pi$$
0.343678 + 0.939088i $$0.388327\pi$$
$$68$$ 0 0
$$69$$ −114.387 −0.199573
$$70$$ 0 0
$$71$$ 1187.60 1.98511 0.992553 0.121810i $$-0.0388698\pi$$
0.992553 + 0.121810i $$0.0388698\pi$$
$$72$$ 0 0
$$73$$ −735.825 −1.17975 −0.589875 0.807494i $$-0.700823\pi$$
−0.589875 + 0.807494i $$0.700823\pi$$
$$74$$ 0 0
$$75$$ 255.774 0.393789
$$76$$ 0 0
$$77$$ 342.387 0.506735
$$78$$ 0 0
$$79$$ 836.774 1.19170 0.595851 0.803095i $$-0.296815\pi$$
0.595851 + 0.803095i $$0.296815\pi$$
$$80$$ 0 0
$$81$$ 81.0000 0.111111
$$82$$ 0 0
$$83$$ −293.732 −0.388450 −0.194225 0.980957i $$-0.562219\pi$$
−0.194225 + 0.980957i $$0.562219\pi$$
$$84$$ 0 0
$$85$$ −862.010 −1.09998
$$86$$ 0 0
$$87$$ −158.350 −0.195137
$$88$$ 0 0
$$89$$ 1298.89 1.54699 0.773496 0.633801i $$-0.218506\pi$$
0.773496 + 0.633801i $$0.218506\pi$$
$$90$$ 0 0
$$91$$ 18.2579 0.0210324
$$92$$ 0 0
$$93$$ −44.3504 −0.0494508
$$94$$ 0 0
$$95$$ 285.051 0.307849
$$96$$ 0 0
$$97$$ −201.041 −0.210440 −0.105220 0.994449i $$-0.533555\pi$$
−0.105220 + 0.994449i $$0.533555\pi$$
$$98$$ 0 0
$$99$$ −440.212 −0.446898
$$100$$ 0 0
$$101$$ −1053.51 −1.03790 −0.518952 0.854804i $$-0.673678\pi$$
−0.518952 + 0.854804i $$0.673678\pi$$
$$102$$ 0 0
$$103$$ −1025.73 −0.981247 −0.490623 0.871372i $$-0.663231\pi$$
−0.490623 + 0.871372i $$0.663231\pi$$
$$104$$ 0 0
$$105$$ −132.387 −0.123044
$$106$$ 0 0
$$107$$ 103.418 0.0934377 0.0467188 0.998908i $$-0.485124\pi$$
0.0467188 + 0.998908i $$0.485124\pi$$
$$108$$ 0 0
$$109$$ −677.134 −0.595024 −0.297512 0.954718i $$-0.596157\pi$$
−0.297512 + 0.954718i $$0.596157\pi$$
$$110$$ 0 0
$$111$$ −1001.72 −0.856570
$$112$$ 0 0
$$113$$ −452.083 −0.376357 −0.188179 0.982135i $$-0.560258\pi$$
−0.188179 + 0.982135i $$0.560258\pi$$
$$114$$ 0 0
$$115$$ −240.370 −0.194910
$$116$$ 0 0
$$117$$ −23.4744 −0.0185488
$$118$$ 0 0
$$119$$ −957.160 −0.737334
$$120$$ 0 0
$$121$$ 1061.42 0.797463
$$122$$ 0 0
$$123$$ −681.788 −0.499795
$$124$$ 0 0
$$125$$ 1325.49 0.948446
$$126$$ 0 0
$$127$$ 1182.88 0.826482 0.413241 0.910622i $$-0.364397\pi$$
0.413241 + 0.910622i $$0.364397\pi$$
$$128$$ 0 0
$$129$$ −1195.30 −0.815816
$$130$$ 0 0
$$131$$ −1257.75 −0.838857 −0.419429 0.907788i $$-0.637770\pi$$
−0.419429 + 0.907788i $$0.637770\pi$$
$$132$$ 0 0
$$133$$ 316.516 0.206356
$$134$$ 0 0
$$135$$ 170.212 0.108515
$$136$$ 0 0
$$137$$ −56.5256 −0.0352504 −0.0176252 0.999845i $$-0.505611\pi$$
−0.0176252 + 0.999845i $$0.505611\pi$$
$$138$$ 0 0
$$139$$ −2113.57 −1.28972 −0.644858 0.764303i $$-0.723083\pi$$
−0.644858 + 0.764303i $$0.723083\pi$$
$$140$$ 0 0
$$141$$ −553.825 −0.330783
$$142$$ 0 0
$$143$$ 127.577 0.0746049
$$144$$ 0 0
$$145$$ −332.754 −0.190577
$$146$$ 0 0
$$147$$ −147.000 −0.0824786
$$148$$ 0 0
$$149$$ 1794.96 0.986904 0.493452 0.869773i $$-0.335735\pi$$
0.493452 + 0.869773i $$0.335735\pi$$
$$150$$ 0 0
$$151$$ −377.032 −0.203195 −0.101597 0.994826i $$-0.532395\pi$$
−0.101597 + 0.994826i $$0.532395\pi$$
$$152$$ 0 0
$$153$$ 1230.63 0.650268
$$154$$ 0 0
$$155$$ −93.1969 −0.0482952
$$156$$ 0 0
$$157$$ −898.701 −0.456842 −0.228421 0.973562i $$-0.573356\pi$$
−0.228421 + 0.973562i $$0.573356\pi$$
$$158$$ 0 0
$$159$$ −1079.47 −0.538414
$$160$$ 0 0
$$161$$ −266.903 −0.130651
$$162$$ 0 0
$$163$$ −3863.52 −1.85653 −0.928264 0.371922i $$-0.878699\pi$$
−0.928264 + 0.371922i $$0.878699\pi$$
$$164$$ 0 0
$$165$$ −925.051 −0.436455
$$166$$ 0 0
$$167$$ 2861.44 1.32590 0.662948 0.748666i $$-0.269305\pi$$
0.662948 + 0.748666i $$0.269305\pi$$
$$168$$ 0 0
$$169$$ −2190.20 −0.996903
$$170$$ 0 0
$$171$$ −406.949 −0.181989
$$172$$ 0 0
$$173$$ 979.005 0.430245 0.215122 0.976587i $$-0.430985\pi$$
0.215122 + 0.976587i $$0.430985\pi$$
$$174$$ 0 0
$$175$$ 596.805 0.257796
$$176$$ 0 0
$$177$$ 299.723 0.127280
$$178$$ 0 0
$$179$$ 1146.27 0.478640 0.239320 0.970941i $$-0.423075\pi$$
0.239320 + 0.970941i $$0.423075\pi$$
$$180$$ 0 0
$$181$$ 3929.33 1.61362 0.806809 0.590812i $$-0.201192\pi$$
0.806809 + 0.590812i $$0.201192\pi$$
$$182$$ 0 0
$$183$$ 2023.55 0.817404
$$184$$ 0 0
$$185$$ −2105.00 −0.836554
$$186$$ 0 0
$$187$$ −6688.15 −2.61543
$$188$$ 0 0
$$189$$ 189.000 0.0727393
$$190$$ 0 0
$$191$$ 2937.36 1.11277 0.556386 0.830924i $$-0.312187\pi$$
0.556386 + 0.830924i $$0.312187\pi$$
$$192$$ 0 0
$$193$$ −3533.17 −1.31774 −0.658868 0.752259i $$-0.728964\pi$$
−0.658868 + 0.752259i $$0.728964\pi$$
$$194$$ 0 0
$$195$$ −49.3286 −0.0181154
$$196$$ 0 0
$$197$$ 584.856 0.211519 0.105760 0.994392i $$-0.466273\pi$$
0.105760 + 0.994392i $$0.466273\pi$$
$$198$$ 0 0
$$199$$ −2158.08 −0.768756 −0.384378 0.923176i $$-0.625584\pi$$
−0.384378 + 0.923176i $$0.625584\pi$$
$$200$$ 0 0
$$201$$ −1130.88 −0.396845
$$202$$ 0 0
$$203$$ −369.484 −0.127747
$$204$$ 0 0
$$205$$ −1432.70 −0.488116
$$206$$ 0 0
$$207$$ 343.160 0.115224
$$208$$ 0 0
$$209$$ 2211.65 0.731976
$$210$$ 0 0
$$211$$ 4290.04 1.39971 0.699855 0.714285i $$-0.253248\pi$$
0.699855 + 0.714285i $$0.253248\pi$$
$$212$$ 0 0
$$213$$ −3562.81 −1.14610
$$214$$ 0 0
$$215$$ −2511.78 −0.796752
$$216$$ 0 0
$$217$$ −103.484 −0.0323731
$$218$$ 0 0
$$219$$ 2207.47 0.681129
$$220$$ 0 0
$$221$$ −356.647 −0.108555
$$222$$ 0 0
$$223$$ 2743.78 0.823932 0.411966 0.911199i $$-0.364842\pi$$
0.411966 + 0.911199i $$0.364842\pi$$
$$224$$ 0 0
$$225$$ −767.321 −0.227354
$$226$$ 0 0
$$227$$ 1724.79 0.504311 0.252155 0.967687i $$-0.418861\pi$$
0.252155 + 0.967687i $$0.418861\pi$$
$$228$$ 0 0
$$229$$ 4201.70 1.21247 0.606237 0.795284i $$-0.292678\pi$$
0.606237 + 0.795284i $$0.292678\pi$$
$$230$$ 0 0
$$231$$ −1027.16 −0.292564
$$232$$ 0 0
$$233$$ 1274.94 0.358472 0.179236 0.983806i $$-0.442637\pi$$
0.179236 + 0.983806i $$0.442637\pi$$
$$234$$ 0 0
$$235$$ −1163.80 −0.323054
$$236$$ 0 0
$$237$$ −2510.32 −0.688029
$$238$$ 0 0
$$239$$ 5967.16 1.61499 0.807497 0.589872i $$-0.200822\pi$$
0.807497 + 0.589872i $$0.200822\pi$$
$$240$$ 0 0
$$241$$ 4881.64 1.30479 0.652395 0.757879i $$-0.273764\pi$$
0.652395 + 0.757879i $$0.273764\pi$$
$$242$$ 0 0
$$243$$ −243.000 −0.0641500
$$244$$ 0 0
$$245$$ −308.903 −0.0805513
$$246$$ 0 0
$$247$$ 117.937 0.0303812
$$248$$ 0 0
$$249$$ 881.197 0.224271
$$250$$ 0 0
$$251$$ −2262.63 −0.568988 −0.284494 0.958678i $$-0.591826\pi$$
−0.284494 + 0.958678i $$0.591826\pi$$
$$252$$ 0 0
$$253$$ −1864.98 −0.463439
$$254$$ 0 0
$$255$$ 2586.03 0.635072
$$256$$ 0 0
$$257$$ −6210.60 −1.50742 −0.753709 0.657209i $$-0.771737\pi$$
−0.753709 + 0.657209i $$0.771737\pi$$
$$258$$ 0 0
$$259$$ −2337.35 −0.560757
$$260$$ 0 0
$$261$$ 475.051 0.112663
$$262$$ 0 0
$$263$$ 2972.69 0.696973 0.348486 0.937314i $$-0.386696\pi$$
0.348486 + 0.937314i $$0.386696\pi$$
$$264$$ 0 0
$$265$$ −2268.38 −0.525833
$$266$$ 0 0
$$267$$ −3896.68 −0.893157
$$268$$ 0 0
$$269$$ −4443.42 −1.00714 −0.503569 0.863955i $$-0.667980\pi$$
−0.503569 + 0.863955i $$0.667980\pi$$
$$270$$ 0 0
$$271$$ 6840.25 1.53327 0.766634 0.642084i $$-0.221930\pi$$
0.766634 + 0.642084i $$0.221930\pi$$
$$272$$ 0 0
$$273$$ −54.7737 −0.0121430
$$274$$ 0 0
$$275$$ 4170.17 0.914439
$$276$$ 0 0
$$277$$ −3228.67 −0.700332 −0.350166 0.936688i $$-0.613875\pi$$
−0.350166 + 0.936688i $$0.613875\pi$$
$$278$$ 0 0
$$279$$ 133.051 0.0285504
$$280$$ 0 0
$$281$$ −6453.83 −1.37012 −0.685059 0.728488i $$-0.740224\pi$$
−0.685059 + 0.728488i $$0.740224\pi$$
$$282$$ 0 0
$$283$$ −3840.72 −0.806739 −0.403369 0.915037i $$-0.632161\pi$$
−0.403369 + 0.915037i $$0.632161\pi$$
$$284$$ 0 0
$$285$$ −855.153 −0.177737
$$286$$ 0 0
$$287$$ −1590.84 −0.327193
$$288$$ 0 0
$$289$$ 13784.1 2.80563
$$290$$ 0 0
$$291$$ 603.124 0.121497
$$292$$ 0 0
$$293$$ 8801.01 1.75481 0.877407 0.479747i $$-0.159272\pi$$
0.877407 + 0.479747i $$0.159272\pi$$
$$294$$ 0 0
$$295$$ 629.830 0.124306
$$296$$ 0 0
$$297$$ 1320.63 0.258017
$$298$$ 0 0
$$299$$ −99.4506 −0.0192354
$$300$$ 0 0
$$301$$ −2789.03 −0.534077
$$302$$ 0 0
$$303$$ 3160.53 0.599234
$$304$$ 0 0
$$305$$ 4252.24 0.798303
$$306$$ 0 0
$$307$$ 3926.72 0.730000 0.365000 0.931008i $$-0.381069\pi$$
0.365000 + 0.931008i $$0.381069\pi$$
$$308$$ 0 0
$$309$$ 3077.20 0.566523
$$310$$ 0 0
$$311$$ 6143.13 1.12008 0.560040 0.828466i $$-0.310786\pi$$
0.560040 + 0.828466i $$0.310786\pi$$
$$312$$ 0 0
$$313$$ 3824.19 0.690594 0.345297 0.938493i $$-0.387778\pi$$
0.345297 + 0.938493i $$0.387778\pi$$
$$314$$ 0 0
$$315$$ 397.160 0.0710396
$$316$$ 0 0
$$317$$ 7949.68 1.40851 0.704257 0.709946i $$-0.251280\pi$$
0.704257 + 0.709946i $$0.251280\pi$$
$$318$$ 0 0
$$319$$ −2581.77 −0.453138
$$320$$ 0 0
$$321$$ −310.255 −0.0539463
$$322$$ 0 0
$$323$$ −6182.78 −1.06508
$$324$$ 0 0
$$325$$ 222.376 0.0379544
$$326$$ 0 0
$$327$$ 2031.40 0.343537
$$328$$ 0 0
$$329$$ −1292.26 −0.216549
$$330$$ 0 0
$$331$$ 11236.4 1.86588 0.932940 0.360032i $$-0.117234\pi$$
0.932940 + 0.360032i $$0.117234\pi$$
$$332$$ 0 0
$$333$$ 3005.17 0.494541
$$334$$ 0 0
$$335$$ −2376.40 −0.387572
$$336$$ 0 0
$$337$$ 8425.41 1.36190 0.680951 0.732329i $$-0.261567\pi$$
0.680951 + 0.732329i $$0.261567\pi$$
$$338$$ 0 0
$$339$$ 1356.25 0.217290
$$340$$ 0 0
$$341$$ −723.095 −0.114832
$$342$$ 0 0
$$343$$ −343.000 −0.0539949
$$344$$ 0 0
$$345$$ 721.110 0.112531
$$346$$ 0 0
$$347$$ −7177.15 −1.11034 −0.555172 0.831735i $$-0.687348\pi$$
−0.555172 + 0.831735i $$0.687348\pi$$
$$348$$ 0 0
$$349$$ 1549.41 0.237644 0.118822 0.992916i $$-0.462088\pi$$
0.118822 + 0.992916i $$0.462088\pi$$
$$350$$ 0 0
$$351$$ 70.4233 0.0107092
$$352$$ 0 0
$$353$$ −566.231 −0.0853752 −0.0426876 0.999088i $$-0.513592\pi$$
−0.0426876 + 0.999088i $$0.513592\pi$$
$$354$$ 0 0
$$355$$ −7486.81 −1.11932
$$356$$ 0 0
$$357$$ 2871.48 0.425700
$$358$$ 0 0
$$359$$ −3848.19 −0.565738 −0.282869 0.959159i $$-0.591286\pi$$
−0.282869 + 0.959159i $$0.591286\pi$$
$$360$$ 0 0
$$361$$ −4814.46 −0.701919
$$362$$ 0 0
$$363$$ −3184.27 −0.460415
$$364$$ 0 0
$$365$$ 4638.74 0.665213
$$366$$ 0 0
$$367$$ −12542.2 −1.78392 −0.891960 0.452113i $$-0.850670\pi$$
−0.891960 + 0.452113i $$0.850670\pi$$
$$368$$ 0 0
$$369$$ 2045.37 0.288557
$$370$$ 0 0
$$371$$ −2518.77 −0.352475
$$372$$ 0 0
$$373$$ 6345.87 0.880903 0.440452 0.897776i $$-0.354818\pi$$
0.440452 + 0.897776i $$0.354818\pi$$
$$374$$ 0 0
$$375$$ −3976.48 −0.547586
$$376$$ 0 0
$$377$$ −137.673 −0.0188078
$$378$$ 0 0
$$379$$ 12200.2 1.65351 0.826757 0.562559i $$-0.190183\pi$$
0.826757 + 0.562559i $$0.190183\pi$$
$$380$$ 0 0
$$381$$ −3548.63 −0.477170
$$382$$ 0 0
$$383$$ 2770.91 0.369679 0.184839 0.982769i $$-0.440824\pi$$
0.184839 + 0.982769i $$0.440824\pi$$
$$384$$ 0 0
$$385$$ −2158.45 −0.285727
$$386$$ 0 0
$$387$$ 3585.90 0.471011
$$388$$ 0 0
$$389$$ 1581.89 0.206183 0.103091 0.994672i $$-0.467127\pi$$
0.103091 + 0.994672i $$0.467127\pi$$
$$390$$ 0 0
$$391$$ 5213.65 0.674336
$$392$$ 0 0
$$393$$ 3773.26 0.484314
$$394$$ 0 0
$$395$$ −5275.13 −0.671951
$$396$$ 0 0
$$397$$ −14235.9 −1.79970 −0.899848 0.436203i $$-0.856323\pi$$
−0.899848 + 0.436203i $$0.856323\pi$$
$$398$$ 0 0
$$399$$ −949.547 −0.119140
$$400$$ 0 0
$$401$$ −9556.18 −1.19006 −0.595028 0.803705i $$-0.702859\pi$$
−0.595028 + 0.803705i $$0.702859\pi$$
$$402$$ 0 0
$$403$$ −38.5592 −0.00476619
$$404$$ 0 0
$$405$$ −510.635 −0.0626510
$$406$$ 0 0
$$407$$ −16332.2 −1.98909
$$408$$ 0 0
$$409$$ 2858.17 0.345544 0.172772 0.984962i $$-0.444728\pi$$
0.172772 + 0.984962i $$0.444728\pi$$
$$410$$ 0 0
$$411$$ 169.577 0.0203518
$$412$$ 0 0
$$413$$ 699.353 0.0833242
$$414$$ 0 0
$$415$$ 1851.73 0.219031
$$416$$ 0 0
$$417$$ 6340.70 0.744617
$$418$$ 0 0
$$419$$ 13333.3 1.55460 0.777299 0.629132i $$-0.216589\pi$$
0.777299 + 0.629132i $$0.216589\pi$$
$$420$$ 0 0
$$421$$ −13567.4 −1.57063 −0.785314 0.619098i $$-0.787499\pi$$
−0.785314 + 0.619098i $$0.787499\pi$$
$$422$$ 0 0
$$423$$ 1661.47 0.190978
$$424$$ 0 0
$$425$$ −11657.9 −1.33057
$$426$$ 0 0
$$427$$ 4721.61 0.535116
$$428$$ 0 0
$$429$$ −382.730 −0.0430732
$$430$$ 0 0
$$431$$ −14207.3 −1.58780 −0.793898 0.608051i $$-0.791952\pi$$
−0.793898 + 0.608051i $$0.791952\pi$$
$$432$$ 0 0
$$433$$ −10530.6 −1.16875 −0.584375 0.811484i $$-0.698660\pi$$
−0.584375 + 0.811484i $$0.698660\pi$$
$$434$$ 0 0
$$435$$ 998.262 0.110030
$$436$$ 0 0
$$437$$ −1724.06 −0.188725
$$438$$ 0 0
$$439$$ 4038.23 0.439030 0.219515 0.975609i $$-0.429553\pi$$
0.219515 + 0.975609i $$0.429553\pi$$
$$440$$ 0 0
$$441$$ 441.000 0.0476190
$$442$$ 0 0
$$443$$ 4574.15 0.490574 0.245287 0.969450i $$-0.421118\pi$$
0.245287 + 0.969450i $$0.421118\pi$$
$$444$$ 0 0
$$445$$ −8188.40 −0.872286
$$446$$ 0 0
$$447$$ −5384.88 −0.569789
$$448$$ 0 0
$$449$$ 14957.1 1.57209 0.786044 0.618171i $$-0.212126\pi$$
0.786044 + 0.618171i $$0.212126\pi$$
$$450$$ 0 0
$$451$$ −11116.0 −1.16060
$$452$$ 0 0
$$453$$ 1131.09 0.117314
$$454$$ 0 0
$$455$$ −115.100 −0.0118593
$$456$$ 0 0
$$457$$ 3027.65 0.309907 0.154954 0.987922i $$-0.450477\pi$$
0.154954 + 0.987922i $$0.450477\pi$$
$$458$$ 0 0
$$459$$ −3691.90 −0.375432
$$460$$ 0 0
$$461$$ −10877.7 −1.09897 −0.549484 0.835504i $$-0.685176\pi$$
−0.549484 + 0.835504i $$0.685176\pi$$
$$462$$ 0 0
$$463$$ 4038.28 0.405345 0.202673 0.979247i $$-0.435037\pi$$
0.202673 + 0.979247i $$0.435037\pi$$
$$464$$ 0 0
$$465$$ 279.591 0.0278833
$$466$$ 0 0
$$467$$ −8411.80 −0.833515 −0.416758 0.909018i $$-0.636834\pi$$
−0.416758 + 0.909018i $$0.636834\pi$$
$$468$$ 0 0
$$469$$ −2638.71 −0.259796
$$470$$ 0 0
$$471$$ 2696.10 0.263758
$$472$$ 0 0
$$473$$ −19488.3 −1.89445
$$474$$ 0 0
$$475$$ 3855.07 0.372384
$$476$$ 0 0
$$477$$ 3238.42 0.310854
$$478$$ 0 0
$$479$$ −7172.70 −0.684194 −0.342097 0.939665i $$-0.611137\pi$$
−0.342097 + 0.939665i $$0.611137\pi$$
$$480$$ 0 0
$$481$$ −870.921 −0.0825584
$$482$$ 0 0
$$483$$ 800.708 0.0754316
$$484$$ 0 0
$$485$$ 1267.39 0.118658
$$486$$ 0 0
$$487$$ 5580.52 0.519256 0.259628 0.965709i $$-0.416400\pi$$
0.259628 + 0.965709i $$0.416400\pi$$
$$488$$ 0 0
$$489$$ 11590.6 1.07187
$$490$$ 0 0
$$491$$ −12489.4 −1.14794 −0.573972 0.818875i $$-0.694598\pi$$
−0.573972 + 0.818875i $$0.694598\pi$$
$$492$$ 0 0
$$493$$ 7217.46 0.659347
$$494$$ 0 0
$$495$$ 2775.15 0.251988
$$496$$ 0 0
$$497$$ −8313.22 −0.750300
$$498$$ 0 0
$$499$$ 15216.6 1.36511 0.682556 0.730834i $$-0.260869\pi$$
0.682556 + 0.730834i $$0.260869\pi$$
$$500$$ 0 0
$$501$$ −8584.31 −0.765506
$$502$$ 0 0
$$503$$ 1814.89 0.160879 0.0804393 0.996760i $$-0.474368\pi$$
0.0804393 + 0.996760i $$0.474368\pi$$
$$504$$ 0 0
$$505$$ 6641.47 0.585231
$$506$$ 0 0
$$507$$ 6570.59 0.575562
$$508$$ 0 0
$$509$$ −4853.68 −0.422663 −0.211332 0.977414i $$-0.567780\pi$$
−0.211332 + 0.977414i $$0.567780\pi$$
$$510$$ 0 0
$$511$$ 5150.77 0.445904
$$512$$ 0 0
$$513$$ 1220.85 0.105072
$$514$$ 0 0
$$515$$ 6466.35 0.553285
$$516$$ 0 0
$$517$$ −9029.63 −0.768129
$$518$$ 0 0
$$519$$ −2937.01 −0.248402
$$520$$ 0 0
$$521$$ −9913.18 −0.833598 −0.416799 0.908999i $$-0.636848\pi$$
−0.416799 + 0.908999i $$0.636848\pi$$
$$522$$ 0 0
$$523$$ 4524.29 0.378267 0.189133 0.981951i $$-0.439432\pi$$
0.189133 + 0.981951i $$0.439432\pi$$
$$524$$ 0 0
$$525$$ −1790.42 −0.148838
$$526$$ 0 0
$$527$$ 2021.45 0.167089
$$528$$ 0 0
$$529$$ −10713.2 −0.880512
$$530$$ 0 0
$$531$$ −899.168 −0.0734850
$$532$$ 0 0
$$533$$ −592.763 −0.0481715
$$534$$ 0 0
$$535$$ −651.963 −0.0526857
$$536$$ 0 0
$$537$$ −3438.82 −0.276343
$$538$$ 0 0
$$539$$ −2396.71 −0.191528
$$540$$ 0 0
$$541$$ 3724.94 0.296022 0.148011 0.988986i $$-0.452713\pi$$
0.148011 + 0.988986i $$0.452713\pi$$
$$542$$ 0 0
$$543$$ −11788.0 −0.931623
$$544$$ 0 0
$$545$$ 4268.74 0.335510
$$546$$ 0 0
$$547$$ −16121.6 −1.26016 −0.630082 0.776528i $$-0.716979\pi$$
−0.630082 + 0.776528i $$0.716979\pi$$
$$548$$ 0 0
$$549$$ −6070.64 −0.471928
$$550$$ 0 0
$$551$$ −2386.69 −0.184530
$$552$$ 0 0
$$553$$ −5857.42 −0.450421
$$554$$ 0 0
$$555$$ 6314.99 0.482985
$$556$$ 0 0
$$557$$ 20451.7 1.55577 0.777887 0.628405i $$-0.216292\pi$$
0.777887 + 0.628405i $$0.216292\pi$$
$$558$$ 0 0
$$559$$ −1039.22 −0.0786303
$$560$$ 0 0
$$561$$ 20064.4 1.51002
$$562$$ 0 0
$$563$$ 10046.7 0.752078 0.376039 0.926604i $$-0.377286\pi$$
0.376039 + 0.926604i $$0.377286\pi$$
$$564$$ 0 0
$$565$$ 2849.99 0.212212
$$566$$ 0 0
$$567$$ −567.000 −0.0419961
$$568$$ 0 0
$$569$$ −15356.4 −1.13141 −0.565705 0.824608i $$-0.691396\pi$$
−0.565705 + 0.824608i $$0.691396\pi$$
$$570$$ 0 0
$$571$$ 19333.6 1.41696 0.708480 0.705731i $$-0.249381\pi$$
0.708480 + 0.705731i $$0.249381\pi$$
$$572$$ 0 0
$$573$$ −8812.07 −0.642460
$$574$$ 0 0
$$575$$ −3250.79 −0.235769
$$576$$ 0 0
$$577$$ 26258.8 1.89458 0.947288 0.320384i $$-0.103812\pi$$
0.947288 + 0.320384i $$0.103812\pi$$
$$578$$ 0 0
$$579$$ 10599.5 0.760795
$$580$$ 0 0
$$581$$ 2056.13 0.146820
$$582$$ 0 0
$$583$$ −17599.9 −1.25028
$$584$$ 0 0
$$585$$ 147.986 0.0104589
$$586$$ 0 0
$$587$$ 4868.98 0.342358 0.171179 0.985240i $$-0.445242\pi$$
0.171179 + 0.985240i $$0.445242\pi$$
$$588$$ 0 0
$$589$$ −668.457 −0.0467628
$$590$$ 0 0
$$591$$ −1754.57 −0.122121
$$592$$ 0 0
$$593$$ −13647.1 −0.945055 −0.472528 0.881316i $$-0.656658\pi$$
−0.472528 + 0.881316i $$0.656658\pi$$
$$594$$ 0 0
$$595$$ 6034.07 0.415752
$$596$$ 0 0
$$597$$ 6474.25 0.443841
$$598$$ 0 0
$$599$$ −7543.11 −0.514529 −0.257265 0.966341i $$-0.582821\pi$$
−0.257265 + 0.966341i $$0.582821\pi$$
$$600$$ 0 0
$$601$$ −19522.4 −1.32501 −0.662507 0.749056i $$-0.730507\pi$$
−0.662507 + 0.749056i $$0.730507\pi$$
$$602$$ 0 0
$$603$$ 3392.63 0.229119
$$604$$ 0 0
$$605$$ −6691.36 −0.449657
$$606$$ 0 0
$$607$$ −13804.5 −0.923079 −0.461539 0.887120i $$-0.652703\pi$$
−0.461539 + 0.887120i $$0.652703\pi$$
$$608$$ 0 0
$$609$$ 1108.45 0.0737550
$$610$$ 0 0
$$611$$ −481.508 −0.0318817
$$612$$ 0 0
$$613$$ 21718.8 1.43102 0.715508 0.698605i $$-0.246195\pi$$
0.715508 + 0.698605i $$0.246195\pi$$
$$614$$ 0 0
$$615$$ 4298.09 0.281814
$$616$$ 0 0
$$617$$ −5183.85 −0.338240 −0.169120 0.985595i $$-0.554093\pi$$
−0.169120 + 0.985595i $$0.554093\pi$$
$$618$$ 0 0
$$619$$ −22003.7 −1.42876 −0.714382 0.699756i $$-0.753292\pi$$
−0.714382 + 0.699756i $$0.753292\pi$$
$$620$$ 0 0
$$621$$ −1029.48 −0.0665244
$$622$$ 0 0
$$623$$ −9092.25 −0.584708
$$624$$ 0 0
$$625$$ 2301.14 0.147273
$$626$$ 0 0
$$627$$ −6634.95 −0.422607
$$628$$ 0 0
$$629$$ 45657.6 2.89426
$$630$$ 0 0
$$631$$ 985.836 0.0621957 0.0310979 0.999516i $$-0.490100\pi$$
0.0310979 + 0.999516i $$0.490100\pi$$
$$632$$ 0 0
$$633$$ −12870.1 −0.808123
$$634$$ 0 0
$$635$$ −7457.01 −0.466020
$$636$$ 0 0
$$637$$ −127.805 −0.00794949
$$638$$ 0 0
$$639$$ 10688.4 0.661702
$$640$$ 0 0
$$641$$ 24282.6 1.49626 0.748131 0.663551i $$-0.230951\pi$$
0.748131 + 0.663551i $$0.230951\pi$$
$$642$$ 0 0
$$643$$ 4743.12 0.290903 0.145451 0.989365i $$-0.453537\pi$$
0.145451 + 0.989365i $$0.453537\pi$$
$$644$$ 0 0
$$645$$ 7535.33 0.460005
$$646$$ 0 0
$$647$$ −29641.1 −1.80110 −0.900549 0.434754i $$-0.856835\pi$$
−0.900549 + 0.434754i $$0.856835\pi$$
$$648$$ 0 0
$$649$$ 4886.72 0.295563
$$650$$ 0 0
$$651$$ 310.453 0.0186906
$$652$$ 0 0
$$653$$ −23046.9 −1.38116 −0.690578 0.723258i $$-0.742644\pi$$
−0.690578 + 0.723258i $$0.742644\pi$$
$$654$$ 0 0
$$655$$ 7929.04 0.472997
$$656$$ 0 0
$$657$$ −6622.42 −0.393250
$$658$$ 0 0
$$659$$ −5795.12 −0.342558 −0.171279 0.985223i $$-0.554790\pi$$
−0.171279 + 0.985223i $$0.554790\pi$$
$$660$$ 0 0
$$661$$ 2592.59 0.152557 0.0762784 0.997087i $$-0.475696\pi$$
0.0762784 + 0.997087i $$0.475696\pi$$
$$662$$ 0 0
$$663$$ 1069.94 0.0626744
$$664$$ 0 0
$$665$$ −1995.36 −0.116356
$$666$$ 0 0
$$667$$ 2012.58 0.116833
$$668$$ 0 0
$$669$$ −8231.33 −0.475697
$$670$$ 0 0
$$671$$ 32992.2 1.89814
$$672$$ 0 0
$$673$$ 28156.0 1.61268 0.806341 0.591451i $$-0.201445\pi$$
0.806341 + 0.591451i $$0.201445\pi$$
$$674$$ 0 0
$$675$$ 2301.96 0.131263
$$676$$ 0 0
$$677$$ 20271.4 1.15080 0.575402 0.817871i $$-0.304846\pi$$
0.575402 + 0.817871i $$0.304846\pi$$
$$678$$ 0 0
$$679$$ 1407.29 0.0795388
$$680$$ 0 0
$$681$$ −5174.38 −0.291164
$$682$$ 0 0
$$683$$ 13267.0 0.743264 0.371632 0.928380i $$-0.378798\pi$$
0.371632 + 0.928380i $$0.378798\pi$$
$$684$$ 0 0
$$685$$ 356.345 0.0198763
$$686$$ 0 0
$$687$$ −12605.1 −0.700022
$$688$$ 0 0
$$689$$ −938.520 −0.0518937
$$690$$ 0 0
$$691$$ 15966.3 0.878995 0.439497 0.898244i $$-0.355157\pi$$
0.439497 + 0.898244i $$0.355157\pi$$
$$692$$ 0 0
$$693$$ 3081.48 0.168912
$$694$$ 0 0
$$695$$ 13324.2 0.727217
$$696$$ 0 0
$$697$$ 31075.3 1.68875
$$698$$ 0 0
$$699$$ −3824.82 −0.206964
$$700$$ 0 0
$$701$$ 7525.29 0.405458 0.202729 0.979235i $$-0.435019\pi$$
0.202729 + 0.979235i $$0.435019\pi$$
$$702$$ 0 0
$$703$$ −15098.1 −0.810010
$$704$$ 0 0
$$705$$ 3491.39 0.186515
$$706$$ 0 0
$$707$$ 7374.58 0.392291
$$708$$ 0 0
$$709$$ 20033.6 1.06118 0.530591 0.847628i $$-0.321970\pi$$
0.530591 + 0.847628i $$0.321970\pi$$
$$710$$ 0 0
$$711$$ 7530.96 0.397234
$$712$$ 0 0
$$713$$ 563.678 0.0296071
$$714$$ 0 0
$$715$$ −804.261 −0.0420666
$$716$$ 0 0
$$717$$ −17901.5 −0.932417
$$718$$ 0 0
$$719$$ 8081.69 0.419188 0.209594 0.977788i $$-0.432786\pi$$
0.209594 + 0.977788i $$0.432786\pi$$
$$720$$ 0 0
$$721$$ 7180.13 0.370876
$$722$$ 0 0
$$723$$ −14644.9 −0.753321
$$724$$ 0 0
$$725$$ −4500.21 −0.230529
$$726$$ 0 0
$$727$$ −34117.8 −1.74052 −0.870262 0.492590i $$-0.836050\pi$$
−0.870262 + 0.492590i $$0.836050\pi$$
$$728$$ 0 0
$$729$$ 729.000 0.0370370
$$730$$ 0 0
$$731$$ 54480.6 2.75655
$$732$$ 0 0
$$733$$ 20048.0 1.01022 0.505110 0.863055i $$-0.331452\pi$$
0.505110 + 0.863055i $$0.331452\pi$$
$$734$$ 0 0
$$735$$ 926.708 0.0465063
$$736$$ 0 0
$$737$$ −18438.0 −0.921534
$$738$$ 0 0
$$739$$ 407.607 0.0202897 0.0101448 0.999949i $$-0.496771\pi$$
0.0101448 + 0.999949i $$0.496771\pi$$
$$740$$ 0 0
$$741$$ −353.811 −0.0175406
$$742$$ 0 0
$$743$$ 128.374 0.00633861 0.00316930 0.999995i $$-0.498991\pi$$
0.00316930 + 0.999995i $$0.498991\pi$$
$$744$$ 0 0
$$745$$ −11315.7 −0.556475
$$746$$ 0 0
$$747$$ −2643.59 −0.129483
$$748$$ 0 0
$$749$$ −723.929 −0.0353161
$$750$$ 0 0
$$751$$ −25076.0 −1.21842 −0.609212 0.793007i $$-0.708514\pi$$
−0.609212 + 0.793007i $$0.708514\pi$$
$$752$$ 0 0
$$753$$ 6787.90 0.328506
$$754$$ 0 0
$$755$$ 2376.86 0.114573
$$756$$ 0 0
$$757$$ −21824.1 −1.04783 −0.523916 0.851770i $$-0.675530\pi$$
−0.523916 + 0.851770i $$0.675530\pi$$
$$758$$ 0 0
$$759$$ 5594.93 0.267567
$$760$$ 0 0
$$761$$ −38939.2 −1.85485 −0.927427 0.374004i $$-0.877985\pi$$
−0.927427 + 0.374004i $$0.877985\pi$$
$$762$$ 0 0
$$763$$ 4739.94 0.224898
$$764$$ 0 0
$$765$$ −7758.09 −0.366659
$$766$$ 0 0
$$767$$ 260.586 0.0122675
$$768$$ 0 0
$$769$$ 3079.42 0.144404 0.0722021 0.997390i $$-0.476997\pi$$
0.0722021 + 0.997390i $$0.476997\pi$$
$$770$$ 0 0
$$771$$ 18631.8 0.870308
$$772$$ 0 0
$$773$$ 31770.9 1.47829 0.739146 0.673545i $$-0.235229\pi$$
0.739146 + 0.673545i $$0.235229\pi$$
$$774$$ 0 0
$$775$$ −1260.41 −0.0584195
$$776$$ 0 0
$$777$$ 7012.06 0.323753
$$778$$ 0 0
$$779$$ −10276.0 −0.472628
$$780$$ 0 0
$$781$$ −58088.5 −2.66142
$$782$$ 0 0
$$783$$ −1425.15 −0.0650458
$$784$$ 0 0
$$785$$ 5665.53 0.257594
$$786$$ 0 0
$$787$$ −6736.02 −0.305099 −0.152550 0.988296i $$-0.548748\pi$$
−0.152550 + 0.988296i $$0.548748\pi$$
$$788$$ 0 0
$$789$$ −8918.06 −0.402397
$$790$$ 0 0
$$791$$ 3164.58 0.142250
$$792$$ 0 0
$$793$$ 1759.32 0.0787834
$$794$$ 0 0
$$795$$ 6805.15 0.303590
$$796$$ 0 0
$$797$$ 13445.1 0.597552 0.298776 0.954323i $$-0.403422\pi$$
0.298776 + 0.954323i $$0.403422\pi$$
$$798$$ 0 0
$$799$$ 25242.8 1.11768
$$800$$ 0 0
$$801$$ 11690.0 0.515664
$$802$$ 0 0
$$803$$ 35991.0 1.58169
$$804$$ 0 0
$$805$$ 1682.59 0.0736689
$$806$$ 0 0
$$807$$ 13330.3 0.581472
$$808$$ 0 0
$$809$$ 20337.7 0.883853 0.441927 0.897051i $$-0.354295\pi$$
0.441927 + 0.897051i $$0.354295\pi$$
$$810$$ 0 0
$$811$$ 23069.9 0.998884 0.499442 0.866347i $$-0.333538\pi$$
0.499442 + 0.866347i $$0.333538\pi$$
$$812$$ 0 0
$$813$$ −20520.8 −0.885233
$$814$$ 0 0
$$815$$ 24356.1 1.04682
$$816$$ 0 0
$$817$$ −18015.8 −0.771471
$$818$$ 0 0
$$819$$ 164.321 0.00701079
$$820$$ 0 0
$$821$$ 8230.74 0.349884 0.174942 0.984579i $$-0.444026\pi$$
0.174942 + 0.984579i $$0.444026\pi$$
$$822$$ 0 0
$$823$$ −17577.9 −0.744506 −0.372253 0.928131i $$-0.621415\pi$$
−0.372253 + 0.928131i $$0.621415\pi$$
$$824$$ 0 0
$$825$$ −12510.5 −0.527951
$$826$$ 0 0
$$827$$ 6440.14 0.270793 0.135396 0.990792i $$-0.456769\pi$$
0.135396 + 0.990792i $$0.456769\pi$$
$$828$$ 0 0
$$829$$ 5084.23 0.213007 0.106503 0.994312i $$-0.466034\pi$$
0.106503 + 0.994312i $$0.466034\pi$$
$$830$$ 0 0
$$831$$ 9686.01 0.404337
$$832$$ 0 0
$$833$$ 6700.12 0.278686
$$834$$ 0 0
$$835$$ −18038.9 −0.747618
$$836$$ 0 0
$$837$$ −399.153 −0.0164836
$$838$$ 0 0
$$839$$ 25150.5 1.03491 0.517456 0.855710i $$-0.326879\pi$$
0.517456 + 0.855710i $$0.326879\pi$$
$$840$$ 0 0
$$841$$ −21602.9 −0.885764
$$842$$ 0 0
$$843$$ 19361.5 0.791038
$$844$$ 0 0
$$845$$ 13807.3 0.562113
$$846$$ 0 0
$$847$$ −7429.96 −0.301413
$$848$$ 0 0
$$849$$ 11522.2 0.465771
$$850$$ 0 0
$$851$$ 12731.5 0.512846
$$852$$ 0 0
$$853$$ −6408.37 −0.257232 −0.128616 0.991694i $$-0.541053\pi$$
−0.128616 + 0.991694i $$0.541053\pi$$
$$854$$ 0 0
$$855$$ 2565.46 0.102616
$$856$$ 0 0
$$857$$ 17248.6 0.687515 0.343758 0.939058i $$-0.388300\pi$$
0.343758 + 0.939058i $$0.388300\pi$$
$$858$$ 0 0
$$859$$ −3159.07 −0.125479 −0.0627393 0.998030i $$-0.519984\pi$$
−0.0627393 + 0.998030i $$0.519984\pi$$
$$860$$ 0 0
$$861$$ 4772.52 0.188905
$$862$$ 0 0
$$863$$ −41071.9 −1.62005 −0.810025 0.586395i $$-0.800547\pi$$
−0.810025 + 0.586395i $$0.800547\pi$$
$$864$$ 0 0
$$865$$ −6171.78 −0.242597
$$866$$ 0 0
$$867$$ −41352.2 −1.61983
$$868$$ 0 0
$$869$$ −40928.6 −1.59771
$$870$$ 0 0
$$871$$ −983.210 −0.0382489
$$872$$ 0 0
$$873$$ −1809.37 −0.0701466
$$874$$ 0 0
$$875$$ −9278.46 −0.358479
$$876$$ 0 0
$$877$$ −1034.90 −0.0398472 −0.0199236 0.999802i $$-0.506342\pi$$
−0.0199236 + 0.999802i $$0.506342\pi$$
$$878$$ 0 0
$$879$$ −26403.0 −1.01314
$$880$$ 0 0
$$881$$ −3109.73 −0.118921 −0.0594606 0.998231i $$-0.518938\pi$$
−0.0594606 + 0.998231i $$0.518938\pi$$
$$882$$ 0 0
$$883$$ 19782.8 0.753955 0.376978 0.926222i $$-0.376963\pi$$
0.376978 + 0.926222i $$0.376963\pi$$
$$884$$ 0 0
$$885$$ −1889.49 −0.0717678
$$886$$ 0 0
$$887$$ −9355.56 −0.354148 −0.177074 0.984198i $$-0.556663\pi$$
−0.177074 + 0.984198i $$0.556663\pi$$
$$888$$ 0 0
$$889$$ −8280.13 −0.312381
$$890$$ 0 0
$$891$$ −3961.90 −0.148966
$$892$$ 0 0
$$893$$ −8347.35 −0.312803
$$894$$ 0 0
$$895$$ −7226.27 −0.269886
$$896$$ 0 0
$$897$$ 298.352 0.0111055
$$898$$ 0 0
$$899$$ 780.322 0.0289491
$$900$$ 0 0
$$901$$ 49201.4 1.81924
$$902$$ 0 0
$$903$$ 8367.09 0.308349
$$904$$ 0 0
$$905$$ −24771.0 −0.909853
$$906$$ 0 0
$$907$$ −16211.9 −0.593505 −0.296752 0.954954i $$-0.595904\pi$$
−0.296752 + 0.954954i $$0.595904\pi$$
$$908$$ 0 0
$$909$$ −9481.60 −0.345968
$$910$$ 0 0
$$911$$ −14540.6 −0.528816 −0.264408 0.964411i $$-0.585177\pi$$
−0.264408 + 0.964411i $$0.585177\pi$$
$$912$$ 0 0
$$913$$ 14367.2 0.520792
$$914$$ 0 0
$$915$$ −12756.7 −0.460901
$$916$$ 0 0
$$917$$ 8804.26 0.317058
$$918$$ 0 0
$$919$$ 37239.2 1.33668 0.668339 0.743857i $$-0.267005\pi$$
0.668339 + 0.743857i $$0.267005\pi$$
$$920$$ 0 0
$$921$$ −11780.2 −0.421466
$$922$$ 0 0
$$923$$ −3097.59 −0.110464
$$924$$ 0 0
$$925$$ −28468.2 −1.01192
$$926$$ 0 0
$$927$$ −9231.59 −0.327082
$$928$$ 0 0
$$929$$ 22062.5 0.779167 0.389583 0.920991i $$-0.372619\pi$$
0.389583 + 0.920991i $$0.372619\pi$$
$$930$$ 0 0
$$931$$ −2215.61 −0.0779954
$$932$$ 0 0
$$933$$ −18429.4 −0.646678
$$934$$ 0 0
$$935$$ 42163.0 1.47473
$$936$$ 0 0
$$937$$ −15597.3 −0.543800 −0.271900 0.962326i $$-0.587652\pi$$
−0.271900 + 0.962326i $$0.587652\pi$$
$$938$$ 0 0
$$939$$ −11472.6 −0.398715
$$940$$ 0 0
$$941$$ −15887.1 −0.550377 −0.275188 0.961390i $$-0.588740\pi$$
−0.275188 + 0.961390i $$0.588740\pi$$
$$942$$ 0 0
$$943$$ 8665.29 0.299237
$$944$$ 0 0
$$945$$ −1191.48 −0.0410147
$$946$$ 0 0
$$947$$ −54754.6 −1.87887 −0.939433 0.342733i $$-0.888648\pi$$
−0.939433 + 0.342733i $$0.888648\pi$$
$$948$$ 0 0
$$949$$ 1919.23 0.0656489
$$950$$ 0 0
$$951$$ −23849.0 −0.813205
$$952$$ 0 0
$$953$$ 12091.3 0.410992 0.205496 0.978658i $$-0.434119\pi$$
0.205496 + 0.978658i $$0.434119\pi$$
$$954$$ 0 0
$$955$$ −18517.5 −0.627447
$$956$$ 0 0
$$957$$ 7745.30 0.261620
$$958$$ 0 0
$$959$$ 395.679 0.0133234
$$960$$ 0 0
$$961$$ −29572.4 −0.992664
$$962$$ 0 0
$$963$$ 930.765 0.0311459
$$964$$ 0 0
$$965$$ 22273.6 0.743017
$$966$$ 0 0
$$967$$ 27415.1 0.911695 0.455848 0.890058i $$-0.349336\pi$$
0.455848 + 0.890058i $$0.349336\pi$$
$$968$$ 0 0
$$969$$ 18548.4 0.614921
$$970$$ 0 0
$$971$$ −55397.0 −1.83087 −0.915435 0.402466i $$-0.868153\pi$$
−0.915435 + 0.402466i $$0.868153\pi$$
$$972$$ 0 0
$$973$$ 14795.0 0.487467
$$974$$ 0 0
$$975$$ −667.127 −0.0219130
$$976$$ 0 0
$$977$$ 18294.3 0.599065 0.299532 0.954086i $$-0.403169\pi$$
0.299532 + 0.954086i $$0.403169\pi$$
$$978$$ 0 0
$$979$$ −63532.0 −2.07405
$$980$$ 0 0
$$981$$ −6094.20 −0.198341
$$982$$ 0 0
$$983$$ 23803.2 0.772334 0.386167 0.922429i $$-0.373799\pi$$
0.386167 + 0.922429i $$0.373799\pi$$
$$984$$ 0 0
$$985$$ −3687.01 −0.119267
$$986$$ 0 0
$$987$$ 3876.77 0.125024
$$988$$ 0 0
$$989$$ 15191.8 0.488445
$$990$$ 0 0
$$991$$ 13624.5 0.436726 0.218363 0.975868i $$-0.429928\pi$$
0.218363 + 0.975868i $$0.429928\pi$$
$$992$$ 0 0
$$993$$ −33709.1 −1.07727
$$994$$ 0 0
$$995$$ 13604.8 0.433470
$$996$$ 0 0
$$997$$ −46834.4 −1.48772 −0.743861 0.668334i $$-0.767008\pi$$
−0.743861 + 0.668334i $$0.767008\pi$$
$$998$$ 0 0
$$999$$ −9015.50 −0.285523
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 336.4.a.n.1.1 2
3.2 odd 2 1008.4.a.y.1.2 2
4.3 odd 2 168.4.a.h.1.1 2
7.6 odd 2 2352.4.a.bv.1.2 2
8.3 odd 2 1344.4.a.bd.1.2 2
8.5 even 2 1344.4.a.bl.1.2 2
12.11 even 2 504.4.a.j.1.2 2
28.27 even 2 1176.4.a.p.1.2 2
By twisted newform
Twist Min Dim Char Parity Ord Type
168.4.a.h.1.1 2 4.3 odd 2
336.4.a.n.1.1 2 1.1 even 1 trivial
504.4.a.j.1.2 2 12.11 even 2
1008.4.a.y.1.2 2 3.2 odd 2
1176.4.a.p.1.2 2 28.27 even 2
1344.4.a.bd.1.2 2 8.3 odd 2
1344.4.a.bl.1.2 2 8.5 even 2
2352.4.a.bv.1.2 2 7.6 odd 2 | 19,389 | 34,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.225009 |
https://www.experts-exchange.com/questions/10305538/Function-Question.html | 1,508,342,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822992.27/warc/CC-MAIN-20171018142658-20171018162658-00628.warc.gz | 923,401,074 | 29,419 | [2 days left] What’s wrong with your cloud strategy? Learn why multicloud solutions matter with Nimble Storage.Register Now
x
Solved
# Function Question
Posted on 2000-03-05
Medium Priority
276 Views
In a Programmer Defined Function , what would the code be to deduct a percentage (ex 20%) from a gross pay. fedTax 20%
0
Question by:nationnon
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Accepted Solution
Shay050799 earned 40 total points
ID: 2584866
x = gross pay
x*0.8 = net pay
0
LVL 22
Expert Comment
ID: 2584867
We cannot provide answers to school assignments. That is grounds for removal from this site. (for both you and the experts involved.) We can provide only limitied help in accademic assignments. We can answer specific (direct) questions, like you might ask your teacher. We can review your work and post suggestions, again, like your teacher might do.
Do you have specific questions?
Do you have any work on this (incomplete even) that we can review?
0
LVL 22
Expert Comment
ID: 2584868
You can stoart by writting the function declaration, that is the name of the function, its parameters and return value.
0
LVL 1
Author Comment
ID: 2584897
FYI, If you were to answer my question it wouldn't be the answer to my school assignment, it would only be the answer to my Programmer Defined Function.
0
LVL 1
Author Comment
ID: 2584908
nietod,
I've already done that.
shay,
I don't understand why it would be
x*0.8 = net pay when 20% should be taken from the gross pay. If 0.8 was the percentage that would mean 80% is taken from the gross pay. Am I missing something?
0
LVL 1
Author Comment
ID: 2584913
Also doesn't * mean multiply?
0
LVL 3
Expert Comment
ID: 2584917
u want to deduct 20% off the gross sallary right ?
that mean that u end up with 80% of ur gross sallary
so multiply ur gross sallary by 80% and thats ur net sallay
0
LVL 1
Author Comment
ID: 2584937
OH! One of the reason's i'm using c++ is to help me learn math.
Thanks!
0
LVL 3
Expert Comment
ID: 2584946
it should be the other way around !!
i think ;-)
0
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Join the community of 500,000 technology professionals and ask your questions. | 911 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-43 | longest | en | 0.928116 |
https://www.enotes.com/homework-help/what-value-x-which-2sin-1-2x-cos-2x-1-252013 | 1,484,972,943,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280900.71/warc/CC-MAIN-20170116095120-00092-ip-10-171-10-70.ec2.internal.warc.gz | 896,728,127 | 12,013 | # what is the value of x in for which 2sin(1/2x)-cos(2x)=1?
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
If the first term is sin(x/2), then we'll use the half angle identity:
sin(x/2) = sqrt[(1-cos x)/2]
We'll use the double angle identity for the 2nd term:
cos 2x = 2(cos x)^2 - 1
2sqrt[(1-cos x)/2] - 2(cos x)^2 + 1 = 1
We'll eliminate 1 both sides:
2sqrt[(1-cos x)/2] - 2(cos x)^2 = 0
We'll divide by 2:
sqrt[(1-cos x)/2] - (cos x)^2 = 0
We'll move (cos x)^2 to the right:
sqrt[(1-cos x)/2] = (cos x)^2
We'll raise to square:
(1-cos x)/2 = (cos x)^4
1-cosx - 2(cos x)^4 = 0
We'll write 2(cos x)^4 = (cos x)^4 + (cos x)^4
(1 - (cos x)^4) - cos x(1 + (cos x)^3) = 0
(1-cos x)(1+cos x)((1 + cos x)^4) - - cos x(1+cos x)(1 + cos x + (cos x)^2) = 0
(1 + cos x)[(1 - cos x)(1 + (cos x)^4) - cos x(1 + cos x + (cos x)^2)] = 0
1 + cos x = 0
cos x = -1
x = pi
The solution of the equation in the interval (0 , 2pi) is x = pi. | 422 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | longest | en | 0.703131 |
https://takenologi.com/qa/question-what-is-the-difference-between-percent-error-and-percent-difference.html | 1,624,407,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00217.warc.gz | 494,128,969 | 7,095 | # Question: What Is The Difference Between Percent Error And Percent Difference?
## What is percentage of error?
Percentage error is a measurement of the discrepancy between an observed and a true, or accepted value.
The absolute error is then divided by the true value, resulting in the relative error, which is multiplied by 100 to obtain the percentage error..
## How do I work out a percentage of two numbers?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage! To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage.
## What does percent difference tell you about an experiment?
The idea is that you take the difference between two numbers and report that difference relative to something by dividing by the something. … If you have an accepted value, like when the object of the lab is to experimentally determine some constant of nature, that formula works.
## What’s a good percent error?
In some cases, the measurement may be so difficult that a 10 % error or even higher may be acceptable. In other cases, a 1 % error may be too high. Most high school and introductory university instructors will accept a 5 % error.
## How do you interpret percent error?
Percent errors tells you how big your errors are when you measure something in an experiment. Smaller percent errors mean that you are close to the accepted or real value. For example, a 1% error means that you got very close to the accepted value, while 45% means that you were quite a long way off from the true value.
## What does percent error tell you about accuracy?
The accuracy is a measure of the degree of closeness of a measured or calculated value to its actual value. The percent error is the ratio of the error to the actual value multiplied by 100. The precision of a measurement is a measure of the reproducibility of a set of measurements. … A systematic error is human error.
## What if percent error is negative?
If the experimental value is less than the accepted value, the error is negative. … If the experimental value is larger than the accepted value, the error is positive. Often, error is reported as the absolute value of the difference in order to avoid the confusion of a negative error.
## What is percentage formula?
If want to find 10% of something, ‘of’ just means ‘times’. So 10% of 150 = 10/100 × 150 = 15. If you have to turn a percentage into a decimal, just divide by 100. For example, 25% = 25/100 = 0.25.
## What is the percentage between two numbers?
The percentage difference between two values is calculated by dividing the absolute value of the difference between two numbers by the average of those two numbers. Multiplying the result by 100 will yield the solution in percent, rather than decimal form.
## What does percent difference mean?
The % difference between two numbers is the absolute value of the difference between the two numbers, divided by the average of those two numbers, multiplied by 100%. That is. therefore. % difference = (approximately) 13.1%.
## How do I calculate percentage difference?
Percentage Change | Increase and DecreaseFirst: work out the difference (increase) between the two numbers you are comparing.Increase = New Number – Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase ÷ Original Number × 100.More items… | 743 | 3,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-25 | latest | en | 0.918653 |
http://mathhelpforum.com/algebra/75877-simple-interest.html | 1,526,927,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00323.warc.gz | 189,499,875 | 9,407 | 1. ## simple interest
P = 9125
R = 0.8% per month
T = ?
I = 511.00
got stuck when i was asked the Rate is 0.8% per month and the T wasn't given
2. Hello Die Hard 4,
Originally Posted by diehardmath4
P = 9125
R = 0.8% per month
T = ?
I = 511.00
got stuck when i was asked the Rate is 0.8% per month and the T wasn't given
when r is in % per month
$\displaystyle I = \frac{P\times t \times \frac{R}{12}}{100}$ | 153 | 411 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-22 | latest | en | 0.92598 |
https://www.cfd-online.com/Forums/fluent/28211-computational-grid-generation-fluent.html | 1,503,032,829,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00663.warc.gz | 886,985,845 | 16,213 | # computational grid generation on fluent
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February 13, 2001, 10:33 computational grid generation on fluent #1 Terry Guest Posts: n/a Sponsored Links Hi! guys, I am a new user of fluent. I am working on a simple aerofoil shape trying to analyse the flow over it when it is near the ground. ie: ground effects any effects if i change the grid definition? i am currently using tri elements and type ave would there be much different if i use other type of elements? I am running it on spalart-allmaras model. what are possible ways of improvements? could some guys who are good in such stuff.. provide so feedback thanks alot really appreciate it terry
February 13, 2001, 10:51 Re: computational grid generation on fluent #2 Jonas Larsson Guest Posts: n/a For airfoil flows you should definitely use quads/hexas (if your geometry allows that). This will resolve the boundary-layers and wakes better since you can assure that the cells are aligned with the flow. The 1-equation SA model is okay for attached flows in this type of applications.
February 13, 2001, 11:18 Re: computational grid generation on fluent #3 Terry Guest Posts: n/a Hi! Jonas Thanks for the response appreciate it, but if we are considering just the forces for life and drag over the foil, what are some of the concerns that we should look out for? its a one dimensional problem and we are using high reynolds number, is it still suitable to use a turbulence model like the SA? its steady state flow... at 50m/s across the aerofoil. thanks again Terry
February 13, 2001, 11:33 Re: computational grid generation on fluent #4 Jonas Larsson Guest Posts: n/a Computing the drag correctly can be a bit tricky since you have to get the skin-friction correct. The lift is easier. I haven't done any validations of the SA model for airfoils so I can't say how well it will perform on the drag. You can always compare it to a more advanced two-equation model (I'd recommend the Realizable k-epsilon model with the Wolfstein 1-eq model in the boundary layers - Fluent calls it "two-layer zonal model"). Lift should be okay as long as you have attached flow. If this is an airfoil in free air make sure you use a very large domain, this is important. Search the forums here and you'll find a lot of discussion about this, including references, if my memory doesn't fail me now. You should of course first compute a wing for which you have data to compare your results with - see the Resources/Reference section for a lot of nice cases to run. I think that both the ERCOFTAC database and the NPARC database includes airfoil cases.
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https://physics.stackexchange.com/questions/296691/where-is-gravity-potential-energy-gone-when-object-has-escape-velocity/296697 | 1,718,608,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00183.warc.gz | 418,819,594 | 39,601 | # Where is gravity "potential energy" gone when object has escape velocity?
If an object leaves Earth with the escape velocity, meaning it'll never fall back on Earth, where does the energy go? You can't say it's converted to "potential energy" anymore because the object will never fall back. Yes, the escaping object velocity will decrease therefore energy will go somewhere else.
The energy is indeed still converted to potential energy. The object doesn't go back on its own, but it still has a potential energy associated with its distance from the other mass.
Similarly, an object perfectly on the top of a hill won't roll down on its own, but it still has a potential energy associated with its being on top of the hill.
But it is converted into potential energy. In non-general relativity terms, it is stored in the gravitational field surrounding and between the two objects (in general relativity terms, the energy is stored in how the curvature of space-time is altered).
In Newtonian gravity, you find the gravitational field of two point-like objects by adding the individual fields of the objects: $$\mathbf{g}_{\mathrm{tot}}(\mathbf{r}) = \mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r}).$$
The energy density, $u_g(\mathbf{r})$ (energy per unit volume of space), stored in the gravitational field is given by $[G / 8\pi] \mathbf{g}_{\mathrm{tot}} \cdot \mathbf{g}_{\mathrm{tot}}$, giving: \begin{align} u_g(\mathbf{r}) &= \frac{G}{8\pi} [\mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r})]\cdot [\mathbf{g}_{1}(\mathbf{r}) + \mathbf{g}_{2}(\mathbf{r})] \\ & = \frac{G}{8\pi}\left[\mathbf{g}_{1}^2 + \mathbf{g}_{2}^2 + 2 \mathbf{g}_{1} \cdot \mathbf{g}_{2} \right] \end{align}
The $\mathbf{g}_{1}^2$ and $\mathbf{g}_{2}^2$ terms are the density of energy needed to create objects 1 and 2, and add up to an energy that is infinite. The energy density of the interaction between 1 and 2 is given by the cross term: $$u_{g\, 1,2} = \frac{G}{4\pi} \mathbf{g}_{1} \cdot \mathbf{g}_{2}.$$ Since only differences in energy are relevant in classical Newtonian physics, the self energies don't contribute to the dynamics. If you integrate $u_{g\, 1,2}$ correctly over all space, you'll find that $$\Delta U = \frac{G m_1 m_2}{|\mathbf{r}_1 - \mathbf{r}_2|},$$ exactly as expected.
There are three important things to note.
One is that you have simplified things by treating the Earth and point mass system as though there is nothing else in the Universe.
"Escape" means that of its own accord the point mass will never return back to the Earth but that does not mean that it ceases to exist.
Lastly you have implied that you can actually travel an infinite distance.
If the system is the mass under consideration and the Earth and there are no external force acting on the system (ie nothing else in the Universe) then as the separation between the mass and the Earth increases the kinetic energy of the system decreases at an every decreasing rate whilst at the same time the gravitational potential energy of the system increases at an ever decreasing rate.
If used in such a context, "infinity" is a word which means "the distance between the mass and the Earth is such that, on the scale set by the example, if there is a change in the distance between the mass and the Earth then the kinetic energy of the system can be sensibly taken to be zero and the gravitational potential energy can be sensibly taken to be constant.
So "mass at/reaches infinity" is a shorthand way of writing "the separation between the mass and the Earth tends to infinity" and this is a very useful idea when one starts to model a system mathematically as the limiting process is implied to have been performed - eg making one of the limits of an integration infinity.
the energy after the object gone out of gravitational field of earth remain constant as it remains in equilibrium in space and does not fall to the earth afterward
1. PE=mgh remain constant if h<=value of g at 'h' height)
• this isnt correct, your formula is an approximation for trajectories small enough for G to be constant. But G isnt constant . Plus my question mentionned escape velocity, meaning the object must moving away and not be "in equilibrium" Commented May 3, 2019 at 20:56 | 1,080 | 4,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.839044 |
http://mathoverflow.net/questions/104783/subset-of-edges-of-graph-touching-all-vertices-such-that-all-paths-consist-of-at | 1,469,508,548,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824624.99/warc/CC-MAIN-20160723071024-00063-ip-10-185-27-174.ec2.internal.warc.gz | 160,443,949 | 20,670 | # Subset of edges of graph touching all vertices such that all paths consist of at most two edges
Let $G=(V,E)$ be a (simple) finite graph such that every vertex has degree at least 1. Then it is easy to see that there is a subset $E'$ of $E$ such every vertex in $G'=(V,E')$ still has degree at least 1 and all paths (with no repeating edges) in $G'$ are of (edge-wise) length at most 2. (I just keep removing middle edges of paths of length 3 until I'm done.) My question is, does this hold for infinite graphs ?
EDITED: tried to make the question more clear, as comments suggested
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Title asks length at most 3, body asks length at most 2. Please edit for consistency. – Gerry Myerson Aug 15 '12 at 23:19
The confusion between 2 and 3 makes it incomprehensible. – Brendan McKay Aug 16 '12 at 3:40
There's also room for confusion as to whether the "length" of a path is the number of vertices or the number of edges. Maybe the two confusions can be made to cancel? I'm pretty sure the OP wants to prohibit paths that have 4 vertices and (therefore) 3 edges. – Andreas Blass Aug 16 '12 at 4:23
Aaron Meyerowitz suggested to try to reduce the problem to trees and, to me, this seems to work. First we can suppose that $G$ is a connected graph, because we can solve the problem separatly for each component. It is easy to see by Zorn's Lemma, that every connected graph contains a spanning tree, i.e. a subgraph which is a tree and which connects all vertices of the original graph. Hence it is enough to solve the problem for a tree.
Put $E_0=\emptyset$. We choose a root $r$ of the tree and denote by $L_n$ the set of vertices which are $n$ edges far from $r$. By hypothesis, $L_1$ is nonempty. If $L_1$ contains at least one vertex of degree 1, we define $E_1$ to be exactly the edges connecting $r$ with the vertices from $L_1$ of degree 1. Otherwise, we pick arbitrary $x_1$ from $L_1$ and define $E_1$ as a singleton consisting just of the edge connecting $r$ and $x_1$. Now we continue inductively by level $n$ of the tree (which is easily well-defined). Let $v \in L_n$, put $E_n=E_{n-1}$:
• If $v$ is leaf, i.e. the tree "under" $v$ has just one vertex, do nothing.
• If there is an edge from $v$ to an element in $L_{n-1}$, add to $E_n$ all edges connecting $v$ with leaves under $v$.
• Otherwise, apply to $v$ the same procudere as to $r$ (if there is a leaf under $v$, add all the edges connecting $v$ with leaves to $E_n$, otherwise pick some edge and add it to $E_n$).
Put $E'=\bigcup E_n$, this (I think) is the desired subset of edges, since:
Let $v$ be a vertex, then $v \in L_n$ for some $n \geq 0$.
• $\operatorname{deg}v \geq 1$: Suppose there is no edge connecting $v$ with any edge from level $n-1$. Then by the construction there must be an edge from $v$ to some vertex in level $n+1$.
• Suppose $v$ has degree 1. Then by the construction, the parent of $v$ is connected only to vertices of degree $1$. Thus there is no path of edge-wise length more then 3.
Thanks for every comment.
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It looks promising. Also, I added some comments to my answer which you might find useful. Thanks for the problem. You might ask if weaker versions of set theory allow you to do this. Gerhard "After This Question, More Await" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 18:39
You should be able to do something similar, at least for graphs with a bijection to the natural numbers. Using the bijection, take the next unprocessed vertex. If it is of degree one, move on. If it has any neighbors of degree one, remove all other edges to neighbors of degree not 1, making a disconnected star. Otherwise remove all edges incident to the vertex except for the edge leading to the smallest numbered vertex. A trans-countable version of this may also work, but you will have to come up with the limit case.
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One thing to watch for in the limit is that a vertex may get degree 0. When that happens, you can add an edge back, but you have to keep track and make sure your path condition is not locally undone. Gerhard "Takes Some And Leaves Some" Paseman, 2012.08.15 – Gerhard Paseman Aug 15 '12 at 20:34
Thanks for input! If I understood your algorithm correctly, I think it could produce a vertex of degree 0 even in the countably infinite case. Say I have a star with $\omega$ rays consisting of two edges and the algorithm processes the "middle" vertices of the rays first. Then I end up with the middle vertex of the star disconnected. For the correcting step, I think that similar star graph with rays of length 3 and suitable numbering of vertices would get lead to situation, where you cannot add any edge connecting the center back without creating a path of length 3. – Fred.Fred Aug 16 '12 at 8:30
Indeed, assuming Choice, there is a transfinite version of the above. I suspect being able to do so is equivalent to AC. Theidea is to add back edges to lower labeled pint in a small way, and remove an edge afterwards in case there is a long path. But I need to sleep. Gerhard "Perhaps Clarity Comes With Morning" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 8:36
Sorry Fred. I see your comment but my eyes are blurry. Will read it sooon. Gerhard " Really Must Go To Sleep" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 8:37
Indeed, morning plus caffeine show my transfinite extension to be flawed. I think Fred has the right take on it, as it seems ugly and problematic to take an approach like mine or like Aaron's initial attempt. Gerhard "Ask Me About System Design" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 18:29
This is not really an answer, but I conjecture that the issues are very closely related to the issues which arise in the case where you are asking for for paths to have lengths not exceeding 1, in other words, a perfect matching. There are fairly subtle issues which come up, but if you read Ron Aharoni's 1991 paper, you will gain enlightenment.
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I think so. here is an incomplete approach which I had thought would answer the question constructively based on distance from a vertex or set of vertices:
Call two vertices connected if there is a finite path between them. The graph consists of one or more connected components. It is enough to prove the result for a connected graph $G$ having at least 3 vertices. So a vertex may have unaccountably many neighbors, however every pair of vertices has a finite shortest path connecting them.
We can label the vertices of $G$ with non-negative integers and possibly delete some edges subject to:
1. There is a vertex labelled $0.$
2. A vertex labelled $k \gt 0$ has at least one neighbor labelled $k-1.$
3. Vertices connected by an edge have labels which differ by $1.$
4. The graph is still connected.
If there is a leaf, a vertex of degree one, the labeling is unique: All leaves are labelled $0$ and each (other) vertex is labelled with the distance to the closest leaf. Then any edges connecting vertices with the same label are deleted
If there are no leaves then label some starting vertex $s=s_G$ with $0$, then label the others according to distance from $s,$ then delete edges connecting vertices with the same label.
Leaf Case Assume first that $G$ has a leaf. Select all edges labelled $2j,2j+1$. If a vertex labelled $2j$ has no neighbors labelled $2j+1,$ then select a single edge on it labelled $2j-1,2j.$ The result is a collection of single edges and stars with an odd label on the center. (We neglected the trivial but very important case that there are two vertices.) So we are done with this Leaf Case in two steps.
Leafless Case If $G$ has no leaves, then delete all but one edge on $s_G$. This creates no isolated vertices but may create several (even uncountably many) components. The component of $s_G$ has at least one leaf (namely $s_G$). Other vertices in that component may need their labels raised or lowered by $1$. However all edges for that component are handled in two steps. Any other components with leaves need all their labels lowered by 1. However all these components are completely handled in two steps. In any component $H$ with no leaves we pick a starting vertex $s_H$ with label $1$. It's label must be lowered to $0\dots$ unfortunately other labels may increase arbitrarily. I had mistakenly thought that Although there can be (countably) infinitely many stages, the selected and discarded edges for a vertex with initial label $m$ are determined by stage $m+1$.
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I am concerned that there will still be an isolated vertex in the limit because it is connected to uncountably many vertices with label 1, and it "never gets its turn." Perhaps you can tell me more about when such a vertex will be processed? (Basically I am worried about the no leaf process continuing infinitely often and affecting a certain unfortunate vertex.) Perhaps this combined with a well ordering argument might work? Gerhard "Ask Me About System Design" Paseman, 2012.08.15 – Gerhard Paseman Aug 16 '12 at 5:13
Yes, that is a problem. I'll think about it. I wonder if one can remove a maximal non-disconnecting set of edges to get a tree and then take it from there. – Aaron Meyerowitz Aug 16 '12 at 7:02
Here is something to think about. Use your same construction iin both cases. In the leafless case, you end up with many stars as well as some with longer branches. The result should be an possibly infinite tree with maximal path length 4, since you added back some edges. Now run your algorithm again on that infinite tree. Gerhard "Hope THIS Makes It Work" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 7:07
Also, in the leaf case, how do we get only stars and not, say bipartite graphs with cycles? Gerhard "Back To The Stick Figures" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 7:13
I think you have a nice reduction of the problem to bipartite graphs. It might be time to read what Igor Rivin suggested. Gerhard "Ask Me About System Design" Paseman, 2012.08.16 – Gerhard Paseman Aug 16 '12 at 7:21 | 2,551 | 9,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-30 | latest | en | 0.955135 |
https://la.mathworks.com/matlabcentral/answers/496633-how-can-i-get-fixed-output-in-ode45 | 1,590,562,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00227.warc.gz | 418,990,346 | 20,212 | MATLAB Answers
How can I get fixed output in ode45
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Hüseyin Cagdas Yatkin on 14 Dec 2019
Answered: Cris LaPierre on 18 Mar 2020
I only get 20 output with this code. How can I get fixed output?
clc
clear
Cmeasured = [0 101 160 239 645 766 766 827 1122 1158 1591 2181 2566 2654 2944 3163 3406 3751 3982 3982 4425 4621 4746 4809 5240 5479 5604 5761 5885 5885 515418 515511 515511 516128 516203 516203 516252 516554 516991 517579 517969 518040 518040 518315 518566 518818 519166 519426 519426 519864 519969 520125 520125 520275 520441 520584 520723 520793 520793 536458 536540 536540 1562585 1562585 1562861 1561528 1561753 1562065 1562947 1562947
];
tspan = 0:1:1562947;
sol = ode45(@mt ,tspan ,[8.850 2] );
plot(sol.x',sol.y')
cagdas = sol.y';
function y = mt(t, C)
Cin = 1.800;
Kd = 0.00125/(24*60); %1/min
KL = 0.29/(24*60); %m/day
kl = 0.29/(24*60); %1/min
hw = 74.75; %depth of lake m
A = 18500*10^6; %Area m^2
L = 311*10^3; %Length m
Qin = 6400*60; %Inflow(niagara) m^3/min
Qout = 6700*60; %Outflow m^3/min
hs = 0.07; %depth of sediment that have DCB concentration m
Vw = hw*A; %Volume of water body m^3
Vs = hs*A*10^6; %Volume of sediment m^3
wt = 106/(365*24*60);
y(1) = ( Qin*Cin -(Qout)*C(1)- KL*A*C(1)-Kd*Vw*C(1)-kl*A*C(1))/Vw;
y(2) = (KL*A*C(1)-KL*A*C(2) - Kd*Vs*C(2))/Vs;
y=y(:);
end
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Answers (1)
Cris LaPierre on 18 Mar 2020
Use the following calling syntax instead. This results in a value in y corresponding to each time point specified in tspan.
tspan = 0:1:1562947;
[t,y] = ode45(@mt ,tspan ,[8.850 2] );
plot(t,y(:,1),t,y(:,2))
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Work-Out Corner has 5 more than 3 times as many exercise bicycles as The Gym. Together they have 21 bicycles. Solve the equation x + 3x + 5 = 21 to find the number of bicycles at Work-Out Corner. A. 4 bicycles B. 17 bicycles C. 7 bicycles D. 25 bicycles
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Not the answer you are looking for? Search for more explanations. | 323 | 1,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-05 | latest | en | 0.360722 |
https://matchmaticians.com/questions/q9hby4/volume-of-the-solid-revolution-for-f-x-sin-x-calculus | 1,709,039,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00429.warc.gz | 394,367,638 | 15,774 | # Volume of the solid of revolution for $f(x)=\sin x$
Find the colume of the solid obtained by rotating the graph of the function $y=\sin x$ around $x$-axis between $x=0$ and $x=\pi$.
$V=\int_0^{\pi}\pi \sin ^2 x dx=\pi \int_0^{\pi}\frac{1-\cos 2x}{2}dx$
$=\pi \left( \frac{x}{2}-\frac{\sin 2x}{4} \right) \bigg |_{x=0}^{x=\pi}=\pi (\frac{\pi}{2}-0)$
$=\frac{\pi^2}{2}.$ | 159 | 372 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | longest | en | 0.565717 |
https://math.stackexchange.com/questions/908056/help-evaluating-lim-x-to-frac-pi2-left-frac1-frac-pi2-x-tan-x/908064 | 1,571,057,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00012.warc.gz | 577,069,827 | 31,652 | # Help Evaluating $\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\frac{\pi}{2}-x}-\tan {x}\right)$
Does anyone know how to evaluate the following limit?
$$\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\frac{\pi}{2}-x}-\tan {x}\right)$$
The answer is 0 , but I want to see a step by step solution if possible.
Here are the steps $$\lim_{x\to \frac{\pi}{2}} \left[\frac{1}{\frac{\pi}{2}-x}-\tan x\right]= \lim_{x\to \frac{\pi}{2}} \left[\frac{2}{\pi-2x}-\frac{\sin x}{\cos x}\right]$$ $$= \lim_{x\to \frac{\pi}{2}} \left[\frac{2\cos x-(\pi-2x)\sin x}{(\pi-2x)\cos x}\right]= \lim_{x\to \frac{\pi}{2}} \left[\frac{\frac{d}{dx}[2\cos x-(\pi-2x)\sin x]}{\frac{d}{dx}[(\pi-2x)\cos x]}\right]$$ $$= \lim_{x\to \frac{\pi}{2}} \left[\frac{-2\sin x-(\pi-2x)\cos x+2\sin x}{-(\pi-2x)\sin x-2\cos x}\right] = \lim_{x\to \frac{\pi}{2}} \left[\frac{\frac{d}{dx}[(\pi-2x)\cos x]}{\frac{d}{dx}[(\pi-2x)\sin x+2\cos x]}\right]$$ $$= \lim_{x\to \frac{\pi}{2}} \left[\frac{-(\pi-2x)\sin x-2\cos x}{(\pi-2x)\cos x-2\sin x-2\sin x}\right] = \lim_{x\to \frac{\pi}{2}} \left[\frac{(\pi-2x)\sin x+2\cos x}{4\sin x -(\pi-2x)\cos x}\right]$$ $$= \frac{0\cdot 1+2\cdot 0}{4\cdot 1 -0\cdot 0}= \frac{0}{4}=0$$
Hints:
1) Put everything over a common denominator.
2) Since $f(\pi/2)$ is undefined (and hence your function $f$ is not continuous at $\pi/2$), you can't just plug in $\pi/2,$ so use L'Hospital's rule (i.e. differentiate the numerator and the denominator with respect to $x$ until you no longer obtain a result in the form $\frac{0}{0}$).
Take $y=\pi/2-x$ and write the limit as $$\lim_{y\rightarrow0}\,\left(\frac{1}{y}-\frac{\cos\,y}{\sin \,y}\right),$$ then use Maclaurin series expansion | 705 | 1,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.50594 |
https://www.jiskha.com/questions/16810/im-really-stuck-on-this-can-someone-please-explain-figure-a-which-i-tried-to | 1,580,055,806,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00021.warc.gz | 930,538,831 | 6,026 | # Physics Phase Constant
..im really stuck on this. can someone please explain?
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Figure A [[which i tried to recreate below]] is a partial graph of the position function x(t) for a simple harmonic oscillator with an angular frequency of 1.1 rad/s.
[a].....x(cm).......
........5-|-........
.........-|-........
=======.3-|-........
.......========.....
........1-|-...=====
----------|---------t
.......-1-|-........
.........-|-........
.......-3-|-........
.........-|-........
.......-5-|-........
Figure B is a partial graph of the corresponding velocity function v(t).
[b]....v(cm/s)......
........5-|-........
.........-|-........
........3-|-........
.........-|-........
........1-|-........
----------|---------t
.......-1-|-........
.........-|-........
=======-3-|-........
.......========.....
.......-5-|-...=====
What is the phase constant of the SHM if the position function x(t) is given the form x = xmcos(ωt + φ)?
sorry that the 'graphs' look distorted on this page that your viewing... i recommend copying the graph and pasting it in the ANSWER box below on this page and it should look right :)
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# Finance Discipline: Using a constant growth rate of 4% pa. (commencing immediately) in the constant dividend growth model, determine a current stock price for your company.
The University of Sydney Business School
Finance Discipline
FINC 2011 Corporate Finance 1
Major Assignment
Due Date: 4pm Wednesday 24 October 2018
ASX- listed Company Valuation Report
This report requires you to undertake a valuation exercise on an Australian Securities
Exchange (ASX) listed company. You are required to use the models presented in
the course to calculate a share price for the company and then critically evaluate the
value you have determined.
Required:
1. Select a publicly-listed company on the ASX that has previously made dividend
payments. The selected company should not be classified by the ASX as a
‘Financial’, ‘Metals and Mining’ or ‘Utilities’ company.
2. Using appropriate historical data calculate a beta for the company you have
selected. Compare your beta to those published by research houses (ie Morningstar
etc) for your chosen company. Are the values different? Why?
3. Using the CAPM model determine an appropriate discount rate for your company.
Use a market risk premium of 6.5% p.a. in your calculations.
4. Using a constant growth rate of 4% pa. (commencing immediately) in the constant
dividend growth model, determine a current stock price for your company.
5. Suppose that the previous dividend growth rate was only expected to apply for 5
years before reverting to a long-term growth rate consistent with the forecast inflation
rate. Demonstrate the impact this would have upon your stock valuation.
6. Now disregard the previous growth forecasts. Using your own variables, determine
a current stock price for the company that you think might represent something close
to its fair value. Why do you believe the variables you have chosen give a more
appropriate value than those used in questions 4 and 5? Discuss.
7. Compare the calculation that you obtained in question 6 with what you would have
obtained for the value of the stock using the ‘method of comparables’ stock valuation
approach.
Formatting and Presentation:
a. The assignment should be no more than 12 pages, with 1.5 line spacing and size
12 Times New Roman or Arial Font. Marks will be reduced by 10% for each page
that the assignment exceeds 12 pages. Hence your mark will be reduced to zero if
the assignment exceeds 22 pages. Brevity and conciseness are key ingredients of a
highly successful report. You will be penalised for inappropriate formatting.
b. Make sure you use proper referencing in your assignment. (Note the Business School’s referencing style that should be used in written reports see http://sydney.edu.au/business/currentstudents/policy ) You will be penalised for inappropriate and/or inconsistent referencing.
c. Pay particular attention to presentation. A significant component of your mark will be based on presentation. Avoid overdoing formatting, and ensure that the assignment is very clear, logical and professional. Pay attention to grammar. Clear and logical presentation is a major challenge in assignment preparation. Preparing a concise assignment is another major challenge. Every part of the assignment should somehow add to the end result otherwise it is superfluous and distracting.
What’s included in the 12 page limit?
– Assignment body
– Tables
– Diagrams
What’s excluded from the 12 page limit?
– Executive summary
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http://threadspodcast.com/margin-of/margin-error-survey-formula.html | 1,518,894,456,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807660.32/warc/CC-MAIN-20180217185905-20180217205905-00651.warc.gz | 371,364,659 | 5,510 | Home > Margin Of > Margin Error Survey Formula
Margin Error Survey Formula
Contents
Typical choices are 90%, 95%, or 99% % The confidence level is the amount of uncertainty you can tolerate. Margin of error is often used in non-survey contexts to indicate observational error in reporting measured quantities. In fact, your survey’s confidence level and margin of error almost solely depends on the number of responses you received. Thanks Reply RickPenwarden says: May 25, 2015 at 2:10 pm Hello Panos! http://threadspodcast.com/margin-of/margin-of-error-in-a-survey.html
If the confidence level is 95%, the z*-value is 1.96. Learn how here: "6 Charts to Create Effective Reports" http://t.co/Dl6ZI5ZJkY #mrx- Wednesday Sep 25 - 7:56pmFluidSurveys's launching Version 5! Sample Size: Margin of Error (%) -- *This margin of error calculator uses a normal distribution (50%) to calculate your optimum margin of error. So you’re probably wondering how to figure out how the Calculator determines what your sample size should be. https://www.surveymonkey.com/mp/margin-of-error-calculator/
Margin Of Error Excel
But, for now, let's assume you can count with 100% accuracy.) Here's the problem: Running elections costs a lot of money. Non-response bias is the difference in responses of those people who complete the survey vs. Sampling error is the only error that can be quantified, but there are many other errors to which surveys are susceptible. This calculation is based on the Normal distribution, and assumes you have more than about 30 samples.
What if Your Sample Size is too High? But how many people do you need to ask to get a representative sample? Like we mentioned earlier, you don’t need to go through this whole formula yourself. Margin Of Error Definition Occasionally you will see surveys with a 99-percent confidence interval, which would correspond to three standard deviations and a much larger margin of error.(End of Math Geek Stuff!) If a poll
Reply aj says: December 5, 2014 at 3:15 am What margin of error and confidence level should I use in order to come up with a product sampling scheme where the What is the population size? The area between each z* value and the negative of that z* value is the confidence percentage (approximately). http://ropercenter.cornell.edu/support/polling-fundamentals-total-survey-error/ This could be expensive, and from a statistical perspective, ultimately frivolous.
The survey results also often provide strong information even when there is not a statistically significant difference. How To Find Margin Of Error On Ti 84 Retrieved on 15 February 2007. About Response distribution: If you ask a random sample of 10 people if they like donuts, and 9 of them say, "Yes", then the prediction that you make about the general At percentages near 50%, the statistical error drops from 7 to 5% as the sample size is increased from 250 to 500.
Confidence Interval Margin Of Error Calculator
To determine the confidence interval for a specific answer your sample has given, you can use the percentage picking that answer and get a smaller interval. http://www.robertniles.com/stats/margin.shtml You can also use a graphing calculator or standard statistical tables (found in the appendix of most introductory statistics texts). Margin Of Error Excel That's because many reporters have no idea what a "margin of error" really represents. Margin Of Error Sample Size According to sampling theory, this assumption is reasonable when the sampling fraction is small.
What confidence level do you need? have a peek at these guys The true p percent confidence interval is the interval [a, b] that contains p percent of the distribution, and where (100 − p)/2 percent of the distribution lies below a, and All rights reserved.About us · Contact us · Careers · Developers · News · Help Center · Privacy · Terms · Copyright | Advertising · Recruiting We use cookies to give you the best possible experience on ResearchGate. If my expected response rate is 10% should I sent an email invitation to 3800 persons to make sure that I will have 380 responses? Margin Of Error In Polls
If the sample size is large, use the z-score. (The central limit theorem provides a useful basis for determining whether a sample is "large".) If the sample size is small, use Most surveys you come across are based on hundreds or even thousands of people, so meeting these two conditions is usually a piece of cake (unless the sample proportion is very An example of such a flaw is to only call people during the day and miss almost everyone who works. http://threadspodcast.com/margin-of/margin-of-error-in-survey.html Stokes, Lynne; Tom Belin (2004). "What is a Margin of Error?" (PDF).
This is due to the Finite Population Correction formula. Margin Of Error Calculator Without Population Size Determine Sample Size Confidence Level: 95% 99% Confidence Interval: Population: Sample size needed: Find Confidence Interval Confidence Level: 95% 99% Sample Size: Population: Percentage: Confidence Interval: Sample To further elaborate, you can say, with 95% confidence red jelly beans make up 30%, {+/- 4% or the range of 26-34%} of the beans in the jar.
For this reason, The Survey System ignores the population size when it is "large" or unknown.
The true answer is the percentage you would get if you exhaustively interviewed everyone. In other words, the more people you ask, the more likely you are to get a representative sample. Introductory Statistics (5th ed.). How To Find Margin Of Error With Confidence Interval Reply Nida Madiha says: March 6, 2015 at 9:40 pm Thanks a lot Rick!
The confidence level score is the standard deviation value that goes along with your confidence level. One example is the percent of people who prefer product A versus product B. Among survey participants, the mean grade-point average (GPA) was 2.7, and the standard deviation was 0.4. http://threadspodcast.com/margin-of/margin-of-error-survey.html What is a Survey?.
A standard survey will usually have a confidence level of 95% and margin of error of 5%. The central limit theorem states that the sampling distribution of a statistic will be nearly normal, if the sample size is large enough. So let's say I conducted a staff survey in 2012 and had a population of 65 people, but in 2013 when the report came out our population was 85. Survey Research Methods Section, American Statistical Association.
Sign up today to join our community of over 11+ million scientific professionals. It holds that the FPC approaches zero as the sample size (n) approaches the population size (N), which has the effect of eliminating the margin of error entirely. If you are not familiar with these terms, click here. Example: You're surveying the attendees to a hockey game, let's say a grand total of 30,000 people, and wanted a margin of error of 5% with a confidence level of 95%.
The following two tabs change content below.BioLatest Posts FluidSurveys Team Latest posts by FluidSurveys Team (see all) It’s All About Timing –When to Send your Survey Email Invites? - April 1, For comparison, let's say you have a giant jar of 200 million jelly beans. This makes intuitive sense because when N = n, the sample becomes a census and sampling error becomes moot. How to Compute the Margin of Error The margin of error can be defined by either of the following equations.
We will describe those computations as they come up. Most surveys report margin of error in a manner such as: "the results of this survey are accurate at the 95% confidence level plus or minus 3 percentage points." That is By calculating your margin of error (also known as a confidence interval), you can tell how much the opinions and behavior of the sample you survey is likely to deviate from You've probably heard that term -- "margin of error" -- a lot before.
Industry standard for marketing research is a 95% confidence level with a margin of error of 5%. I gave you the math up above. Reply RickPenwarden says: March 5, 2015 at 11:41 am Hi Wisdom, The more of your population that respond to your survey the more confident you can be in your findings. Maximum and specific margins of error While the margin of error typically reported in the media is a poll-wide figure that reflects the maximum sampling variation of any percentage based on | 1,851 | 8,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-09 | latest | en | 0.88731 |
https://goudryan.com/the-rbpr/the-numbers/ | 1,610,845,231,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00257.warc.gz | 370,133,072 | 9,777 | ## The Neolithic Numbers
Introduction
These are the bare data of major works of the British Isles, Maeshowe, Ring of Brodgar, Stones of Stenness and Stonehenge, all translated into the units of measurement as employed on this website. This is the system I have created on the basis of already known units which I have crafted in such a way that they have values which stand in a mathematical order. [differences of thousandths (like 0.009 etc.) and less are considered ‘exact’ matches throughout these pages, it is an integral aspect of the model of approximation]
The units are the following, respectively: Megalithic Ell (ME), -Remen (MR), -Yard (MY), -Foot (MF), these are all based on human bone lengths of the average Neolithic skeleton, the Ell on 2 male ulna (fore-arm), here called Orkney Ulna (OU), the Remen on 1 female fibula (calf) (OF), the Yard on 2 male tibia (shin) (OT), the Foot on 1 female humerus (upper-arm) (OH).
Ell (ME) = 52.360cm —– Ulna (OU) = 26.18cm
Remen (MR) = 37.025cm —– = Fibula (OF)
Yard (MY) = 82.280cm —– Tibia (OT) = 41.14cm
Foot (MF) = 29.620cm —– = Humerus (OH)
These values relate as follows:
11 ME = 7 MY
9 MY = 20 MR = 25 MF
99 ME = 140 MR
11 OU = 7 OT
9 OT = 10 OF
4 OF = 5 OH
the upper row works out as:
63 MY = 99 ME = 140 MR = 175 MF
We see that this is a fully integrated calculator, that it is in principle simple, but can be confusing because of the doubles and the halves and the ensuing ratios because for instance:
11 ME = 7 MY, but as a ratio it is 1 ME : 1 MY = 7 : 11, which looks like a mistake but is not, since
52.36 : 82.28 = 7 : 11 (52.36/7 = 7.48, this x 11 = 82.28)
So use a calculator when you think it is wrong, I have made mistakes in the past myself, but I think I’ve taken out most mistakes. If there are still mistakes I apologize for that, it does though not affect the consistency of the model as it stands. Remember it are all ‘approximations’ in terms of accepted mathematics
The practical explanation and applications of these numbers are given elsewhere (The Neolithic Numbers explained). This ordering is to show the recurrence of the same numbers and ratios in all the different buildings.
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$R(A,B,C,D)$ is a relation. Which of the following does not have a lossless join, dependency preserving $BCNF$ decomposition?
1. $A \rightarrow B, B \rightarrow CD$
2. $A \rightarrow B, B \rightarrow C, C \rightarrow D$
3. $AB \rightarrow C, C \rightarrow AD$
4. $A \rightarrow BCD$
taking up option A first :
We have, R(A, B, C, D) and the Functional Dependency set = {A→B, B→CD}.
Now we will try to decompose it such that the decomposition is a Lossless Join, Dependency Preserving and new relations thus formed are in BCNF.
We decomposed it to R1(A, B) and R2(B, C, D). This decomposition satisfies all three properties we mentioned prior.
taking up option B :
we have, R(A, B, C, D) and the Functional Dependency set = {A→B, B→C, C→D}.
we decomposed it as R1(A, B), R2(B, C) and R3(C, D). This decomposition too satisfies all properties as decomposition in option A.
taking up option D :
we have, R(A, B, C, D) and the Functional Dependency set = {A→BCD}.
This set of FDs is equivalent to set = {A→B, A→C, A→D} on applying decomposition rule which is derived from Armstrong's Axioms
we decomposed it as R1(A, B), R2(A, C) and R3(A, D). This decomposition also satisfies all properties as required.
taking up option C :
we have, R(A, B, C, D) and the Functional Dependency set = {AB→C, C→AD}.
we decompose it as R1(A, B, C) and R2(C, D). This preserves all dependencies and the join is lossless too, but the relation R1 is not in BCNF. In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a superkey. Condition that all relations formed after decomposition should be in BCNF is not satisfied here.
We need to identify the INCORRECT, Hence mark option C.
thanks .Well explained :)
Great explanation.
Just one thing:
Do we need to decompose in case of option D {A → BCD}?
edited
I have also same question. Do we need to decompose option D? It is already in BCNF.
Also, can you post BCNF decomposition algorithm?
I am using algorithm 11.3 of Navathe book. But getting different decomposed relation in option C. Keys are AB and BC. So C -> AD not in BCNF. So, decomposition result as BC and CAD. So, decomposition result as BC and CAD. Please comment.
Algorithm 11.3: Relational Decomposition into BCNF with Nonadditive Join Property
Input: A universal relation Rand a set of functional dependencies F on the attributes of R.
1. Set D := {R};
2. While there is a relation schema Q in D that is not in BCNFdo
{
choose a relation schema Q in D that is not in BCNF;
find a functional dependency X ~ Y in Qthat violates BCNFj
replace Q in D by two relation schemas (Q - Y) and (X U Y);
};
No, we don't need to decompose R(A, B, C, D) for option D.
And the algorithm you posted is the correct one, and your decomposition for option C is also correct, that follows lossless join property but doesn't preservere dependency.
Should I take decomposed table as R1(BC) and R2(CAD)? what will be the situation now?
@Rajesh Raj , If we do so, then what should we take FDs for R1 ??
here Lossless is okay but dependecy(AB -> C) would not be preserved any more...
am I right ??
yup thats why we could also choose the option C because the question is asking "Which of the following does not have a lossless join, dependency preserving BCNF decomposition? ". whats say ??
perfect explanation :)
that's the beauty of gate overflow answer
@vijay @Rajesh raj Although it's very appropriate explanation i understood it very well,i have a small query
what if i decompose (c) in the table like this
R1(ABC) R2(ACD)
AB->C C->AD // here also doesn't have lossless join dependency preserving decomposition
AC common part which doesn't imply any key of both tables
m i correct ??
AC common part which doesn't imply any key of both tables.
First of all, the common part should be key for any table ( R1 or R2 ).( Both not necessary)
And No, It is not so. If C is alone a key of R2(C -> AD), then why not AC. ?
@vijay i was meant "ANY" of table
yeah AC would be super key bcoz union with candidate key would give super key
Okay : )
In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a superkey.
Why can't we decompose like R1 like (A,B,C) and R2 like (A,C,D)? Then both tables will be BCNF right?
Because for lossless decomposition common attribute has to be work as a key for any of the decomposed relation.
If we decompose as R1(ABC) and R2(CAD),then C is the common attribute and is key for R2. Why are we decomposing as R1(ABC),R2(CD)?
@amitqy
Both will give the same result.
Since given, AB$\rightarrow$C and C$\rightarrow$AD
So rewriting them, AB$\rightarrow$C, C$\rightarrow$A, C$\rightarrow$D
Though R2(CAD) is in BCNF as C$\rightarrow$AD is the only FD present but
R1(ABC) will have FDs AB$\rightarrow$C and C$\rightarrow$A
C$\rightarrow$A violates the BCNF property
There is one more decomposition possible for option B i.e R1(A,B), R2(A,D) , R3(B,C) which is also lossless but not dependency preserving .Now, which one to choose between option B and option C and why?
If we decompose as R1(ABC) and R2(CAD),then AC is the common attribute and AC is not a key in any of the two decomposed relations as the FD for R1 is AB->C and for R2 C->AD.
We can't have AC as a key to get lossless decomposition.
If a given relation is decomposed in more than 2 relations then how we can conclude that it is loseless join decomposition.??????
(How to identify and tackle such problem like option (B) in the above question)
so using this approach I can also state that option C is incorrect, right???
To get lossless decomposition, one of the common attribute must be candidate key in any one table. So i think, arguing "AC is common and AC is not a candidate key in any relation so decomposition does not have lossless join " is not correct.
We can take C as common attribute and C is candidate key in the relation $R_{2}$ (CAD) and it is enough to say that decompostion to $R_{1}$ (ABC) and $R_{2}$ (CAD) is lossless. But $R_{1}$ (ABC) is not in BCNF, refer Minipanda's explanation.
We can break C->AD as C->A and C->D, here only C-> D will get break in the process because it is not in 2NF and we will get decomposition as 1.(ABC) and 2.(CD). Please correct me if I am wrong.
(C) is the answer. Because of AB $\to$ C and C $\to$ A, we cannot have A, B and C together in any BCNF relation- in relation ABC, C is not a super key and C$\to$ A exists violating BCNF condition. So, we cannot preserve AB $\to$ C dependency in any decomposition of ABCD.
For (A) we can have AB, BCD, A and B the respective keys
For (B) we can have AB, BC, CD, A, B and C the respective keys
For (D) we can have ABCD, A is key
by
We are not given the candidate keys here so how can we determine that which decomposition is in BCNF....?Please explain....
explanation is not clear.
@dhingrak candidate keys won't be given. We have to find them from the FDs.
@Tamojit-Chatterjee I have edited the answer. Is it clear now?
hmmm thanx
@Arjun Sir ..I have a doubt regarding dependency preserving- if all attributes of a particular Functional Dependency are not present in any decomposition then we apply the dependency preserving algorithm...
http://www.cs.uakron.edu/~chanc/cs475/Fall2000/ExampleDp%20decomposition.htm
But its application is too lengthy ....is there any other short method..?
The steps of that algorithm is lengthy. But if you correctly apply it, it is very easy. I mean for most of the questions, the algorithm stops very fast.
That means we need to apply this algo everytime we need to check for functional dependency?
Yes. But it can be skipped if you are sure, we cannot check for any dependency without a join of the relations.
edited
Do we need to decompose option D? It is already in BCNF.
Also, can you post BCNF decomposition algorithm?
I am using algorithm 11.3 of Navathe book. But getting different decomposed relation in option C. Keys are AB and BC. So C -> AD not in BCNF. So, decomposition result as BC and CAD. Please comment.
Algorithm 11.3: Relational Decomposition into BCNF with Nonadditive Join Property
Input: A universal relation Rand a set of functional dependencies F on the attributes of R.
1. Set D := {R};
2. While there is a relation schema Q in D that is not in BCNFdo
{
choose a relation schema Q in D that is not in BCNF;
find a functional dependency X ~ Y in Qthat violates BCNFj
replace Q in D by two relation schemas (Q - Y) and (X U Y);
};
@Arjun sir can you give the intuition with this algorithm on the link. Will be helpful? not able to get it
@Arjun Sir, I think it is silly to ask but why can't we do like:
R(A,B,C,D)
R1(ABC) R2(CD)
AB->C, C->A //NOT IN BCNF C->D //IN BCNF
R11(ABC) R12(AC)
AB->C // IN BCNF C->A //IN BCNF
the decomposition you have ABC AC and CD lossless join and dependency preserving, but it is still not in BCNF. because in ABC $C\rightarrow A$ will still be implied.
@Shivansh Gupta thanks!!
edited
@Arjun sir @Bikram sir pls look at this once. I have a doubt on c option $AB\rightarrow C, C\rightarrow AD$.
We have 2 algorithms for decomposition,
1. BCNF decompositoin algorithm, according to this we will get two relations R1(CAD) with $C\rightarrow AD$ and R2(BC) , where $AB\rightarrow C$ is not preserved. So its not possible this way.
But given Schema is also not in 3NF so we can try 3NF atleast.
2. 3NF synthesis algorithm, which makes a relation for each dependency in the minimal cover of given relation A. This guarantees lossless and dependency preserving decompositoin. If we use this in C option then we will get 2 relations R1(ABC) with FD $AB\rightarrow C$ and R2(CAD) with FD $C\rightarrow AD$. So this is in 3NF according to the algorithm, But this also happens to be in BCNF.
I use this because sometimes through 3NF decomposition we get a BCNF decomposition. Also for $AB\rightarrowC, C\rightarrowA$ the 3NF algorithm gives no decomposition as expected.
Since given that C is wrong, does it mean that we must use only BCNF decompostition algorithm in this case.
Sir means it is possible that relation can never be in BCNF?
(A) A->B, B->CD
AB and BCD, B is the key of second and hence decomposition is lossless.
(B) A->B, B->C, C->D
AB, BC, CD B is the key of second and C is the key of third, hence lossless.
ABC, CD. C is key of second, but C->A violates BCNF condition in ABC as C is not a key. We cannot decompose ABC further as AB->C dependency would be lost. Hence the ANSWER.
(D) A ->BCD
by
Arjun suresh sir, in third option (CAD) nd (BC) lossless decomposition is also possible ???
Yes, that is lossless but AB->C, dependency is not preserved.
@ARJUN sir we have, R(A, B, C, D) and the Functional Dependency set = {AB→C, C→AD}.
we decompose it as R1(A, B, C) and R2(C, D). This preserves all dependencies and the join is lossless too, but the relation Ris not in BCNF. In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a superkey. Condition that all relations formed after decomposition should be in BCNF is not satisfied here..
why i can not do this R1(ABC) and R2(ACD) ??
here AC is common and it is not the key any of the R1 and R2 that's why it is not lossless.
Sir,can you share link for how to convert dependency to any normal form ??
R1(ABC) and R2(ACD) is also dependency preserving because AC is common and c is candidate key for R2
ans C)
Explain ur reason ...
### @Puja... i have a question... may be this is really silly but somehow i am not getting this question....
if possible help me..
R(A,B,C,D)
OPTION A : A ->B, B->CD which can be further written in to B->C , C->D
clearly from here in right hand side we dont have A, so the p.k. will surely have A
so lets find (A)+ --> ABCD ... so we get A is the only primary key right?
and according to the rule B,C,D are non-prime attributes. but in relation's F.D. we have B->C & B-->D where a non prime determines a non prime attribute, which clearly says the relation is not in 3NF also. Then how the hell the relation can be in BCNF?
@Puja ... Sorry to bother... got this :-P | 3,432 | 12,491 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-39 | latest | en | 0.865476 |
http://econpage.com/202/questions/daythirteen.html | 1,508,653,140,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825147.83/warc/CC-MAIN-20171022060353-20171022080353-00546.warc.gz | 105,574,944 | 1,290 | ## Question of the Day: Day Thirteen
Assume we have an economy where the Price Level is constant. We can describe that economy with the following equations (below).
C = 0.8(DI) + 2000 C = Consumption Expenditure, DI = Disposable Income I = 1000 I = Investment Expenditure G = 1200 G = Government Expenditure X = 500 X = Expenditure on Exports M = 500 M = Expenditure on Imports T = 1000 T = Tax Revenue DI = Y - T Y = real GDP
a. Using the equations above, how does the equilibrium change if G increases by 1000?
Assume that this country adopts a 10% flat income tax. The net result is that we replace the Tax equation with the following: T = 1000 + 0.1Y (where Y = real GDP). Find the new equilibrium real GDP and then answer the question below.
b. Using the new Tax equation and the remaining equations above, how does the new equilibrium change if G increases by 1000 now? | 231 | 881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | longest | en | 0.88945 |
https://www.aqua-calc.com/one-to-one/acceleration/microinch-per-hour-squared/thou-per-second-squared/1 | 1,606,253,818,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00569.warc.gz | 579,854,938 | 7,535 | # 1 microinch per hour squared [µin/h²] in thous per second squared
## microinches/hour² to thou/second² unit converter of acceleration
1 microinch per hour squared [µin/h²] = 7.716049394 × 10-11 thou per second squared [thou/s²]
### microinches per hour squared to thous per second squared acceleration conversion cards
• 1
through
25
microinches per hour squared
• 1 µin/h² to thou/s² = 7.716049394 × 10-11 thou/s²
• 2 µin/h² to thou/s² = 1.543209879 × 10-10 thou/s²
• 3 µin/h² to thou/s² = 2.314814818 × 10-10 thou/s²
• 4 µin/h² to thou/s² = 3.086419758 × 10-10 thou/s²
• 5 µin/h² to thou/s² = 3.858024697 × 10-10 thou/s²
• 6 µin/h² to thou/s² = 4.629629637 × 10-10 thou/s²
• 7 µin/h² to thou/s² = 5.401234576 × 10-10 thou/s²
• 8 µin/h² to thou/s² = 6.172839516 × 10-10 thou/s²
• 9 µin/h² to thou/s² = 6.944444455 × 10-10 thou/s²
• 10 µin/h² to thou/s² = 7.716049394 × 10-10 thou/s²
• 11 µin/h² to thou/s² = 8.487654334 × 10-10 thou/s²
• 12 µin/h² to thou/s² = 9.259259273 × 10-10 thou/s²
• 13 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 14 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 15 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 16 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 17 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 18 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 19 µin/h² to thou/s² = 1 × 10-9 thou/s²
• 20 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 21 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 22 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 23 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 24 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 25 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 26
through
50
microinches per hour squared
• 26 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 27 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 28 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 29 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 30 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 31 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 32 µin/h² to thou/s² = 2 × 10-9 thou/s²
• 33 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 34 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 35 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 36 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 37 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 38 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 39 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 40 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 41 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 42 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 43 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 44 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 45 µin/h² to thou/s² = 3 × 10-9 thou/s²
• 46 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 47 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 48 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 49 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 50 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 51
through
75
microinches per hour squared
• 51 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 52 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 53 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 54 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 55 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 56 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 57 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 58 µin/h² to thou/s² = 4 × 10-9 thou/s²
• 59 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 60 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 61 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 62 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 63 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 64 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 65 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 66 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 67 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 68 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 69 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 70 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 71 µin/h² to thou/s² = 5 × 10-9 thou/s²
• 72 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 73 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 74 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 75 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 76
through
100
microinches per hour squared
• 76 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 77 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 78 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 79 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 80 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 81 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 82 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 83 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 84 µin/h² to thou/s² = 6 × 10-9 thou/s²
• 85 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 86 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 87 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 88 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 89 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 90 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 91 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 92 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 93 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 94 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 95 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 96 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 97 µin/h² to thou/s² = 7 × 10-9 thou/s²
• 98 µin/h² to thou/s² = 8 × 10-9 thou/s²
• 99 µin/h² to thou/s² = 8 × 10-9 thou/s²
• 100 µin/h² to thou/s² = 8 × 10-9 thou/s²
#### Foods, Nutrients and Calories
MUSHROOM STEMS and PIECES, UPC: 075450126884 contain(s) 31 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Bermuda Pink weighs 1 281.5 kg/m³ (80.00143 lb/ft³) with specific gravity of 1.2815 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Propane [CH3CH2CH3] weighs 493 kg/m³ (30.77698 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Crambe oil with temperature in the range of 23.9°C (75.02°F) to 110°C (230°F)
#### Weights and Measurements
Imperial (UK) quart is a non-metric liquid measurement unit of volume [ Imperial quart ]
The density of a material or substance is defined as its mass per unit of volume.
sl/US c to t/l conversion table, sl/US c to t/l unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume, area and surface to volume ratio of a torus | 2,917 | 6,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.335242 |
https://investsense.com/2013/09/03/the-active-management-value-ratio-worksheet/ | 1,686,180,742,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00576.warc.gz | 352,017,156 | 25,874 | ## The Active Management Value Ratio 2.0 Worksheet
I have had a lot of people ask me if could provide a simple worksheet that would allow them to quickly calculate the AMVR. Ask and ye shall receive. I have also provided an example of a completed worksheet. All you have to do is obtain the necessary data from an online source such as morningstar.com or yahoo.finance.com and you’re good to go.
The Active Management Value Ratio™ allows investors to evaluate an actively managed mutual funds in terms of its efficency, both in terms of cost and performance. WordPress will not let me insert a table, but the steps are simple:
(1) Obtain the current expense ratios and the turnover ratios for the actively managed fund in question and an appropriate benchmark, such as the Vanguard S&P 500 Index.
(2) Calculate each fund’s trading costs (See note below). Add each fund’s trading costs to its annual expense ratio to get the funds total expenses
(3) Subtract the index fund’s total expense ratio from the fund’s total expense ratio to obtain the actively managed fund’s incremental cost, the added cost of the actively managed fund.
(4) Obtain the 5-year annualized return for the actively managed fund in question and the benchmark used in Step 1.
(5) Subtract the index fund’s 5-year annualized return from the fund’s 5-year annualized return to obtain the actively managed fund’s incremental benefit, the added benefit, if any, provided by the actively managed fund. If the fund does not provide any incremental benefit, the fund should be excluded from consideration.
(6) Calculate the actively managed fund’s AMVR™ score by dividing the actively managed fund’s incremental costs by the actively managed fund’s incremental benefit.
Example:
Actively managed fund: Annual Expense Ratio – 1.00%, Annual Turnover Ratio 100%, and 5-Year Annualized Rtn – 20%
Benchmark Fund: Annual Expense Ratio – 0.17%; Annual Turnover Ratio 3%, and 5-Year Annualized Rtn – 18%
Incremental Cost = Active (1.00 + 1.20) – Passive (0.17 + .03)= 2.00
Incremental Return = 20.00 – 18.00 = 2.00
AMVR™ Score = 2.00/2.00 = 1.00
Therefore, an investor is effectively paying an expense ratio of 100% for the active management component of the actively managed fund.
Another way of looking at the analysis is by an analogy. Which is more prudent, paying \$20 for an annual return of 18 percent, or paying \$200 dollars for an annual return of 2 percent?
Yet another way to analyze the AMVR™ Score is to compare cost to relative contribution. In our example, 90 percent of the actively managed fund’s cost is only providing 10 percent of the fund’s return.
Bottom line, as to these two investments, it would be hard to justify an investment in the actively managed fund as a prudent investment in comparison to the passively managed fund.
Note: Mutual funds are not required to provide the actual trading costs of the fund. To calculate trading costs for the purposes of the AMVR™, we used a formula created by John Bogle. Bogle’s formula is (Turnover Ratio x 2) x 0.60. In our example, (100 x 2) x 0.60 gives us the 1.20 number. Another method of computing trading costs would be to simply take 1 percent of the stated turnover ratio. The calculation would simply involve multiplying the stated turnover ratio by .o1. In our example, 100 x 0.01 would result in a trading cost of 1.00 percent
Annual Fees Total Annual Fees Annual Returns Actively Managed Fund 20.00 Annual Expense Ratio 1.00 Annual Trading Expenses 1.20 Total Active Expenses 2.20 Passively Managed Fund 18.00 Annual Expense Ratio 0.17 Annual Trading Expenses 0.03 Total Passive Expenses 0.20 Incremental Costs/Expenses 2.00 2.00
This article is neither designed nor intended to provide legal, investment, or other professional advice since such advice always requires consideration of individual circumstances. If legal, investment, or other professional assistance is needed, the services of an attorney or other professional advisor should be sought. | 929 | 4,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-23 | latest | en | 0.890823 |
http://www.fractii.ro/fractions-compared-results-explained.php?lang=en&compared_fractions=1/3vs.3/8&sorted_fractions=1/3%3C3/8 | 1,550,845,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247518425.87/warc/CC-MAIN-20190222135147-20190222161147-00076.warc.gz | 353,881,790 | 9,396 | # Compare, what fraction is larger? 1/3 vs. 3/8. Fractions sorted in ascending order: 1/3 < 3/8. Ordinary math fractions compared, result explained below
## Latest fractions compared or sorted in ascending order
1/3 < 3/8 Feb 22 14:30 UTC (GMT) 9/14 < 5/7 Feb 22 14:30 UTC (GMT) 8/9 < 9/10 Feb 22 14:30 UTC (GMT) 7/16 < 1/2 Feb 22 14:30 UTC (GMT) 2/3 < 7/10 < 4/5 Feb 22 14:30 UTC (GMT) 3/8 < 1/2 Feb 22 14:30 UTC (GMT) 1/3 < 3/2 < 9 < 27 Feb 22 14:30 UTC (GMT) 1/36 < 1/6 Feb 22 14:30 UTC (GMT) 5/8 < 3/4 < 5/6 Feb 22 14:30 UTC (GMT) 15/6 < 17/6 Feb 22 14:30 UTC (GMT) 1/11 < 1/10 < 1/9 Feb 22 14:30 UTC (GMT) 2/3 < 15/20 Feb 22 14:30 UTC (GMT) 19/25 < 4/5 Feb 22 14:30 UTC (GMT) see more... compared fractions see more... sorted fractions
## How to compare two fractions?
### 1. EQUAL DENOMINATORS but unlike numerators fractions
• a) To compare two positive fractions that have EQUAL DENOMINATORS (like denominators) but different numerators (unlike numerators), simply compare the numerators: the larger fraction is the one with the larger numerator, ie: 24/25 > 19/25
• b) To compare two negative fractions that have EQUAL DENOMINATORS (like denominators) but different numerators (unlike numerators), simply compare the numerators: the larger fraction is the one with the smaller numerator, ie: -19/25 < -17/25
• c) To compare two fractions of different signs (one positive and one negative) that have EQUAL DENOMINATORS (like denominators) but different numerators (unlike numerators), the rule is that any positive fraction is larger than any negative fraction, ie: 2/25 > -1/25
### 2. EQUAL NUMERATORS but unlike denominators fractions
• a) To compare two positive fractions that have EQUAL NUMERATORS (like numerators) but different denominators (unlike denominators), simply compare the denominators: the larger fraction is the one with the smaller denominator, ie: 24/25 > 24/26
• b) To compare two negative fractions that have EQUAL NUMERATORS (like numerators) but different denominators (unlike denominators), simply compare the denominators: the larger fraction is the one with the larger denominator, ie: -17/25 < -17/29
• c) To compare two fractions of different signs (one positive and one negative) that have EQUAL NUMERATORS (like numerators) but different denominators (unlike denominators), the rule is that any negative fraction is smaller than any positive fraction, ie: -1/25 < 1/200
### 3. Different denominators and numerators (unlike denominators and unlike numerators) fractions
• a) To compare two fractions of the same sign (both positive or both negative) that have different denominators and numerators (unlike denominators and unlike numerators), fractions should be brought to the same denominator (or if it's easier, to the same numerators). Please see the next paragraph, 3.a)
• b) To compare two fractions of different signs (one positive and one negative) that have different denominators and numerators (unlike denominators and unlike numerators), the rule is that any negative fraction is smaller than any positive fraction, ie: -11/24 < 10/13 | 896 | 3,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-09 | latest | en | 0.785545 |
https://homework.cpm.org/category/CC/textbook/cca/chapter/10/lesson/10.2.1/problem/10-32 | 1,620,451,003,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00255.warc.gz | 316,495,578 | 14,937 | ### Home > CCA > Chapter 10 > Lesson 10.2.1 > Problem10-32
10-32.
Jessica has three fewer candies than twice the number Dante has.
1. If Dante has $d$ candies, write an expression to represent how many candies Jessica has.
$2d-3$
2. If Jessica has $19$ candies, write and solve an equation to find out how many candies Dante has.
$11=d$ | 95 | 342 | {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | longest | en | 0.828104 |
https://mathematica.stackexchange.com/questions/221985/how-can-we-generate-this-more-general-form-of-the-continued-fraction | 1,632,113,559,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057018.8/warc/CC-MAIN-20210920040604-20210920070604-00249.warc.gz | 435,055,349 | 33,199 | # How can we generate this more general form of the continued fraction? [duplicate]
How can I write Mathematica code for this continued fraction with alternating terms?
How can we generate this more general form of the above-continued fraction?
$$x+\cfrac{a}{y+\cfrac{a^2}{x+\cfrac{a^3}{y+\cfrac{a^4}{x+\cfrac{a^5}{y+\cfrac{a^6}{x+\cdots}}}}}}$$
• Yes thank you so much for your help. May 16 '20 at 20:26
Once again, you can use ContinuedFractionK[]:
With[{n = 7},
x + ContinuedFractionK[a^k, Piecewise[{{y, Mod[k, 2] == 1}}, x], {k, 1, n}]]
$$x+\cfrac{a}{y+\cfrac{a^2}{x+\cfrac{a^3}{y+\cfrac{a^4}{x+\cfrac{a^5}{y+\cfrac{a^6}{x+\cfrac{a^7}{y}}}}}}}$$
z = {x, y};
i = 1;
n = 6;
Style[x + Nest[a^(n - i++)/(z[[Mod[i, 2, 1]]] + #) &, …, n - 1], 32,
ScriptSizeMultipliers -> 1]
• Thank you very much. May 16 '20 at 20:27 | 332 | 827 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.6741 |
https://de.scribd.com/document/130870878/2b365Ant-Tute | 1,563,657,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526670.1/warc/CC-MAIN-20190720194009-20190720220009-00051.warc.gz | 368,289,417 | 59,407 | You are on page 1of 13
# TUTORIAL SHEET -1
Q1) Point charges 1 mC and -2mC are located at (3.2,-1) and (-1,-1, 4) respectively. Calculate electric force on a -10 nC charge located at (0, 3, 1) and electric field at that point. Q2) Point charges 5nC and -2nC are located at (2, 0, 4) and (-3, 0, 5) respectively. a) Determine the force on 1nC point charge located at (1,-3, 7). b) Find electric field E at (1,-3, 7). Q3) Two charges Q1=2C and Q2=5C are located at (-3,7,-4) and (2,4,-1) respectively. Determine the force on Q2 due to Q1 and force on Q1 due to Q2. Q4) A point charge Q=10nC is at the origin in free space. Find electric field at P(1,0,1).Also find electric flux density at P.
2 Q5) If electric flux density D is given by D = [(2 y + z )a x + 4 xya y + xa z ] C / m 2 .Find volume charge density at (0,0,0)and (-1,0,4).
Q6) Find the potential and volume charge density at P(0.5,1.5,1) in free space given the potential field a) V = 2 x 2 y 2 z 2 volts b) V=6z volts c) V= 5(2r 2 7) coscos volts d) V=3x-y volts Q7) Let D = 2 z 2 a +cos 2 a z .Evaluate a) D.ds
S
b) Ddv
V
over the region defined by 0 5,1 z 1,0 2 . Q8) electric field is given 2 D = 2 ( z + 1) cos a ( z + 1) sin a + cos a z C/m a) Find charge density b) Calculate charge enclosed by volume 0<<2, 0<</2,0<z<4. c) Confirm gauss law by finding the net flux through the surface of volume in b.
2
In
certain
region
the
by
Q9) In free space V = x 2 y ( z + 3) volts. Find a) E at (3, 4, -6) b) the charge within the cube 0<x, y ,z<1.
TUTORIAL SHEET -2
Q1) Find the total charge inside each of the volume indicated 2 0.1 x sin y,0 y 1,1 x 2,3 z 3.6 a) V = 10 z e b) V = 4 xyz 2 ,0 2,0 / 2,0 z 3 Q2) In Free space let Q1=10nC be at P1(0,-4,0) and Q2=20nC be at P2(0,0,4). a) Find E at the origin. b) Where should a 30nC point charge be located so that E is zero at the origin. Q3) A point charge QA=1C is at A(0,0,1) and QB=-1C is at B (0,0,-1). Find E r, E and E at P (1,2,3). nC/m 2 in free space. D = a r Q4. Let a) Find E at r=0.2m b) find total charge with in sphere r=0.2m. pC/m2 in free space. D = y 2 z 3 a x + 2 xyz 3 a y + 3 xy 2 z 2 a z Q5. Let a) Find total electric flux passing the surface defined by x=3, 0 y 2 , 0 z 1 in the direction away from origin. b) Find | E | at P (3, 2, 1). c) Find the total charge enclosed in an incremental sphere having radius of 2 m centered at P (3, 2, 1). Q6. If = 0 and V= 8x2yz find a) V at P(2,-1,3) b) v at P c) E at P. d) Does V satisfy Laplaces equation? Q7. A homogenous dielectric ( r = 2.5 ) fills region 1( x 0 ) while region-2( x 0 ) is a free space. a) If D1=12ax -10ay+4az nC/m2 find D2 and 2 b) if E2=12 volt/m and 2 =600 find E1 and 1. Take 1 and 2 as the angles which E1 and E2 makes with normal to the interface. Q8) The vector magnetic potential A due to direct current in a conductor in free space is given by A = ( x 2 + y 2 )a z Wb/m2.Determine the magnetic filed produced by current element at (1,2,3).
20( xa x + ya y ) Q9) A certain magnetic field intensity is given by H = A/m in free space. x2 + y2
r 3
## c) Find A (x, y, z) if Ax =Ay = 0 and Az = 0 at P(1,1,1). Q10) In a certain conducting region
H = yz ( x 2 + y 2 )a x y 2 xza y + 4 x 2 y 2 a z A/m.
a) Determine J at (5,2,-3) 1,0 <y, z<2. b) Find current passing through x = c) Show that B = 0.
TUTORIAL SHEET -3
Q1) Given the magnetic flux density B = 6 cos 10 6 t sin 0.01xa z mT. Find a) the magnetic flux passing through the surface defined z=0,0<x<20m,0<y<3m at t=1microseconds. by
b)
The value of closed line integral of E around the perimeter of the surface specified above at t=1microseconds.
Q2) In free space E = 20 cos( wt 50 x)a y V/m calculate a) JD , displacement current density b) H i.e Magnetic field c) B i.e Magnetic flux Density d) , frequency of elctromagnetic field wave. Q3) material of infinite extent with 9 = 2 10 F / m, =1.25 10 H / m, and =0.Let E = 400 cos(10 t kz ) a y V/m. If all the fields vary sinusoidally use maxwells equation to find D, B, H and k.
10 5
Assume
homogenous
Q4) A certain material has =0 and r=1.If E = 800 sin(10 6 t 0.01z ) a y V/m. Make use of the maxwells equation to find a) r b) H ( z , t ) Q5) In free space E = 20 cos(t 50 x)a y V/m. Calculate a) JD b)H c)
7 Q6) In air E = sin cos(6 10 t r )a V/m. Find and H. r Q7) In a charge free region for which =0, = r and = , H = 5 cos(1011 t 4 y ) a z A/m. Find a) J D and D b) r
Q8) In a certain Region with =0,= and = 6.25 the magnetic field of wave is 8 H = 0.6 cos x cos10 ta z A / m . Find and corresponding E using Maxwells equation. Q9) In free space
H = (sin a + 2 cos a ) cos(4 10 6 t ) A/m. Find JD and E.
r
## Find corresponding E in terms of r and .
TUTORIAL SHEET -4
1. The radiation resistance of an antenna is 80 ohm and loss resistance is 10ohm.What is the Directivity, if the power gain is 20? 2. Calculate the radiation Resistance of /20 dipole in free space.
3. Calculate the maximum Effective aperture of an antenna which is operating at wavelength of two meters and a directivity of 100. 4. Find out radiation resistance of l/8 wire dipole in free space. 5. An antenna has a loss resistance 10ohm, power gain of 20 and directivity of 22.Calculate its radiation resistance. 6. Calculate the maximum effective aperture of microwave antenna which has Directivity of 900. 7. Two vertically oriented half wave dipoles are spaced 1.5 apart to form an array .Calculate half power bandwidth of the major lobes of the array in horizontal plane, for the following two cases, when the dipoles are fed with: 8. a)equal and in phase current (broadside array) and 9. b)equal current but with a phase difference of 540 between the two current(End fire Array) 10. In the radiation pattern of a 3 element array of isotropic radiators equally spaced at distances l/4; it is required to place a null at an angle 33.56 degree of the endfire direction .Calculate the progressive phase shifts to be applied to the elements. Also calculate the angle at which the mail beam is placed for this phase distribution.
TUTORIAL SHEET -5
1. Calculate the radiation resistance of / 10 wire dipole in free space. 2. A maximum current carried by / 40 antenna is 125 amp. Calculate the power radiated by this antenna and its radiation resistance.
3. In Hertzian dipole, the radiation and induction fields have equal amplitude at distance. Justify the statement. 4. The power radiated from the dipole antenna is maximum at right angle to the axis of the antenna. 5. Explain the effective length of an antenna as a radiator of electromagnetic energy. 6. Drive an expression for the vector potential Az at a point P located at a large distance r from a half wave dipole placed along the Z- axis. 7. What do you mean by the radiation resistance of an antenna? What is the nature of the current distribution in a base fed half wave vertical antenna erected just above a perfect earth? 8. Calculate efficiency and total resistance of / 16 dipole antenna if the loss resistance is 1.8. 9. The total resistance of an antenna having effective height of 62 meters is 48.5. The rms current of antenna is 40 amperes at 550 kHz. Calculate: I. Radiation resistance of antenna II. III. Radiation power from this antenna Efficiency of antenna
## 10. Show that the directivity of isotropic antenna is unity.
TUTORIAL SHEET -6
1. Compute the effective length of a half wave dipole. 2. Establish that
where the terms have their usual meaning. 3. Show that the effective area and effective length of an antenna are related by . 4. The power delivered to an isotropic radiator is 1KW and antenna efficiency is 90%. Find the electric field intensity at a distance of 100 km. 5. A directional antenna has an effective radiated power of 1.1kW, when it is fed with a terminal input power of 90 watts. The radiation resistance is 74 at resonance and the measured antenna current is 1.088 amp rms. Find a) The antenna efficiency b) The antenna power loss c) The directive gain in decibels over an isotropic radiator. 6. What is Radiation Resistance of an antenna? Show that the radiation resistance of Half Wave dipole is 73ohms. 7. A transmitting antenna with an effective height of 100 meters has a current at the base 100 amperes (rms) at the frequency of 300 KHz. a) Find the Field strength at a distance of 10Km. b) Find the Power Radiated.
TUTORIAL SHEET -7
1. The electric field in free space is given by E = 50 cos(108 t + x)a y V/m a. Find the direction of wave propagation. b. Calculate and time it takes to travel a distance of /2. c. Sketch the wave at 0, T/4& T/2 sec.
30 at a particular frequency. 2. A lossy dielectric has an intrinsic impedance of 20 If at that frequency, the plane wave propagating through the dielectric has the magnetic field component
H= 10e-x cos ( t x/2) ay A/m Find E and . Determine the skin depth and wave polarization. 3. In a lossless medium for which = 60, r =1 and H= - 0.1cos ( t-z)ax +0.5 sin ( t-z) ay A/m. Calculate r, and E. 4. In free space (z 0), a plane with H = 10 cos (10 8t-x) ax mA/m is incident normally on a lossless medium ( = 2 , = 8 ) in region (z 0). Determine the reflected wave Hr Er and the transmitted wave Ht, Et. 5. The plane wave E = 50 sin (t - 5 x ) a y V/m in a lossless medium ( = 4 , = ) encounters a lossy medium( = , = , = 0.1mho / m ) normal to x-axis at x=0. Find (a) Reflection and transmission coefficient (b) E r and H r (c) E t and H t 6. A lossy dielectric is characterized by r = 2.5, r = 4 and =10 -3 MHz. Let E s = 20 e y z a x V/m then find (a) attenuation constant (b) Phase constant (c) Wave velocity u (d) Wavelength (e) Intrinsic impedance (f) H s (g) E (2,3,4,10ns) mho/m at 10
7. A uniform plane wave in free space is propagation in the -ay direction at a frequency of 10 MHz. If E = 400cos t ax V/m at y=0.write expressions for (a) E (x,y,z,t) (b) E s(x,y,z) (c) H s(x,y,z) (d) H (x,y,z,t) 8. A plane wave propagating through a medium with r = 8, r = 2 has E =0.5e-z/3 sin(108 t- z)ax V/m. Determine (a) (b)Loss tangent (c) wave impedance
(d) wave velocity (e) H field 9. The magnetic field component of a plane wave in a lossless dielectric is H=30 sin (2 108t-5x)az mA/m (a) If r =1 find r (b) Calculate the wavelength and wave velocity (c) Determine the polarization of wave. (d) Find the corresponding electric field component. (e) Find displacement current density. 10. In a certain medium E=10 cos(2 107t- x)(ay+ az) V/m If =50 , = 2 and = 0 find and H.
TUTORIAL SHEET -8
1. Calculate the maximum Aperture of a /2 (Half wave) Antenna. 2. Calculate the Effective length of an Antenna. Given Rr= 73 ohms, Ae(max)= 0.13 ^2 and =120. 3. Calculate the maximum effective aperture of an antenna which is operating at a wavelength 0f 2 meters and has a directivity of 100.
4. Calculate the gain of an Antenna with a circular aperture of diameter 3 meters at a frequency of 5 GHz. 5. A thin dipole antenna is /15 long. If its loss resistance is 1.5 ohms, find radiation resistance and the efficiency. 6. Calculate the radiation resistance of an antenna which is radiating 1000 watts and drawing a current of 5 amps. 7. Calculate the FBR (Front to Back) ratio of an antenna in DB which radiates 3 KW in its most optimum direction and 500 Watts in the Opposite direction. 8. An antenna has loss resistance 10 ohms, power gain of 20 and directivity 22. Calculate its radiation resistance. 9. Find the gain, Beamwidth and capture area for a parabolic antenna with a 6 meters diameter dish and dipole feed at a frequency of 10 GHz. 10. The Noise figure of an amplifier at room temperature (T=290 K) is 0.2 DB. Find the equivalent temperature.
TUTORIAL SHEET -9
1. Calculate the radiation resistance of 2. A maximum current carried by wire dipole in free space. antenna is 125 amp. Calculate the power
radiated by this antenna and its radiation resistance. 3. In Hertzian dipole, the radiation and induction fields have equal amplitude at distance. Justify the statement. 4. The power radiated from the dipole antenna is maximum at right angle to the axis of the antenna. 5. Explain the effective length of an antenna as a radiator of electromagnetic energy. 6. Drive an expression for the vector potential Az at a point P located at a large distance r from a half wave dipole placed along the Z- axis. 7. What do you mean by the radiation resistance of an antenna? What is the nature of the current distribution in a base fed half wave vertical antenna erected just above a perfect earth? 8. Calculate effeciency and total resistance of is 1.8. 9. The total resistance of an atenna having effective height of 62 metres is 48.5. The rms current of antenna is 40 amp. at 550 kHz. Calculate: a) Radiation resistance of antenna b) Radiation power from this antenna c) Efficiency of antenna 10. Show that the directivity of isotropic antenna is unity. dipole antenna if the loss resistance
## TUTORIAL SHEET -10
1. Compute the effective length of a half wave dipole. 2. Establish that Where the terms have their usual meaning. 3. Show that the effective area and effective length of an antenna are related by . 4. The power delivered to an isotropic radiator is 1KW and antenna effeiciency is 90%. Find the electric field intensity at a distance of 100 km. 5. A directional antenna has an effective radiated power of 1.1kW, when it is fed with a terminal input power of 90 watts. The radiation resistance is 74 at resonance and the measured antenna current is 1.088 amp rms. Find a) The antenna efficiency b) The antenna power loss c) The directive gain in decibels over an isotropic radiator. 6. (a) What is Radiation Resistance of an antenna ? (b) Show that the radiation resistance of Half Wave dipole is 73ohms. 7. A transmitting antenna with an effective height of 100 meters has a current at the base 8. 100 amp(rms) at the frequency of 300KHz. 9. Find the Field strength at a distance of 10Km. 10. Find the Power Radiated. | 4,255 | 13,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-30 | latest | en | 0.83514 |
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# How to Generate mesh in Python with Gmsh module?
In this article, we will cover how to Generate meshes using Gmsh module in Python.
## What is Mesh?
A connected 2D, 3D, or multi-dimension structure that is made up of points, lines, and curves is called mesh. In real-world meshes are made up of many different materials such as metals, fibers, and ductile materials. For example, the image is shown below:
3D mesh:
### Methods:
After understanding mesh and functions it’s time to create the above mesh. So without further delay, let’s jump right in.
## Module required:
Gmsh library: It is a script of promotion and API Python wrapper for gmsh.
`pip install gmsh`
Sys library: It is a script of promotion and API Python wrapper for sys.
`pip install sys`
## Stepwise implementation
Step1: Import modules, initialize gmsh.
## Python3
`# Import modules:``import` `gmsh``import` `sys` `# Initialize gmsh:``gmsh.initialize()`
Step2: The above mesh is made up of three shapes a cube and two pentagons. So first we create a cube then both pentagons. To create a cube we need to create 8 points, 12 edges/lines, 6 faces, and their surfaces. To create points use the code given below.
## Python3
`# Import modules:``import` `gmsh``import` `sys` `# Initialize gmsh:``gmsh.initialize()` `# cube points:``lc ``=` `1e``-``2``point1 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``0``, lc)``point2 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``0``, lc)``point3 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``0``, lc)``point4 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``0``, lc)``point5 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``1``, lc)``point6 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``1``, lc)``point7 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``1``, lc)``point8 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``1``, lc)` `# Create the relevant Gmsh data structures``# from Gmsh model.``gmsh.model.geo.synchronize()` `# Generate mesh:``gmsh.model.mesh.generate()` `# Write mesh data:``gmsh.write(``"GFG.msh"``)` `# Creates graphical user interface``if` `'close'` `not` `in` `sys.argv:`` ``gmsh.fltk.run()` `# It finalize the Gmsh API``gmsh.finalize()`
Output:
Step3: After creating points, we are able to create lines from these points using the add_line() method. To create lines use the code given below.
## Python3
`# Import modules:``import` `gmsh``import` `sys` `# Initialize gmsh:``gmsh.initialize()` `# cube points:``lc ``=` `1e``-``2``point1 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``0``, lc)``point2 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``0``, lc)``point3 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``0``, lc)``point4 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``0``, lc)``point5 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``1``, lc)``point6 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``1``, lc)``point7 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``1``, lc)``point8 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``1``, lc)` `# Edge of cube:``line1 ``=` `gmsh.model.geo.add_line(point1, point2)``line2 ``=` `gmsh.model.geo.add_line(point2, point3)``line3 ``=` `gmsh.model.geo.add_line(point3, point4)``line4 ``=` `gmsh.model.geo.add_line(point4, point1)``line5 ``=` `gmsh.model.geo.add_line(point5, point6)``line6 ``=` `gmsh.model.geo.add_line(point6, point7)``line7 ``=` `gmsh.model.geo.add_line(point7, point8)``line8 ``=` `gmsh.model.geo.add_line(point8, point5)``line9 ``=` `gmsh.model.geo.add_line(point4, point5)``line10 ``=` `gmsh.model.geo.add_line(point6, point1)``line11 ``=` `gmsh.model.geo.add_line(point7, point2)``line12 ``=` `gmsh.model.geo.add_line(point3, point8)` `# Create the relevant Gmsh data structures``# from Gmsh model.``gmsh.model.geo.synchronize()` `# Generate mesh:``gmsh.model.mesh.generate()` `# Write mesh data:``gmsh.write(``"GFG.msh"``)` `# Creates graphical user interface``if` `'close'` `not` `in` `sys.argv:`` ``gmsh.fltk.run()` `# It finalize the Gmsh API``gmsh.finalize()`
Output:
Step 4: Next we create faces and surfaces but before that one should understand two methods more clearly that are add_curve_loop() and add_plane_surface(). In the above output, it looks like faces are generated but it is not true. So to create solid faces we use the add_curve_loop() then we use the add_plane_surface() method so that we get a solid face with different colors.
• add_curve_loop([line1, line2, line3, line4]): It required a list of lines in close loop format. For example: [line1(point1 to point2) -> line2(point2 to point3) -> line3(point3 to point4) -> line4(point4 to point1)], In this way point1 is connected to point4. So a face of the above square is produced. If we are moving in opposite directions of line i.e line1 = (point1 to point2) and if we are moving from (point2 to point1) then line1 is written as “-line1” in add_curve_loop() function because of the opposite direction.
• add_plane_surface(): If only one face is passed as a parameter then the function creates an only surface on the face. If two faces are passed as parameters in the list then the function connects faces if it is possible i.e if the faces lie in the same plane.
After understanding both methods cube can be created using the given code below.
## Python3
`# Import modules:``import` `gmsh``import` `sys` `# Initialize gmsh:``gmsh.initialize()` `# cube points:``lc ``=` `1e``-``2``point1 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``0``, lc)``point2 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``0``, lc)``point3 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``0``, lc)``point4 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``0``, lc)``point5 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``1``, lc)``point6 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``1``, lc)``point7 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``1``, lc)``point8 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``1``, lc)` `# Edge of cube:``line1 ``=` `gmsh.model.geo.add_line(point1, point2)``line2 ``=` `gmsh.model.geo.add_line(point2, point3)``line3 ``=` `gmsh.model.geo.add_line(point3, point4)``line4 ``=` `gmsh.model.geo.add_line(point4, point1)``line5 ``=` `gmsh.model.geo.add_line(point5, point6)``line6 ``=` `gmsh.model.geo.add_line(point6, point7)``line7 ``=` `gmsh.model.geo.add_line(point7, point8)``line8 ``=` `gmsh.model.geo.add_line(point8, point5)``line9 ``=` `gmsh.model.geo.add_line(point4, point5)``line10 ``=` `gmsh.model.geo.add_line(point6, point1)``line11 ``=` `gmsh.model.geo.add_line(point7, point2)``line12 ``=` `gmsh.model.geo.add_line(point3, point8)` `# faces of cube:``face1 ``=` `gmsh.model.geo.add_curve_loop([line1, line2, line3, line4])``face2 ``=` `gmsh.model.geo.add_curve_loop([line5, line6, line7, line8])``face3 ``=` `gmsh.model.geo.add_curve_loop([line9, line5, line10, ``-``line4])``face4 ``=` `gmsh.model.geo.add_curve_loop([line9, ``-``line8, ``-``line12, line3])``face5 ``=` `gmsh.model.geo.add_curve_loop([line6, line11, ``-``line1, ``-``line10])``face6 ``=` `gmsh.model.geo.add_curve_loop([line11, line2, line12, ``-``line7])` `# surfaces of cube:``gmsh.model.geo.add_plane_surface([face1])``gmsh.model.geo.add_plane_surface([face2])``gmsh.model.geo.add_plane_surface([face3])``gmsh.model.geo.add_plane_surface([face4])``gmsh.model.geo.add_plane_surface([face5])``gmsh.model.geo.add_plane_surface([face6])` `# Create the relevant Gmsh data structures``# from Gmsh model.``gmsh.model.geo.synchronize()` `# Generate mesh:``gmsh.model.mesh.generate()` `# Write mesh data:``gmsh.write(``"GFG.msh"``)` `# Creates graphical user interface``if` `'close'` `not` `in` `sys.argv:`` ``gmsh.fltk.run()` `# It finalize the Gmsh API``gmsh.finalize()`
Output:
Step5: Now the final step is to create both pentagons and connect them. To do so one should need to create their points and lines using proper measurement it is better to draw figures on paper first then code points according. Here we create two pentagons one bigger and one smaller using add_point(), add_line() functions. After that, we will create surfaces of both the pentagons using add_curve_loop() and then connect the bigger and smaller pentagon face with the cube face using add_plane_surface. At last, we will connect both pentagon’s faces with each other using the add_plane_surface() function.
## Python3
`# Import modules:``import` `gmsh``import` `sys` `# Initialize gmsh:``gmsh.initialize()` `# cube points:``lc ``=` `1e``-``2``point1 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``0``, lc)``point2 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``0``, lc)``point3 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``0``, lc)``point4 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``0``, lc)``point5 ``=` `gmsh.model.geo.add_point(``0``, ``1``, ``1``, lc)``point6 ``=` `gmsh.model.geo.add_point(``0``, ``0``, ``1``, lc)``point7 ``=` `gmsh.model.geo.add_point(``1``, ``0``, ``1``, lc)``point8 ``=` `gmsh.model.geo.add_point(``1``, ``1``, ``1``, lc)` `# Edge of cube:``line1 ``=` `gmsh.model.geo.add_line(point1, point2)``line2 ``=` `gmsh.model.geo.add_line(point2, point3)``line3 ``=` `gmsh.model.geo.add_line(point3, point4)``line4 ``=` `gmsh.model.geo.add_line(point4, point1)``line5 ``=` `gmsh.model.geo.add_line(point5, point6)``line6 ``=` `gmsh.model.geo.add_line(point6, point7)``line7 ``=` `gmsh.model.geo.add_line(point7, point8)``line8 ``=` `gmsh.model.geo.add_line(point8, point5)``line9 ``=` `gmsh.model.geo.add_line(point4, point5)``line10 ``=` `gmsh.model.geo.add_line(point6, point1)``line11 ``=` `gmsh.model.geo.add_line(point7, point2)``line12 ``=` `gmsh.model.geo.add_line(point3, point8)` `# faces of cube:``face1 ``=` `gmsh.model.geo.add_curve_loop([line1, line2, line3, line4])``face2 ``=` `gmsh.model.geo.add_curve_loop([line5, line6, line7, line8])``face3 ``=` `gmsh.model.geo.add_curve_loop([line9, line5, line10, ``-``line4])``face4 ``=` `gmsh.model.geo.add_curve_loop([line9, ``-``line8, ``-``line12, line3])``face5 ``=` `gmsh.model.geo.add_curve_loop([line6, line11, ``-``line1, ``-``line10])``face6 ``=` `gmsh.model.geo.add_curve_loop([line11, line2, line12, ``-``line7])` `# surfaces of cube:``gmsh.model.geo.add_plane_surface([face1])``gmsh.model.geo.add_plane_surface([face2])``gmsh.model.geo.add_plane_surface([face3])``gmsh.model.geo.add_plane_surface([face4])``gmsh.model.geo.add_plane_surface([face5])``gmsh.model.geo.add_plane_surface([face6])` `# Points of bigger petagon:``point9 ``=` `gmsh.model.geo.add_point(``0.3``, ``0.3``, ``-``2``, lc)``point10 ``=` `gmsh.model.geo.add_point(``0.7``, ``0.3``, ``-``2``, lc)``point11 ``=` `gmsh.model.geo.add_point(``0.7``, ``0.5``, ``-``2``, lc)``point12 ``=` `gmsh.model.geo.add_point(``0.5``, ``0.7``, ``-``2``, lc)``point13 ``=` `gmsh.model.geo.add_point(``0.3``, ``0.5``, ``-``2``, lc)` `# Points of smaller petagon:``point14 ``=` `gmsh.model.geo.add_point(``0.4``, ``0.4``, ``2``, lc)``point15 ``=` `gmsh.model.geo.add_point(``0.6``, ``0.4``, ``2``, lc)``point16 ``=` `gmsh.model.geo.add_point(``0.6``, ``0.5``, ``2``, lc)``point17 ``=` `gmsh.model.geo.add_point(``0.5``, ``0.6``, ``2``, lc)``point18 ``=` `gmsh.model.geo.add_point(``0.4``, ``0.5``, ``2``, lc)` `# lines of bigger pentagon:``line13 ``=` `gmsh.model.geo.add_line(point9, point10)``line14 ``=` `gmsh.model.geo.add_line(point10, point11)``line15 ``=` `gmsh.model.geo.add_line(point11, point12)``line16 ``=` `gmsh.model.geo.add_line(point12, point13)``line17 ``=` `gmsh.model.geo.add_line(point13, point9)` `# lines of smaller pentagon:``line18 ``=` `gmsh.model.geo.add_line(point14, point15)``line19 ``=` `gmsh.model.geo.add_line(point15, point16)``line20 ``=` `gmsh.model.geo.add_line(point16, point17)``line21 ``=` `gmsh.model.geo.add_line(point17, point18)``line22 ``=` `gmsh.model.geo.add_line(point18, point14)` `# face of bigger pentagon.``face7 ``=` `gmsh.model.geo.add_curve_loop([line13, line14, line15, line16, line17])` `# face of smaller pentagon.``face8 ``=` `gmsh.model.geo.add_curve_loop([line18, line19, line20, line21, line22])` `# connection of cube faces with pentagon``# and bigger pentagon with smaller.``gmsh.model.geo.add_plane_surface([face1, face7])``gmsh.model.geo.add_plane_surface([face2, face8])``gmsh.model.geo.add_plane_surface([face7, face8])` `# Create the relevant Gmsh data structures``# from Gmsh model.``gmsh.model.geo.synchronize()` `# Generate mesh:``gmsh.model.mesh.generate()` `# Write mesh data:``gmsh.write(``"GFG.msh"``)` `# Creates graphical user interface``if` `'close'` `not` `in` `sys.argv:`` ``gmsh.fltk.run()` `# It finalize the Gmsh API``gmsh.finalize()`
Output:
My Personal Notes arrow_drop_up | 4,050 | 12,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-23 | latest | en | 0.688858 |
https://www.cpalms.org/Public/PreviewCourse/Preview/13049?isShowCurrent=false' | 1,603,792,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00208.warc.gz | 668,600,760 | 41,966 | # Algebra 1-B for Credit Recovery (#1200385)
## General Course Information and Notes
### Version Description
Special Notes: Credit Recovery courses are credit bearing courses with specific content requirements defined by Next Generation Sunshine State Standards and/or Florida Standards. Students enrolled in a Credit Recovery course must have previously attempted the corresponding course (and/or End-of-Course assessment) since the course requirements for the Credit Recovery course is exactly the same as the previously attempted corresponding course. For example, Geometry (1206310) and Geometry for Credit Recovery (1206315) have identical content requirements. It is important to note that Credit Recovery courses are not bound by Section 1003.436(1)(a), Florida Statutes, requiring a minimum of 135 hours of bona fide instruction (120 hours in a school/district implementing block scheduling) in a designed course of study that contains student performance standards, since the students have previously attempted successful completion of the corresponding course. Additionally, Credit Recovery courses should ONLY be used for credit recovery, grade forgiveness, or remediation for students needing to prepare for an End-of-Course assessment retake.
### General Notes
The fundamental purpose of this course is to formalize and extend the mathematics that students learned in the middle grades. The critical areas, called units, deepen and extend understanding of linear and exponential relationships by contrasting them with each other and by applying linear models to data that exhibit a linear trend, and students engage in methods for analyzing, solving, and using quadratic functions. The Standards for Mathematical Practice apply throughout each course and, together with the content standards, prescribe that students experience mathematics as a coherent, useful, and logical subject that makes use of their ability to make sense of problem situations.
Algebra 1A (Year 1)
Unit 1- Relationships Between Questions and Reasoning with Equations: By the end of eighth grade, students have learned to solve linear equations in one variable and have applied graphical and algebraic methods to analyze and solve systems of linear equations in two variables. Now, students analyze and explain the process of solving an equation. Students develop fluency writing, interpreting, and translating between various forms of linear equations and inequalities, and using them to solve problems. They master the solution of linear equations and apply related solution techniques and the laws of exponents to the creation and solution of simple exponential equations.
Unit 2- Linear and Exponential Relationships: In earlier grades, students define, evaluate, and compare functions, and use them to model relationships between quantities. In this unit, students will learn function notation and develop the concepts of domain and range. They explore many examples of functions, including sequences; they interpret functions given graphically, numerically, symbolically, and verbally, translate between representations, and understand the limitations of various representations. Students build on and informally extend their understanding of integer exponents to consider exponential functions. They compare and contrast linear and exponential functions, distinguishing between additive and multiplicative change. Students explore systems of equations and inequalities, and they find and interpret their solutions. They interpret arithmetic sequences as linear functions and geometric sequences as exponential functions.
Algebra 1B (Year 2)
Unit 3- Descriptive Statistics: This unit builds upon students prior experiences with data, providing students with more formal means of assessing how a model fits data. Students use regression techniques to describe and approximate linear relationships between quantities. They use graphical representations and knowledge of the context to make judgments about the appropriateness of linear models. With linear models, they look at residuals to analyze the goodness of fit.
Unit 4- Expressions and Equations: In this unit, students build on their knowledge from unit 2, where they extended the laws of exponents to rational exponents. Students apply this new understanding of number and strengthen their ability to see structure in and create quadratic and exponential expressions. They create and solve equations, inequalities, and systems of equations involving quadratic expressions.
Unit 5- Quadratic Functions and Modeling: In this unit, students consider quadratic functions, comparing the key characteristics of quadratic functions to those of linear and exponential functions. They select from among these functions to model phenomena. Students learn to anticipate the graph of a quadratic function by interpreting various forms of quadratic expressions. In particular, they identify the real solutions of a quadratic equation as the zeros of a related quadratic function. Students expand their experience with functions to include more specialized functions absolute value, step, and those that are piece wise-defined.
English Language Development ELD Standards Special Notes Section:
Teachers are required to provide listening, speaking, reading and writing instruction that allows English language learners (ELL) to communicate information, ideas and concepts for academic success in the content area of Mathematics. For the given level of English language proficiency and with visual, graphic, or interactive support, students will interact with grade level words, expressions, sentences and discourse to process or produce language necessary for academic success. The ELD standard should specify a relevant content area concept or topic of study chosen by curriculum developers and teachers which maximizes an ELL’s need for communication and social skills. To access an ELL supporting document which delineates performance definitions and descriptors, please click on the following link:
### Version Requirements
Fluency Recommendations
A/G- Algebra I students become fluent in solving characteristic problems involving the analytic geometry of lines, such as writing down the equation of a line given a point and a slope. Such fluency can support them in solving less routine mathematical problems involving linearity, as well as in modeling linear phenomena (including modeling using systems of linear inequalities in two variables).
A-APR.1- Fluency in adding, subtracting, and multiplying polynomials supports students throughout their work in Algebra, as well as in their symbolic work with functions. Manipulation can be more mindful when it is fluent.
A-SSE.1b- Fluency in transforming expressions and chunking (seeing parts of an expression as a single object) is essential in factoring, completing the square, and other mindful algebraic calculations.
### General Information
Course Number: 1200385
Course Path:
Abbreviated Title: ALG 1-B CR
Number of Credits: One (1) credit
Course Length: Credit Recovery (R)
Course Type: Elective Course
Course Level: 2
Course Status: Course Approved
## Educator Certifications
One of these educator certification options is required to teach this course.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this course.
## Original Student Tutorials
Exponential Functions Part 1:
Learn about exponential functions and how they are different from linear functions by examining real world situations, their graphs and their tables in this interactive tutorial.
Type: Original Student Tutorial
Solving Rational Equations: Using Common Denominators:
Learn how to solve rational functions by getting common denominators in this interactive tutorial.
Type: Original Student Tutorial
Dilations...The Effect of k on a Graph:
Visualize the effect of using a value of k in both kf(x) or f(kx) when k is greater than zero in this interactive tutorial.
Type: Original Student Tutorial
Solving Rational Equations: Cross Multiplying:
Learn how to solve rational linear and quadratic equations using cross multiplication in this interactive tutorial.
Type: Original Student Tutorial
Solving Inequalities and Graphing Solutions Part 2:
Learn how to solve and graph compound inequalities and determine if solutions are viable in part 2 of this interactive tutorial series.
Type: Original Student Tutorial
Writing Equations in Two Variables:
Learn how to write equations in two variables in this interactive tutorial.
Type: Original Student Tutorial
Solving Inequalities and Graphing Solutions: Part 1:
Learn how to solve and graph one variable inequalities, including compound inequalities, in part 1 of this interactive tutorial series.
Type: Original Student Tutorial
Reflections...The Effect of k on a Graph:
Learn how reflections of a function are created and tied to the value of k in the mapping of f(x) to -1f(x) in this interactive tutorial.
Type: Original Student Tutorial
Translations...The Effect of k on the Graph:
Explore translations of functions on a graph that are caused by k in this interactive tutorial. GeoGebra and interactive practice items are used to investigate linear, quadratic, and exponential functions and their graphs, and the effect of a translation on a table of values.
Type: Original Student Tutorial
The Year-Round School Debate: Identifying Faulty Reasoning — Part Two:
Practice identifying faulty reasoning in this two-part, interactive, English Language Arts tutorial. You'll learn what some experts say about year-round schools, what research has been conducted about their effectiveness, and how arguments can be made for and against year-round education. Then, you'll read a speech in favor of year-round schools and identify faulty reasoning within the argument, specifically the use of hasty generalizations.
Make sure to complete Part One before Part Two! Click HERE to launch Part One.
Type: Original Student Tutorial
The Year-Round School Debate: Identifying Faulty Reasoning – Part One:
Learn to identify faulty reasoning in this two-part interactive English Language Arts tutorial. You'll learn what some experts say about year-round schools, what research has been conducted about their effectiveness, and how arguments can be made for and against year-round education. Then, you'll read a speech in favor of year-round schools and identify faulty reasoning within the argument, specifically the use of hasty generalizations.
Make sure to complete both parts of this series! Click HERE to open Part Two.
Type: Original Student Tutorial
Evaluating an Argument – Part Four: JFK’s Inaugural Address:
Examine President John F. Kennedy's inaugural address in this interactive tutorial. You will examine Kennedy's argument, main claim, smaller claims, reasons, and evidence.
In Part Four, you'll use what you've learned throughout this series to evaluate Kennedy's overall argument.
Make sure to complete the previous parts of this series before beginning Part 4.
• Click HERE to launch Part One.
• Click HERE to launch Part Two.
• Click HERE to launch Part Three.
Type: Original Student Tutorial
Evaluating an Argument – Part Three: JFK’s Inaugural Address:
Examine President John F. Kennedy's inaugural address in this interactive tutorial. You will examine Kennedy's argument, main claim, smaller claims, reasons, and evidence. By the end of this four-part series, you should be able to evaluate his overall argument.
In Part Three, you will read more of Kennedy's speech and identify a smaller claim in this section of his speech. You will also evaluate this smaller claim's relevancy to the main claim and evaluate Kennedy's reasons and evidence.
Make sure to complete all four parts of this series!
• Click HERE to launch Part One.
• Click HERE to launch Part Two.
Type: Original Student Tutorial
It's a Slippery Slope!:
Learn what slope is in mathematics and how to calculate it on a graph and with the slope formula in this interactive tutorial.
Type: Original Student Tutorial
Finding the Zeros of Quadratic Functions:
Quadratic functions can be used to model real-world phenomena. Key features of quadratic functions such as maximum values and zeros can often reveal important qualities of these phenomena. By the end of this tutorial, you should be able to find the zeros of a quadratic function and interpret their meaning in real-world contexts.
Type: Original Student Tutorial
Introduction to Polynomials, Part 2 - Adding and Subtracting:
Learn how to add and subtract polynomials in this online tutorial. You will learn how to combine like terms and then use the distribute property to subtract polynomials.
This is part 2 of a two-part lesson. Click below to open part 1.
Type: Original Student Tutorial
Introduction to Polynomials, Part 1:
Learn how to identify monomials and polynomials and determine their degree in this interactive tutorial.
This is part 1 in a two-part series. Click HERE to open Part 2.
Type: Original Student Tutorial
Ready for Takeoff! -- Part Two:
Want to learn about Amelia Earhart, one of the most famous female aviators of all time? If so, then this interactive tutorial is for YOU! This tutorial is Part Two of a two-part series. In this series, you will study a speech by Amelia Earhart. You will practice identifying the purpose of her speech and practice identifying her use of rhetorical appeals (ethos, logos, pathos, Kairos). You will also evaluate the effectiveness of Earhart's rhetorical choices based on the purpose of her speech.
Please complete Part One before beginning Part Two. Click HERE to view Part One.
Type: Original Student Tutorial
Ready for Takeoff! -- Part One:
Want to learn about Amelia Earhart, one of the most famous female aviators of all time? If so, then this interactive tutorial is for YOU! This tutorial is Part One of a two-part series. In this series, you will study a speech by Amelia Earhart. You will practice identifying the purpose of her speech and practice identifying her use of rhetorical appeals (ethos, logos, pathos, Kairos). You will also evaluate the effectiveness of Earhart's rhetorical choices based on the purpose of her speech.
Type: Original Student Tutorial
Untangling Food Webs:
Learn how living organisms can be organized into food webs and how energy is transferred through a food web from producers to consumers to decomposers. This interactive tutorial also includes interactive knowledge checks.
Type: Original Student Tutorial
Data and Frequencies:
Learn to define, calculate, and interpret marginal frequencies, joint frequencies, and conditional frequencies in the context of the data with this interactive tutorial.
Type: Original Student Tutorial
Finding the Maximum or Minimum of a Quadratic Function:
Learn to complete the square of a quadratic expression and identify the maximum or minimum value of the quadratic function it defines. In this interactive tutorial, you'll also interpret the meaning of the maximum and minimum of a quadratic function in a real world context.
Type: Original Student Tutorial
Comparing Mitosis and Meiosis:
Compare and contrast mitosis and meiosis in this interactive tutorial. You'll also relate them to the processes of sexual and asexual reproduction and their consequences for genetic variation.
Type: Original Student Tutorial
Changing Rates:
Learn how to calculate and interpret an average rate of change over a specific interval on a graph.
Type: Original Student Tutorial
The graph of a quadratic equation is called a parabola [puh-ra-bow-luh]. The key features we will focus on in this tutorial are the vertex (a maximum or minimum extreme) and the direction of its opening. You will learn how to examine a quadratic equation written in vertex form in order to distinguish each of these key features.
Type: Original Student Tutorial
Cancer: Mutated Cells Gone Wild!:
Explore the relationship between mutations, the cell cycle, and uncontrolled cell growth which may result in cancer with this interactive tutorial.
Type: Original Student Tutorial
Climbing Around the Hominin Family Tree:
Learn to identify basic trends in the evolutionary history of humans, including walking upright, brain size, jaw size, and tool use in "Climbing Around the Hominin Family Tree" online tutorial.
Type: Original Student Tutorial
Writing Inequalities with Money, Money, Money:
Write linear inequalities for different money situations in this interactive tutorial.
Type: Original Student Tutorial
## Educational Games
Timed Algebra Quiz:
In this timed activity, students solve linear equations (one- and two-step) or quadratic equations of varying difficulty depending on the initial conditions they select. This activity allows students to practice solving equations while the activity records their score, so they can track their progress. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Educational Game
Algebra Four:
In this activity, two students play a simulated game of Connect Four, but in order to place a piece on the board, they must correctly solve an algebraic equation. This activity allows students to practice solving equations of varying difficulty: one-step, two-step, or quadratic equations and using the distributive property if desired. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the Java applet.
Type: Educational Game
## Educational Software / Tool
This Excel spreadsheet allows the educator to input data into a two way frequency table and have the resulting relative frequency charts calculated automatically on the second sheet. This resource will assist the educator in checking student calculations on student-generated data quickly and easily.
Steps to add data: All data is input on the first spreadsheet; all tables are calculated on the second spreadsheet
2. Input joint frequency data.
3. Click the second tab at the bottom of the window to see the automatic calculations.
Type: Educational Software / Tool
## Lesson Plan
Do Credit Cards Make You Gain Weight? What is Correlation, and How to Distinguish It from Causation:
This lesson introduces the students to the concepts of correlation and causation, and the difference between the two. The main learning objective is to encourage students to think critically about various possible explanations for a correlation, and to evaluate their plausibility, rather than passively taking presented information on faith. To give students the right tools for such analysis, the lesson covers most common reasons behind a correlation, and different possible types of causation.
Type: Lesson Plan
## Perspectives Video: Experts
Type: Perspectives Video: Expert
Mathematically Exploring the Wakulla Caves:
The tide is high! How can we statistically prove there is a relationship between the tides on the Gulf Coast and in a fresh water spring 20 miles from each other?
Type: Perspectives Video: Expert
MicroGravity Sensors & Statistics:
Statistical analysis played an essential role in using microgravity sensors to determine location of caves in Wakulla County.
Type: Perspectives Video: Expert
## Perspectives Video: Professional/Enthusiasts
Base 16 Notation in Computing:
Listen in as a computing enthusiast describes how hexadecimal notation is used to express big numbers in just a little space.
Type: Perspectives Video: Professional/Enthusiast
Correlation and Causation in a Scientific Study:
Watching this video will cause your critical thinking skills to improve. You might also have a great day, but that's just correlation.
Type: Perspectives Video: Professional/Enthusiast
Students explore the structure of the operation s/(vn). This question provides students with an opportunity to see expressions as constructed out of a sequence of operations: first taking the square root of n, then dividing the result of that operation into s.
Speed Trap:
The purpose of this task is to allow students to demonstrate an ability to construct boxplots and to use boxplots as the basis for comparing distributions.
Musical Preferences:
This problem solving task asks students to make deductions about what kind of music students like by examining a table with data.
SAT Scores:
This problem solving task challenges students to answer probability questions about SAT scores, using distribution and mean to solve the problem.
Haircut Costs:
This problem could be used as an introductory lesson to introduce group comparisons and to engage students in a question they may find amusing and interesting.
Golf and Divorce:
This is a simple task addressing the distinction between correlation and causation. Students are given information indicating a correlation between two variables, and are asked to reason out whether or not a causation can be inferred.
The purpose of this task is to assess ability to interpret the slope and intercept of the least squares regression line in context.
Coffee and Crime:
This problem solving task asks students to examine the relationship between shops and crimes by using a correlation coefficient.
Should We Send Out a Certificate?:
The purpose of this task is to have students complete normal distribution calculations and to use properties of normal distributions to draw conclusions.
Do You Fit in This Car?:
This task requires students to use the normal distribution as a model for a data distribution. Students must use given means and standard deviations to approximate population percentages.
Random Walk III:
The task provides a context to calculate discrete probabilities and represent them on a bar graph.
As the Wheel Turns:
In this task, students use trigonometric functions to model the movement of a point around a wheel and, through space. Students also interpret features of graphs in terms of the given real-world context.
Population and Food Supply:
In this task students use verbal descriptions to construct and compare linear and exponential functions and to find where the two functions intersect (F-LE.2, F-LE.3, A-REI.11).
Braking Distance:
This task provides an exploration of a quadratic equation by descriptive, numerical, graphical, and algebraic techniques. Based on its real-world applicability, teachers could use the task as a way to introduce and motivate algebraic techniques like completing the square, en route to a derivation of the quadratic formula.
Cash Box:
The given solutions for this task involve the creation and solving of a system of two equations and two unknowns, with the caveat that the context of the problem implies that we are interested only in non-negative integer solutions. Indeed, in the first solution, we must also restrict our attention to the case that one of the variables is further even. This aspect of the task is illustrative of mathematical practice standard MP4 (Model with mathematics), and crucial as the system has an integer solution for both situations, that is, whether or not we include the dollar on the floor in the cash box or not.
A Cubic Identity:
Solving this problem with algebra requires factoring a particular cubic equation (the difference of two cubes) as well as a quadratic equation. An alternative solution using prime numbers and arithmetic is presented.
Two Squares are Equal:
This classroom task is meant to elicit a variety of different methods of solving a quadratic equation (A-REI.4). Some are straightforward (for example, expanding the square on the right and rearranging the equation so that we can use the quadratic formula); some are simple but clever (reasoning from the fact that x and (2x - 9) have the same square); some use tools (using a graphing calculator to graph the functions f(x) = x^2 and g(x) = (2x-90)^2 and looking for values of x at which the two functions intersect). Some solution methods will work on an arbitrary quadratic equation, while others (such as the last three) may have difficulty or fail if the quadratic equation is not given in a particular form, or if the solutions are not rational numbers.
Exponential growth versus linear growth I:
The purpose of this task it to have students discover how (and how quickly) an exponentially increasing quantity eventually surpasses a linearly increasing quantity. Students' intuitions will probably have them favoring Option A for much longer than is actually the case, especially if they are new to the phenomenon of exponential growth. Teachers might use this surprise as leverage to segue into a more involved task comparing linear and exponential growth.
Finding Parabolas through Two Points:
This problem-solving task challenges students to find all quadratic functions described by given equation and coordinates, and describe how the graphs of those functions are related to one another.
Exponential growth versus polynomial growth:
This problem solving task shows that an exponential function takes larger values than a cubic polynomial function provided the input is sufficiently large. This resource also includes standards alignment commentary and annotated solutions.
Exponential growth versus linear growth II:
This task asks students to calculate exponential functions with a base larger than one.
Which Function?:
The task addresses knowledge related to interpreting forms of functions derived by factoring or completing the square. It requires students to pay special attention to the information provided by the way the equation is represented as well as the sign of the leading coefficient, which is not written out explicitly, and then to connect this information to the important features of the graph.
Warming and Cooling:
This task is meant to be a straight-forward assessment task of graph reading and interpreting skills. This task helps reinforce the idea that when a variable represents time, t = 0 is chosen as an arbitrary point in time and positive times are interpreted as times that happen after that.
Throwing Baseballs:
This task could be used for assessment or for practice. It allows students to compare characteristics of two quadratic functions that are each represented differently, one as the graph of a quadratic function and one written out algebraically. Specifically, students are asked to determine which function has the greatest maximum and the greatest non-negative root.
Average Cost:
This task asks students to find the average, write an equation, find the domain, and create a graph of the cost of producing DVDs.
Springboard Dive:
The problem presents a context where a quadratic function arises. Careful analysis, including graphing of the function, is closely related to the context. The student will gain valuable experience applying the quadratic formula and the exercise also gives a possible implementation of completing the square.
The High School Gym:
This task asks students to consider functions in regard to temperatures in a high school gym.
Telling a Story with Graphs:
In this task students are given graphs of quantities related to weather. The purpose of the task is to show that graphs are more than a collection of coordinate points; they can tell a story about the variables that are involved, and together they can paint a very complete picture of a situation, in this case the weather. Features in one graph, like maximum and minimum points, correspond to features in another graph. For example, on a rainy day, the solar radiation is very low, and the cumulative rainfall graph is increasing with a large slope.
Oakland Coliseum:
This deceptively simple task asks students to find the domain and range of a function from a given context. The function is linear and if simply looked at from a formulaic point of view, students might find the formula for the line and say that the domain and range are all real numbers.
Logistic Growth Model, Explicit Version:
This problem introduces a logistic growth model in the concrete settings of estimating the population of the U.S. The model gives a surprisingly accurate estimate and this should be contrasted with linear and exponential models.
Logistic Growth Model, Abstract Version:
This task is for instructional purposes only and students should already be familiar with some specific examples of logistic growth functions. The goal of this task is to have students appreciate how different constants influence the shape of a graph.
How Is the Weather?:
This task can be used as a quick assessment to see if students can make sense of a graph in the context of a real world situation. Students also have to pay attention to the scale on the vertical axis to find the correct match. The first and third graphs look very similar at first glance, but the function values are very different since the scales on the vertical axes are very different. The task could also be used to generate a group discussion on interpreting functions given by graphs.
Equations and Formulas:
In this task, students will use inverse operations to solve the equations for the unknown variable or for the designated variable if there is more than one.
Bernardo and Sylvia Play a Game:
This task presents a simple but mathematically interesting game whose solution is a challenging exercise in creating and reasoning with algebraic inequalities. The core of the task involves converting a verbal statement into a mathematical inequality in a context in which the inequality is not obviously presented, and then repeatedly using the inequality to deduce information about the structure of the game.
Regular Tessellations of the Plane:
This task examines the ways in which the plane can be covered by regular polygons in a very strict arrangement called a regular tessellation. These tessellations are studied here using algebra, which enters the picture via the formula for the measure of the interior angles of a regular polygon (which should therefore be introduced or reviewed before beginning the task). The goal of the task is to use algebra in order to understand which tessellations of the plane with regular polygons are possible.
Compounding with a 100% Interest Rate:
This task provides an approximation, and definition, of e, in the context of more and more frequent compounding of interest in a bank account. The approach is computational.
Building a quadratic function from f(x) = x^2:
This task aims for students to understand the quadratic formula in a geometric way in terms of the graph of a quadratic function.
Building an explicit quadratic function by composition:
This task is intended for instruction and to motivate "Building a general quadratic function." This task assumes that the students are familiar with the process of completing the square.
A Sum of Functions:
In this example, students are given the graph of two functions and are asked to sketch the graph of the function that is their sum. The intent is that students develop a conceptual understanding of function addition.
Forms of Exponential Expressions:
There are many different ways to write exponential expressions that describe the same quantity, in this task the amount of a radioactive substance after t years. Depending on what aspect of the context we need to investigate, one expression of the quantity may be more useful than another. This task contrasts the usefulness of four equivalent expressions. Students first have to confirm that the given expressions for the radioactive substance are equivalent. Then they have to explain the significance of each expression in the context of the situation.
Harvesting the Fields:
This is a challenging task, suitable for extended work, and reaching into a deep understanding of units. Students are given a scenario and asked to determine the number of people required to complete the amount of work in the time described. The task requires students to exhibit MAFS.K12.MP.1.1, Make sense of problems and persevere in solving them. An algebraic solution is possible but complicated; a numerical solution is both simpler and more sophisticated, requiring skilled use of units and quantitative reasoning. Thus the task aligns with either MAFS.912.A-CED.1.1 or MAFS.912.N-Q.1.1, depending on the approach.
Calculating the Square Root of 2:
This task is intended for instructional purposes so that students can become familiar and confident with using a calculator and understanding what it can and cannot do. This task gives an opportunity to work on the notion of place value (in parts [b] and [c]) and also to understand part of an argument for why the square root of 2 is not a rational number.
Throwing a Ball:
Students manipulate a given equation to find specified information.
Paying the Rent:
Students solve problems tracking the balance of a checking account used only to pay rent. This simple conceptual task focuses on what it means for a number to be a solution to an equation, rather than on the process of solving equations.
Students extrapolate the list price of a car given a total amount paid in states with different tax rates. The emphasis in this task is not on complex solution procedures. Rather, the progression of equations, from two that involve different values of the sales tax, to one that involves the sales tax as a parameter, is designed to foster the habit of looking for regularity in solution procedures, so that students don't approach every equation as a new problem but learn to notice familiar types.
Graphs of Compositions:
This task addresses an important issue about inverse functions. In this case the function f is the inverse of the function g but g is not the inverse of f unless the domain of f is restricted.
Planes and Wheat:
In this resource, students refer to given information which defines 5 variables in the context of real world government expenses. They are then asked to write equations based upon specific known values for some of the variables. The emphasis is on setting up, rather than solving, the equations.
Crude Oil and Gas Mileage:
This task asks students to write expressions for various problems involving distance per units of volume.
Flu on Campus:
The context of this example is the spread of a flu virus on campus and the related sale of tissue boxes sold. Students interpret the composite function and determine values simply by using the tables of values.
In this resource, a method of deriving the quadratic formula from a theoretical standpoint is demonstrated. This task is for instructional purposes only and builds on "Building an explicit quadratic function."
Compounding with a 5% Interest Rate:
This task develops reasoning behind the general formula for balances under continuously compounded interest. While this task itself specifically address the standard (F-BF), building functions from a context, an auxiliary purpose is to introduce and motivate the number e, which plays a significant role in the (F-LE) domain of tasks.
Profit of a Company:
This task compares the usefulness of different forms of a quadratic expression. Students have to choose which form most easily provides information about the maximum value, the zeros and the vertical intercept of a quadratic expression in the context of a real world situation. Rather than just manipulating one form into the other, students can make sense out of the structure of the expressions.
(From Algebra: Form and Function, McCallum et al., Wiley 2010)
Increasing or Decreasing? Variation 2:
The purpose of this task is to help students see manipulation of expressions as an activity undertaken for a purpose.
Variation 1 of this task presents a related more complex expression already in the correct form to answer the question.
The expression arises in physics as the reciprocal of the combined resistance of two resistors in parallel. However, the context is not explicitly considered here.
Downhill:
This task would be especially well-suited for instructional purposes. Students will benefit from a class discussion about the slope, y-intercept, x-intercept, and implications of the restricted domain for interpreting more precisely what the equation is modeling.
Ice Cream:
This task illustrates the process of rearranging the terms of an expression to reveal different aspects about the quantity it represents, precisely the language being used in standard MAFS.912.A-SSE.2.3. Students are provided with an expression giving the temperature of a container at a time t, and have to use simple inequalities (e.g., that 2t>0 for all t) to reduce the complexity of an expression to a form where bounds on the temperature of a container of ice cream are made apparent.
Sum of Even and Odd:
Students explore and manipulate expressions based on the following statement:
A function f defined for -a < x < a is even if f(-x)=f(x) and is odd if f(-x)=-f(x) when -a < x < a. In this task we assume f is defined on such an interval, which might be the full real line (i.e., a=8).
Students compare graphs of different quadratic functions, then produce equations of their own to satisfy given conditions.
This exploration can be done in class near the beginning of a unit on graphing parabolas. Students need to be familiar with intercepts, and need to know what the vertex is. It is effective after students have graphed parabolas in vertex form (y=a(x–h)2+k), but have not yet explored graphing other forms.
Equivalent Expressions:
This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see a, the coefficient of the x2 term; k, the leading coefficient of the x term; and n, the constant term.
Students are asked to interpret the effect on the value of an expression given a change in value of one of the variables.
Mixing Fertilizer:
Students examine and answer questions related to a scenario similar to a "mixture" problem involving two different mixtures of fertilizer. In this example, students determine and then compare expressions that correspond to concentrations of various mixtures. Ultimately, students generalize the problem and verify conclusions using algebraic rather than numerical expressions.
Mixing Candies:
Students are asked to interpret expressions and equations within the context of the amounts of caramels and truffles in a box of candy.
Kitchen Floor Tiles:
This problem asks students to consider algebraic expressions calculating the number of floor tiles in given patterns. The purpose of this task is to give students practice in reading, analyzing, and constructing algebraic expressions, attending to the relationship between the form of an expression and the context from which it arises. The context here is intentionally thin; the point is not to provide a practical application to kitchen floors, but to give a framework that imbues the expressions with an external meaning.
Delivery Trucks:
This resource describes a simple scenario which can be represented by the use of variables. Students are asked to examine several variable expressions, interpret their meaning, and describe what quantities they each represent in the given context.
Animal Populations:
In this task students interpret the relative size of variable expressions involving two variables in the context of a real world situation. All given expressions can be interpreted as quantities that one might study when looking at two animal populations.
Computations with Complex Numbers:
This resource involves simplifying algebraic expressions that involve complex numbers and various algebraic operations.
Operations with Rational and Irrational Numbers:
This task has students experiment with the operations of addition and multiplication, as they relate to the notions of rationality and irrationality.
Seeing Dots:
The purpose of this task is to identify the structure in the two algebraic expressions by interpreting them in terms of a geometric context. Students will have likely seen this type of process before, so the principal source of challenge in this task is to encourage a multitude and variety of approaches, both in terms of the geometric argument and in terms of the algebraic manipulation.
Graphs of Power Functions:
This task requires students to recognize the graphs of different (positive) powers of x.
Transforming the Graph of a Function:
This problem solving task examines, in a graphical setting, the impact of adding a scalar, multiplying by a scalar, and making a linear substitution of variables on the graph of the function f. This resource also includes standards alignment commentary and annotated solutions.
Temperature Conversions:
Unit conversion problems provide a rich source of examples both for composition of functions (when several successive conversions are required) and inverses (units can always be converted in either of two directions).
Susita's Account:
This task asks students to determine a recursive process from a context. Students who study computer programming will make regular use of recursive processes.
Summer Intern:
This task asks students to use proportions of mass and volume to create ideal brine for saltwater fish tanks. It also asks students to compare graphs.
Skeleton Tower:
This problem is a quadratic function example. The other tasks in this set illustrate MAFS.912.F.BF.1.1.a in the context of linear, exponential, and rational functions.
Medieval Archer:
The task addresses the first part of standard MAFS.912.F-BF.2.3: "Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative)."
Lake Algae:
The purpose of this task is to introduce students to exponential growth. While the context presents a classic example of exponential growth, it approaches it from a non-standard point of view.
Kimi and Jordan:
In the middle grades, students have lots of experience analyzing and comparing linear functions using graphs, table, symbolic expressions, and verbal descriptions. In this task, students may choose a representation that suits them and then reason from within that representation.
The Canoe Trip, Variation 2:
The primary purpose of this task is to lead students to a numerical and graphical understanding of the behavior of a rational function near a vertical asymptote, in terms of the expression defining the function.
The Canoe Trip, Variation 1:
The purpose of this task is to give students practice constructing functions that represent a quantity of interest in a context, and then interpreting features of the function in the light of the context. It can be used as either an assessment or a teaching task.
Identifying Even and Odd Functions:
This task asks students to determine whether a the set of given functions is odd, even, or neither.
## Student Center Activity
Method to Multiplying Polynomials:
This video will demonstrate how to multiply polynomials.
Type: Student Center Activity
## Tutorials
You will learn in this video how to solve Quadratic Equations using the Quadratic Formula.
Type: Tutorial
Graphs and Solutions of Functions in Quadratic Equations:
You will learn how the parent function for a quadratic function is affected when f(x) = x2.
Type: Tutorial
Learning How to Complete the Square:
You will learn int his video how to solve the Quadratic Equation by Completing the Square.
Type: Tutorial
Graphs of second degree polynomials:
In this tutorial, students will look at input and output values of quadratic functions to help them understand why the graph of a second degree polynomial curves.
Type: Tutorial
This video tutorial shows students: the standard form of a polynomial, how to identify polynomials, how to determine the degree of a polynomial, how to add and subtract polynomials, and how to represent the area of a shape as an addition or subtraction of polynomials.
Type: Tutorial
Solving Quadratic Equations by Square Roots:
In this video tutorial students will learn how to solve quadratic equations by square roots.
Type: Tutorial
Subtracting Polynomials with Multiple Variables:
This video explains how to subtract polynomials with multiple variables and reinforces how to distribute a negative number.
Type: Tutorial
Squaring a Binomial:
This video covers squaring a binomial with two variables. Students will be given the area of a square.
Type: Tutorial
Constructing an Equations with Two Variables - Yoga Plan:
This video provides a real-world scenario and step-by-step instructions to constructing equations using two variables. Possible follow-up videos include Plotting System of Equations - Yoga Plan, Solving System of Equations with Substitution - Yoga Plan, and Solving System of Equations with Elimination - Yoga Plan.
Type: Tutorial
Graphing Quadractic Functions in Vertex Form:
This tutorial will help the students to identify the vertex of a parabola from the equation, and then graph the parabola.
Type: Tutorial
This tutorial will help the learners to graph the equation of the quadratic function using the coordinates of the vertex of a parabola adn its x- intercepts.
Type: Tutorial
Example: Evaluating expressions with 2 variables:
Evaluating Expressions with Two Variables
Type: Tutorial
Graphing Exponential Equations:
This tutorial will help you to learn about the exponential functions by graphing various equations representing exponential growth and decay.
Type: Tutorial
How to evaluate an expression using substitution:
In this example we have a formula for converting Celsius temperature to Fahrenheit. Let's substitute the variable with a value (Celsius temp) to get the degrees in Fahrenheit. Great problem to practice with us!
Type: Tutorial
How to evaluate an expression with variables:
Learn how to evaluate an expression with variables using a technique called substitution (or "plugging in").
Type: Tutorial
Why aren't we using the multiplication sign?:
Great question. In algebra, we do indeed avoid using the multiplication sign. We'll explain it for you here.
Type: Tutorial
What is a variable?:
Our focus here is understanding that a variable is just a letter or symbol (usually a lower case letter) that can represent different values in an expression. We got this. Just watch.
Type: Tutorial
Power of a Power Property:
This tutorial demonstrates how to use the power of a power property with both numerals and variables.
Type: Tutorial
Special Products of Binomials:
The video tutorial discusses about two typical polynomial multiplications. First, squaring a binomial and second, product of a sum and difference.
Type: Tutorial
Dividing Polynomials:
This tutorial will help the learners practice division of polynomials. Students will recognize that dividing polynomials is similar to simplifying fractions.
Type: Tutorial
Multiplying Polynomials:
This tutorial will help the learners practice multiplication of polynomials. Learners will understand that when they multiply expressions with more than two terms, they need to make sure each term in the first expression multiplies every term in the second expression.
Type: Tutorial
Multiplying Bionomials:
Binomials are the polynomials with two terms. This tutorial will help the students learn about the multiplication of binomials. In multiplication, we need to make sure that each term in the first set of parenthesis multiplies each term in the second set.
Type: Tutorial
## Video/Audio/Animations
Solving Quadratic Equations using Square Roots:
This video will demonstrate how to solve a quadratic equation using square roots.
Type: Video/Audio/Animation
Relations and Functions:
This video demonstrates how to determine if a relation is a function and how to identify the domain.
Type: Video/Audio/Animation
Solving Mixture Problems with Linear Equations:
Mixture problems can involve mixtures of things other than liquids. This video shows how Algebra can be used to solve problems involving mixtures of different types of items.
Type: Video/Audio/Animation
Using Systems of Equations Versus One Equation:
When should a system of equations with multiple variables be used to solve an Algebra problem, instead of using a single equation with a single variable?
Type: Video/Audio/Animation
Systems of Linear Equations in Two Variables:
The points of intersection of two graphs represent common solutions to both equations. Finding these intersection points is an important tool in analyzing physical and mathematical systems.
Type: Video/Audio/Animation
Slope:
"Slope" is a fundamental concept in mathematics. Slope is often defined as " the rise over the run"....but why?
Type: Video/Audio/Animation
Point-Slope Form:
Th point-slope form of the equation for a line can describe any non-vertical line in the Cartesian plane, given the slope and the coordinates of a single point which lies on the line.
Type: Video/Audio/Animation
Two Point Form:
The two point form of the equation for a line can describe any non-vertical line in the Cartesian plane, given the coordinates of two points which lie on the line.
Type: Video/Audio/Animation
Linear Equations in the Real World:
Linear equations can be used to solve many types of real-word problems. In this episode, the water depth of a pool is shown to be a linear function of time and an equation is developed to model its behavior. Unfortunately, ace Algebra student A. V. Geekman ends up in hot water anyway.
Type: Video/Audio/Animation
Solving Literal Equations:
Literal equations are formulas for calculating the value of one unknown quantity from one or more known quantities. Variables in the formula are replaced by the actual or 'literal' values corresponding to a specific instance of the relationship.
Type: Video/Audio/Animation
Example of Solving for a Variable - Khan Academy:
This video takes a look at rearranging a formula to highlight a quantity of interest.
Type: Video/Audio/Animation
Basic Linear Function:
This video demonstrates writing a function that represents a real-life scenario.
Type: Video/Audio/Animation
This video gives a more in-depth look at graphing quadratic functions than previously offered in Quadratic Functions 1.
Type: Video/Audio/Animation
Graphing Lines 1:
Khan Academy video tutorial on graphing linear equations: "Algebra: Graphing Lines 1"
Type: Video/Audio/Animation
Fitting a Line to Data:
Khan Academy tutorial video that demonstrates with real-world data the use of Excel spreadsheet to fit a line to data and make predictions using that line.
Type: Video/Audio/Animation
Averages:
This Khan Academy video tutorial introduces averages and algebra problems involving averages.
Type: Video/Audio/Animation
## Virtual Manipulatives
This resource will assess students' understanding of addition and subtraction of polynomials.
Type: Virtual Manipulative
Solving Quadratics By Taking The Square Root:
This resource can be used to assess students' understanding of solving quadratic equation by taking the square root. A great resource to view prior to this is "Solving quadratic equations by square root' by Khan Academy.
Type: Virtual Manipulative
Algebra Tiles (Multiplying Binomials):
This virtual manipulative is intended to allow the student to practice multiplication of binomials. The student should understand how to use algebra tiles before using this tool.
Type: Virtual Manipulative
Histogram vs. Box Plot:
This simulation allows the student to create a box plot and a histogram for the same set of data and toggle between the two displays. Maximum, minimum, median and mean are shown for the data set. The student can change the cell width to explore how the histogram is affected.
Type: Virtual Manipulative
Scatterplot:
This manipulative will help students in understanding scatter plots which are particularly useful when investigating whether there is a relationship between two variables. Students could develop a systematic plan for collecting and entering data into the scatter plot manipulative and set appropriate ranges for the x and y scales.
Type: Virtual Manipulative
Slope Slider:
In this activity, students adjust slider bars which adjust the coefficients and constants of a linear function and examine how their changes affect the graph. The equation of the line can be in slope-intercept form or standard form. This activity allows students to explore linear equations, slopes, and y-intercepts and their visual representation on a graph. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Graphing Equations Using Intercepts:
This resource provides linear functions in standard form and asks the user to graph it using intercepts on an interactive graph below the problem. Immediate feedback is provided, and for incorrect responses, each step of the solution is thoroughly modeled.
Type: Virtual Manipulative
Graphing Lines:
Allows students access to a Cartesian Coordinate System where linear equations can be graphed and details of the line and the slope can be observed.
Type: Virtual Manipulative
Box Plot:
In this activity, students use preset data or enter in their own data to be represented in a box plot. This activity allows students to explore single as well as side-by-side box plots of different data. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the Java applet.
Type: Virtual Manipulative
Data Flyer:
Using this virtual manipulative, students are able to graph a function and a set of ordered pairs on the same coordinate plane. The constants, coefficients, and exponents can be adjusted using slider bars, so the student can explore the affect on the graph as the function parameters are changed. Students can also examine the deviation of the data from the function. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Function Matching:
This is a graphing tool/activity for students to deepen their understanding of polynomial functions and their corresponding graphs. This tool is to be used in conjunction with a full lesson on graphing polynomial functions; it can be used either before an in depth lesson to prompt students to make inferences and connections between the coefficients in polynomial functions and their corresponding graphs, or as a practice tool after a lesson in graphing the polynomial functions.
Type: Virtual Manipulative
Normal Distribution Interactive Activity:
With this online tool, students adjust the standard deviation and sample size of a normal distribution to see how it will affect a histogram of that distribution. This activity allows students to explore the effect of changing the sample size in an experiment and the effect of changing the standard deviation of a normal distribution. Tabs at the top of the page provide access to supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Function Flyer:
In this online tool, students input a function to create a graph where the constants, coefficients, and exponents can be adjusted by slider bars. This tool allows students to explore graphs of functions and how adjusting the numbers in the function affect the graph. Using tabs at the top of the page you can also access supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
This is an online graphing utility that can be used to create box plots, bubble graphs, scatterplots, histograms, and stem-and-leaf plots.
Type: Virtual Manipulative
Number Cruncher:
In this activity, students enter inputs into a function machine. Then, by examining the outputs, they must determine what function the machine is performing. This activity allows students to explore functions and what inputs are most useful for determining the function rule. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Curve Fitting:
With a mouse, students will drag data points (with their error bars) and watch the best-fit polynomial curve form instantly. Students can choose the type of fit: linear, quadratic, cubic, or quartic. Best fit or adjustable fit can be displayed.
Type: Virtual Manipulative
Equation Grapher:
This interactive simulation investigates graphing linear and quadratic equations. Users are given the ability to define and change the coefficients and constants in order to observe resulting changes in the graph(s).
Type: Virtual Manipulative
Line of Best Fit:
This manipulative allows the user to enter multiple coordinates on a grid, estimate a line of best fit, and then determine the equation for a line of best fit.
Type: Virtual Manipulative
Histogram Tool:
This virtual manipulative histogram tool can aid in analyzing the distribution of a dataset. It has 6 preset datasets and a function to add your own data for analysis.
Type: Virtual Manipulative
Multi Bar Graph:
This activity allows the user to graph data sets in multiple bar graphs. The color, thickness, and scale of the graph are adjustable which may produce graphs that are misleading. Users may input their own data, or use or alter pre-made data sets. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Histogram:
In this activity, students can create and view a histogram using existing data sets or original data entered. Students can adjust the interval size using a slider bar, and they can also adjust the other scales on the graph. This activity allows students to explore histograms as a way to represent data as well as the concepts of mean, standard deviation, and scale. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this course. | 11,399 | 59,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-45 | latest | en | 0.919297 |
https://www.hellovaia.com/textbooks/math/the-practice-of-statistics-for-ap-4th/exploring-data/q-10-spam-email-spam-is-the-curse-of-the-internet-here-is-a-/ | 1,695,814,022,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00608.warc.gz | 867,094,373 | 19,375 | Suggested languages for you:
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Q 10.
Expert-verified
Found in: Page 22
### The Practice of Statistics for AP
Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339
# Spam Email spam is the curse of the Internet. Here is a compilation of the most common types of spam:(a) What percent of spam would fall in the “Other” category?(b) Display these data in a bar graph. Be sure to label your axes and title your graph.(c) Would it be appropriate to make a pie chart of these data? Explain.
Part (a) The percentage of other spam is $7%$
Part (b) The following is a bar chart.
Part (c) Pie chart can also be drawn.
See the step by step solution
## Part (a) Step 1: Given Information
Given data on the proportion of spam and their type
Type of spam Percent Adult 19 Financial 20 Health 7 Internet 7 Leisure 6 Products 25 Scams 9 Other ?
## Part (a) Step 2: Concept
The distribution of a category variable is displayed using pie charts and bar graphs. Any set of quantities measured in the same units can be compared using bar graphs. It's a good idea to ask yourself, "What do I see?" when looking at any graph.
## Part (a) Step 3: Explanation
The above information is listed in the following table.
Since the total percentage will add up to $100$, the percentage of other spam is $100-19+20+7+7+6+25+9=100-93=7$
Hence, the percentage of other spam is $7%$
Type of spam Percent Adult 19 Financial 20 Health 7 Internet 7 Leisure 6 Products 25 Scams 9 Other ?
## Part (b) Step 1: Explanation
The bar chart is drawn in excel by selecting the two columns of the above table.
## Part (c) Step 1: Explanation
When a phenomenon is broken down into sub-events and each event is represented in the diagram by a sub-component, the pie chart is utilized. Because there is no missing data, each category's number of spam is represented as a sector in a pie graphic. | 493 | 1,930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-40 | latest | en | 0.860472 |
https://leanprover-community.github.io/mathlib4_docs/Mathlib/Data/Vector/Defs.html | 1,720,998,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00318.warc.gz | 329,302,887 | 7,573 | # Documentation
## Mathlib.Data.Vector.Defs
The type Vector represents lists with fixed length.
def Mathlib.Vector (α : Type u) (n : ) :
Vector α n is the type of lists of length n with elements of type α.
Equations
• = { l : List α // l.length = n }
Instances For
instance Mathlib.Vector.instDecidableEq {α : Type u} {n : } [] :
Equations
@[match_pattern]
def Mathlib.Vector.nil {α : Type u} :
The empty vector with elements of type α
Equations
• Mathlib.Vector.nil = [],
Instances For
@[match_pattern]
def Mathlib.Vector.cons {α : Type u} {n : } :
αMathlib.Vector α n.succ
If a : α and l : Vector α n, then cons a l, is the vector of length n + 1 whose first element is a and with l as the rest of the list.
Equations
• = match x✝, x with | a, v, h => a :: v,
Instances For
@[reducible]
def Mathlib.Vector.length {α : Type u} {n : } :
The length of a vector.
Equations
• x.length = n
Instances For
def Mathlib.Vector.head {α : Type u} {n : } :
Mathlib.Vector α n.succα
The first element of a vector with length at least 1.
Equations
• x.head = match x with | a :: tail, property => a
Instances For
theorem Mathlib.Vector.head_cons {α : Type u} {n : } (a : α) (v : ) :
The head of a vector obtained by prepending is the element prepended.
def Mathlib.Vector.tail {α : Type u} {n : } :
Mathlib.Vector α (n - 1)
The tail of a vector, with an empty vector having empty tail.
Equations
• x.tail = match x with | [], h => [], | head :: v, h => v,
Instances For
theorem Mathlib.Vector.tail_cons {α : Type u} {n : } (a : α) (v : ) :
().tail = v
The tail of a vector obtained by prepending is the vector prepended. to
@[simp]
theorem Mathlib.Vector.cons_head_tail {α : Type u} {n : } (v : Mathlib.Vector α n.succ) :
Prepending the head of a vector to its tail gives the vector.
def Mathlib.Vector.toList {α : Type u} {n : } (v : ) :
List α
The list obtained from a vector.
Equations
• v.toList = v
Instances For
def Mathlib.Vector.get {α : Type u} {n : } (l : ) (i : Fin n) :
α
nth element of a vector, indexed by a Fin type.
Equations
Instances For
def Mathlib.Vector.append {α : Type u} {n : } {m : } :
Mathlib.Vector α (n + m)
Appending a vector to another.
Equations
• x✝.append x = match x✝, x with | l₁, h₁, l₂, h₂ => l₁ ++ l₂,
Instances For
def Mathlib.Vector.elim {α : Type u_1} {C : {n : } → Sort u} (H : (l : List α) → C l, ) {n : } (v : ) :
C v
Elimination rule for Vector.
Equations
• = match x with | l, h => match n, h with | .(l.length), => H l
Instances For
def Mathlib.Vector.map {α : Type u} {β : Type v} {n : } (f : αβ) :
Map a vector under a function.
Equations
• = match x with | l, h => List.map f l,
Instances For
@[simp]
theorem Mathlib.Vector.map_nil {α : Type u} {β : Type v} (f : αβ) :
Mathlib.Vector.map f Mathlib.Vector.nil = Mathlib.Vector.nil
A nil vector maps to a nil vector.
@[simp]
theorem Mathlib.Vector.map_cons {α : Type u} {β : Type v} {n : } (f : αβ) (a : α) (v : ) :
map is natural with respect to cons.
def Mathlib.Vector.map₂ {α : Type u} {β : Type v} {φ : Type w} {n : } (f : αβφ) :
Mapping two vectors under a curried function of two variables.
Equations
• = match x✝, x with | x, property, y, property_1 => List.zipWith f x y,
Instances For
def Mathlib.Vector.replicate {α : Type u} (n : ) (a : α) :
Vector obtained by repeating an element.
Equations
• = ⟨,
Instances For
def Mathlib.Vector.drop {α : Type u} {n : } (i : ) :
Mathlib.Vector α (n - i)
Drop i elements from a vector of length n; we can have i > n.
Equations
• = match x with | l, p => ⟨,
Instances For
def Mathlib.Vector.take {α : Type u} {n : } (i : ) :
Mathlib.Vector α (min i n)
Take i elements from a vector of length n; we can have i > n.
Equations
• = match x with | l, p => ⟨,
Instances For
def Mathlib.Vector.eraseIdx {α : Type u} {n : } (i : Fin n) :
Mathlib.Vector α (n - 1)
Remove the element at position i from a vector of length n.
Equations
• = match x with | l, p => l.eraseIdx i,
Instances For
@[deprecated Mathlib.Vector.eraseIdx]
def Mathlib.Vector.removeNth {α : Type u} {n : } (i : Fin n) :
Mathlib.Vector α (n - 1)
Alias of Mathlib.Vector.eraseIdx.
Remove the element at position i from a vector of length n.
Equations
Instances For
def Mathlib.Vector.ofFn {α : Type u} {n : } :
(Fin nα)
Vector of length n from a function on Fin n.
Equations
Instances For
def Mathlib.Vector.congr {α : Type u} {n : } {m : } (h : n = m) :
Create a vector from another with a provably equal length.
Equations
• = match x with | x, p => x,
Instances For
def Mathlib.Vector.mapAccumr {α : Type u} {β : Type v} {n : } {σ : Type} (f : ασσ × β) :
σσ ×
Runs a function over a vector returning the intermediate results and a final result.
Equations
• = match x✝, x with | x, px, c => let res := ; (res.fst, res.snd, )
Instances For
def Mathlib.Vector.mapAccumr₂ {n : } {α : Type} {β : Type} {σ : Type} {φ : Type} (f : αβσσ × φ) :
σσ ×
Runs a function over a pair of vectors returning the intermediate results and a final result.
Equations
Instances For
### Shift Primitives #
def Mathlib.Vector.shiftLeftFill {α : Type u} {n : } (v : ) (i : ) (fill : α) :
shiftLeftFill v i is the vector obtained by left-shifting v i times and padding with the fill argument. If v.length < i then this will return replicate n fill.
Equations
Instances For
def Mathlib.Vector.shiftRightFill {α : Type u} {n : } (v : ) (i : ) (fill : α) :
shiftRightFill v i is the vector obtained by right-shifting v i times and padding with the fill argument. If v.length < i then this will return replicate n fill.
Equations
Instances For
### Basic Theorems #
theorem Mathlib.Vector.eq {α : Type u} {n : } (a1 : ) (a2 : ) :
a1.toList = a2.toLista1 = a2
Vector is determined by the underlying list.
theorem Mathlib.Vector.eq_nil {α : Type u} (v : ) :
v = Mathlib.Vector.nil
A vector of length 0 is a nil vector.
@[simp]
theorem Mathlib.Vector.toList_mk {α : Type u} {n : } (v : List α) (P : v.length = n) :
Vector of length from a list v with witness that v has length n maps to v under toList.
@[simp]
theorem Mathlib.Vector.toList_nil {α : Type u} :
Mathlib.Vector.nil.toList = []
A nil vector maps to a nil list.
@[simp]
theorem Mathlib.Vector.toList_length {α : Type u} {n : } (v : ) :
v.toList.length = n
The length of the list to which a vector of length n maps is n.
@[simp]
theorem Mathlib.Vector.toList_cons {α : Type u} {n : } (a : α) (v : ) :
().toList = a :: v.toList
toList of cons of a vector and an element is the cons of the list obtained by toList and the element
@[simp]
theorem Mathlib.Vector.toList_append {α : Type u} {n : } {m : } (v : ) (w : ) :
(v.append w).toList = v.toList ++ w.toList
Appending of vectors corresponds under toList to appending of lists.
@[simp]
theorem Mathlib.Vector.toList_drop {α : Type u} {n : } {m : } (v : ) :
().toList = List.drop n v.toList
drop of vectors corresponds under toList to drop of lists.
@[simp]
theorem Mathlib.Vector.toList_take {α : Type u} {n : } {m : } (v : ) :
().toList = List.take n v.toList
take of vectors corresponds under toList to take of lists.
instance Mathlib.Vector.instGetElemNatLt {α : Type u} {n : } :
GetElem () α fun (x : ) (i : ) => i < n
Equations
• Mathlib.Vector.instGetElemNatLt = { getElem := fun (x : ) (i : ) (h : i < n) => x.get i, h } | 2,242 | 7,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-30 | latest | en | 0.630808 |
https://www.flashcardmachine.com/two-parallellinesintersectedbyatransversal.html | 1,718,670,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00223.warc.gz | 670,860,299 | 5,391 | # Shared Flashcard Set
## Details
Two Parallel Lines Intersected by a Transversal
Describing the relationship between angles
4
Mathematics
10th Grade
03/06/2011
## Cards Return to Set Details
Term
Alternate Exterior Angles
Definition
If two parallel lines are crossed by a transversal, then these angles are 1. both outside the two parallel lines2. on opposite sides of the transversal3. congruent to each other
Term
Alternative Interior Angles
Definition
If two parallel lines are crossed by a transversal, then these angles are 1. both inside the two parallel lines2. on opposite sides of the transversal3. congruent to each other
Term
Same Side Interior Angles(Consecutive Interior Angles)
Definition
If two parallel lines are crossed by a transversal, then these types of angles are 1. both inside the two parallel lines2. on the same side of the transversal3. supplementary
Term
Corresponding Angles
Definition
If two parallel lines are crossed by a transversal, then these angles are 1. arranged such that one angle is on the inside of the two parallel lines and the other angle is on the outside of the two parallel lines2. on the same side of the transversal3. congruent
Supporting users have an ad free experience! | 294 | 1,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.865674 |
https://answers.yahoo.com/question/index?qid=20070818064926AAmOLXj | 1,579,721,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607407.48/warc/CC-MAIN-20200122191620-20200122220620-00343.warc.gz | 319,919,131 | 31,351 | # Which is the lethal element in a strong electrical current?
Volts or amperes?
Relevance
• goring
Lv 6
An Ampere is the electro motive power that causes electrons to move inside a conductor.
It is the product of an electrical force(Volt) and conductance(which has the Units of velocity).
What affects the heart muscles is the amount of electronic Power(amperes)that the Human body receives from an electric shock.
The amount of power that would stop the heart is one hundred milliamperes. The victim would die if the Heart is not restarted within 5 minutes.
Whether a person survices a shock or not depends on the amount of conductivity the body had during the Shock at that particular voltage.
Electricity can be very dangerous it is not something to fool around with.
And people who dont understand it should not play with it.
What makes the difference is the amount of current flowing through you. Touch your fingers to 200 volts and you'll likely get nothing more than a nasty shock. Connect 12 volts to a low-resistance path like pins penetrating the skin, and it's likely to be fatal.
Most fatal electrocution results when the current interferes with the control mechanism of the heart. At higher currents, the damage is due to burning.
Skin effect is significant only at higher frequencies. At 60 Hz, skin depth is over 8mm.
• Anonymous
Current (amps).
AND....AC is more dangerous than an equal voltage of DC.
Allow me to elaborate a bit.
Current is measured in a unit call Amperes (or just Amps). Current is a flow of electrons. They flow on the *outside of bare wiring and metal (conducting) objects. This is why insulation is on wiring, so you can't touch the wire and get shocked from it.
When you do happen to touch something that is 'hot' it is this flow of electrons that flows thru your body, depending on how much current and voltage there is, and on what your body chemistry is also. Your body may have more or less of what is needed to conduct electricity in it, such as water and salts and so on.
If the current is sever enough and/or passes close enough to vital organs, such as your heart, brain, or lungs, etc it can cause them physical damage or cause them to convulse or seize. Obviously this won't be a happy thing for you.......so people have learned to be careful around electricity.
Amperes, a cars ignition is between 30 and 40 thousand volts but only 2 amps, you can get a nasty kick off the spark leads but it wont kill.
Amps. However there needs to be sufficient voltage for the current to flow at all; how much depending on variable factors like exactly how damp your skin is at the point of contact. Once you have the voltage, the energy burning you is proportional to the current.
Its the ampage that kills. They only put voltage on the signs cos it sounds more leathal like 30000 volts. When the reality is less than 1 amp can kill you
• Anonymous
Volts and amps in combination. One is basically how much flows and the other is how fast. I've heard that amps are the most deadly however.
• emiss
Lv 5
if the Resistance of the body will allow upwards of a quarter amp it can be lethal | 694 | 3,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.946311 |
https://shikshaway.com/class-6-maths-decimals-exercise-8-5/ | 1,726,806,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00673.warc.gz | 472,993,633 | 29,476 | Categories:
## NCERT Solutions For Class 6 Maths Decimals Exercise 8.5
• Decimals Exercise 8.1
• Decimals Exercise 8.2
• Decimals Exercise 8.3
• Decimals Exercise 8.4
• Decimals Exercise 8.6
Exercise 8.5
Ex 8.5 Class 6 Maths Question 1.
Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Solution:
(a) 0.007 + 8.5 + 30.08
= 0.007 + 8.500 + 30.080 (making like decimals)
= 38.587
(b) 15 + 0.632 + 13.8
= 15.000 + 0.632 + 13.800 (making like decimals)
= 29.432
(c) 27.076 + 0.55 + 0.004
= 27.076 + 0.550 + 0.004 (making like decimals)
= 27.630
(d) 25.65 + 9.005 + 3.7
= 25.650 + 9.005 + 3.700 (making like decimals)
= 38.355
(e) 0.75 + 10.425 + 2
= 0.750 + 10.425 + 2.000 (making like decimals)
= 13.175
(f) 280.69 + 25.2 + 38
= 280.69 + 25.20 + 38.00 (making like decimals)
= 343.89
Ex 8.5 Class 6 Maths Question 2.
Rashid spent ₹35.75 for Maths book and ₹32.60 for Science book. Find the total amount spent by Rashid.
Solution:
Money spent by Rashid for Maths book = ₹35.75
Money spent by Rashid for Science book = ₹32.60
∴ Total money spent by Rashid on both books = ₹35.75 + ₹32.60 = ₹68.35
Ex 8.5 Class 6 Maths Question 3.
Radhika’s mother gave her ₹10.50 and her father gave her ₹15.80, find the total amount given to Radhika by her parents.
Solution:
Money given by Radhika’s mother = ₹10.50
Money given by her father = ₹15.80
∴ Total money given to her by her parents
= ₹10.50 + ₹15.80 = ₹26.30
Ex 8.5 Class 6 Maths Question 4.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solution:
Length of cloth bought by Nasreen for her shirt = 3 m 20 cm = 3.20 m
Length of cloth brought by her for her trouser = 2 m 5 cm = 2.05 m
Total length of cloth bought by her = 3.20 m + 2.05 m = 5.25 m
Ex 8.5 Class 6 Maths Question 5.
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution:
Distance walked by Naresh in the morning = 2 km 35 m = (2 + (frac { 35 }{ 1000 })) km = 2.035 km.
Distance walked by him in the evening = 1 km 7 m = (1 + (frac { 7 }{ 1000 })) km = 1.007 km 1000)
∴ Total distance walked by him in all
= (2.035 + 1.007) km
= 3.042 km
Ex 8.5 Class 6 Maths Question 6.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solution:
Distance travelled by Sunita by bus
= 15 km 268 m = ( 15 + (frac { 268 }{ 1000 })) = 15.268 km
Distance travelled by her by car
= 7 km 7 m = (7 + (frac { 7 }{ 1000 })) km
= 7.007km
<!– –> | 1,068 | 2,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.886882 |
http://www.physics.udel.edu/~watson/phys208/exercises/answers/29-54E.html | 1,518,897,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807660.32/warc/CC-MAIN-20180217185905-20180217205905-00089.warc.gz | 514,266,306 | 1,027 | ## PHYS208 Fundamentals of Physics II
### Answer for Ch. 29, 54E
Let B = B i
Fa = -(0.138 N) k
Fb = (0.138 N) k
Fc = 0
Extension:
Each of the forces Fa and Fa contribute a torque of 0.045 N-m up.
The magnetic dipole moment is 1.2 A-m2. | 96 | 241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-09 | longest | en | 0.474921 |
https://e-hipokrates.com/delta-and-singapore-airlines-case/ | 1,726,249,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00007.warc.gz | 205,047,528 | 12,371 | # Delta and Singapore Airlines Case:
1. a. Delta Airlines Depreciation Method Depreciation MethodSalvage ValueFor every \$100 milDepreciatedAnnual Depreciation Prior to 1986Straight-line, 10 years10%100-(. 1*100)=9090/10=9\$9 mil 1968 – 1993Straight-line, 15 years10%100-(. 1*100)=9090/15=6\$6 mil After 1993Straight-line, 20 years5%100-(. 05*100)=9595/20=4. 75\$4. 75 mil b. Singapore Airlines Depreciation Method Depreciation MethodSalvage ValueFor every \$100 milDepreciatedAnnual Depreciation Prior to 1989Straight-line, 8 years10%=\$10100-(. 1*100)=9090/8=11. 5\$9 mil After 1989Straight-line, 10 years20%=\$20100-(. 2*100)=8080/10=8\$6 mil 2. Both use a straight line depreciation method to depreciate the value of the fleet on the Balance Sheet. However, as noted in the computations above, Singapore depreciates the value of its fleet twice as fast. Additionally, they assume a much higher salvage value of the total purchase price of the asset. The estimation of depreciation and salvage value are both assumptions of the company’s management. There are several reasons to drive the company’s choice of depreciation schedule.
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Delta may have wanted to reduce depreciation expense recorded on the Balance Sheet and therefore increase the depreciation period to reduce the annual expenditure. As noted in the case text, Delta was reporting losses in other areas and wanted to improve their overall earnings through lessening the affect of fleet depreciation expense. Currently Delta depreciates over 20 years and Singapore depreciates over 10; a significant difference. Some of the differences in the method chosen for depreciation can be based on the use and maintenance schedule of the fleet.
In the case, it states that Delta gleaned 21% of its revenues from international travel. In contrast, 56% of Singapore’s revenues are for travel outside of Asia, and it can also be inferred that while 44% of its revenue come from Asia, not all of those are inner continental. If you assume that longer distance flights lead to a reduction in serviceable life of an aircraft, this explanation may account for some of the difference in depreciation schedule between Delta and Singapore. Additionally, it may be inferred that climate and weather may affect the life of an aircraft and this too may account for the difference between Delta and Singapore.
Here Singapore’s fleet is presumably exposed to a wider array of climate vis-a-vis its extensive international routes. Taken these assumptions as given, it would be proper to have different depreciation schedules for the two companies. 3. Difference in Annual Depreciation between Delta and Singapore Total AssetsAssets including Capital LeaseCalculations Avg value of flight equipment in ’93 avg of the past 2 years9216. 00=Expense Singapore-Delta Annual depreciation expense before ‘86 542. 58 552. 96 =Asset Value/100*6. 00 Annual depreciation expense after ‘93 429. 54 33. 8 =Asset Value/100*4. 75 Expense reduction in ’93 113. 04 519. 78 =Expense before ’86 + after ‘93 Using Singapore’s method 723. 44 737. 28 =Asset Value/100*6. 00 Savings compared to Singapore 293. 90 704. 10 =Expense Singapore-Delta 4. Singapore Gains: There is a tax benefit to depreciating assets more rapidly through lowering the net income annually. Lower tax liability is a benefit of the more aggressive depreciation schedule. If one assumes planes are sold or retired after they are fully depreciated, then Singapore can increase their next income upon sale of its retied fleet.
This salvage value is more significant that Delta’s. If this is not true and the planes stay in service, the company can also expect a future increase in net income as the asset has been fully expensed. Singapore Losses: The aggressive depreciation lowers the net income which is typically not ideal for stock value. It is noted in the case that Singapore’s stock value is highly monitored since it is partially government owned. The rapid depreciation schedule and relatively high salvage value may be due to ???
Analysis: The overall strategy of Singapore Airlines may be related to their commitment to maintaining their high rankings for customer service amongst business commuters. Here having current technology in the airplanes may be relevant to Singapore Airlines in implementing this strategy. To maintain current technologies in aircrafts, it may be more cost effective to truncate the serviceable life of aircrafts and purchase new aircrafts versus allowing the fleet’s amenities – i. e. telecommunication capability for passengers or flat screen televisions in every seat – to become outdated or choose to retrofit the aging aircrafts. . Average age can be computed as a ratio of accumulated depreciation divided by the annual depreciation expense. Here age is a function of the depreciation rules in place by the company. As stated in the case, it is difficult to compare over airlines within the industry due to different depreciable lives and fleet ages. Most specifically when analyzing average age of a fleet as relative to annual depreciation expenses the case states: “There is no necessary connection, however, between the average age of an airline’s fleet and the assumption it made regarding the fleet’s depreciable life. In the chart below, you will see the amount of total asset value that has been depreciated as of 1993. This value is similar for both companies. However, this result is deceiving because it only considers the total assets relative to that year. All things held the same in 1994 Singapore will have a much higher percentage depreciated than Delta since their annual depreciation expense is much greater relative to their total assets. AirlineAvg AgeDepreciationTotal AssetsCalculations% of Total Assets Depreciated ‘93 Delta8. 8 years20 years, 5%9043=9043*(1-. 05)*(8. 8/20)41. 8% Singapore5. 1 years10 years, 20%9224=9224*(1-. 2)*(5. 1/10)40. 8% | 1,341 | 5,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.878844 |
https://courses.lumenlearning.com/chemistryformajors/chapter/assignment-thermochemistry/ | 1,618,662,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00595.warc.gz | 296,740,974 | 10,055 | ## Assignment: Enthalpy, Energy, and Temperature Change
Complete the following thermochemistry problems – show all of your work as you set up equations and make calculations.
1. The Travis Scott promotional meal at McDonald’s had $1180$ Calories (note the capital letter- this indicates nutritional Calories). Find the amount of energy in Joules.
2. A $9.00$ g nugget of pure gold absorbed $276$ J of heat. What was the final temperature of the gold if the initial temperature was $25.0$° C? The specific heat of gold is $0.129$ J/(g °C).
3. A sample of silver with a mass of $63.3$ g is heated to a temperature of $384.4$ K and placed in a container of water at $290.0^{\circ}$K. The final (equilibrium) temperature of the silver and water is $292.4^{\circ}$K. What volume of water is in the container? The specific heat of silver is $0.235$ J/(g °C).
4. NSU Chemistry 101L class conducted an experiment in which a measured amount of $\text{NaOH}$ is dissolved in water inside a foam cup calorimeter. A student collected data throughout the experiment. Calculate $\Delta\text{H}$ for the process in kJ/mol.
• Data table from the experiment:
Mass of NaOH + weigh boat 14.13 g Mass of weigh boat 0.70 g Mass of NaOH Mass of foam cup+water 109.85 g Mass of foam cup 10.25 g Mass of water Initial temp of water 24.5°C Final temp of water 36.8°C Change in temperature (ΔT)
5. You can light a charcoal grill in about a minute. If the grill is prepared using one $20$ pound bag of charcoal (primarily carbon) and two $1$ gallon buckets of liquid oxygen (the density of LOX (liquid oxygen) is $1142$ kg/m3, CRC Handbook of Chemistry and Physics.)
• Write a balanced chemical equation describing this reaction.
• Determine the limiting reagent for this reaction.
• Calculate the mass of each product and any remaining reactant.
• Calculate the amount of heat released by this reaction.
• If this heat is used to warm a $500$ gallon tank of water at $15^{\circ}$C, what is the final temperature?
• If this heat is used to warm $40$ kg of ice at $200^{\circ}$K, what is the final temperature? | 537 | 2,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-17 | latest | en | 0.880571 |
https://community.qlik.com/thread/249078 | 1,537,806,422,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160568.87/warc/CC-MAIN-20180924145620-20180924170020-00149.warc.gz | 478,759,197 | 20,249 | 0 Replies Latest reply: Feb 8, 2017 3:36 AM by Mohammad Sarif
# Cumulative Chart is not working when Denominator is dynamic
Hi All,
I have a Cumulative Chart. Which is average of last few same days value. However the denominator is dynamic depending on no of holidays.
Below is the expression:
(RangeSum(Above(Sum({\$<Z_DATE=, Holiday_HD = {'N'},M_DATE={'\$(=(Date(Date#('\$(vDATE)','M/D/YYYY') - 7)))'}>} [ORD_LINE_CNT]),0,RowNo()))
+RangeSum(Above(Sum({\$<Z_DATE=,Holiday_HD = {'N'}, M_DATE={'\$(=(Date(Date#('\$(vDATE)','M/D/YYYY') - 14)))'}>} [ORD_LINE_CNT]),0,RowNo()))
+RangeSum(Above(Sum({\$<Z_DATE=,Holiday_HD = {'N'}, M_DATE={'\$(=(Date(Date#('\$(vDATE)','M/D/YYYY') - 21)))'}>} [ORD_LINE_CNT]),0,RowNo()))
+RangeSum(Above(Sum({\$<Z_DATE=,Holiday_HD = {'N'}, M_DATE={'\$(=(Date(Date#('\$(vDATE)','M/D/YYYY') - 28)))'}>} [ORD_LINE_CNT]),0,RowNo()))) /
((4*Count (distinct PLANT)) - Count({<Dates1 = {"\$(=Date(Date#('\$(vDATE)','M/D/YYYY') - 7))","\$(=Date(Date#('\$(vDATE)','M/D/YYYY') - 14))","\$(=Date(Date#('\$(vDATE)','M/D/YYYY') - 21))","\$(=Date(Date#('\$(vDATE)','M/D/YYYY') - 28))"} >}Dates))
The denominator is dynamic here. If we put the same expression in single KPI. It is returning correct value. And if we put that numeric value in expression denominator, the cumulative chart is behaving wrongly. | 432 | 1,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-39 | latest | en | 0.526277 |
http://nrich.maths.org/public/leg.php?code=-93&cl=4&cldcmpid=418 | 1,485,132,385,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281659.81/warc/CC-MAIN-20170116095121-00067-ip-10-171-10-70.ec2.internal.warc.gz | 204,290,505 | 10,150 | # Search by Topic
#### Resources tagged with Making and proving conjectures similar to Target Six:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### Recent Developments on S.P. Numbers
##### Stage: 5
Take a number, add its digits then multiply the digits together, then multiply these two results. If you get the same number it is an SP number.
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### The Clue Is in the Question
##### Stage: 5 Challenge Level:
This problem is a sequence of linked mini-challenges leading up to the proof of a difficult final challenge, encouraging you to think mathematically. Starting with one of the mini-challenges, how. . . .
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Polite Numbers
##### Stage: 5 Challenge Level:
A polite number can be written as the sum of two or more consecutive positive integers. Find the consecutive sums giving the polite numbers 544 and 424. What characterizes impolite numbers?
### Plus or Minus
##### Stage: 5 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Fibonacci Factors
##### Stage: 5 Challenge Level:
For which values of n is the Fibonacci number fn even? Which Fibonnaci numbers are divisible by 3?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Conjugate Tracker
##### Stage: 5 Challenge Level:
Make a conjecture about the curved track taken by the complex roots of a quadratic equation and use complex conjugates to prove your conjecture.
### Cyclic Triangles
##### Stage: 5 Challenge Level:
Make and prove a conjecture about the cyclic quadrilateral inscribed in a circle of radius r that has the maximum perimeter and the maximum area.
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Integral Sandwich
##### Stage: 5 Challenge Level:
Generalise this inequality involving integrals.
### Why Stop at Three by One
##### Stage: 5
Beautiful mathematics. Two 18 year old students gave eight different proofs of one result then generalised it from the 3 by 1 case to the n by 1 case and proved the general result.
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Discrete Trends
##### Stage: 5 Challenge Level:
Find the maximum value of n to the power 1/n and prove that it is a maximum.
### 2^n -n Numbers
##### Stage: 5
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Fibonacci Fashion
##### Stage: 5 Challenge Level:
What have Fibonacci numbers to do with solutions of the quadratic equation x^2 - x - 1 = 0 ?
### Steve's Mapping
##### Stage: 5 Challenge Level:
Steve has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### Multiplication Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Thebault's Theorem
##### Stage: 5 Challenge Level:
Take any parallelogram and draw squares on the sides of the parallelogram. What can you prove about the quadrilateral formed by joining the centres of these squares?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Pythagorean Fibs
##### Stage: 5 Challenge Level:
What have Fibonacci numbers got to do with Pythagorean triples?
### Poly Fibs
##### Stage: 5 Challenge Level:
A sequence of polynomials starts 0, 1 and each poly is given by combining the two polys in the sequence just before it. Investigate and prove results about the roots of the polys.
### Vecten
##### Stage: 5 Challenge Level:
Join in this ongoing research. Build squares on the sides of a triangle, join the outer vertices forming hexagons, build further rings of squares and quadrilaterals, investigate.
### Prime Sequences
##### Stage: 5 Challenge Level:
This group tasks allows you to search for arithmetic progressions in the prime numbers. How many of the challenges will you discover for yourself?
### Center Path
##### Stage: 3 and 4 Challenge Level:
Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . .
### Trig Rules OK
##### Stage: 5 Challenge Level:
Change the squares in this diagram and spot the property that stays the same for the triangles. Explain...
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Close to Triangular
##### Stage: 4 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Painting by Numbers
##### Stage: 5 Challenge Level:
How many different colours of paint would be needed to paint these pictures by numbers?
### Alison's Mapping
##### Stage: 4 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
##### Stage: 4 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
### Least of All
##### Stage: 5 Challenge Level:
A point moves on a line segment. A function depends on the position of the point. Where do you expect the point to be for a minimum of this function to occur.
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
### Sixty-seven Squared
##### Stage: 5 Challenge Level:
Evaluate these powers of 67. What do you notice? Can you convince someone what the answer would be to (a million sixes followed by a 7) squared?
### OK! Now Prove It
##### Stage: 5 Challenge Level:
Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
##### Stage: 4 Challenge Level:
Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles?
### Summats Clear
##### Stage: 5 Challenge Level:
Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1, 2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a - b) = ab.
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
##### Stage: 5 Challenge Level:
With red and blue beads on a circular wire; 'put a red bead between any two of the same colour and a blue between different colours then remove the original beads'. Keep repeating this. What happens?
### Binary Squares
##### Stage: 5 Challenge Level:
If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number. | 2,228 | 9,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2017-04 | longest | en | 0.821133 |
https://www.airmilescalculator.com/distance/nan-to-trw/ | 1,718,532,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00679.warc.gz | 582,389,271 | 11,138 | # How far is Tarawa from Nadi?
The distance between Nadi (Nadi International Airport) and Tarawa (Bonriki International Airport) is 1347 miles / 2168 kilometers / 1171 nautical miles.
1347
Miles
2168
Kilometers
1171
Nautical miles
3 h 3 min
170 kg
## Distance from Nadi to Tarawa
There are several ways to calculate the distance from Nadi to Tarawa. Here are two standard methods:
Vincenty's formula (applied above)
• 1347.421 miles
• 2168.464 kilometers
• 1170.877 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 1354.175 miles
• 2179.334 kilometers
• 1176.746 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Nadi to Tarawa?
The estimated flight time from Nadi International Airport to Bonriki International Airport is 3 hours and 3 minutes.
## Flight carbon footprint between Nadi International Airport (NAN) and Bonriki International Airport (TRW)
On average, flying from Nadi to Tarawa generates about 170 kg of CO2 per passenger, and 170 kilograms equals 375 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path from Nadi to Tarawa
See the map of the shortest flight path between Nadi International Airport (NAN) and Bonriki International Airport (TRW). | 382 | 1,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.808976 |
https://www.jiskha.com/questions/1824705/calculate-the-potential-of-a-platinum-electrode-immersed-in-a-solution-that-is-0-1996-m-in | 1,591,385,098,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00470.warc.gz | 771,554,688 | 5,550 | # Chemistry
Calculate the potential of a platinum electrode immersed in a solution that is 0.1996 M in V(OH)4 ^+ . 0.0789 M in VO^2+ and 0.0800 M in HClO4
1. 👍 0
2. 👎 0
3. 👁 50
1. I'm a little confused about exactly what you want but I'll guess this.
V(OH)4^+ + e + 2H^+ ==> VO^2+ + 3H2O
E = Eo - 0.0591/1 log[V(OH)4]^+[H^+)^2/[VO]^2[H2O]^3
You will need to look up the Eo value for [V(OH)4]^+ to [VO]^2+
1. 👍 0
2. 👎 0
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Consider the cell: (Pt) H2/H+ || (Pt) H+/H2. In the anode half-cell, hydrogen gas at 1.0 atm is bubbled over a platinum electrode dipping into a solution that has a pH of 7.0. The other half-cell is identical to the first except
asked by Anonymous on March 27, 2012
6. ### chemistry
A galvanic cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode composed of silver in a 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential
asked by katie on February 27, 2012 | 1,015 | 3,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.918439 |
https://www.takshilalearning.com/ncert-solutions/ncert-solutions-for-class-10-economics-chapter-1-development/ | 1,680,231,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00362.warc.gz | 1,106,505,941 | 75,505 | NCERT Solutions for Class 10 Economics Chapter 1 Development
Economics Class 10 NCERT Solutions Chapter 1
NCERT Books Solutions for Class 10 Economics Chapter 1
NCERT Solutions for Class 10 Economics Chapter 1 Development is the first stepping stone for a student in the competitive world. With the introduction of the CBSE Board Exam for class 10 a few years back, this has become an important gateway for a student. Based on the results of class 10th a student selects his future stream of Science, Commerce or Arts suiting his interest.
Takshila Learning provides you with detailed and well explained NCERT Solutions for Class 10 Social Science of each chapter of each subject for NCERT Class 10. These NCERT Solutions help you to easily understand every concept so that you can score high in your CBSE Class 10 Board Exams.
Below you can find the NCERT Solutions for Class 10 Economics Chapter 1. You can get a Solution for the all-important question of “Development”
Q1. Development of a country can generally be determined by
1. its per capita income
2. its average literacy level
3. health status of its people
4. all the above
Q2. Which of the following neighbouring countries has better performance in terms of human development than India?
2. Sri Lanka
3. Nepal
4. Pakistan
Q3. Assume there are four families in a country. The average per capita income of these families is Rs 5000. If the income of three families is Rs 4000, Rs 7000 and Rs 3000 respectively, what is the income of the fourth family?
1. Rs 7500
2. Rs 3000
3. Rs 2000
4. Rs 6000
(4000+7000+3000+x) ÷ 4 = 5000
14000+x = 5000 × 4
x = 20000-14000
x = 6000
Q4. What is the main criterion used by the World Bank in classifying different countries? What are the limitations of this criterion, if any?
Answer: The World Bank uses per capita income to classify various countries. Per capita income is calculated by dividing the total income of the country by the population of the country. For the year 2017, countries with a per capita income of US \$ 12,056 per year are declared as rich countries and countries with per capita income of US \$ 955 or less are called low income countries. The limitations of the criterion are:1. While classifying countries, literacy rate, infant mortality, health care and other important factors are ignored.2. Information about uneven distribution of income has not been mentioned by the World Bank3. The economy of the country cannot determine the development of the country.
Q5. In what respects is the criterion used by the UNDP for measuring development different from the one used by the World Bank?
Answer: The criterion used by UNDP is different from that used by the World Bank becauseUNDP compares countries based on people’s educational level, their health status, and per capita income. This is contrary to the method used by the World Bank because the World Bank only calculates per capita income to measure growth.
Q6. Why do we use averages? Are there any limitations to their use? Illustrate with your own examples related to development.
Answer: Different countries have a different population so calculating the average helps to get an approximate answer that can be used to compare different things at different levels. Average calculation has limitations because we cannot know the difference in income of people and improper distribution of income in a country or state. For example, if we calculate the per capita income of two countries A and B with 5 people, then the salary of five people of country A is Rs. 2,3,000, Rs. 2,2,000, Rs. 2,3,500, Rs. 2,8,000 and Rs. 25,000 and the income of people living in Country B is 1 lakh 50 thousand rupees. 22,000, Rs.50,000, Rs.4000, Rs.2500. The average income of Country A is Rs.24,300 and the average income of Country B is Rs. 45,700. This proves that the average of Country B is higher than that of Country A, yet there is a disparity in the income distribution of Country B and that the income is distributed evenly for Country A.
Q7. Kerala, with lower per capita income has a better human development ranking than Haryana. Hence, per capita income is not a useful criterion at all and should not be used to compare states. Do you agree? Discuss.
Answer: Kerala with lower per capita income has a better human development ranking than Haryana. Therefore, per capita income is not a useful criterion and should not be used to compare states. This is true because Kerala has better literacy rate, infant mortality, health facilities, etc. than Haryana. Per capita income is calculated by simply calculating the state’s average income, regardless of any other factor.
Q8. Find out the present sources of energy that are used by the people in India. What could be the other possibilities fifty years from now?
Answer: Current sources of energy used by people in India include firewood, coal, petroleum, crude oil and natural gas. Fifty years from now, other possibilities can use solar energy and wind energy as a source for various energy forms. This is because the current use of sources of energy can cause the loss of these resources for future generations.
Q9. Why is the issue of sustainability important for development?
Answer: Sustainable development refers to the use of natural resources in a manner so that they can be used by present and future generations. The issue of sustainability is important for development because if natural resources are not used carefully, they may not be available to future generations. The dwindling resources of a country can eventually result in a decrease in the country’s development.
Q10. “The Earth has enough resources to meet the needs of all but not enough to satisfy the greed of even one person”. How is this statement relevant to the discussion of development? Discuss.
Answer:Development depends not only on the economic factors of a country, but also on the resources that are available for the use of the people of a country. Statement: “Earth has enough resources to meet the needs of all, but not enough to satisfy the greed of one person” is completely relevant in the context of a country’s development because natural resources are non-renewable. There are resources and it is the responsibility of the people to fulfill only their needs and their greed. If natural resources are not used wisely, generations to come may not use them for their own needs, which will not result in the development of a country.
Q11. List a few examples of environmental degradation that you may have observed around you.
Answer:Some examples of environmental degradation that we can see around us are:1. Pollution due to vehicles and excessive use of fuel in vehicles.2. Industrial waste is collected in residential areas and released into water bodies3. Deforestation4. Mining5. Soil degradationIncreased pollution in the environment has resulted in global warming and the decay of glaciers and atmospheric conditions.
Q12. For each of the items given in Table 1.6, find out which country is at the top and which is at the bottom.
Answer:According to Table 1.6, Sri Lanka topped all four categories. It has the highest gross national income, life expectancy at birth, mean year of schooling for people age 25 and older and HDI rank in the world. Nepal has the lowest gross national income among the given countries. Pakistan has the lowest life expectancy at birth and the lowest HDI rank in the world among given countries. Schooling for people 25 years and older is the lowest for Myanmar and Nepal.
Q13. The following table shows the proportion of adults (aged 15-49 years) whose BMI is below normal (BMI <18.5 kg/m2) in India. It is based on a survey of various states for the year 2015-16. Look at the table and answer the following questions.
State Male Female Kerala Karnataka Madhya Pradesh 8.5 17 28 10 21 28 All States 20 23
A) Compare the nutritional level of people in Kerala and Madhya Pradesh.
Answer: The nutritional level of people in Kerala is higher than the nutritional level of people in Madhya Pradesh.
1. B) Can you guess why around one-fifth of people in the country are undernourished even though it is argued that there is enough food in the country? Describe in your own words.
Answer: One-fifth of the population in the country is undernourished, although it is argued that there is enough food in the country for the following reasons:1. Inequality in distribution of food grains through Public Distribution System (PDS)2. Nutritious food cannot be taken by the poor population in the country.3. Unemployment occurs due to educational backwardness of the people, due to which people are not able to meet the basic requirement of food.4. There is no proper distribution of ration at fixed price shops.
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0Shares | 1,990 | 9,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-14 | latest | en | 0.907829 |
http://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-12th-edition/chapter-12-analysis-of-variance-review-chapter-quick-quiz-page-627/9 | 1,524,407,193,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00528.warc.gz | 421,011,575 | 11,864 | # Chapter 12 - Analysis of Variance - Review - Chapter Quick Quiz: 9
No effect.
#### Work Step by Step
The given P-value: P=0.395. If the P-value is less than the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.1$, hence we fail reject the null hypothesis. Hence we can say that there is no effect from the row factor.
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. | 128 | 539 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-17 | latest | en | 0.873737 |
https://byjus.com/question-answer/in-a-parallelogram-abcd-diagonals-ac-and-bd-intersects-at-o-and-ac-12-8-11/ | 1,686,285,899,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00157.warc.gz | 188,101,271 | 23,075 | Question
# In a parallelogram ABCD diagonals AC and BD intersects at O and AC =12.8 cm and BD=7.6 cm, then the measure of OC and OD respectively equal to
A
1.9 cm and 6.4 cm
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B
3.8 cm and 3.2 cm
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C
3.8 cm, 3.8 cm
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D
6.4 cm and 3.8 cm
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Solution
## The correct option is C 6.4 cm and 3.8 cmSince diagonals of a parallelogram bisect each other. ∴OC=12×AC =12×12.8 =6.4 cm Now,OD=12×BD =12×7.6 =3.8 cm
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Watch in App | 275 | 765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-23 | latest | en | 0.827812 |
https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/12/lesson/12.4.1/problem/12-169 | 1,726,469,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00820.warc.gz | 267,636,901 | 16,217 | ### Home > A2C > Chapter 12 > Lesson 12.4.1 > Problem12-169
12-169.
Explain each transformation below in terms of the parent graph $y = \operatorname{tan}θ$
1. $y = \operatorname{tan}θ + 1$
1. $y = \operatorname { tan } ( \theta + \frac { \pi } { 4 } )$
1. $y = −\operatorname{tan}\left(θ\right)$
1. $y = 4 \operatorname{tan}θ$
For the following questions, compare the equations to the parent $y = tan\left(x\right)$.
Use the eTool below to explore the transformations.
Click the link at right for the full version of the eTool: A2C 12-169 HW eTool | 184 | 556 | {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.534037 |
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Explore the properties of perspective drawing. | 431 | 1,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | latest | en | 0.788522 |
https://www.vedantu.com/question-answer/find-the-range-of-log-esin-x-is-a-infty-infty-b-class-12-maths-cbse-5edcbb2b4d8add132469cb6c | 1,620,578,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00080.warc.gz | 1,119,350,215 | 82,916 | Question
# Find the range of ${\log _e}(\sin x)$ isA)$( - \infty ,\infty )$B)$( - \infty ,1)$C)$( - \infty ,0]$D)$( - \infty ,0)$
Hint: Here we will proceed with the solution as we know the range of $\sin x = [1, - 1]$ which is required to solve this problem.
Here we need to find the range of given value that is ${\log _e}(\sin x)$
As we know that the domain of logarithmic functions are of positive value only
Since we know the range of $\sin x$ is $[ - 1,1]$
Then the domain value of above log function would be $(0,1]$
So, now to get the range of ${\log _e}(\sin x)$
Let us substitute $x = 0$ in the given function ${\log _e}(\sin x)$
i.e.
For $x \to 0 \Rightarrow {\log _e}x = - \infty$
Now let us substitute $x = 1$ in the given function ${\log _e}(\sin x)$
i.e.
For$x \to 1 \Rightarrow {\log _e}x = 0$
Hence from this we can say that the range of given function ${\log _e}(\sin x)$=$( - \infty ,0]$
NOTE: In this particular problem we know the range of $\sin x$ is $[ - 1,1]$ and domain of log function is $(0,1]$ so by substituting the domain values ($x$ values) in the given function i.e. ${\log _e}(\sin x)$ . We will get the range of the given function. | 392 | 1,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-21 | longest | en | 0.732342 |
https://engineering.stackexchange.com/questions/25136/how-does-stiffness-rigidity-affect-the-bending-moment-of-a-beam | 1,653,554,686,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00667.warc.gz | 298,255,403 | 70,776 | # How does stiffness/rigidity affect the bending moment of a beam
My question may sound silly as it’s mainly good sense but I may be mixing things up.
Let’s consider a flexible cantilever beam of a given length and material is fixed at one end and free at the other end. Let’s suppose a dead weight is put at the free end. This results in a bending moment at the fixed end of say 10 Nm and a given deflection at the free end of say 1 cm.
Now let’s suppose the beam thickness gets increased and as a result stiffer. The deflection at the tip will be reduced for sure, but what about the bending moment at the fixed end? Is making the beam more rigid going to reduce the bending moment?
Looking forward to your thoughts! Many thanks
Thanks for you answers, much appreciated.
I get that the bending moment caused by a static load applied at the free-end does not depend on the elasticity of the beam. Only the deflection does. I would have thought that the greater the deflection the higher the bending moment... Can you confirm this isn't true? Imagine a 100 m long vertical monopile support structure looking like a slender cylinder anchored in the seabed and subjected to dynamic wave loadings exciting all the cylinder modes. The thinner the cylinder wall is, the more elastic the cylinder gets and the greater the deflections at the top get. I would think that the maximum bending moment would be reached when the deflection is the most significant and therefore making the cylinder more rigid with thicker wall would both reduce the deflections and the max bending moment, wouldn't it?
• The bending moment is independent of the geometry of beam, however the curvature and vertical deflection are inversely proportional to second area of moment, how thicker how bigger the latter one. Dec 17, 2018 at 22:01
• For large displacements, linear beam theory is no longer valid, and second order effects can increase or decrease the loads. Dec 19, 2018 at 15:08
There are two basic types of structure.
Statically determinate structures are those where you can calculate the forces at the restraints without knowing anything about the flexibility of the structure itself (you assume it is strong enough to carry the loads without breaking, of course).
Statically indeterminate or redundant structures are those where the way the loads are transferred to the restraints depend on the relative stiffness (or flexibility) of the different parts of the structure. A simple example of a statically indeterminate structure would be a heavy rectangular box resting on a table. The total force exerted on the table top is simply the weight of the box, but the way the force is distributed over the area the box is resting on depends on the flexibility of both the box and the table.
A cantilever beam fixed at one end and loaded at the other end is statically determinate. For most beam structures you can construct a shear force and bending moment diagram which shows the forces and moments everywhere along the structure.
The size of the deflection of the beam depends on its flexibility, but the distribution of the internal forces on it does not, and the forces don't change for different "shapes" of the beam, e.g. changes in cross section along it lengths, or sections made from different materials.
Statically determinate structures are very convenient for making engineering designs, because the design process splits into two parts: first you calculate the internal forces everywhere in the structure, and the you can select the size of the structure to support those loads with acceptable stress levels (to avoid failure) and sufficiently small deformations. For example structures like the trusses which support the roof of buildings are not "perfectly" statically determinate, but they are close enough that they can be designed assuming they are.
In your example the change in the cross section of the beam doesn't have any effect on the end moment even if the beam is a hollow section such as a pipe as long as the ratio of length to depth is greater than 10.
When the beam is very deep and this ratio is less than 10, shear deformation and web warping effect could change the picture. We follow Euler- Bernoulli beam theory which is a great simplification of linear elasticity albeit a genius one. And that theory is very effective predictive of stresses when the beam is slender.
See a brief history of the theory here.
Actually it's the other way around. Imagine a load on your beam. The integration of the loading is the shear force. The integration of the shear force is the moment. The angle of deflection is the integration of the moment divided by E*I (this is where the material kicks in, E is Young's modulus and I is the second moment of area of beam's section). Then you integrate the angle of deflection to find the vertical deflection. (This may be helpful: https://en.m.wikipedia.org/wiki/Direct_integration_of_a_beam)
In a statically determined system, axial & shear forces along with bending and torsional moments are all unrelated to your material or your section (considering been in the elastic region, small deflections etc).
The biggest moment is not at the point where the biggest displacement is, in a double fixed beam, the supports have the biggest moment but the deflections there are 0.
As mentioned by other answers, when dealing with a statically determinate structure, the stiffness of each element is irrelevant when calculating the bending moment, but a key variable when calculating the deflection. Meanwhile, for statically indeterminate structures, even the calculation of bending moment requires the stiffness.
In simplified terms, this is because in statically determinate structures one can determine how the load is transferred through the structure to the supports without caring about the stiffness itself. In indeterminate structures, however, the stiffness directly impacts how the load is shared by the supports.
So let's try a variation on your cantilever example:
That's just a cantilever with a force applied at the free end. There's also a dangling vertical beam at the end (which is unsupported at the bottom).
Now, I don't need to know anything about the stiffness of each of those beam segments to determine how the load will be divided between them. We trivially know that dangling beam won't do anything, and all the load will be resisted by the cantilever. After all, that dangling beam might be infinitely stiff, but it's a dead end.
Now imagine you get the thinnest possible steel wire and use it to connect the bottom of the dangling beam to the support (with fixed connections, not pinned). Your structure diagram then becomes something like this:
Now, given how weak that new diagonal "beam" is (just a tiny wire), we can probably approximate its stiffness as equal to zero (especially when compared to the stiffness of an actual steel beam). Which basically means we can pretend the diagonal doesn't exist. Therefore, the structure will actually behave exactly as it did in the original case: the entire load will be resisted by the horizontal beam.
So, that was easy enough.
But now imagine the diagonal and horizontal members were reversed: the diagonal is the steel beam and the horizontal is just a pitiful wire. In this case, we can obviously state that the load will travel down the vertical (previously dangling) beam and then up the diagonal towards the support, with the tiny wire not really doing much of anything.
But what if both the horizontal and diagonal beams are steel beams, and therefore both contribute to resisting the load? Well, then we can't trivially figure this out anymore.1
And that's why calculating the shear force and bending moment of statically indeterminate structures depends on the elements' stiffness: each element's stiffness defines how much load it must support. And the proportion of the load going to each element is directly proportional to its stiffness: the stiffer the beam (as compared to others sharing the load), the more load it must support.
As for your follow-up question, the answer is simply "no". Just think of the fundamental beam equation:
$$\dfrac{\partial^2}{\partial x^2}\left(EI\dfrac{\partial^2 w}{\partial x^2}\right) = q$$
This tells us that the first integral of the loading is the shear force, the second integral is the bending moment, the third integral is the angle of rotation times the stiffness, and the fourth integral is the deflection times the stiffness.
Obviously, the greater the load applied on a beam, the greater the shear force and bending moment will be and therefore, the greater the deflection. But if you change the beam's stiffness (and the applied loading remains the same), the bending moment will remain the same, but the deflection will change.
For example, if you double a beam's stiffness, how do you expect its deflection to behave?
The correct answer is that if you double the beam's stiffness, then the deflection will be halved. This is consistent with what I've said: the loading remained the same, therefore so did the shear force (first integral of loading) and bending moment (second integral of loading) diagrams. However, since the third and fourth integrals of loading are equal to the product of stiffness and rotation or deflection, the resulting rotation/deflection is halved if the stiffness is doubled (basically, $$EIw = (2EI)(w/2)$$).
If your idea were correct, then doubling the stiffness would lead to another answer: the bending moment diagram itself would change (say, decrease by half), and then the deflection due to that reduced moment would itself be reduced by half due to the stiffness coefficient when integrating from moment to rotation and deflection.
1 This is a bit off-topic, so putting this as a footnote. To actually calculate this case, you need to use compatibility equations which effectively guarantee that the diagonal's deflection at the free end is equal to the horizontal's deflection at its free end plus the vertical's compressive deflection. Basically equations which guarantee that everyone agrees where (and at which angle of rotation) the nodes end up in the structure's deflected shape. | 2,080 | 10,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-21 | longest | en | 0.94376 |
https://vitez.me/parity-clarity | 1,713,958,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00111.warc.gz | 554,234,338 | 6,698 | # Parity Clarity
Sometimes good performance seems to be in direct conflict with programmer-understandable types. This post is about ways to get fast parity checking on natural number types, while also maintaining as many invariants as possible automatically. The example here is definitely contrived—if you wanted fast numeric parity calculations you could just check the lowest bit. However, it’s a pretty good stand-in for more complicated encodings, and how we can play with them to make them both performant and easy to understand. Along the way we’ll briefly touch on some cool Haskell language extensions: type families, GADTs, and data kinds.
Let’s start with a tiny subset of the Prelude, and we’ll also define a Bool type as well as the not operation on it which we can use later.
import Prelude((++), Show(..))
data Bool = True | False
deriving Show
not :: Bool -> Bool
not False = True
not True = False
Hopefully this is straightforward. Here’s our stand-in for an existing type with a focus on understandability by human programmers. We’re encoding natural numbers as Peano numerals, where Z means 0, and S means successor. Let’s also throw in a helpful succ function for finding successors.
data Nat
= Z
| S Nat
deriving Show
succ :: Nat -> Nat
succ = S
Our version of succ here only needs to provide the S constructor, which means it’s simple to write and easy to understand.
For whatever reason, the other constraint on our type is that we want fast parity checking, i.e. a way to tell whether a number is even or odd. Let’s write a version of that now.
even :: Nat -> Bool
even Z = True
even (S Z) = False
even (S (S n)) = even n
odd :: Nat -> Bool
odd n = not (even n)
Here we’re deconstructing our Nat and recursively finding out whether it’s even by reducing the number by two, and checking the parity of that. Intuitively, this works, because we always end at one of the base cases and are making the number smaller at each step. However, it’s rather slow. We’re taking around $$n/2$$ steps to check the parity of a number $$n$$, which means the runtime of this function grows linearly with the size of its input. Let’s see if we can do better.
Here’s another idea: we can encode odd and even numbers separately. This way, there’s an easy way to check a number’s parity from its encoding in a single step. Maybe our new data type looks like this:
data ParityNat
= ZZ
| SO Parity Nat
| SE ParityNat
deriving Show
ZZ is the equivalent of Z in Nat, but now we have both an SO (“odd successor”) and an SE (“even successor”). This makes the succ function a bit more complicated now:
succPar :: ParityNat -> ParityNat
succPar ZZ = SO ZZ
succPar n@(SO _) = SE n
succPar n@(SE _) = SO n
We want the SEs and SOs to alternate, which we can do as shown. However, note that now there’s a higher burden on people writing functions dealing with ParityNat as opposed to just plain Nats. Along with additional cases to handle, there are more places for implementations to contain errors. We’re also relying on people to use succ now, rather than just the plain data constructors. At least our new parity checking functions are speedy:
evenPar :: ParityNat -> Bool
evenPar ZZ = True
evenPar (SO _) = False
evenPar (SE _) = True
oddPar :: ParityNat -> Bool
oddPar n = not (evenPar n)
There’s another encoding option, which puts more burden on the author of the encoding, but hopefully automatically enforces more constraints for later users of the library. We can push parity checking into the types. First, let’s make a new type for the parity itself.
data Parity = Even | Odd
deriving Show
If this seems weirdly familiar, that’s because it’s isomorphic to Bool which we defined earlier. We can also define a “not” operation on Paritys, but first let’s take one step back. If we want to enforce parity in types, having Even and Odd terms isn’t that helpful. We need to promote Parity to a kind, and promote Even and Odd to types. While we’re at it, let’s also add the type families extension so we can define the equivalent to our not function, but at the type level instead of the term level. All in all, we’ll want the following extensions enabled:
{-# LANGUAGE GADTs #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE TypeFamilies #-}
Now we can define the type-level equivalent of not, but for Parity instead of Bool. Let’s call it “Opp”. Here’s its definition:
type family Opp (n :: Parity) :: Parity
type instance Opp 'Even = 'Odd
type instance Opp 'Odd = 'Even
This may look bizarre, but it’s more or less just a different syntax for defining a function. Opp takes a Parity called n and returns a Parity. The opposite of even is odd, and the opposite of odd is even. The ticks in front of 'Even and 'Odd remind us that these have been promoted from terms to types.
We can now define our constructors analogous to Z and S. This stuff here is the reason we needed GADTs.
data Natural :: Parity -> * where
Zero :: Natural 'Even
Succ :: Natural p -> Natural (Opp p)
instance Show (Natural p) where
show Zero = "Zero"
show (Succ n) = "(Succ " ++ show n ++ ")"
Breaking this down a bit further, a Natural is a type which takes something of kind Parity and gives us back a normal type (something of kind *). Zero is even, and the Succ of any Natural has the opposite parity as that Natural.
The successor function is again trivial to write.
succNat :: Natural p -> Natural (Opp p)
succNat = Succ
Weirdly, it now almost doesn’t make sense to have even and odd functions. Because this information is encoded in the types, it’s already sort of carried along with every Natural. However, just for the sake of completeness we can write something like this:
evenNat :: Natural 'Even -> Bool
evenNat _ = True
oddNat :: Natural 'Odd -> Bool
oddNat _ = True
This was a brief tour of a few possible encodings for a small bit of data with additional outside constraints on it. The first encoding was very simple, but runtime for operations we care about a lot (even and odd) was too slow. We switched to the more performant ParityNat, but this came at the cost of ease of use when writing functions using that type. Finally, we sort of pushed the problem up to the type level so that anything of type Natural 'Even has even parity, and likewise any Natural 'Odd has odd parity. This did away with the need for parity checking as functions, but comes at the cost of a more complicated type system encoding of our desired result.
Like so many places in engineering, this provides an interesting example of multiple tradeoffs that have to be balanced. Normally I lean towards “clarity at most costs”, sacrificing performance to make programmers’ jobs more manageable. However, occasionally understandability has to be traded away for enhancements of the details of how our programs actually run, and this tradeoff is more common as systems become more heavily relied upon by others. | 1,595 | 6,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-18 | latest | en | 0.924951 |
https://community.victronenergy.com/questions/114813/do-solar-panels-only-provide-wattage-required-by-l.html | 1,723,135,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00115.warc.gz | 140,928,718 | 35,568 | question
Do solar panels only provide wattage required by load?
I have two 350 watt 24 volt panels in parallel connected to two 140 amp hour agm batteries through a Victron smart solar 100/30 MPPT. Just put the system together. Main load is a 12 volt 45 watt fridge. In ten days of good sunshine the MPPT unit has never displayed solar power above 269 watts. It’s usually showng Pmax around 160 watts. Batteries don’t drop lower than 12.8 volts at night and quickly get up to 13 volts or more and fridge running fine so all seems good. Will the panels only produce close to their maximum when the load is larger? Thanks
matthewmiles
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Yes.
But you are also limited by that 30A maximum current which your 100/30 charger can output.
Since you are using a 12V battery, this means maximum power output is about 12Vx30A = 360W (slightly higher in practice - up to about 430W or so, because the battery voltage can go higher than 12V; I used the nominal voltage for simplicity).
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·
Thanks. Maybe I should have chosen the 100/50 unit to get the most out of my panels.. But my load will never be very large - a few LEDs and the fridge is all I need.
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matthewmiles ·
Although at first look it might seem as a complete waste, you still benefit from your oversized PV array during cloudy weather, when the PV panels can produce only a fraction of their rated power.
Also, if you have no shadow issues and if the two panels are placed in the same plane (oriented the same), you could wire them in series instead.
Your 100/30 probably can take that configuration (string voltage should be under the maximum limit of 100V for your charger, even in the coldest weather).
The benefit is higher PV array voltage (the charging will start sooner), lower current (less loses on the PV array wires).
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seb71 ·
As you can probably see I’m a complete newbie. I was surprised how cheap the panels were (and they’re Australian made with good reviews) so bought two.
I had thought that the 30 amp figure on the Victron unit was referring to input from the panels
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matthewmiles ·
Technically there is also a maximum allowed limit for the Isc from the PV array connected to the MPPT charger, but the parameter you must pay close attention to is the maximum voltage of your PV array.
If you will consider changing from panels in parallel to panels in series, better check the actual Voc and the Voc temperature coefficient (combined with the minimum temperature) for your panels and location.
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seb71 ·
Oversized arrays to help during cloudy weather are such a myth and waste of money.
The reduction in power is so enormous, you'd need to erect 10X the panels to compensate.
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nickdb ♦♦ ·
When there are very dark (thick) clouds or panels with a thick snow layer on them, yes.
But not all cloudy days are pitch black.
Sure, if you have stable grid power and/or just a few cloudy days each year, it's less tempting to oversize the PV array.
It also helps even in sunny winter days, if the panel angle is not optimal for the low winter Sun.
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seb71 ·
Thanks again for your help. I will look at a series configuration. At present the panels get a bit of shade in the morning but I’ll deal with that soon. The data sheet states Voc as 45 at standard test conditions and 41.5 at nominal module operating temperature. My location is usually very sunny and never gets snow. Minimum temperature for last few years was around 4 degrees or 39 in Fahrenheit. It’s usually too hot, hence the new fridge. I’m twenty miles from the grid and there’s no mobile phone or internet or tv. Perfect!
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matthewmiles ·
The data sheet states Voc as 45 at standard test conditions
In this case, DO NOT WIRE THEM IN SERIES.
At 4°C, panel Voc would probably be close to 48V. Two in series would be too close to the 100V maximum allowed for the MPPT charger.
Keep them as they are, in parallel.
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seb71 ·
I had no idea that temperature is so critical. Thanks again. As I said it’s only a simple shack in the bush that I visit on weekends so no need for lots of power. I will take your advice and leave it as is
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matthewmiles ·
In your case, most of the time the voltage will be much lower, because of the high temperatures. But one chilly day is enough to damage the charger.
If you want to take a closer look about how much the panel temperature influences the voltage, you can read this reply I recently wrote for a similar situation.
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seb71 ·
You are most helpful! Using your calculations and factoring in the increasingly erratic weather in my part of the world- crazy hot droughts and wildfires followed immediately by cold conditions and gigantic floods- I can readily see that the 100 volt limit of my charger would be inadequate in a series configuration. | 1,207 | 5,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.929269 |
https://mcqslearn.com/business-statistics/confidence-interval-estimation.php | 1,719,298,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00629.warc.gz | 318,351,923 | 16,475 | BBA Management Courses
Books:
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## Confidence Interval Estimation MCQs: Questions and Answers
MCQ 1:
If the point estimate is 8 and the margin of error is 5 then the confidence interval is
1. 3 to 13
2. 4 to 14
3. 5 to 15
4. 6 to 16
MCQ 2:
In confidence interval estimation, the confidence efficient is denoted by
1. 1 + β
2. 1 - β
3. 1 - α
4. 1 + α
MCQ 3:
In confidence interval estimation, the interval estimate is also classified as
1. confidence efficient
2. confidence deviation
3. confidence mean
4. marginal coefficient
MCQ 4:
In confidence interval estimation, the formula of calculating confidence interval is
1. point estimate * margin of error
2. point estimate ± margin of error
3. point estimate - margin of error
4. point estimate + margin of error | 394 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.689801 |
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# Causality For Someone who Doesn't Get it!
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24 minutes ago, StrictlyLogical said:
The law of causation is not a definition of causation. It is a limitation or a condition for it.
Entities ACT in accordance with their nature.
Then what is the definition of causation?
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37 minutes ago, StrictlyLogical said:
I think I know what your problem is.
The law of causation is not a definition of causation. It is a limitation or a condition for it.
Entities ACT in accordance with their nature.
This means that in the context of an entity reacting in an interaction with some thing acting on it, the action (reaction) it takes is not arbitrary, is not self-contradictory, it is in accordance with its nature. Meaning, it is of a nature such that when some thing else acts on it in a specific way it reacts in accordance with its nature.
Rand's law of causation does not in any way imply entities do not react or act when acted upon by other entities ... they obviously do, and Rand knew this.
Ok, so change happens.
The law of causation is what prevents really weird things from happening.
Without the concept of cause and effect, things would happen/change in ways that are not imaginable. A cat turns into a dog into a
table that burns and becomes a flower. This would be the world that does not limit change by some laws.
So causality implies that things don't happen arbitrarily, without any reason.
But it does not tell you what "cause" means!
So, to find the reason for something happening, you have look at the changes and identify the entities involved?
That is not how I do it intuitively.
I look at the change and through experimentation find the one factor that without it, the change does not occur.
That problem is at first glance, and maybe final glance, the one factor can be an attribute or an action.
"pressing" causes the "shining" in the light - if I don't press the light does not come on
"throwing" the ball caused the dog to "yelp" - if I don't throw, the dog does not yelp
I think that is how most people identify cause intuitively. If "the factor that is necessary" is the definition of cause, It is hard for me to make a case that the action is not the cause. Mainly because from a practical standpoint people don't need to identify the entity behind it. It is only when you need to prove that there is no necessity for a "first mover" and highly philosophical arguments that this precision is necessary.
I suppose I could make the case that the way that you (event causation believer) see it would not explain why the world is orderly and not arbitrary. There is no connection between an event and another except saying there is probability or likelihood that the sun will come up unless you look at entities and their identities and limitations on their actions.
I expect them to retort by saying: Without telescopes and science, as primitives in the jungle, what should someone say? "I am not sure why the sun comes up?" or based on the experience of having seen it before, there is the likelihood of seeing it again.
I think I am getting closer, the conversation has helped.
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1 hour ago, Eiuol said:
You are making it more complicated than it is. Start off with the idea that all actions are embodied by some entity. Sometimes, two or more entities interact which leads to all of them producing an action as a group. So BOTH the bubbles act, they BOTH do something. However, their relationship to one another is not identical. The important point is that there is not -only- a one way relationship, and that there is no such thing as a "pure" action.
Then I could take the billiard table as a group of balls (plural) and everything that happened on the table is the action (singular).
I have to go through this tedious drilling down to come up with a simple example. This is in order to get every step of the way right.
So, it is not that an entity causes an action but also that entities (plural) cause an action(singular).
Edited by Easy Truth
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5 hours ago, Easy Truth said:
The problem is being able to explain to the layperson that cause and effect, looked at in terms "the law of identity applied to action".
I have seen objectivists say that to lay people and I can see that it is not understood. So then you have to give examples and there are no good ones. I am looking for good, explanatory, educational examples!
There are no disembodied actions. Actions are verbs enacted by things which are nouns. "Jumping" cannot exist without the thing which jumps. Why would this be hard to explain or understand?
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1 hour ago, StrictlyLogical said:
It is common knowledge that entities act on one another. An entity can act in isolation (I can wave my hand in outer space) but they act on each other as well. I can grab a stick and push it. Without resorting to Newton or physics (mass, momentum etc) a consequence of the action is the effect on the stick.
Entities act in accordance with their nature. Actions of entities affect, i.e. interact with other entities.
(Leonard Peikoff does this bit about the abstraction "number"... someone says "I saw 9 running in the lobby")
Eiuol said
The important point is that there is not -only- a one way relationship and that there is no such thing as a "pure" action.
So you can't wave your arm in a vacuum. Granted, but we are speaking abstractly here, how else can we get to the bottom of this.
I would agree that entities act on each other and that it is common knowledge. But, I think the point that the objectivism wants to make is that entities don't cause other entities, philosophically speaking.
Is the argument against it that If that if entities could cause other entities, then something could cause the existence of something that did not exist, out of nothing?
My understanding is that the absolute necessary factor which is required is for an action to be "determined" by an entity, the entity's state at that moment. By state I mean is qualities, properties, attributes, basically its identity.
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6 minutes ago, Easy Truth said:
I would agree that entities act on each other and that it is common knowledge. But, I think the point that the objectivism wants to make is that entities don't cause other entities, philosophically speaking.
More from page 16 of OPAR:
An entity may be said to have a cause only if it is the kind of entity that is noneternal; and then what one actually explains causally is a process, the fact of its coming into being or another thing's passing away. Action is the crux of the law of cause and effect: it is action that is caused—by entities.
What is your source of the point(s) that Objectivism is making as you are trying to summarize?
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34 minutes ago, Grames said:
There are no disembodied actions. Actions are verbs enacted by things which are nouns. "Jumping" cannot exist without the thing which jumps. Why would this be hard to explain or understand?
Because it is far too abstract and the context has to be carefully explained first. If I say, it the way you said it, I could see people say "Why are you wasting my time with something that obvious?".
For instance, I have seen objectivists say "The law of causality is the law of identity applied to action." in midst of one of those meetings and the people's eyes glaze over. It is meaningless in a conversation.
I have to bring it to life, make it educational, make it simple.
In the context of explaining why the universe has a rhyme or reason and is not all chaos, being the way it is, the Objectivist explanation of causality is great. But when it comes to explaining what caused something in a person's daily life, either it is not applicable or the application is hard.
It is like being given a field theory and trying to explain a particular interaction. Like knowing the laws of economics and expecting to determine if the stock market will go up or down the next day. I am starting to think that I have to preface my explanation with the fact that it is not as applicable as thinking that events cause events but that it is what is real and it explains why entities don't have to come from non existence.
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1 hour ago, Easy Truth said:
Ok, so change happens.
The law of causation is what prevents really weird things from happening.
Without the concept of cause and effect, things would happen/change in ways that are not imaginable. A cat turns into a dog into a
table that burns and becomes a flower. This would be the world that does not limit change by some laws.
So causality implies that things don't happen arbitrarily, without any reason.
But it does not tell you what "cause" means!
So, to find the reason for something happening, you have look at the changes and identify the entities involved?
That is not how I do it intuitively.
I look at the change and through experimentation find the one factor that without it, the change does not occur.
That problem is at first glance, and maybe final glance, the one factor can be an attribute or an action.
"pressing" causes the "shining" in the light - if I don't press the light does not come on
"throwing" the ball caused the dog to "yelp" - if I don't throw, the dog does not yelp
I think that is how most people identify cause intuitively. If "the factor that is necessary" is the definition of cause, It is hard for me to make a case that the action is not the cause. Mainly because from a practical standpoint people don't need to identify the entity behind it. It is only when you need to prove that there is no necessity for a "first mover" and highly philosophical arguments that this precision is necessary.
I suppose I could make the case that the way that you (event causation believer) see it would not explain why the world is orderly and not arbitrary. There is no connection between an event and another except saying there is probability or likelihood that the sun will come up unless you look at entities and their identities and limitations on their actions.
I expect them to retort by saying: Without telescopes and science, as primitives in the jungle, what should someone say? "I am not sure why the sun comes up?" or based on the experience of having seen it before, there is the likelihood of seeing it again.
I think I am getting closer, the conversation has helped.
Take some time to keep chewing on these ideas ... also remember Ocam's razor and Rand's razor. Some philosophers thrive on inventing distinctions, subtleties, explananda, and floating concepts where none exist and none are needed. When you see Objectivism dispensing entirely with some concept there's a good reason for it: complete absence for the need of the concept on the basis of a complete absence of evidence of the senses to the contrary.
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6 minutes ago, dream_weaver said:
More from page 16 of OPAR:
An entity may be said to have a cause only if it is the kind of entity that is noneternal; and then what one actually explains causally is a process, the fact of its coming into being or another thing's passing away. Action is the crux of the law of cause and effect: it is action that is caused—by entities.
What is your source of the point(s) that Objectivism is making as you are trying to summarize?
2
Ah, so "some" entities can cause other entities. That is a huge misunderstanding on my part. Let me go through that then.
I thought that since there is a causal link between entity and action, there was none between entity and entity. It turns out that the universe is a special case, eternal don't have a cause.
Sources are the standard ones, the books, the lexicon and Peikoff's course on the history of philosophy.
I also saw the formal proof that is used to prove the direct linkage between entity and action, Kelly brought that up. I did not see a proof regarding entities and other entities.
Okay, I have to go back and reformulate.
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10 minutes ago, Easy Truth said:
I also saw the formal proof that is used to prove the direct linkage between entity and action
Formal proof, or validation? Peikoff classified causality as a corollary of identity, where [a] "corollary"
is a self-evident implication of already established knowledge. A corollary of an axiom is not itself an axiom; it is not self-evident apart from the principle(s) at its root (an axiom, by contrast, does not depend on an antecedent context). Nor is a corollary a theorem; it does not permit or require a process of proof; like an axiom, it is self-evident (once its context has been grasped).
(. . . . still more from page 16 of OPAR)
When you identify "sources" as "the books", do you consider Leonard Peikoff's "Objectivism: The Philosophy of Ayn Rand" one of them?
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18 minutes ago, dream_weaver said:
When you identify "sources" as "the books", do you consider Leonard Peikoff's "Objectivism: The Philosophy of Ayn Rand" one of them?
Yes I do.
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20 minutes ago, dream_weaver said:
Formal proof, or validation? Peikoff classified causality as a corollary of identity, where [a] "corollary"
is a self-evident implication of already established knowledge. A corollary of an axiom is not itself an axiom; it is not self-evident apart from the principle(s) at its root (an axiom, by contrast, does not depend on an antecedent context). Nor is a corollary a theorem; it does not permit or require a process of proof; like an axiom, it is self-evident (once its context has been grasped).
(. . . . still more from page 16 of OPAR)
The proof that David Kelly mentioned in his talk about Causality is this .. It links entity with action of the entity ...
If a thing a under conditions c produces a change x in subject s_the way
in which it acts must be regarded as a partial expression of what it is. It could
only act differently, if it were different. As long therefore as it is a, and stands
related under conditions c to a subject that is s, no other effect than x can be
produced; and to say that the same thing acting on the same thing may yet
produce a different effect, is to say that a thing need not be what it is. But this
is in flat contradiction to the Law of Identity. A thing, to be at all, must be
something, and can only be what it is. To assert a causal connexion between
a and x implies that a acts as it does because of what it is: because, in fact, it
is a. So long therefore as it is a, it must act thus; and to assert that it may act
otherwise on a subsequent occasion is to assert that what is a is something
else than the a which it is declared to be.
It is from a book "An Introduction to Logic" by someone name Joseph
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An Introduction to Logic by Horace William Brindley Joseph. It is in the public domain and available freely via http://books.google.com/.
While it is on my Kindle, it is what I've considered a dry, tough read. For an introduction to logic, I found Peikoff's Introduction to Logic helpful.
While reading OPAR this afternoon, I noticed that a point correlating the concepts of validation and volition share an etymological root was mentioned. Page 69 of OPAR:
The concept of "volition" is one of the roots of the concept of "validation" (and of its subdivisions, such as "proof"). A validation of ideas is necessary and possible only because man's consciousness is volitional. This applies to any idea, including the advocacy of free will: to ask for its proof is to presuppose the reality of free will.
Causality in conscious beings differs from causality in inanimate objects, a distinction that is eloquently captured in Miss Rand's identification of the need to distinguish between the metaphysical and the man-made.
Even in actions predicated by dogs and/or cats, adding the attribute of consciousness (not conceptual) adds a distinction between actions of conscious entities and inanimate entities. (How far down the animate/inanimate chain this goes is an investigation in, and of itself.)
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In the passage you quote (p 70 in my copy), Peikoff says that volition and validation are hierarchically connected but not that they are etymologically connected. "Volition" comes from a verb for willing or wishing, and "validation" from an adjective for strength or power.
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I saw the term roots, and just thought it was etymological. And yes, it is page 70. Depending on which side of page 70 it appears, sometimes I need to subtract 1, as I erroneously did when I typed that.
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I have a question about the concept action ...
Is "Existing" an action?
Because actions are aspects of the entities that act, the actions are part
of the identity of the entity
That implies that not all entities act.
But if "being" or "existing" is an action, that would imply that ALL entities act.
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Easy Truth, welcome to Objectivism Online.
I liked Aristotle’s bit about truth being in one way hard and in another way easy, “like the proverbial door one cannot avoid bumping into.”
Your insight that if existing were itself an action, it would imply that all entities act is right, provided we keep constant the sort of action we are talking about throughout that if-then statement. That kind of action would be something more inert than when, in science or in everyday experience, we say that such-and-such thing acted in such-and-such way. Nevertheless, for those who love not only easy truth, but hard truth, the question of whether existing is itself an action, or activity, is a good one. Aristotle, Leibniz, and Lotze affirmed. Russell denied, arguing against Lotze. The case that Aristotle is an affirmer on this question is made out by Aryeh Kosman in his book The Activity of Being (2013).
I’m a denier on the conception of existing as necessarily being a sort of acting. I affirm that all concrete existence is temporal, but for that, it suffices that some concrete part in the whole of concrete existence is acting in our ordinary and scientific types of acting. To be clear, I’m talking about any sort of acting that has been connected in a necessary way with existence per se in the history of philosophic reflection or is thusly connected by us in our philosophic reflection today. What we know from science (e.g. that mass is convertible to energy, that mass-energy has some dynamical relations with spacetime, and that the vacuum has energy) concerns other sorts of activity than the one that has been claimed by philosophers for existence per se, and these activities we learn in the physical sciences were conceived and discovered by means necessarily beyond philosophic reflection.
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4 hours ago, Easy Truth said:
Is "Existing" an action?
No. If 'existing' was a type of action, actions would hold metaphysical primacy. In other words, first there would be the platonic form of action, from which its numerous manifestations (including 'existing') would spring.
But actions can't have metaphysical primacy. Let's say for a moment that existing is a type of action. But what is the most fundamental thing you can say about action? That it exists
Entities that exist, act. Entities that do not exist, do not act.
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6 hours ago, Easy Truth said:
But if "being" or "existing" is an action, that would imply that ALL entities act.
Your logic is correct when you state that if being is an action then all entities act.
But your conclusion, that what Rand says "implies that not all entities act", is not entirely correct. What is correctly drawn from Rand, is that it would be false to conclude that simply by virtue of every entity existing, every entity is always acting.
It may be a fact of existence, that no entity may be 100% isolated from every other entity in existence and hence, it is always interacting with something else, i.e. it is actually acting. But this does not mean it is its being as such which IS the action. Its being makes the actions possible. Gravitational attraction would be an example kind of action an entity would perform which might not be capable of isolation.
That said, if one were to try to consider an entity which literally never acted, the entity necessarily never would have interacted (a kind of action) with anything else in the universe, the entity could not directly or indirectly affect our sensory apparatus or anything else in universe for that matter which could form evidence of the senses from which we could build knowledge of it. Such a thing is a ghost of no consequence whatever... wandering through a universe (if one could even intelligibly say this) with which it has no connection and being utterly unknowable. In fact, such a thing, which here I hypothesized in absence of evidence (which evidence by definition could not exist in any case) is simply a bald assertion of the arbitrary. An empty insanity dreamt up on a whim.
All entities that exist, act in the sense that they participate with other entities and are ultimately open to our direct or indirect perception.
Edited by StrictlyLogical
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5 hours ago, KyaryPamyu said:
No. If 'existing' was a type of action, actions would hold metaphysical primacy. In other words, first there would be the platonic form of action, from which its numerous manifestations (including 'existing') would spring.
But actions can't have metaphysical primacy. Let's say for a moment that existing is a type of action. But what is the most fundamental thing you can say about action? That it exists
Entities that exist, act. Entities that do not exist, do not act.
So a table is not "acting" when it is "preventing" a glass from hitting the floor. After all, the table is just being. But it has an effect on things around it. It has this stopping power that it is exerting. It seems to be interacting with the things on it.
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2 hours ago, StrictlyLogical said:
That said, if one were to try to consider an entity which literally never acted, the entity necessarily never would have interacted (a kind of action) with anything else in the universe, the entity could not directly or indirectly affect our sensory apparatus or anything else in universe for that matter which could form evidence of the senses from which we could build knowledge of it. Such a thing is a ghost of no consequence whatever... wandering through a universe (if one could even intelligibly say this) with which it has no connection and being utterly unknowable. In fact, such a thing, which here I hypothesized in absence of evidence (which evidence by definition could not exist in any case) is simply a bald assertion of the arbitrary. An empty insanity dreamt up on a whim.
All entities that exist, act in the sense that they participate with other entities and are ultimately open to our direct or indirect perception.
1
I have several questions based on your post but I will start by one. It seems the proper thing to say is that "all entities can act", they have that potential. There is no entity that is missing that potential.
I am leaning toward thinking that the action is not the best word, that I should use the word "interact" instead of "act". Because actions don't happen in a vacuum in a sense.
An entity causes the action and the action seems to have a linkage to a "subject of the action".
The dog eats (eats what?)
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To exist at all is to act, "act" in the ordinary active sense not a contrived passive sense. A table which holds up a glass is acting in accord with Newton's Third Law of Motion (for every action there is an equal and opposite reaction) as applied to static entities. Analyzing the same table-and-glass with the latest state-of-the-art physics does not change the conclusion that there are multiple entities acting and interacting as their attributes dictate..
A passive sense of action attributed to merely existing is an empty concept because it cannot refer to anything which is solely inherent to existing. Any possible candidate referent of such a concept would have to be an attribute that necessarily interacted with other entities before we could become aware of it. To return to Rand's short style phrases, "Existence is Identity" strictly contradicts "Existence has Identity".
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10 hours ago, StrictlyLogical said:
if one were to try to consider an entity which literally never acted, the entity necessarily never would have interacted (a kind of action) with anything else in the universe, the entity could not directly or indirectly affect our sensory apparatus or anything else in universe for that matter which could form evidence of the senses from which we could build knowledge of it. Such a thing is a ghost of no consequence whatever... wandering through a universe (if one could even intelligibly say this) with which it has no connection and being utterly unknowable. In fact, such a thing, which here I hypothesized in absence of evidence (which evidence by definition could not exist in any case) is simply a bald assertion of the arbitrary. An empty insanity dreamt up on a whim.
All entities that exist, act in the sense that they participate with other entities and are ultimately open to our direct or indirect perception.
1
If all entities act, and all actions are interactions, then all entities interact.
Is this right?
and the next one
If existing is an action, then action is not always a change and all changes are actions but not all actions are changes.
If existing is not an action, then action is change.
Is this right?
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12 hours ago, KyaryPamyu said:
No. If 'existing' was a type of action, actions would hold metaphysical primacy. In other words, first there would be the platonic form of action, from which its numerous manifestations (including 'existing') would spring.
But actions can't have metaphysical primacy. Let's say for a moment that existing is a type of action. But what is the most fundamental thing you can say about action? That it exists
Entities that exist, act. Entities that do not exist, do not act.
1
I tried to understand what "metaphysical primacy" means, I still don't get it.
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In the case where a rock is thrown at a window ...
Shattering is something the window does when it is struck by a rock. The glass (entity / cause) is shattering (action / effect).
Can I say
The glass (entity / cause) with attribute (collision by rock) shatters (action / effect).
Is there a causal connection with the person throwing the rock? If so, how does the person put that into words? It seems it is not correct to say the person caused the shattering. And why not?
It seems as if the person throwing the rock is changing an attribute of the glass window. The person is changing the attribute (free from collisions) to (collision by rock). And of course, based on the rule above, a glass with that attribute will act/shatter.
I can't seem to say the throwing caused a collision that changed the attribute. That would be action to action causation.
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What is Run Rate?
Definition: The concept of run rate is used to extrapolate financial results to estimate future revenues. For example, a new company may have sold goods of a value of \$1 million in its first six months. Using this run rate, the company could say that it expects its revenues to be \$2 million in its first year of operations.
What does Run Rate mean?
At times, it may be necessary for a company to make a prediction about its financial results. A startup may need to convince prospective investors about the viability of its business model. A new division within a company may need to prove to its top management that the product that it has recently launched has gained the acceptance of customers.
In situations like these, it is useful to talk about the results that are expected to be achieved in a full year of operations. But this is not possible if the firm or the division within the company has been in existence for a shorter period, say, three months.
The idea of a run rate is useful here. The company could take its revenues in a certain period, say, three months, and convert it into an annual figure. It could do this by simply multiplying its sales for three months by four to get a projected yearly sales figure.
However, this method of calculating future revenues has its limitations. Circumstances could change in subsequent months. A company may have enjoyed high sales in the initial period because customers may have wanted to try out its product. They could be dissatisfied with the quality and could refuse to make additional purchases.
The run rate could also give you an inaccurate figure of future sales if the product is seasonal. In addition to this, there are other factors that make this an inexact tool. The competition that a company faces could suddenly increase. A key set of employees may leave. All these factors could lead to the estimate arrived at using the run rate being widely off the mark.
Example of Run Rate
The Financial Times, a leading newspaper that specializes in business and economic news, recently carried a report headlined: “Zendesk Surpasses \$500M Annual Revenue Run Rate….”
The news report goes on to explain that in 2014, the company’s revenues stood at \$100 million. But, over the years, the sales figure increased as the company added new customers and increased revenues from existing clients.
At the beginning of April 2018, Zendesk announced that it expects the year’s revenues to exceed \$500 million based on its run rate in the first quarter.
Summary
A company can use its run rate to estimate future revenues. But the method should be used with caution as there may be circumstances that prevent a firm from keeping up the pace of sales that it has registered in an earlier period.
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In Week 3, you ran the independent-samples t-testto compare the mean of two groups. However, there may be circumstances where you need to compare more than two group means. An ANOVA is a statistical method used to compare the means of two or more groups. For example, an IT manager wants to determine whether the mean (average) times required to complete a certain IT task differ based on the three types of employee training. The IT manager randomly selects 10 employees who have undergone the three types of training (three groups). Using an ANOVA, the IT manager could analyze data to verify whether the mean times required to complete the same IT task varies based on the three types of training.
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Submit a synthesis of statistical findings derived from ANOVA that follows the Week 4 Assignment Template. Your paper must include the following:
• A description and justification for using the one-way ANOVA
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• An APA-formatted “Results” section for the one-way ANOVA
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• Identification of independent and dependent variables, including the identification of the number of levels for the independent variable
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## Presentation on theme: "Atomic, molecular, and formula masses"— Presentation transcript:
Atomic, molecular, and formula masses
Atomic weight? So what is the unit for atomic mass? An analogy for atomic mass An "atomic weight" experiment Avogadro's Hypothesis So then, why isn't the atomic mass of Hydrogen exactly 1? Isotopes Who picked hydrogen, anyway? What, you say? Not a molecule? Calculating formula masses Of Moles and other furry creatures How many is that? Does it matter? It's enough! Where'd they get that number? Summary - Molar Mass
The Mole Concept It is often more convenient for us to group objects into defined units of measure for calculations. For example, roses and eggs are commonly sold in units of 12, a dozen. Fruit at a farmer’s market is often sold by the bushel. Chemists use the unit of a mole to group atoms, molecules, or other things into workable units. If you have a mole of something, that means that you have 602,204,500,000,000,000,000,000 of it. To make it easier to write we usually use scientific notation and say that we have x 1023 things. The reason why this strange unit was chosen is historical, but as a consequence, the mole concept makes our life much easier by simplifying complex calculations. The number, x 1023, is also known as Avogadro’s number after its discoverer.
The Mole Concept The molar scale was set up so that a mole of any substance will be equal to its atomic (or molecular) mass expressed in grams. For example, twelve grams of carbon-12 is equal to one mole of carbon-12. Therefore, if we count the number of atoms in grams of carbon-12, we will find x 1023 of them. Restated, the mass of a mole of carbon-12 is 12.00g. The mass of a mole of a substance is defined as the molar mass and is given in units of g/mol. (The abbreviation mol is used to indicate mole(s) – NOT molecules.) Based on this equivalence, we can easily convert from atoms (or molecules) to grams using the molar mass and the molar definition as conversion factors.
How to do calculations between moles, atoms or molecules, and grams of a substance One of the main problems that beginning chemistry students have is in doing conversions between grams, moles, and molecules (or atoms). Usually, a question will be asked of you in the following form: How many moles are in 22 grams of copper metal? If you're confused by this problem, don't worry. Most people are when they start doing this kind of problem. To make life easier for you, I put together a "road map" which tells you exactly what you need to do to convert between atoms (or molecules), grams, and moles.
The Four Steps to Solving Mole Problems:
Step 1: Figure out how many parts your calculation will have by using the diagram Looking at the diagram above, we can see that we are going between grams and moles, which is a one-step conversion. Furthermore, we can see that we need to use the atomic mass of copper as our conversion factor. Step 2: Make a T-chart, and put whatever information the problem gave you in the top left. After that, put the units of whatever you were given in the bottom right of the T, and the units of what you want to find in the top right. In this case, the problem gave you "22 grams of copper" as the starting information. Because this is what you were given, put "22 grams of copper" in the top left of the T. Since "grams of copper" is the unit of what you were given, put this in the bottom right of the T. Since you want to find out how many moles of copper are going to be made, put "moles of copper" as your unit in the top right. When you've done this, your calculation should look like this:
Step 3: Put the conversion factors into the T-chart in front of the units on the right.
As we saw from the "map", the conversion factor between grams and moles is the atomic mass of copper. Because we measure atomic mass in grams, you need to put the atomic mass in front of the unit "grams of copper". What do you put in front of moles? Whenever you do a calculation of this kind, you need to put "1" in front of moles, like you see here: Step 4: Cancel out the units from the top left and bottom right, then find the answer by multiplying all the stuff on the top together and dividing it by the stuff on the bottom. In this case, you'd multiply 22 by one and divide the result by Your answer, 0.35 moles of copper:
Solving Two-Step Mole Calculation Problems:
What happens if we need to solve a problem that requires we not just go from one box in the next in our diagram, but across the entire diagram? Well, it means that we need to do two steps in our calculation. Let's see that "map" again to see what I mean: If we were asked to convert 22 grams of copper to atoms of copper, we'd have to go from one end of the map to the other. Instead of doing a simple one step calculation, we'd need to do a two-step calculation, with the first step going from grams to moles and the second step going from moles to atoms. How can we solve this kind of problem? Well, we start off by doing the same thing that we did in our last example: We had to convert grams to moles before, and we can see from the map that we have to convert grams to moles now, too. To refresh your memory, here's the calculation from last time:
In the next step, we do the same thing over again, except that we need to add another T to the T-chart. When you do this, take the units of the thing at the new top left and put them on the bottom right (in this case, moles). Then take the units of what you want (in this case, atoms) and put it in the top right. Finally, put in your conversion factors, which from the chart above is Avogadro's number, or 6.02E23. Since this number refers to the number of atoms in a mole of a substance, we put this in front of "atoms of copper". Again, put the number "1" in front of moles, because we're saying that there are 6.02E23 atoms in ONE mole of an element. When we add all these terms in, we can cross out the units that cancel out, as shown. To get the answer, multiply all the numbers on the top together and divide by the numbers on the bottom. Your answer should then be set up like this: | 1,486 | 6,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | longest | en | 0.927428 |
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The flashcards below were created by user insouci on FreezingBlue Flashcards.
1. What are the two types of reasoning we use in everyday life?
• 1. Inductive reasoning
• 2. Deductive reasoning
2. inductive reasoning
Inductive reasoning is the process of reasoning that arrives at a general conclusion or conjecture based on the observation of specific examples.
**Using specific examples to --- arrive --- at a general conclusion
3. deductive reasoning
Deductive reasoning is the process of reasoning that arrives at a specific conclusion based on previously accepted general statements.
** Arrive at a specific conclusion --- via --- previously accepted statements.
4. George Polya
FOUR major steps of problem solving
• 1. Understand the problem
• 2. Devise a plan
• 3. Carry out the plan
• 4. Look back, check the work
5. estimation
Estimation is the process of finding an approximate answer to a mathematical problem.
In many cases, it is not necessary to find the exact answer to a problem. When only an approximate answer is needed, you can use estimation. This is accopmplished by rounding the numbers used in the problem, then performing the necessary operatioin or operations.
6. define a set
A set is a well-defined collection of objects/elements.
7. What is meant when a set is well defined?
A set is said to be well defined when there is no misunderstanding as to whether or not an element belongs to a set.
Ex:the set of "letters of the English alphabet" is a well-defined set since it consists of the 26 symbols.
• The set of Great Lakes is {Ontario, Erie, Huron, Michigan, Superior}
• The days of the week are {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
• Write the set of natural numbers less than 6 {1, 2, 3, 4, 5}
8. Name the 3 ways to designate sets
• 1. list or roster method
• 2. descriptive method
• 3. set-builder method
9. Each object of a set is called...
an element or member of the set.
10. Write the set of natural numbers less than 8.
{1, 2, 3, 4, 5, 6, 7}
11. Write the descriptive of the set containing 2, 4, 6, 8, ...
Since the elements in the set are called the even natural numbers, the anser is:
E = even natural numbers
12. Use set-builder notation to designate the set {2, 4, 6}
{ x | x E and x < 7 }
Reads: "The set of all x such that x is an even natural number and x is less than seven."
13. Use set-builder notation to designate the set {red, yellow, blue}
{x | x is a primary color}
Reads: "The set of all x such that x is a primary color."
14. Designate the set 32, 33, 34, 35, ... using the roster method.
{32, 33, 34, 35, ...}
15. Designate the set 32, 33, 34, 35, ... using the descriptive method.
Natural numbers greater than 31.
16. Designate the set 32, 33, 34, 35, ... using the set-builder notation.
{ x | x N and x > 31 }
17. finite set
A set is said to be a finite set if the number of elements contained in the set is either 0 or a natural number.
18. infinite set
A set is said to be an infinite set if it has an unlimited number of elements.
19. finite? or infinte?
{the natural numbers that are multiples of 6}
infinite
20. finite? or infinte?
{ x | x is a member of the U.S. Senate}
finite
21. finite? or infinte?
{3, 6, 9, ..., 24}
finite
22. null set
A set with no elements is called an empty set or null set. The symbols used to represent the null set are { } or circle w/line
23. Two sets are equal if...
they have exactly the same members or elements.
24. Two sets have a one-to-one correspondence if and only if...
it is possible to pair the elements of one set with the elements of the other set in such a way that for each element in the first set there exists one and only one element in the second set.
25. What is a subset?
When all, some, or none of the elements of one set are used in another set, the second set is called a subset of the original set. Formally defined.
Ex: Subsets of {bacon, egg}
• {bacon, egg}
• {bacon}
• {egg}
• { }
26. What is a proper subset?
If a subset of a given set is NOT equal to the original set, then the subset is called a proper subset of the original set.
Original set: {bacon, egg}
{bacon, egg} ⊄ {bacon, egg} **Because it is equal to the original
{bacon}⊂ {bacon, egg}
27. What is the union of two sets?
All the elements of each set. Duplicates are not written twice.
Ex: A={10, 12, 14, 15} and B={13, 14, 15, 16, 17}
A ∪ B={10, 12, 13, 14, 15, 16, 17}
28. What is the intersection of two sets?
The set of elements that are common to both sets.
Ex: A={10, 12, 14, 15} and B={13, 14, 15, 16, 17}
A∩ B = {14, 15}
29. What is the complement of a set?
• It is the set of elements contained in the universal set that are NOT contained in the set noted.
• _
• A (the line over the set name denotes the compliment of that set)
• Basically everything outside of the mentioned set.
30. What is a Venn Diagram?
When a set or sets are represented pictorially using Venn Diagrams.
Author: insouci ID: 4347 Card Set: MATC1100 Updated: 2010-01-17T05:29:31Z Folders: Description: Ch. 1&2 Show Answers: | 1,410 | 5,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-40 | latest | en | 0.924476 |
http://school.eckovation.com/tips-tricks-simplification/ | 1,527,154,231,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866201.72/warc/CC-MAIN-20180524092814-20180524112814-00122.warc.gz | 253,818,320 | 18,666 | # Tips and Tricks on Simplification related problems in Aptitude
1. ‘BODMAS’ Rule:This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of given expression.Here B – Bracket,
O – of,
D – Division,
M – Multiplication,
S – Subtraction
Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (), {} and ||.
After removing the brackets, we must use the following operations strictly in the order:
(i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.
2. Modulus of a Real Number:Modulus of a real number a is defined as
|a| = a, if a > 0 –a, if a < 0
Thus, |5| = 5 and |-5| = -(-5) = 5.
3. Virnaculum (or Bar):When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum
A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A. 45 B. 60 C. 75 D. 90
Explanation:
Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
16x = 480
x = 30.
Hence, total number of notes = 3x = 90.
2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
A. 20 B. 80 C. 100 D. 200
Explanation:
Let the number of students in rooms A and B be x and y respectively.
Then, x – 10 = y + 10 x – y = 20 …. (i)
and x + 20 = 2(y – 20) x – 2y = -60 …. (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100. | 576 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2018-22 | latest | en | 0.830494 |
https://www.aqua-calc.com/what-is/surface-density/grain-per-square-nanometer | 1,652,829,101,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00007.warc.gz | 730,648,264 | 12,980 | # What is a grain per square nanometer (unit)
## Grain per square nanometer is a unit of measurement of surface density
The grain per square nanometer surface density measurement unit is used to measure area in square nanometers in order to estimate weight or mass in grains. The surface density is used to measure the thickness of paper, fabric and other thin materials.
• What is surface densityInstant conversionsConversion tables
• 1 gr/nm² = 0.06479891 µg/pm²
• 1 gr/nm² = 647.9891 µg/Ų
• 1 gr/nm² = 64 798.91 µg/nm²
• 1 gr/nm² = 64 798 910 000 µg/µ²
• 1 gr/nm² = 64 798 910 000 µg/µm²
• 1 gr/nm² = 6.479891×10+16 µg/mm²
• 1 gr/nm² = 6.479891×10+18 µg/cm²
• 1 gr/nm² = 6.479891×10+22 µg/m²
• 1 gr/nm² = 64.79891 µg/ha
• 1 gr/nm² = 4.18056647756×10+19 µg/inch²
• 1 gr/nm² = 6.02001572769×10+21 µg/ft²
• 1 gr/nm² = 5.41801415492×10+22 µg/yd²
• 1 gr/nm² = 6.479891×10-5 mg/pm²
• 1 gr/nm² = 0.6479891 mg/Ų
• 1 gr/nm² = 64.79891 mg/nm²
• 1 gr/nm² = 64 798 910 mg/µ²
• 1 gr/nm² = 64 798 910 mg/µm²
• 1 gr/nm² = 64 798 910 000 000 mg/mm²
• 1 gr/nm² = 6.479891×10+15 mg/cm²
• 1 gr/nm² = 6.479891×10+19 mg/m²
• 1 gr/nm² = 4.18056647756×10+16 mg/in²
• 1 gr/nm² = 6.02001572769×10+18 mg/ft²
• 1 gr/nm² = 5.41801415492×10+19 mg/yd²
• 1 gr/nm² = 6.479891×10-8 g/pm²
• 1 gr/nm² = 0.0006479891 g/Ų
• 1 gr/nm² = 0.06479891 g/nm²
• 1 gr/nm² = 64 798.91 g/µ²
• 1 gr/nm² = 64 798.91 g/µm²
• 1 gr/nm² = 64 798 910 000 g/mm²
• 1 gr/nm² = 6 479 891 000 000 g/cm²
• 1 gr/nm² = 6.479891×10+16 g/m²
• 1 gr/nm² = 41 805 664 775 600 g/in²
• 1 gr/nm² = 6.02001572769×10+15 g/ft²
• 1 gr/nm² = 5.41801415492×10+16 g/yd²
• 1 gr/nm² = 6.479891×10-11 kg/pm²
• 1 gr/nm² = 6.479891×10-7 kg/Ų
• 1 gr/nm² = 6.479891×10-5 kg/nm²
• 1 gr/nm² = 64.79891 kg/µ²
• 1 gr/nm² = 64.79891 kg/µm²
• 1 gr/nm² = 64 798 910 kg/mm²
• 1 gr/nm² = 6 479 891 000 kg/cm²
• 1 gr/nm² = 64 798 910 000 000 kg/m²
• 1 gr/nm² = 6.479891×10+17 kg/ha
• 1 gr/nm² = 41 805 664 776 kg/in²
• 1 gr/nm² = 6 020 015 727 690 kg/ft²
• 1 gr/nm² = 54 180 141 549 200 kg/yd²
• 1 gr/nm² = 6.479891×10+15 centner/ha
• 1 gr/nm² = 6.479891×10-14 t/pm²
• 1 gr/nm² = 6.479891×10-10 t/Ų
• 1 gr/nm² = 6.479891×10-8 t/nm²
• 1 gr/nm² = 0.06479891 t/µ²
• 1 gr/nm² = 0.06479891 t/µm²
• 1 gr/nm² = 64 798.91 t/mm²
• 1 gr/nm² = 6 479 891 t/cm²
• 1 gr/nm² = 64 798 910 000 t/m²
• 1 gr/nm² = 41 805 664.8 t/in²
• 1 gr/nm² = 6 020 015 728 t/ft²
• 1 gr/nm² = 54 180 141 549 t/yd²
• 1 gr/nm² = 2.285714288×10-9 oz/pm²
• 1 gr/nm² = 2.285714288×10-5 oz/Ų
• 1 gr/nm² = 0.002285714288 oz/nm²
• 1 gr/nm² = 2 285.71429 oz/µ²
• 1 gr/nm² = 2 285.71429 oz/µm²
• 1 gr/nm² = 2 285 714 283 oz/mm²
• 1 gr/nm² = 228 571 428 250 oz/cm²
• 1 gr/nm² = 2.2857142825×10+15 oz/m²
• 1 gr/nm² = 1 474 651 429 130 oz/in²
• 1 gr/nm² = 2.12349805198×10+14 oz/ft²
• 1 gr/nm² = 1.91114824679×10+15 oz/yd²
• 1 gr/nm² = 1.42857142867×10-10 lb/pm²
• 1 gr/nm² = 1.428571429×10-6 lb/Ų
• 1 gr/nm² = 0.0001428571429 lb/nm²
• 1 gr/nm² = 142.857143 lb/µ²
• 1 gr/nm² = 142.857143 lb/µm²
• 1 gr/nm² = 142 857 143 lb/mm²
• 1 gr/nm² = 14 285 714 274 lb/cm²
• 1 gr/nm² = 1.42857142737×10+14 lb/m²
• 1 gr/nm² = 92 165 714 240 lb/in²
• 1 gr/nm² = 13 271 862 816 800 lb/ft²
• 1 gr/nm² = 1.19446765351×10+14 lb/yd²
• 1 gr/nm² = 9.999999998×10-7 gr/pm²
• 1 gr/nm² = 0.009999999998 gr/Ų
• 1 gr/nm² = 1 000 000 gr/µ²
• 1 gr/nm² = 1 000 000 gr/µm²
• 1 gr/nm² = 1 000 000 000 000 gr/mm²
• 1 gr/nm² = 1.0×10+14 gr/cm²
• 1 gr/nm² = 1.0×10+18 gr/m²
• 1 gr/nm² = 15.4323584 gr/ha
• 1 gr/nm² = 6.4516×10+14 gr/inch²
• 1 gr/nm² = 9.29030400001×10+16 gr/ft²
• 1 gr/nm² = 8.3612736×10+17 gr/yd²
• 1 gr/nm² = 7.14285713687×10-14 short tn/pm²
• 1 gr/nm² = 7.14285713687×10-10 short tn/Ų
• 1 gr/nm² = 7.142857137×10-8 short tn/nm²
• 1 gr/nm² = 0.07142857137 short tn/µ²
• 1 gr/nm² = 0.07142857137 short tn/µm²
• 1 gr/nm² = 71 428.5714 short tn/mm²
• 1 gr/nm² = 7 142 857.14 short tn/cm²
• 1 gr/nm² = 71 428 571 369 short tn/m²
• 1 gr/nm² = 46 082 857.1 short tn/in²
• 1 gr/nm² = 6 635 931 441 short tn/ft²
• 1 gr/nm² = 59 723 382 838 short tn/yd²
• 1 gr/nm² = 6.37755102034×10-14 long tn/pm²
• 1 gr/nm² = 6.37755102034×10-10 long tn/Ų
• 1 gr/nm² = 6.37755102×10-8 long tn/nm²
• 1 gr/nm² = 0.0637755102 long tn/µ²
• 1 gr/nm² = 0.0637755102 long tn/µm²
• 1 gr/nm² = 63 775.5102 long tn/mm²
• 1 gr/nm² = 6 377 551.02 long tn/cm²
• 1 gr/nm² = 63 775 510 203 long tn/m²
• 1 gr/nm² = 41 145 408.1 long tn/in²
• 1 gr/nm² = 5 924 938 775 long tn/ft²
• 1 gr/nm² = 53 324 449 004 long tn/yd²
• 1 gr/nm² = 1.02040816334×10-11 st/pm²
• 1 gr/nm² = 1.020408163×10-7 st/Ų
• 1 gr/nm² = 1.020408163×10-5 st/nm²
• 1 gr/nm² = 10.2040816 st/µ²
• 1 gr/nm² = 10.2040816 st/µm²
• 1 gr/nm² = 10 204 081.6 st/mm²
• 1 gr/nm² = 1 020 408 162 st/cm²
• 1 gr/nm² = 10 204 081 624 100 st/m²
• 1 gr/nm² = 0.004628493572 st/ha
• 1 gr/nm² = 6 583 265 303 st/inch²
• 1 gr/nm² = 947 990 201 200 st/ft²
• 1 gr/nm² = 8 531 911 810 800 st/yd²
• 1 gr/nm² = 2.083333333×10-9 oz t/pm²
• 1 gr/nm² = 2.083333333×10-5 oz t/Ų
• 1 gr/nm² = 0.002083333333 oz t/nm²
• 1 gr/nm² = 2 083.33333 oz t/µ²
• 1 gr/nm² = 2 083.33333 oz t/µm²
• 1 gr/nm² = 2 083 333 333 oz t/mm²
• 1 gr/nm² = 208 333 333 333 oz t/cm²
• 1 gr/nm² = 2.08333333333×10+15 oz t/m²
• 1 gr/nm² = 0.03215074667 oz t/ha
• 1 gr/nm² = 1 344 083 333 330 oz t/inch²
• 1 gr/nm² = 1.93548×10+14 oz t/ft²
• 1 gr/nm² = 1.741932×10+15 oz t/yd²
• 1 gr/nm² = 1.73611111076×10-10 troy/pm²
• 1 gr/nm² = 1.736111111×10-6 troy/Ų
• 1 gr/nm² = 0.0001736111111 troy/nm²
• 1 gr/nm² = 173.611111 troy/µ²
• 1 gr/nm² = 173.611111 troy/µm²
• 1 gr/nm² = 173 611 111 troy/mm²
• 1 gr/nm² = 17 361 111 111 troy/cm²
• 1 gr/nm² = 1.73611111111×10+14 troy/m²
• 1 gr/nm² = 0.002679228889 troy/ha
• 1 gr/nm² = 112 006 944 444 troy/inch²
• 1 gr/nm² = 16 129 000 000 000 troy/ft²
• 1 gr/nm² = 1.45161×10+14 troy/yd²
• 1 gr/nm² = 4.166666666×10-8 dwt/pm²
• 1 gr/nm² = 0.0004166666666 dwt/Ų
• 1 gr/nm² = 0.04166666667 dwt/nm²
• 1 gr/nm² = 41 666.6667 dwt/µ²
• 1 gr/nm² = 41 666.6667 dwt/µm²
• 1 gr/nm² = 41 666 666 667 dwt/mm²
• 1 gr/nm² = 4 166 666 666 670 dwt/cm²
• 1 gr/nm² = 4.16666666667×10+16 dwt/m²
• 1 gr/nm² = 0.6430149333 dwt/ha
• 1 gr/nm² = 26 881 666 666 700 dwt/inch²
• 1 gr/nm² = 3.87096×10+15 dwt/ft²
• 1 gr/nm² = 3.483864×10+16 dwt/yd²
#### Foods, Nutrients and Calories
PECAN PIE SNACK MIX, UPC: 708820049406 weigh(s) 169 grams per metric cup or 5.6 ounces per US cup, and contain(s) 475 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
5050 foods that contain Lysine. List of these foods starting with the highest contents of Lysine and the lowest contents of Lysine
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, West Caribbean Reef weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Tar weighs 1 153 kg/m³ (71.97944 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-407C, liquid (R407C) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
gram per foot (g/ft) is a non-metric measurement unit of linear or linear mass density.
The surface density of a two-dimensional object, also known as area or areal density, is defined as the mass of the object per unit of area.
µg/fl.oz to t/cm³ conversion table, µg/fl.oz to t/cm³ unit converter or convert between all units of density measurement.
#### Calculators
Volume of a rectangular box calculator. Definition and formulas. | 3,933 | 7,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-21 | latest | en | 0.236146 |
http://www.peachpit.com/articles/article.aspx?p=784304&seqNum=5 | 1,563,523,470,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00239.warc.gz | 253,052,570 | 10,779 | Publishers of technology books, eBooks, and videos for creative people
# Unlock the True Power of Illustrator By Mastering Vectors
• Print
This chapter is from the book
## Exploring the Pathfinder Panel
In Chapter 2, Vectors 101, you learned about using the basic drawing tools such as the Rectangle tool and the Ellipse tool. Those tools are great on their own, but you'll often need to create shapes that are a bit more complex. Although you can use a variety of the tools we've mentioned so far in this chapter to create and edit paths of any shape, many times it's far easier to combine simple shapes to create more complex ones. It can also be easier to edit existing shapes using other shapes rather than trying to adjust the anchor points of individual paths.
Illustrator's Pathfinder panel, which you can open by choosing Window > Pathfinder, contains a wellspring of functions that you can perform with at least two selected paths.
### Combining Shapes with Shape Modes
The top row of the Pathfinder panel contains four functions, called shape modes, which are used to combine multiple selected shapes in different ways. Once a shape mode is applied, the resulting shape is referred to as a compound shape.
When you create a compound shape from multiple selected objects, the resulting shape appears as a single object and takes on the attributes of the topmost object (Figure 4.62). Using the Direct Selection tool, you can select the individual objects in the compound shape and edit them. See the sidebar "Illustrator Shape Modes and Photoshop Shape Layers" for additional functionality that you can take advantage of when using compound shapes.
The following are the four shape modes you can choose from in the Pathfinder panel:
• Add. The Add shape mode combines all the selected shapes and gives the appearance as if they were all joined together. This function replaces the Unite pathfinder, which you can find in older versions of Illustrator.
• Subtract. The Subtract shape mode combines all the selected shapes and takes the top objects and removes them from the bottommost object. This function replaces the Minus Front pathfinder, which was found in older versions of Illustrator.
• Intersect. The Intersect shape mode combines all the selected shapes and displays only the areas in which all the objects overlap with each other.
• Exclude. The Exclude shape mode combines all the selected shapes and removes the areas in which the objects overlap with each other.
It is certainly useful to be able to select the individual objects of a compound shape, but many times you just want to create a new shape that combines all the selected shapes. To do so, you can expand a compound shape by clicking the Expand button in the Pathfinder panel. If, when you're creating a compound shape, you know that you want to expand it, you can hold the Option (Alt) key while clicking the Add, Subtract, Intersect, or Exclude button. This applies the function and expands the shape in one step.
Additionally, you can release a compound shape by choosing Release Compound Shape from the Pathfinder panel menu. Releasing compound shapes returns the objects to their individual states and appearances.
### Changing Paths with Pathfinders
The functions in the second row of the Pathfinder panel are called pathfinders, and unlike with compound shapes, when you use pathfinders, they do not retain their original objects. Once you apply a pathfinder function, the paths are changed permanently (Figure 4.64). The following are the six pathfinder functions in the Pathfinder panel:
• Divide. One of the most often-used pathfinders, Divide takes all selected objects and breaks them apart into individual shapes based on their overlapping parts. Open paths act like knives and slice paths that intersect with them.
• Trim. The Trim pathfinder removes all overlapping areas from the selected paths.
• Merge. The Merge pathfinder removes all overlapping areas from the selected paths and joins all areas of the same color.
• Crop. The Crop pathfinder takes the topmost selected object and removes all objects and areas beneath it that fall outside its path. Unfortunately, this pathfinder works on vector objects only, and you can't use it to crop a raster image (you'll need Photoshop for that). This function ignores strokes on objects, so it's best to perform an Outline Paths function before applying the Crop pathfinder.
• Outline. The Outline pathfinder converts the selected shapes to outlines and divides the lines where they intersect.
• Minus Back. The Minus Back pathfinder is similar to the Subtract shape mode, but instead of using the top object to define the subtracted area, the function uses the bottom object.
Once you've applied a pathfinder function, you can choose Repeat Pathfinder from the Pathfinder panel menu to apply the same effect again. In reality, it takes longer to access the panel menu than it does to just click the icon in the panel, but by having this function available, you can assign a keyboard shortcut to it in the Keyboard Shortcuts dialog if you find you use these functions often.
From the Pathfinder panel menu, you can also choose Pathfinder Options, where you can set the level of precision to use when applying pathfinder functions (lower numbers may result in more complex paths). You can also specify that Illustrator should remove redundant points (always a good idea) and unpainted artwork when performing Divide or Outline functions. | 1,080 | 5,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-30 | latest | en | 0.938689 |
https://fbsymbols.net/half-symbol/ | 1,708,529,708,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00286.warc.gz | 252,923,894 | 22,514 | # Half Symbol [Meaning, Copy and Paste]
## Half Symbol For You To Copy and Paste is ½
The ½ Symbol, also known as the fraction one-half, is a universally recognized mathematical symbol representing dividing something into two equal parts. This seemingly simple Symbol carries a rich history and has found its way into various fields beyond mathematics, such as typography, design, and everyday life. In this blog post, we will explore the origins, significance, and usage of the ½ Symbol and provide you with the essential alt codes for easy access.
Please scroll down if you want to copy the Emoji/Symbol
## Half Symbol 2024:
#### Copy & Paste
½
`½`
1/2
`1/2`
1️⃣↗️2️⃣
`1️⃣↗️2️⃣`
½, 1/2, 1️⃣↗️2️⃣
## How to Use These Symbols?
Copy and paste the Half Symbol in just one click. Just click on the Half symbol copy button next to it and insert it anywhere.
## A Brief History Of The ½ Symbol
The concept of dividing a whole into equal parts dates back to ancient civilizations, but the ½ Symbol has roots in the medieval era. The Latin word ‘dimidius,’ meaning ‘half,’ was abbreviated as ‘di’ and eventually transformed into the familiar ½ Symbol we use today. Its unique design has withstood the test of time and remains an essential part of mathematical notation.
## The Significance Of The ½ Symbol
1. Mathematics: The ½ Symbol is a cornerstone of fractions, a fundamental mathematical concept. It represents dividing a whole into two equal parts and is crucial for understanding ratios, proportions, and percentages.
2. Typography and Design: In the world of typography, the ½ Symbol is a unique character that adds visual interest and clarity to the text. It is used in various contexts, such as recipes, measurements, and pricing, to convey the idea of ‘half efficiently.’
3. Everyday Life: The ½ Symbol is a versatile and easily recognizable shorthand for ‘half’ appearing in everyday situations. From clothing sizes to street addresses, the ½ Symbol has become essential to our daily lives.
## Alt Codes For The ½ Symbol
To access the ½ Symbol on your keyboard, you can use alt codes – a series of numerical combinations that enable you to type special characters. Here are the alt codes for the ½ Symbol on different operating systems:
1. Windows: To type the ½ Symbol on a Windows computer, press and hold the ‘Alt’ key while typing the code ‘0189’ on your numeric keypad. Ensure that Num Lock is enabled before attempting this.
2. Mac: On a Mac, press and hold the ‘Option’ key, then type the code ’00BD’ to create the ½ Symbol.
3. HTML: For web designers and developers, you can use the HTML entity code ‘½’ or ‘½’ to display the ½ Symbol on a webpage.
## Conclusion
The ½ Symbol has a rich history and plays a significant role. Whether using it in mathematics, typography, or everyday situations, this versatile Symbol is functional and visually appealing. By learning the alt codes for the ½ Symbol, you can effortlessly incorporate it into your writing and designs, enhancing your work’s clarity and aesthetics. | 710 | 3,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.907438 |
https://askfilo.com/user-question-answers-mathematics/the-mast-of-the-actual-ship-is-high-if-the-length-of-the-33303933303637 | 1,723,019,666,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00375.warc.gz | 83,732,703 | 33,944 | Question
# the mast of the actual ship is high. If the length of the ship is , how long is the model ship? 7. Suppose 2 kg of sugar contains crystals. How many sugar crystals are there in (i) of sugar? (ii) of sugar? 8. Rashmi has a road map with a scale of representing . She drives on a road for . What would be her distance covered in the map? 9. A high vertical pole casts a shadow long. Find at the same time (i) the length of the shadow cast by another pole high (ii) the height of a pole which casts a shadow long. 10. A loaded truck travels in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
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Question Text the mast of the actual ship is high. If the length of the ship is , how long is the model ship? 7. Suppose 2 kg of sugar contains crystals. How many sugar crystals are there in (i) of sugar? (ii) of sugar? 8. Rashmi has a road map with a scale of representing . She drives on a road for . What would be her distance covered in the map? 9. A high vertical pole casts a shadow long. Find at the same time (i) the length of the shadow cast by another pole high (ii) the height of a pole which casts a shadow long. 10. A loaded truck travels in 25 minutes. If the speed remains the same, how far can it travel in 5 hours? Updated On Nov 9, 2022 Topic All topics Subject Mathematics Class Class 11 Answer Type Video solution: 1 Upvotes 119 Avg. Video Duration 22 min | 522 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.905313 |
https://www.reference.com/math/24kgb-mean-2d34619d737ed27f | 1,472,006,186,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982290765.41/warc/CC-MAIN-20160823195810-00278-ip-10-153-172-175.ec2.internal.warc.gz | 896,548,482 | 20,400 | Q:
What does "24KGB" mean?
A:
"24KGB" is short for 24-karat gold bonding. This is a technique in which base layers of 24-karat gold are covered with layers of 14- or 18-karat gold to create a more affordable replica.
Keep Learning
Gold bonding generally starts with a base material such as bronze. The base material is then covered in layers of 24-karat gold, and these are then covered in layers of 14- or 18-karat gold. The resulting material is very similar in appearance, weight and flexibility to 14- or 18-karat gold but does not cost as much. Gold bonding may also be referred to as "gold filled" or "gold overlay."
Sources:
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https://math.stackexchange.com/questions/3761261/why-cant-wolfram-alpha-calculate-int-02-pi-sqrta-cos-theta2-sin2-th | 1,660,273,433,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00024.warc.gz | 361,321,093 | 66,289 | # Why can't Wolfram Alpha calculate $\int_0^{2\pi}\sqrt{(a-\cos\theta)^2+\sin^2\theta}\ d\theta$?
In this answer to How is the average distance between 2 objects orbiting around a third object calculated? I had to integrate
$$\int_0^{2 \pi}\sqrt{(a-\cos \theta)^2 + \sin^2 \theta} \ d\theta.$$
I tried to find the integral analytically with Wolfram Alpha but it returned an error message:
Standard computation time exceeded...
which surprised me; I'd figured that this was known and easily looked-up by the site.
Does this mean that there is no known analytical form for this definite integral? Or for some reason is it particularly challenging?
• The result is nasty! Assuming $a$ is real, we have $$\fbox{2 \left(\sqrt{(a-1)^2} E\left(-\frac{4 a}{(a-1)^2}\right)+\sqrt{(a+1)^2} E\left(\frac{4 a}{(a+1)^2}\right)\right)\text{ if }\Re((a-2) a)>-1\land \Re(a (a+2))>-1}$$
– Moo
Jul 18, 2020 at 13:57
• If you need approximation formulae, let me know. Jul 18, 2020 at 13:58
• @Moo "The result is nasty" is essentially the answer, thanks! ;-)
– uhoh
Jul 18, 2020 at 14:12
• @uhoh. The result is not nasty at all. Jul 18, 2020 at 14:34
• wolframalpha.com/input/… gives the result immediately for the antiderivative. Jul 18, 2020 at 15:24
It is not an error message but just "Standard computation time exceeded".
What you should have obtained is $$I=\int_0^{2 \pi}\sqrt{(a-\cos (\theta))^2 + \sin^2 (\theta)} \ d\theta=$$ $$I=2 \left(\sqrt{(a-1)^2} E\left(-\frac{4 a}{(a-1)^2}\right)+\sqrt{(a+1)^2} E\left(\frac{4 a}{(a+1)^2}\right)\right)$$ provided, if $$a$$ is a real, that $$\Re(a (a+2))>-1\land \Re((a-2) a)>-1$$ where appear ellptic integrals of the second kind. In fact, this reduces to $$I=4(a+1)E\left(\frac{4 a}{(a+1)^2}\right)$$
• Wolfram Alpha can certainly handle lengthy expressions, so is the reason that getting to this result is a challenge, requiring a large number of steps, substitutions, or dead-ends from which it must back-track and then try something else? I'm really after the "Why...?" here, thanks!
– uhoh
Jul 18, 2020 at 14:15
• @uhoh. Take the Pro version. It is just because you have limited ressources with the standard version. It is a quite long calculation. Jul 18, 2020 at 14:19
• okay then "It is a quite long calculation" would help round out an answer to my question, ideally supported in some way. That it is long is probably well known to some, but not to others.
– uhoh
Jul 18, 2020 at 14:29
• – uhoh
Jul 19, 2020 at 7:43 | 812 | 2,467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-33 | latest | en | 0.8321 |
https://aakashdigitalsrv1.meritnation.com/cbse-class-9/maths/math-ncert-solutions/circles/ncert-solutions/11_1_1308_107_171_2427 | 1,638,133,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00183.warc.gz | 155,156,920 | 21,610 | Mathematics NCERT Grade 9, Chapter 10: Circles- We all see many objects in our daily life which are round in shape.
In this chapter, students will study circles, its related terms and some properties of a circle
• A circle is a collection of all points in a plane, which is equidistant from a fixed point in the plane
• In this chapter we will study some terms related to circles like centre of circle, radius, diameter, arcs and its types- major arc and minor arc, semicircle, segments and its two types major segment and minor segment.
• The next section is about Angle Subtended by a Chord at a point. 2 theorems are discussed in this section, equal chords of a circle subtend equal angles at the centre and its converse
This is followed by the theorems that perpendicular from the centre of a circle to a chord bisects the chord and its converse.
Next we learn about circumcircle. According to which through three given non-collinear points there is one circle that can be drawn.The centre and radius are respectively circumcentre and circumradius respectively.
The next section covers the topics equal chords of a circle are equidistant from the centre and its converse.
This topic is followed by Angle Subtended by an Arc of a Circle where three theorems are discussed.
• The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
• Angles in the same segment of a circle are equal.
• If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
Cyclic Quadrilaterals is another important topic from this chapter.
• A quadrilateral is cyclic if all the vertices of the quadrilateral lie on a circle.
• The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
• If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.
The chapter contains 5 unsolved exercises, and also an optional exercise is given.
The summary of the chapter-Circles is provided in the end.
#### Question 1:
Fill in the blanks
(i) The centre of a circle lies in __________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a __________ of the circle.
(iv) An arc is a __________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and __________ of the circle.
(vi) A circle divides the plane, on which it lies, in __________ parts.
(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semi-circle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.
##### Video Solution for circles (Page: 171 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 171 , Question 1
#### Question 2:
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
(i) True. All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle.
(ii) False. There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.
(iii) False. Consider three arcs of same length as AB, BC, and CA. It can be observed that for minor arc BDC, CAB is a major arc. Therefore, AB, BC, and CA are minor arcs of the circle.
(iv) True. Let AB be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle. Therefore, it will be the diameter of the circle.
(v) False. Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example, in the given figure, OAB is the sector of the circle.
(vi) True. A circle is a two-dimensional figure and it can also be referred to as a plane figure.
##### Video Solution for circles (Page: 171 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 171 , Question 2
#### Question 1:
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius.
Consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths.
In ΔAOB and ΔCO'D,
AB = CD (Chords of same length)
OA = O'C (Radii of congruent circles)
OB = O'D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO'D (SSS congruence rule)
⇒ ∠AOB = ∠CO'D (By CPCT)
Hence, equal chords of congruent circles subtend equal angles at their centres.
##### Video Solution for circles (Page: 173 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 173 , Question 1
#### Question 2:
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Let us consider two congruent circles (circles of same radius) with centres as O and O'.
In ΔAOB and ΔCO'D,
∠AOB = ∠CO'D (Given)
OA = O'C (Radii of congruent circles)
OB = O'D (Radii of congruent circles)
∴ ΔAOB ≅ ΔCO'D (SAS congruence rule)
⇒ AB = CD (By CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
##### Video Solution for circles (Page: 173 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 173 , Question 2
#### Question 1:
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Consider the following pair of circles.
The above circles do not intersect each other at any point. Therefore, they do not have any point in common.
The above circles touch each other only at one point Y. Therefore, there is 1 point in common.
The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.
These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.
##### Video Solution for circles (Page: 176 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 176 , Question 1
#### Question 2:
Suppose you are given a circle. Give a construction to find its centre.
The below given steps will be followed to find the centre of the given circle.
Step1. Take the given circle.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords.
Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.
##### Video Solution for circles (Page: 176 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 176 , Question 2
#### Question 3:
If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.
Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.
Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O.
Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will also pass through O’.
Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.
##### Video Solution for circles (Page: 176 , Q.No.: 3)
NCERT Solution for Class 9 maths - circles 176 , Question 3
#### Question 1:
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
∴ AC = CB
It is given that, OO' = 4 cm
Let OC be x. Therefore, O'C will be x − 4
In ΔOAC,
OA2 = AC2 + OC2
⇒ 52 = AC2 + x2
⇒ 25 − x2 = AC2 ... (1)
In ΔO'AC,
O'A2 = AC2 + O'C2
⇒ 32 = AC2 + (x − 4)2
⇒ 9 = AC2 + x2 + 16 − 8x
⇒ AC2 = − x2 − 7 + 8x ... (2)
From equations (1) and (2), we obtain
25 − x2 = − x2 − 7 + 8x
8x = 32
x = 4
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.
Length of the common chord AB = 2 O'A = (2 × 3) cm = 6 cm
##### Video Solution for circles (Page: 179 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 179 , Question 1
#### Question 2:
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ VT = UT (By CPCT) ... (1)
It is given that,
PQ = RS ... (2)
⇒ PV = RU ... (3)
On adding equations (1) and (3), we obtain
PV + VT = RU + UT
⇒ PT = RT ... (4)
On subtracting equation (4) from equation (2), we obtain
PQ − PT = RS − RT
⇒ QT = ST ... (5)
Equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.
##### Video Solution for circles (Page: 179 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 179 , Question 2
#### Question 3:
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ ∠OTV = ∠OTU (By CPCT)
Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.
##### Video Solution for circles (Page: 179 , Q.No.: 3)
NCERT Solution for Class 9 maths - circles 179 , Question 3
#### Question 4:
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure 10.25).
Let us draw a perpendicular OM on line AD.
It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
∴ BM = MC ... (1)
And, AM = MD ... (2)
On subtracting equation (2) from (1), we obtain
AM − BM = MD − MC
⇒ AB = CD
##### Video Solution for circles (Page: 179 , Q.No.: 4)
NCERT Solution for Class 9 maths - circles 179 , Question 4
#### Question 5:
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Draw perpendiculars OA and OB on RS and SM respectively.
OR = OS = OM = 5 m. (Radii of the circle)
In ΔOAR,
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 − 9) m2 = 16 m2
OA = 4 m
ORSM will be a kite (OR = OM and RS = SM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
∴∠RCS will be of 90° and RC = CM
Area of ΔORS =
Therefore, the distance between Reshma and Mandip is 9.6 m.
##### Video Solution for circles (Page: 179 , Q.No.: 5)
NCERT Solution for Class 9 maths - circles 179 , Question 5
#### Question 6:
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
It is given that AS = SD = DA
Therefore, ΔASD is an equilateral triangle.
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write
∴ AB = OA + OB = (20 + 10) m = 30 m
In ΔABD,
Therefore, the length of the string of each phone will be m.
##### Video Solution for circles (Page: 179 , Q.No.: 6)
NCERT Solution for Class 9 maths - circles 179 , Question 6
#### Question 1:
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
##### Video Solution for circles (Page: 184 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 184 , Question 1
#### Question 2:
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60°
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.
##### Video Solution for circles (Page: 185 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 185 , Question 2
#### Question 3:
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°
##### Video Solution for circles (Page: 185 , Q.No.: 3)
NCERT Solution for Class 9 maths - circles 185 , Question 3
#### Question 4:
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)
##### Video Solution for circles (Page: 185 , Q.No.: 4)
NCERT Solution for Class 9 maths - circles 185 , Question 4
#### Question 5:
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
In ΔCDE,
∠CDE + ∠DCE = ∠CEB (Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE (Angles in the same segment of a circle)
⇒ ∠BAC = 110°
##### Video Solution for circles (Page: 185 , Q.No.: 5)
NCERT Solution for Class 9 maths - circles 185 , Question 5
#### Question 6:
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
For chord CD,
∠CBD = ∠CAD (Angles in the same segment)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
⇒ ∠BCA = 30°
We have, ∠BCD = 80°
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°
##### Video Solution for circles (Page: 185 , Q.No.: 6)
NCERT Solution for Class 9 maths - circles 185 , Question 6
#### Question 7:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as a chord)
∠BCD = 180° − 90° = 90°
(Considering AC as a chord)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.
##### Video Solution for circles (Page: 185 , Q.No.: 7)
NCERT Solution for Class 9 maths - circles 185 , Question 7
#### Question 8:
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
∠AMD = ∠BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
##### Video Solution for circles (Page: 185 , Q.No.: 8)
NCERT Solution for Class 9 maths - circles 185 , Question 8
#### Question 9:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Join chords AP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) ... (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) ... (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD
##### Video Solution for circles (Page: 186 , Q.No.: 9)
NCERT Solution for Class 9 maths - circles 186 , Question 9
#### Question 10:
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
∠ADB = 90° (Angle subtended by semi-circle)
∠ADC = 90° (Angle subtended by semi-circle)
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC.
##### Video Solution for circles (Page: 186 , Q.No.: 10)
NCERT Solution for Class 9 maths - circles 186 , Question 10
#### Question 11:
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
∠CDA + ∠ACD + ∠DAC = 180° (Angle sum property of a triangle)
⇒ 90° + ∠ACD + ∠DAC = 180°
⇒ ∠ACD + ∠DAC = 90° ... (2)
Adding equations (1) and (2), we obtain
∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°
⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°
∠BCD + ∠DAB = 180° ... (3)
However, it is given that
∠B + ∠D = 90° + 90° = 180° ... (4)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠CAD = ∠CBD (Angles in the same segment)
##### Video Solution for circles (Page: 186 , Q.No.: 11)
NCERT Solution for Class 9 maths - circles 186 , Question 11
#### Question 12:
Prove that a cyclic parallelogram is a rectangle.
Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ... (1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.
##### Video Solution for circles (Page: 186 , Q.No.: 12)
NCERT Solution for Class 9 maths - circles 186 , Question 12
#### Question 1:
Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Let two circles having their centres as O and intersect each other at point A and B respectively. Let us join O.
In ΔAO and BO,
OA = OB (Radius of circle 1)
A = B (Radius of circle 2)
O = O (Common)
ΔAO ≅ ΔBO (By SSS congruence rule)
∠OA = ∠OB (By CPCT)
Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
##### Video Solution for circles (Page: 186 , Q.No.: 1)
NCERT Solution for Class 9 maths - circles 186 , Question 1
#### Question 2:
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
(Perpendicular from the centre bisects the chord)
Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
In ΔNOD,
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
From equation (2),
Therefore, the radius of the circle iscm.
##### Video Solution for circles (Page: 186 , Q.No.: 2)
NCERT Solution for Class 9 maths - circles 186 , Question 2
#### Question 3:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB =
In ΔOMB,
In ΔOND,
Therefore, the distance of the bigger chord from the centre is 3 cm.
##### Video Solution for circles (Page: 186 , Q.No.: 3)
NCERT Solution for Class 9 maths - circles 186 , Question 3
#### Question 4:
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
In ΔAOD and ΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) ... (1)
∠ODA = ∠OEC (By CPCT) ... (2)
Also,
∠OAD = ∠ODA (As OA = OD) ... (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180º − 2xa ... (4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2y = 2a − 2 (180º − 2xa)
= 4a + 4x − 360° ... (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)
Similarly, ∠ACB = 180º − (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − ax) − (180º − ax)
= 2a + 2x − 180º
= [4a + 4x − 360°]
∠ABC = [∠DOE − ∠ AOC] [Using equation (5)]
##### Video Solution for circles (Page: 186 , Q.No.: 4)
NCERT Solution for Class 9 maths - circles 186 , Question 4
#### Question 5:
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Clearly, point O has to lie on the circle.
##### Video Solution for circles (Page: 186 , Q.No.: 5)
NCERT Solution for Class 9 maths - circles 186 , Question 5
#### Question 6:
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA ... (1)
For a parallelogram, opposite angles are equal.
From (1) and (2),
AD = AE (Angles opposite to equal sides of a triangle)
##### Video Solution for circles (Page: 186 , Q.No.: 6)
NCERT Solution for Class 9 maths - circles 186 , Question 6
#### Question 7:
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
OA = OC (Given)
OB = OD (Given)
∠AOB = ∠COD (Vertically opposite angles)
ΔAOB ≅ ΔCOD (SAS congruence rule)
AB = CD (By CPCT)
Similarly, it can be proved that ΔAOD ≅ ΔCOB
∴ AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C
However, ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.
∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
##### Video Solution for circles (Page: 186 , Q.No.: 7)
NCERT Solution for Class 9 maths - circles 186 , Question 7
#### Question 8:
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° .
It is given that BE is the bisector of ∠B.
∴ ∠ABE =
However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)
Similarly, ∠ACF = ∠ADF = (Angle in the same segment for chord AF)
Similarly, it can be proved that
##### Video Solution for circles (Page: 186 , Q.No.: 8)
NCERT Solution for Class 9 maths - circles 186 , Question 8
#### Question 9:
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
AB is the common chord in both the congruent circles.
∴ ∠APB = ∠AQB
In ΔBPQ,
∠APB = ∠AQB
∴ BQ = BP (Angles opposite to equal sides of a triangle)
##### Video Solution for circles (Page: 187 , Q.No.: 9)
NCERT Solution for Class 9 maths - circles 187 , Question 9
#### Question 10:
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.
Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.
Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 ∠BAC = 2 ∠A ... (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHS congruence rule)
∠BOE = ∠COE (By CPCT) ... (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A ... (3)
Since AD is the bisector of angle ∠A,
⇒ 2 ∠BAD = ∠A ... (4)
From equations (3) and (4), we obtain | 8,549 | 28,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-49 | latest | en | 0.915723 |
https://sinews.siam.org/Details-Page/stochastic-modeling-for-weather-and-climate-prediction | 1,719,161,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00845.warc.gz | 453,879,949 | 38,207 | SIAM News Blog
SIAM News
Stochastic Modeling for Weather and Climate Prediction
Regardless of the time or location, people seemingly always want to know the future weather. As late as the mid-20th century, the favored approach for weather forecasting involved analogues that were based on a large historical dataset of past weather reports. One would simply examine the record to find a day that was similar to the present day in question, then issue the historical evolution of the atmosphere as the forecast for the coming week. However, this method does not work in practice because the atmosphere is chaotic — its evolution is very sensitive to small details in the initial state. This was the central message of meteorologist Edward Lorenz’s groundbreaking 1963 paper: analogue forecasting is doomed to fail since one simply cannot find a historical match to the current weather with sufficient accuracy [3].
Instead of using analogues, meteorologists now generate forecasts by combining the Navier-Stokes equations with equations that describe radiation, thermodynamics, water phase changes, and other phenomena in order to build a computer model of the atmosphere. Numerically solving these equations involves setting a discretization scale, which should be as fine as possible. However, we must also produce weather forecasts in a timely manner. Despite access to some of the world’s largest supercomputers, this stipulation puts a hard limit on how fine the discretization scale can be. For weather forecasts that are out one or two weeks, this scale is around 10 kilometers. We must include the effects of all processes that occur below the discretization scale—including clouds, convection, and turbulence—in the model, but can only do so in an approximate manner via so-called “parametrization schemes.” A key assumption is that one can successfully approximate the unresolved scales’ impact on the resolved scale flow with a deterministic function of the resolved scales.
Two problems are immediately apparent. First, the Navier-Stokes equations show strong evidence of scaling symmetries [5]. In other words, if $$\mathbf{u}(\mathbf{x}, t)$$, $$p(\mathbf{x}, t)$$ is a solution to the Navier-Stokes equation, then $$\mathbf{u}_r(\mathbf{x},t)=r\mathbf{u}(r\mathbf{x}, r^2t)$$, $$p_r(\mathbf{x},t)=r^2p(r\mathbf{x}, r^2t)$$ is also a solution for any scaling parameter $$r>0$$. This scaling symmetry is consistent with the power-law behavior that is evident in atmospheric observations [6]. However, truncating the equations of motion at the discretization scale and replacing the unresolved scales in computer models with a deterministic parametrization scheme violate these scaling symmetries. Deterministic parametrizations essentially assume the presence of a spectral gap between resolved and unresolved scales, which does not exist in reality. The parametrization process is therefore a source of error in our forecasts. The second problem is that small-scale forecast errors will not remain confined to the smallest scales. Instead, they will exponentially grow in time and cascade upscale in space, thus causing our forecasts to diverge from the atmosphere’s true evolution.
One solution to these two issues is to replace conventional, deterministic parametrizations with stochastic parametrization schemes [7]. We recognize that the grid-scale variables cannot fully constrain subgrid motions without a spectral gap. We therefore choose to describe the subgrid in terms of a probability density function (PDF) that is constrained by the resolved scale flow, then randomly draw from this evolving PDF to step our computer model forward. For example, instead of including the effects of the most likely arrangement of clouds, we include the effect of just one possible cloud field on the forecast’s evolution.
To derive an appropriate form for the stochastic parametrization, we can characterize small-scale variability using high-resolution simulations that resolve the small-scale phenomena of interest. We do this by coarse graining these simulations before comparing them to a low-resolution forecasting model. Measurements of the “true” PDF of subgrid motions that are conditioned on the large-scale state not only provide further evidence that parametrization schemes should be stochastic, but also motivate the form of the stochastic parametrizations themselves [1] (see Figure 1).
Figure 1. Coarse-graining studies motivate and constrain stochastic parametrizations. 1a. The coarse-graining approach. 1b. True subgrid temperature tendency distribution (blue) compared to the estimate from a deterministic parametrization (grey rectangle and panel titles). Figure adapted from [1].
To address the second aforementioned problem—the fact that small-scale forecast errors will not remain confined to the smallest scales—we transition from making a single forecast for an upcoming period to making a set of forecasts. The forecasts originate from different but equally likely starting conditions, which we estimate based on our measurements of the atmosphere. Each forecast also utilizes different random numbers in the stochastic parametrization scheme, thereby indicating various possible realizations of the small-scale processes. By skillfully accounting for all sources of error in our forecasts, we can ensure that they are reliable — i.e., statistically consistent with the observed evolution of the atmosphere. For instance, if we collect all of the days for which we predicted a 10 percent chance of rain, it should rain on 10 percent of those days. Stochastic parametrizations are clearly necessary; we cannot produce reliable forecasts without them.
Nowadays, however, we are not only interested in predicting the weather. Climate prediction is extremely important because it provides guidance for policymakers and enables a range of sectors to prepare for the future. But predicting the climate is a different problem than predicting the weather. In fact, Lorenz referred to weather prediction as a “prediction of the first kind” [4]. Such problems are initial value problems — the skill in the forecast comes primarily from accurate specification of the starting conditions and the system’s resulting evolution away from these conditions. Climate prediction, on the other hand, is a “prediction of the second kind.” In this context, we are interested in predicting a system’s response to an external forcing. We cannot hope to predict the specifics of the weather on any given day, but rather seek to predict the weather’s changing statistics.
Despite these differences, we produce climate predictions much like weather forecasts — though now we use a computer model of the entire Earth system, including the atmosphere, oceans, biosphere, and cryosphere, among other components. We also incorporate an estimate of how anthropogenic greenhouse gases and other emissions will evolve in the future—based on a range of policy-driven “emission pathways”—to assess possible forthcoming changes to the Earth’s climate. The added complexity of a climate model, coupled with the need to produce predictions on century-long timescales, means that we must substantially coarsen the discretization scale to the order of 100 kilometers.
While the weather forecasting community has readily adopted stochastic parametrizations because of their measurable positive impact on forecast skill, the climate modeling community generally still uses deterministic models. However, our recent work demonstrates the potential of stochastic parametrizations to transform climate modeling much like they have transformed weather prediction. We show that the presence of stochasticity in climate models can alleviate long-standing systematic biases, such as mean state biases—like the distribution of precipitation [8]—and biases in modes of variability, like the El Niño–Southern Oscillation [2] (see Figure 2). Despite concerted efforts from the community, these stubborn biases in deterministic models have long resisted improvement.
Figure 2. Power spectra of modeled (black) and observed (grey) El Niño–Southern Oscillation time series in three climate models—the Community Climate System Model (CCSM), EC-Earth, and Met Office Unified Model (MetUM)—with and without the stochastically perturbed parameterization tendencies (SPPT). The power spectrum for each model with stochastic parametrization better matches the observed data. Figure adapted from [2] and [9].
Unpicking the way in which stochasticity leads to such dramatic improvements is nontrivial, and we generally must assess the mechanism for each phenomenon of interest. For example, while researchers can understand El Niño’s basic existence as a deterministic coupling between atmosphere and ocean, its variability stems from high-frequency atmospheric wind stress forcing on the ocean surface. Simulations that include a stochastic parametrization reveal an improved distribution of atmospheric winds. In the Community Climate System Model (CCSM) and Met Office Unified Model (MetUM), the parametrization dampens an excessively active El Niño. But in the EC-Earth climate model, it enhances a too-weak El Niño (see Figure 2). If we assume that the underlying coupling strength between atmosphere and ocean differs among the various climate models, then a very simple delayed oscillator model of El Niño predicts this extraordinary result [9].
By improving the statistics of the fast “weather” in climate models, we enable the simulated Earth system to explore its attractor in a more realistic way and thus improve the model’s fidelity. As I write this article, the International Panel on Climate Change is producing its Sixth Assessment Report. This report will collate the state of the art in climate prediction and compare coordinated climate change experiments that were created with the world’s leading climate models. For the first time, one of these models includes stochastic parametrization schemes. This is an exciting development, and I trust that many climate centers will soon adopt these techniques.
References
[1] Christensen, H.M. (2020). Constraining stochastic parametrisation schemes using high-resolution simulations. Quart. J. Roy. Meteorol. Soc., 146(727), 938-962.
[2] Christensen, H.M., Berner, J., Coleman, D.R.B., & Palmer, T.N. (2017). Stochastic parameterization and the El Niño–Southern Oscillation. J. Clim., 30(1), 17-38.
[3] Lorenz, E.N. (1963). Deterministic nonperiodic flow. J. Atmos. Sci., 20(2), 130-141.
[4] Lorenz, E.N. (1975). Climatic predictability. In The physical basis of climate and climate modelling (GARP Publication Series No. 16) (pp. 132-136). Stockholm, Sweden.
[5] Lovejoy, S., & Schertzer, D. (2013). The weather and climate: Emergent laws and multifractal cascades. Cambridge, U.K.: Cambridge University Press.
[6] Nastrom, G.D., & Gage, K.S. (1985). A climatology of atmospheric wavenumber spectra observed by commercial aircraft. J. Atmos. Sci., 42(9), 950-960.
[7] Palmer, T.N. (2001). A nonlinear dynamical perspective on model error: A proposal for non-local stochastic-dynamic parameterization in weather and climate prediction. Quart. J. Roy. Meteorol. Soc., 127(572), 279-304.
[8] Strommen, K., Christensen, H.M., MacLeod, D., Juricke, S., & Palmer, T.N. (2019). Progress towards a probabilistic Earth system model: Examining the impact of stochasticity in the atmosphere and land component of EC-Earth v3.2. Geosci. Model Dev., 12(7), 3099-3118.
[9] Yang, C., Christensen, H.M., Corti, S., von Hardenberg, J., & Davini, P. (2019). The impact of stochastic physics on the El Niño Southern Oscillation in the EC-Earth coupled model. Clim. Dyn., 53, 2843-2859.
Hannah Christensen is an associate professor in the Department of Physics at the University of Oxford and the David Richards Fellow at Oxford’s Wadham College. Her research focuses on the role of fast atmospheric processes in the climate system, including convective storms, clouds, and turbulence. | 2,489 | 12,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.95071 |
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Posted: Fri Mar 13, 2009 2:39 pm Post subject: Count occuprance of | in a row Hi, I have a requirement where I need to find the count for number of occurance of | in a single row. Ex. "asdsf"|1234|2009-09-01|"asdr" The result of count should be: 3 Can Parse be used for this requirement.
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Posted: Fri Mar 13, 2009 4:24 pm Post subject: Re: Reply to: Count occuprance of | in a row
Assuming there are no numbers in the characters for which you want to count x's, you can use this DFSORT job to do what you asked for:
Code: //S1 EXEC PGM=ICEMAN //SYSOUT DD SYSOUT=* //SORTIN DD * aaaaxxaaaaaxxxx aaaaaaffdfxmm dddddxxxdddddd //SORTOUT DD SYSOUT=* //SYSIN DD * OPTION COPY * Change 'x' to '1' ALTSEQ CODE=(A7F1) * Copy 1-15 to 21-35 and change each 'x' to a '1' in 21-35. INREC FIELDS=(1,15,21:1,15,TRAN=ALTSEQ) * Get total of 1s in 21-35 (with FS, letters will be treated as 0). OUTREC FIELDS=(1,15,X, 21,1,FS,ADD,22,1,FS,ADD,23,1,FS,ADD,24,1,FS,ADD,25,1,FS,ADD, 26,1,FS,ADD,27,1,FS,ADD,28,1,FS,ADD,29,1,FS,ADD,30,1,FS,ADD, 31,1,FS,ADD,32,1,FS,ADD,33,1,FS,ADD,34,1,FS,ADD,35,1,FS, TO=FS,LENGTH=3) /*
SORTOUT will have:
Code: aaaxxaaaaaxxxx 6 aaaaaffdfxmm 1 ddddxxxdddddd 3
If there are numbers in the characters for which you want to count x's, change the ALTSEQ statement to:
Code: ALTSEQ CODE=(A7F1,F1F0,F2F0,F3F0,F4F0,F5F0,F6F0,F7F0,F8F0,F9F0)
This will ensure that numbers are counted as 0s and only x's are counted as 1s
Skolusu
Senior Member
Joined: 07 Dec 2007
Posts: 2205
Location: San Jose
Posted: Sat Mar 14, 2009 1:26 am Post subject: Reply to: Count occuprance of | in a row
shuklas,
Here is a DFSORT JCL which will give you the desired results. Using FINDREP we eliminate all character A thru Z and 0 thru 9 and ", - . By doing this we are only left with the character which you want to count. And now we convert this to VB file so that we get the accurate count of the characters. It also eliminates records without the | character
FINDREP function of DFSORT is available with z/OS DFSORT V1R5 PTF UK90013 (July, 2008)
Code: //STEP0100 EXEC PGM=SORT //SYSOUT DD SYSOUT=* //SORTIN DD * "ASDSF"|1234|2009-09-01|"ASDR" "ASDSF"|1234 "ASDSF" //SORTOUT DD DSN=&&T1,DISP=(,PASS),SPACE=(CYL,(1,1),RLSE) //SYSIN DD * SORT FIELDS=COPY INREC FINDREP=(IN=(C'A',C'B',C'C',C'D',C'E',C'F',C'G',C'H',C'I', C'J',C'K',C'L',C'M',C'N',C'O',C'P',C'Q',C'R', C'S',C'T',C'U',C'V',C'W',C'X',C'Y',C'Z',C'0', C'1',C'2',C'3',C'4',C'5',C'6',C'7',C'8',C'9', C'"',C'-'),OUT=C'') OUTFIL FTOV,VLTRIM=C' ',OMIT=(1,1,CH,NE,C'|') //* //STEP0200 EXEC PGM=SORT //SYSOUT DD SYSOUT=* //SORTIN DD DSN=&&T1,DISP=SHR //SORTOUT DD SYSOUT=* //SYSIN DD * SORT FIELDS=COPY OUTFIL VTOF,BUILD=(5,1,C' COUNT = ',1,2,BI,SUB,+4,M10,LENGTH=5) /*
The output from this job is
Code: | COUNT = 3 | COUNT = 1
If you don't have the July, 2008 PTF installed, ask your System Programmer to install it (it's free).
For complete details on the new FINDREP and the other new functions available with PTF UK90013, see:
www.ibm.com/systems/support/storage/software/sort/mvs/ugpf/
dick scherrer
Site Director
Joined: 23 Nov 2006
Posts: 19270
Location: Inside the Matrix
Posted: Sat Mar 14, 2009 1:37 am Post subject: Hello, Suggest if the file has 100 million records some other solution would be a far better use of the system. To simply count the |'s, there will be 100 million "writes" and then another 100 million reads to actually get the count. . . This to get around 5 minutes of writing code? (read, inspect/tallying, add to total - at eof display the total) . . .
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Job Vacancies | Forum Rules | Bookmarks | Subscriptions | FAQ | Polls | Contact Us | 1,755 | 5,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-17 | latest | en | 0.754222 |
http://mathmisery.com/wp/2020/02/03/playful-math-education-carnival-134/ | 1,603,529,909,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00140.warc.gz | 78,890,584 | 9,197 | # Playful Math Education Carnival: #134
It’s the 134th (wow!) installment of the Playful Math Education Carnival!
What numerology is there in the number 134?
• $$1 + 3 = 4$$
• In honor of Super Groundhog Day (02/02/2020 … a palindromic date!) we have $$134 + 431 = 565$$
• $$134 = (9-7)\times(2^{6} + 8-5) + 0$$ … a pandigital expression!
Ok, now on to the blogs! And in this one, we’ll also extend to videos!
Super Bowl! Check out this activity with Roman numerals.
Now, I know this is a little old, but I recently found this beautiful video explaining the 2016 Nobel Prize. It’s a must watch!!
Now moving to a slightly less complicated topic, we have this submission from
Tom Bennison. Do check it out and give him some feedback on the five questions he has.
Earlier in January, Jim Doherty, had this write-up about an inspired approach by one of his students on polynomial division.
Staying in integration land, we have some advice from Colin Beveridge about inverse trig integrals.
In Twitter puzzle-land, I found this cute one:
I have a solution. Do you?
Next, we have a stellar write-up by Evelyn Lamb on measurement.
$$2,3,5,7,11,12,14,16$$ | 312 | 1,160 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-45 | latest | en | 0.923025 |
http://www.catb.org/~esr/writings/taouu/html/ch04s03.html | 1,652,833,801,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00495.warc.gz | 75,564,034 | 3,204 | ## Hick's Law and Fitts's Law
Analysis of timings on real-world interfaces sometimes yields suggestive curves. That is, suggestive to a statistician, who may be able to recognize a powerful qualitative principle in the numbers. One such regularity is Hick's Law,[26] which for our purposes we can paraphrase as: The time M(n) required to make a choice from a menu of n items rises with the log to the base two of n.
The key fact here is that the rise of M(n) is sublinear. Thus, the Rule of Large Menus: one large menu is more time-efficient than several small submenus supporting the same choices, even if we ignore the time overhead of moving among submenus.
A related result is Fitts's Law, which predicts that the time T(D,S) to move the mouse pointer to a region on a screen, when is initially D units away from the region border and the region is S units deep in the direction of motion, rises with the log to the base two of D/S.
Since you can't control D — the user will start his mouse movements from unpredictable locations — it follows that the way to reduce the D/S ratio that predicts select time in a point-and-click interface is to make the target larger. But not huge; the log to the base two in the formula means that, as with Hick's Law, the efficiency gains are sublinear and fall off as the ratio rises, and that gain has to be traded off against the value of other uses for the screen space.
Still, we get from this the Rule of Target Size: The size of a button should be proportional to its expected frequency of use.
Macintosh fans like to point out that Fitts's Law implies a very large advantage for Mac-style edge-of-screen menus with no borders, because they effectively extend the depth of the target area offscreen. This prediction is borne out by experiment. Under Unix. we can capture this benefit with a (borderless) taskbar adjacent to any screen edge, but the standard Unix toolkits don't give us a way to harness it within applications (because application windows have borders and anyway don't typically appear nestled up to a sctreen edge).[27]
We can get from this the Rule of the Infinite Edge: The easiest target rectangles on the screen are those adjacent to its edges.
Hick's Law and Fitts's Law come from a place even deeper than evolved human instinctual wiring. They're related to the Shannon-Hartley Theorem in information theory and would probably just hold as true for intelligent squids, robots, or anything else with an eye-brain-hand loop that has to check whether the mouse pointer has landed in the right spot by tracking progress against a visual boundary.
For more on the application of Fitts's Law, including some marvelously detailed case studies, see Bruce Tognazzini's February 1999 Ask Tog[28].
[26] For the precise mathematical statements of both Hick's Law and Fitts's Law, see the discussion in [Raskin].
[27] In fact Fitts's Law tells us that the most easily targeted areas would be the (unbordered) four corners of the screen, which have offscreen landing zones on two sides. No GUI toolkit in history has actually used this fact. | 684 | 3,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.950926 |
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1. !****************************************************
2. ! 単回帰曲線（2次回帰）計算
3. ! : y = a + b * x + c * x^2
4. ! : 連立方程式を ガウスの消去法で解く方法
5.
6. ! date name version
7. ! 2019.03.17 mk-mode.com 1.00 新規作成
8. !
10. !****************************************************
11. !
12. module const
13. ! SP: 単精度(4), DP: 倍精度(8)
14. integer, parameter :: SP = kind(1.0)
15. integer(SP), parameter :: DP = selected_real_kind(2 * precision(1.0_SP))
16. end module const
17.
18. module comp
19. use const
20. implicit none
21. private
22. public :: calc_reg_curve
23.
24. contains
25. ! 単回帰曲線（2次回帰）計算
26. !
27. ! :param(in) real(8) x(:): 説明変数配列
28. ! :param(in) real(8) y(:): 目的変数配列
29. ! :param(out) real(8) a: 係数 a
30. ! :param(out) real(8) b: 係数 b
31. ! :param(out) real(8) b: 係数 c
32. subroutine calc_reg_curve(x, y, a, b, c)
33. implicit none
34. real(DP), intent(in) :: x(:), y(:)
35. real(DP), intent(out) :: a, b, c
36. integer(SP) :: size_x, size_y, i
37. real(DP) :: sum_x, sum_x2, sum_x3, sum_x4
38. real(DP) :: sum_y, sum_xy, sum_x2y
39. real(DP) :: mtx(3, 4)
40.
41. size_x = size(x)
42. size_y = size(y)
43. if (size_x == 0 .or. size_y == 0) then
44. print *, "[ERROR] array size == 0"
45. stop
46. end if
47. if (size_x /= size_y) then
48. print *, "[ERROR] size(X) != size(Y)"
49. stop
50. end if
51.
52. sum_x = sum(x)
53. sum_x2 = sum(x * x)
54. sum_x3 = sum(x * x * x)
55. sum_x4 = sum(x * x * x * x)
56. sum_y = sum(y)
57. sum_xy = sum(x * y)
58. sum_x2y = sum(x * x * y)
59. mtx(1, :) = (/real(size_x, DP), sum_x, sum_x2, sum_y/)
60. mtx(2, :) = (/ sum_x, sum_x2, sum_x3, sum_xy/)
61. mtx(3, :) = (/ sum_x2, sum_x3, sum_x4, sum_x2y/)
62. call solve_ge(3, mtx)
63. a = mtx(1, 4)
64. b = mtx(2, 4)
65. c = mtx(3, 4)
66. end subroutine calc_reg_curve
67.
68. ! 連立方程式を解く（ガウスの消去法）
69. !
70. ! :param(in) integer(4) n: 元数
71. ! :param(inout) real(8) a(n,n+1): 係数配列
72. subroutine solve_ge(n, a)
73. implicit none
74. integer(SP), intent(in) :: n
75. real(DP), intent(inout) :: a(n, n + 1)
76. integer(SP) :: i, j
77. real(DP) :: d
78.
79. ! 前進消去
80. do j = 1, n - 1
81. do i = j + 1, n
82. d = a(i, j) / a(j, j)
83. a(i, j+1:n+1) = a(i, j+1:n+1) - a(j, j+1:n+1) * d
84. end do
85. end do
86.
87. ! 後退代入
88. do i = n, 1, -1
89. d = a(i, n + 1)
90. do j = i + 1, n
91. d = d - a(i, j) * a(j, n + 1)
92. end do
93. a(i, n + 1) = d / a(i, i)
94. end do
95. end subroutine solve_ge
96. end module comp
97.
98. program regression_line
99. use const
100. use comp
101. implicit none
102. real(DP) :: a, b, c
103. integer(SP) :: n, i
104. character(20) :: f
105. real(DP), allocatable :: x(:), y(:)
106.
108. allocate(x(n))
109. allocate(y(n))
110. do i = 1, n
112. end do
113. write (f, '("(A, ", I0, "F8.2, A)")') n
114. print f, "説明変数 X = (", x, ")"
115. print f, "目的変数 Y = (", y, ")"
116. print '(A)', "---"
117. call calc_reg_curve(x, y, a, b, c)
118. print '(A, F12.8)', "a = ", a
119. print '(A, F12.8)', "b = ", b
120. print '(A, F12.8)', "c = ", c
121. deallocate(x)
122. deallocate(y)
123. end program regression_line
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perplexus dot info
Equal Temperatures (Posted on 2016-10-31)
Find all pairs (A,B) of nonzero integers such that:
(80/A)o Fahrenheit = (80/B)o Celsius.
*** F = 9*C/5 + 32, where F and C respectively represents Fahrenheit and Celsius scales of temperatures.
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
re: No Subject | Comment 3 of 6 |
(In reply to No Subject by xdog)
"By terms of the problem only (2A-5) can be negative"
Why is that?
Posted by Charlie on 2016-10-31 11:54:31
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## Seize This Chance To Win 2,000 Sapphires!
1264 Replies
User
13 February, 2017, 5:09 PM UTC
25
UTC +8:00
0
User
13 February, 2017, 6:45 PM UTC
i say again it is 25!!!!
there are 1102 replies and about the half saying it is 25. So if all right results will get a piece of the cake you can split
it... so everyone could get 2 or 3 saphieres ;-DD
UTC +7:00
0
User
13 February, 2017, 6:46 PM UTC
26
UTC +2:00
0
User
13 February, 2017, 6:49 PM UTC
25
UTC +6:00
0
User
13 February, 2017, 7:06 PM UTC
25 if an half ax means 1 ...
UTC +1:00
0
User
13 February, 2017, 7:07 PM UTC
25
UTC +8:00
0
User
13 February, 2017, 7:30 PM UTC
25
UTC +1:00
0
User
13 February, 2017, 8:34 PM UTC
it's 25 :)
UTC +2:00
0
User
13 February, 2017, 9:15 PM UTC
25
UTC +5:00
0
User
13 February, 2017, 9:16 PM UTC
25
UTC +5:00
0
User
13 February, 2017, 10:09 PM UTC
26 is the answer, i hope :)
UTC +2:00
0
User
13 February, 2017, 10:46 PM UTC
25
UTC +8:00
0
User
13 February, 2017, 11:01 PM UTC
50
UTC +2:00
0
User
13 February, 2017, 11:42 PM UTC
25
UTC -6:00
0
User
14 February, 2017, 12:19 AM UTC
26
UTC +8:00
0
User
14 February, 2017, 2:01 AM UTC
25
UTC +6:00
0
User
14 February, 2017, 2:19 AM UTC
Lord Oberon said:
Can you think fast under pressure?
Simply solve the puzzle for the chance to win 2,000 Sapphires for your Castle!
25
UTC +7:00
0
User
14 February, 2017, 3:13 AM UTC
UTC +7:00
0
User
14 February, 2017, 4:14 AM UTC | 617 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | latest | en | 0.799937 |
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e-math IV: Advanced Algebra and Trigonometry (Revised edition 2010)
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# How many of the integers from 1 to 16, inclusive, have 3
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How many of the integers from 1 to 16, inclusive, have 3 different factors?
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20 Sep 2008, 12:53
vivektripathi wrote:
How many of the integers from 1 to 16, inclusive, have 3 different factors?
1 thru 16 inclusive are 16 numbers.
Any number that is prime will have that number and 1 as factors.
Numbers we need= 16 - (number of primes between 1 & 16)
primes between 1 & 16= 1,2,3,5,7,11,13 = 7
So numbers we need =9
All these numbers have at least 3 different factors. For example 8 has 1, 2, 4 & 8
9 has 1, 3 & 9
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20 Sep 2008, 12:54
stallone wrote:
i think 3 , as 1 is also a factor
only squares of prime numbers have 3 factors..so based on that only 2 numbers from 1 to 16 have exactly 3 factors..4 and 9..
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20 Sep 2008, 12:57
fresinha12 wrote:
stallone wrote:
i think 3 , as 1 is also a factor
only squares of prime numbers have 3 factors..so based on that only 2 numbers from 1 to 16 have exactly 3 factors..4 and 9..
Why are you looking for exactly 3?? The Q says 3 different factors ( 3 or more ) not exactly 3 different factors.
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20 Sep 2008, 23:55
icandy wrote:
fresinha12 wrote:
stallone wrote:
i think 3 , as 1 is also a factor
only squares of prime numbers have 3 factors..so based on that only 2 numbers from 1 to 16 have exactly 3 factors..4 and 9..
Why are you looking for exactly 3?? The Q says 3 different factors ( 3 or more ) not exactly 3 different factors.
Eve i find difficulty in getting the Quant questions !!!meaning is vague!!!
How to prac on this !!!
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21 Sep 2008, 09:20
I took the question to mean 'have 3 different factors' as only 3... not more
Answer is 2 (9, 4)
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Re: PS [#permalink] 21 Sep 2008, 09:20
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# How many of the integers from 1 to 16, inclusive, have 3
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http://us.metamath.org/mpeuni/dchrisumn0.html | 1,653,465,678,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00376.warc.gz | 56,250,499 | 9,610 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dchrisumn0 Structured version Visualization version GIF version
Theorem dchrisumn0 25255
Description: The sum Σ𝑛 ∈ ℕ, 𝑋(𝑛) / 𝑛 is nonzero for all non-principal Dirichlet characters (i.e. the assumption 𝑋 ∈ 𝑊 is contradictory). This is the key result that allows us to eliminate the conditionals from dchrmusum2 25228 and dchrvmasumif 25237. Lemma 9.4.4 of [Shapiro], p. 382. (Contributed by Mario Carneiro, 12-May-2016.)
Hypotheses
Ref Expression
rpvmasum.z 𝑍 = (ℤ/nℤ‘𝑁)
rpvmasum.l 𝐿 = (ℤRHom‘𝑍)
rpvmasum.a (𝜑𝑁 ∈ ℕ)
dchrmusum.g 𝐺 = (DChr‘𝑁)
dchrmusum.d 𝐷 = (Base‘𝐺)
dchrmusum.1 1 = (0g𝐺)
dchrmusum.b (𝜑𝑋𝐷)
dchrmusum.n1 (𝜑𝑋1 )
dchrmusum.f 𝐹 = (𝑎 ∈ ℕ ↦ ((𝑋‘(𝐿𝑎)) / 𝑎))
dchrmusum.c (𝜑𝐶 ∈ (0[,)+∞))
dchrmusum.t (𝜑 → seq1( + , 𝐹) ⇝ 𝑇)
dchrmusum.2 (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘((seq1( + , 𝐹)‘(⌊‘𝑦)) − 𝑇)) ≤ (𝐶 / 𝑦))
Assertion
Ref Expression
dchrisumn0 (𝜑𝑇 ≠ 0)
Distinct variable groups: 𝑦, 1 𝑦,𝐶 𝑦,𝐹 𝑦,𝑎 𝑦,𝑁 𝑦,𝑇 𝑦,𝑍 𝑦,𝐷 𝐿,𝑎,𝑦 𝑋,𝑎,𝑦
Allowed substitution hints: 𝜑(𝑦,𝑎) 𝐶(𝑎) 𝐷(𝑎) 𝑇(𝑎) 1 (𝑎) 𝐹(𝑎) 𝐺(𝑦,𝑎) 𝑁(𝑎) 𝑍(𝑎)
Proof of Theorem dchrisumn0
Dummy variable 𝑚 is distinct from all other variables.
StepHypRef Expression
1 rpvmasum.z . . . 4 𝑍 = (ℤ/nℤ‘𝑁)
2 rpvmasum.l . . . 4 𝐿 = (ℤRHom‘𝑍)
3 rpvmasum.a . . . . 5 (𝜑𝑁 ∈ ℕ)
43adantr 480 . . . 4 ((𝜑𝑇 = 0) → 𝑁 ∈ ℕ)
5 dchrmusum.g . . . 4 𝐺 = (DChr‘𝑁)
6 dchrmusum.d . . . 4 𝐷 = (Base‘𝐺)
7 dchrmusum.1 . . . 4 1 = (0g𝐺)
8 eqid 2651 . . . 4 {𝑦 ∈ (𝐷 ∖ { 1 }) ∣ Σ𝑚 ∈ ℕ ((𝑦‘(𝐿𝑚)) / 𝑚) = 0} = {𝑦 ∈ (𝐷 ∖ { 1 }) ∣ Σ𝑚 ∈ ℕ ((𝑦‘(𝐿𝑚)) / 𝑚) = 0}
9 dchrmusum.b . . . . . 6 (𝜑𝑋𝐷)
10 dchrmusum.n1 . . . . . 6 (𝜑𝑋1 )
11 dchrmusum.f . . . . . 6 𝐹 = (𝑎 ∈ ℕ ↦ ((𝑋‘(𝐿𝑎)) / 𝑎))
12 dchrmusum.c . . . . . 6 (𝜑𝐶 ∈ (0[,)+∞))
13 dchrmusum.t . . . . . 6 (𝜑 → seq1( + , 𝐹) ⇝ 𝑇)
14 dchrmusum.2 . . . . . 6 (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘((seq1( + , 𝐹)‘(⌊‘𝑦)) − 𝑇)) ≤ (𝐶 / 𝑦))
151, 2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 8dchrvmaeq0 25238 . . . . 5 (𝜑 → (𝑋 ∈ {𝑦 ∈ (𝐷 ∖ { 1 }) ∣ Σ𝑚 ∈ ℕ ((𝑦‘(𝐿𝑚)) / 𝑚) = 0} ↔ 𝑇 = 0))
1615biimpar 501 . . . 4 ((𝜑𝑇 = 0) → 𝑋 ∈ {𝑦 ∈ (𝐷 ∖ { 1 }) ∣ Σ𝑚 ∈ ℕ ((𝑦‘(𝐿𝑚)) / 𝑚) = 0})
171, 2, 4, 5, 6, 7, 8, 16dchrisum0 25254 . . 3 ¬ (𝜑𝑇 = 0)
1817imnani 438 . 2 (𝜑 → ¬ 𝑇 = 0)
1918neqned 2830 1 (𝜑𝑇 ≠ 0)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1523 ∈ wcel 2030 ≠ wne 2823 ∀wral 2941 {crab 2945 ∖ cdif 3604 {csn 4210 class class class wbr 4685 ↦ cmpt 4762 ‘cfv 5926 (class class class)co 6690 0cc0 9974 1c1 9975 + caddc 9977 +∞cpnf 10109 ≤ cle 10113 − cmin 10304 / cdiv 10722 ℕcn 11058 [,)cico 12215 ⌊cfl 12631 seqcseq 12841 abscabs 14018 ⇝ cli 14259 Σcsu 14460 Basecbs 15904 0gc0g 16147 ℤRHomczrh 19896 ℤ/nℤczn 19899 DChrcdchr 25002 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 ax-inf2 8576 ax-cnex 10030 ax-resscn 10031 ax-1cn 10032 ax-icn 10033 ax-addcl 10034 ax-addrcl 10035 ax-mulcl 10036 ax-mulrcl 10037 ax-mulcom 10038 ax-addass 10039 ax-mulass 10040 ax-distr 10041 ax-i2m1 10042 ax-1ne0 10043 ax-1rid 10044 ax-rnegex 10045 ax-rrecex 10046 ax-cnre 10047 ax-pre-lttri 10048 ax-pre-lttrn 10049 ax-pre-ltadd 10050 ax-pre-mulgt0 10051 ax-pre-sup 10052 ax-addf 10053 ax-mulf 10054 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-fal 1529 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-nel 2927 df-ral 2946 df-rex 2947 df-reu 2948 df-rmo 2949 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-int 4508 df-iun 4554 df-iin 4555 df-disj 4653 df-br 4686 df-opab 4746 df-mpt 4763 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-se 5103 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-pred 5718 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-isom 5935 df-riota 6651 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-of 6939 df-rpss 6979 df-om 7108 df-1st 7210 df-2nd 7211 df-supp 7341 df-tpos 7397 df-wrecs 7452 df-recs 7513 df-rdg 7551 df-1o 7605 df-2o 7606 df-oadd 7609 df-omul 7610 df-er 7787 df-ec 7789 df-qs 7793 df-map 7901 df-pm 7902 df-ixp 7951 df-en 7998 df-dom 7999 df-sdom 8000 df-fin 8001 df-fsupp 8317 df-fi 8358 df-sup 8389 df-inf 8390 df-oi 8456 df-card 8803 df-acn 8806 df-cda 9028 df-pnf 10114 df-mnf 10115 df-xr 10116 df-ltxr 10117 df-le 10118 df-sub 10306 df-neg 10307 df-div 10723 df-nn 11059 df-2 11117 df-3 11118 df-4 11119 df-5 11120 df-6 11121 df-7 11122 df-8 11123 df-9 11124 df-n0 11331 df-xnn0 11402 df-z 11416 df-dec 11532 df-uz 11726 df-q 11827 df-rp 11871 df-xneg 11984 df-xadd 11985 df-xmul 11986 df-ioo 12217 df-ioc 12218 df-ico 12219 df-icc 12220 df-fz 12365 df-fzo 12505 df-fl 12633 df-mod 12709 df-seq 12842 df-exp 12901 df-fac 13101 df-bc 13130 df-hash 13158 df-word 13331 df-concat 13333 df-s1 13334 df-shft 13851 df-cj 13883 df-re 13884 df-im 13885 df-sqrt 14019 df-abs 14020 df-limsup 14246 df-clim 14263 df-rlim 14264 df-o1 14265 df-lo1 14266 df-sum 14461 df-ef 14842 df-e 14843 df-sin 14844 df-cos 14845 df-pi 14847 df-dvds 15028 df-gcd 15264 df-prm 15433 df-numer 15490 df-denom 15491 df-phi 15518 df-pc 15589 df-struct 15906 df-ndx 15907 df-slot 15908 df-base 15910 df-sets 15911 df-ress 15912 df-plusg 16001 df-mulr 16002 df-starv 16003 df-sca 16004 df-vsca 16005 df-ip 16006 df-tset 16007 df-ple 16008 df-ds 16011 df-unif 16012 df-hom 16013 df-cco 16014 df-rest 16130 df-topn 16131 df-0g 16149 df-gsum 16150 df-topgen 16151 df-pt 16152 df-prds 16155 df-xrs 16209 df-qtop 16214 df-imas 16215 df-qus 16216 df-xps 16217 df-mre 16293 df-mrc 16294 df-acs 16296 df-mgm 17289 df-sgrp 17331 df-mnd 17342 df-mhm 17382 df-submnd 17383 df-grp 17472 df-minusg 17473 df-sbg 17474 df-mulg 17588 df-subg 17638 df-nsg 17639 df-eqg 17640 df-ghm 17705 df-gim 17748 df-ga 17769 df-cntz 17796 df-oppg 17822 df-od 17994 df-gex 17995 df-pgp 17996 df-lsm 18097 df-pj1 18098 df-cmn 18241 df-abl 18242 df-cyg 18326 df-dprd 18440 df-dpj 18441 df-mgp 18536 df-ur 18548 df-ring 18595 df-cring 18596 df-oppr 18669 df-dvdsr 18687 df-unit 18688 df-invr 18718 df-dvr 18729 df-rnghom 18763 df-drng 18797 df-subrg 18826 df-lmod 18913 df-lss 18981 df-lsp 19020 df-sra 19220 df-rgmod 19221 df-lidl 19222 df-rsp 19223 df-2idl 19280 df-psmet 19786 df-xmet 19787 df-met 19788 df-bl 19789 df-mopn 19790 df-fbas 19791 df-fg 19792 df-cnfld 19795 df-zring 19867 df-zrh 19900 df-zn 19903 df-top 20747 df-topon 20764 df-topsp 20785 df-bases 20798 df-cld 20871 df-ntr 20872 df-cls 20873 df-nei 20950 df-lp 20988 df-perf 20989 df-cn 21079 df-cnp 21080 df-haus 21167 df-cmp 21238 df-tx 21413 df-hmeo 21606 df-fil 21697 df-fm 21789 df-flim 21790 df-flf 21791 df-xms 22172 df-ms 22173 df-tms 22174 df-cncf 22728 df-0p 23482 df-limc 23675 df-dv 23676 df-ply 23989 df-idp 23990 df-coe 23991 df-dgr 23992 df-quot 24091 df-log 24348 df-cxp 24349 df-em 24764 df-cht 24868 df-vma 24869 df-chp 24870 df-ppi 24871 df-mu 24872 df-dchr 25003 This theorem is referenced by: dchrmusumlem 25256 dchrvmasumlem 25257
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## Math Vocabulary Crossword
Type
Crossword
Description
the act or breaking up numbers division
the act of calculating the total of 2 or more numbers addition
the act of combining sets of numbers multiplication
solving a math problem where the answer is close to the real answer estimating
finding a number that is nearest to an event 10, 100, 1,000, etc. rounding
rules for dividing numbers to help with factoring rules of divisibility
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the answer to a multiplication problem product
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## 5th Grade Math Word Search
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## 5th grade Math Vocabulary Word Search
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only one number can make a equation true one solution
700/4 divided by x/22 what is this called? Cross Multiply
____ lie in the same plane and never intersect parallel lines
the set of all the input/ output values in a function domain
to find a __ that is neither horizontal nor vertical on the coordinate gird , you can apply the Pythagorean Theorem Distance
any nonzero number rasied to the power of zero is 1 power of zero
to raise a power to a power, multipy the exponents power of a power
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the measure of a right angle is ____ degrees ninety
an ___ number can be expressed as an infinite, non-repeating decimal irrational
any number that can be expressed as the ratiom of two integers excluding division by zero rational
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when a fatcor is multiplied by the sum/ difference of two numbers,you can multiply each two numbers by the factor and then add/ subtract the products distributive property
the longest side of a right triangle , opposite the right triangle hypotenuse
## Exponent crossword
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Crossword
Description
Apply the exponent to both numbers of a quotient Power of a quotient law
Anything to the power of zero is one Zero rule
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Subtract the exponents if they have the same base Division law
Multiply two exponents to get a single power Power of a power law
Apply the exponent to both numbers of a product Power of a product law
The order of operations acronym Bedmas
The number of times you multiply the base by its self Exponent
Number that gets multiplied Base
Base & Exponent put together Power
Make something simple or easier to understand simplify
4x4x4x4x4x4 Repeated multiplication
## Math Power! Crossword
Type
Crossword
Description
What is the number that divides the divider in division? DIVISOR
What is the answer to a Multiplication problem? PRODUCT
What is the answer to the Division problem? QUOTIENT
What is the answer to the Subtraction problem? DIFFERENECE
What is the number that times another number? FACTOR
What is the whole of a question? SUM
What is the great as, or same of to a question? EQUALS
What is Equal to another number? EQUIVALENT
What is the bring together two or more numbers together to make a new total? ADD
What is finding the total or sum, by combining two or more numbers? ADDITION
What is it when one number can be divided by another and the results is an exact whole number? Divisible
What is the amount left over after division called? REMAINDER
What is a symbol used to make numerals? DIGIT
What is exactly the same amount or value of a number? EQUAL
What is a close guess of the actually value? ESTIMATE
What is another word for mathematics? MATH
What is the number that you multiple by called? MULTIPLIER
What is the term used to take away? Minus
What is a mathematical process used in math? (Add, subtract, multiple and divide) Operation
What is taking one number away from another called? Subtraction
## Math Word Search
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## Math Vocabulary Word Search
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brackets
braces
axis
attribute
array
area
algorithm
acute triangle
## Math Terms Crossword Puzzle
Type
Crossword
Description
A two-termed polynomial is a __________. binomial
In a term, the sum of the exponents on the variables is the _______ __ __ ______ degreeofaterm
A _________ is an algebraic expression made by adding or subtracting terms. polynomial
The number that a variable is being multiplied with is called the __________. coefficient
An __________ is a mathematical statement that says two expressions are equal. equation
Numbers, variables, and the products of numbers and variables are all ______. terms
A term with no variables, that's value remains the same is a ________ _____ constantterm
A mathematical phrase with operators connecting numbers and variables is an _________. expression
A three-termed polynomial is a _________. trinomial
The degree of the term with the greatest degree is the _______ __ __ ________ degreeofapolynomial
Terms that have the same variable(s) raised to the same exponent(s) are ______ ______. liketerms
Terms with different variables, or the same variables raised to different exponents are ________ ______. unliketerms
A one-termed polynomial is a _________. monomial
A letter representing a number or value that can change or vary is a _________. variable
When the terms of a polynomial are ordered from highest to lowest degree is known as __________ ______. standardform
Numbers or variables that form a product are ________. factors | 1,767 | 7,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.762298 |
http://heasarc.gsfc.nasa.gov/xanadu/xspec/manual/XSmethod.html | 1,371,625,762,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708143620/warc/CC-MAIN-20130516124223-00033-ip-10-60-113-184.ec2.internal.warc.gz | 114,418,428 | 3,835 | Next: statistic Up: Setting Commands Previous: cosmo
### method: change the fitting method
Set the minimization method.
Syntax: method <algorithm> [<# of trials/evaluations> [<critical delta>] [method-specific options]]
where <algorithm> is the method in use and the other arguments are control values for the minimization. Their meanings are explained under the individual methods. Note that all but leven require the MINUIT library from CERN to be linked into XSPEC. If any of the MINUIT library methods are set, then the error command will use the MINUIT MINOS command to find the confidence regions.
XSPEC is required to calculate derivates of the fit statistic with respect to the model parameters for both the leven and migrad methods. leven further requires second derivatives. By default, XSPEC calculates these using an analytic expression which assumes that partial 2nd derivatives of the model with respect to its parameters may be ignored. This may be changed by setting the USE_NUMERICAL_DIFFERENTIATION flag to “true” in the user’s startup Xspec.init initialization file. XSPEC will then calculate all derivatives numerically, which can be noticeably slower.
#### leven
method leven [<# of eval> [<crit delta>] [<crit beta>]]
The default XSPEC minimization method using the modified Levenberg-Marquardt based on the CURFIT routine from Bevington. <# of eval> is the number of trial vectors before the user is prompted to say whether they want to continue fitting. <crit delta> is the convergence criterion, which is the (absolute, not fractional) difference in fit statistic between successive iterations, less than which the fit is determined to have converged.
<crit beta> refers to the |beta|/N value reported during a fit. This is the norm of the vector of derivatives of the statistic with respect to the parameters divided by the number of parameters. At the best fit this should be zero, and so provides another measure of how well the fit is converging. When this is set to a positive value, it will provide another fit stopping criterion in addition to that of the <crit delta> setting.
<# of eval>, <crit delta>, and <crit beta> may also be changed through the fit command.
method migrad [<# of eval> [<crit delta>]]
The MINUIT MIGRAD method. <# of eval> is the number of function evaluations to perform before giving up and <crit delta> is the convergence criterion.
XSPEC12.0 includes version94.1 of the CERN MINUIT library – dated August 1998. The manual for the library is included with the XSPEC12 documentation and can be accessed by
XSPEC12>help minuit
When minuit is used, the output from the fitting procedure is different from xspec’s normal behavior. It is written to the file mn_output.log in the current directory. For uncertainty calculations (the error command), XSPEC calls the equivalent MINUIT implementation (MNERRS).
Following the advice in section 5 of the MINUIT manual, instead of providing the full range of MINUIT methods, most of which are said to be inferior, we have chosen to give access to the robust migrad algorithm.
migrad uses only first derivatives of models, and part of its operation is to approximate the Hessian, or second derivative matrix. The Levenberg-Marquadt assumes that the model is twice [numerically] differentiable, and calculates the Hessian explicitly. Thus the latter is the method of choice for analytical models.
#### minim
method minim [<# of evaluations> [<critical delta>] ]
The MINUIT MINIMIZE method, used MIGRAD unless it gets into trouble in which case it switches to SIMPLEX. <# of evaluations> is the number of function evaluations to perform before giving up and <critical delta> is the convergence criterion.
#### monte
method monte [<# of evaluations> [<critical delta>] ]
The MINUIT SEEK method, a simple random sampling of parameter space (not recommended!). <# of evaluations> is the number of function evaluations to perform before giving up and <critical delta> is the convergence criterion.
#### simplex
method simplex [<# of evaluations> [<critical delta>] ]
The MINUIT SIMPLEX method. <# of evaluations> is the number of function evaluations to perform before giving up and <critical delta> is the convergence criterion.
Next: statistic Up: Setting Commands Previous: cosmo | 929 | 4,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-20 | longest | en | 0.844428 |
https://economics.stackexchange.com/questions/39944/indifference-curve | 1,716,051,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00159.warc.gz | 188,989,594 | 38,720 | # indifference curve
Suppose my preferences are such that I like more of both goods, but only up to a point. After I have 5 units of both goods, that’s as good as it gets, and I’m indifferent if I get more. how do u draw the indifference curves? it seems that any bundle with more than five units of both goods are the same, so any line in that area is useless
• Perhaps the "curve" is not a line, but an area then? Sep 28, 2020 at 5:32
The area from $$(0, 0)$$ to $$(5, 5)$$ would be just like any other two-good indifference curve. Then since the area $$\{(x, y): x \geq 5, y \geq 5\}$$ has the same utility, they're all on the same "indifference curve" (so that's more like an "indifference area").
Not sure what the indifference curves would look like, say, at $$(80, 2)$$ though since you didn't specify if that's better or worse than $$(5, 5)$$.
This is an example of a satiation point or bliss point. Basically you want a particular point and getting further away from this point decreases your utility. The indifference curves look like concentric circles around a specific point, in your case (5,5) is the bliss point.
• I think the bliss point is a different situation, in this question the point (5,5) is indifferent to the point (6,6). as more isn't better in this situation. But in a bliss point case, more is actually worse as u get away from your bliss point, so the point (6,6) would be worse than (5,5) Oct 2, 2020 at 5:07 | 390 | 1,445 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.966828 |
https://www.printablemultiplication.com/worksheets/0-And-1-Multiplication-Worksheets | 1,606,280,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00425.warc.gz | 840,828,771 | 76,726 | ## Printable Multiplication Facts Worksheets
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…by posts one particular through 12. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly. Now we will improve the amount of… | 604 | 2,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-50 | longest | en | 0.840999 |
https://www.qb365.in/materials/stateboard/post/11th-standard/business-maths-188?medium=english | 1,590,962,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00462.warc.gz | 859,427,784 | 41,223 | ### 11th Standard Business Maths Study material & Free Online Practice Tests - View Model Question Papers with Solutions for Class 11 Session 2019 - 2020 TN Stateboard [ Chapter , Marks , Book Back, Creative & Term Based Questions Papers - Syllabus, Study Materials, MCQ's Practice Tests etc..]
#### 11th Standard Business maths English Medium One Mark important Questions Book back and Creative - 2020 - by Banumathi - Nilgiris - View & Read
• 1)
If A is a square matrix of order 3, then |kA| is
• 2)
The number of Hawkins-Simon conditions for the viability of an input - output analysis is
• 3)
The value of $\begin{vmatrix} x & x^2 & -yz & 1 \\ y & y^2 & -zx & 1 \\ z & z^2 & -xy &1 \end{vmatrix}$ is
• 4)
If $\begin{vmatrix} x & 2 \\ 8 &5 \end{vmatrix}=0$ then the value of x is
• 5)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
#### 11th Standard Business Mathematics English Medium All Chapter Book Back and Creative One Marks Questions 2020 - by Banumathi - Nilgiris - View & Read
• 1)
If A and B are non-singular matrices then, which of the following is incorrect?
• 2)
If $\begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix}=-5$ then value of $\begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix}$ is
• 3)
If nC3 = nC2, then the value of nC4 is
• 4)
The possible out comes when a coin is tossed five times
• 5)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
#### 11th Standard Business Mathematics English Medium All Chapter Book Back and Creative Two Marks Questions 2020 - by Banumathi - Nilgiris - View & Read
• 1)
Suppose the inter-industry flow of the product of two sectors X and Yare given as under.
Production Sector Consumption Sector Domestic demand Gross output
X Y
X 15 10 10 35
Y 20 30 15 65
Find the gross output when the domestic demand changes to 12 for X and 18 for Y.
• 2)
If $A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ then show that |2A| = 4 |A|.
• 3)
Find the values of x if $\begin{vmatrix} 2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4\\6 & x \end{vmatrix}.$
• 4)
Show that $\left[ \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right]$is a singular matrix.
• 5)
Resolve into partial fractions for the following
$\frac { 3x+7 }{ { x }^{ 2 }-3x+2 }$
#### 11th Standard Business Mathematics English Medium All Chapter Book Back and Creative Three Marks Questions 2020 - by Banumathi - Nilgiris - View & Read
• 1)
Evaluate: $\begin{bmatrix} 3&-2&4\\2&0&1\\1&2&3 \end{bmatrix}$
• 2)
Show that $\begin{vmatrix}x+a &b&c \\a &x+b&c\\a&b&x+c \end{vmatrix}=x^2(x+a+b+c)$
• 3)
Using the properties of determinants, show that $\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right|$=0
• 4)
If $A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right]$then, find A -1.
• 5)
How many number lesser than 1000 can be formed using the digits 5,6,7,8 and 9 if no digit is repeated?
#### 11th Standard Business Mathematics English Medium All Chapter Book Back and Creative Five Marks Questions 2020 - by Banumathi - Nilgiris - View & Read
• 1)
Prove that $\begin{vmatrix} {1\over a}&bc&b+c\\{1\over b}&ca&c+a\\{1\over c}&ab&a+b \end{vmatrix}=0$
• 2)
The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it, we get 46. By using matrix inversion method find the numbers.
• 3)
Use the product $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} \right]$to solve the system of equations x-2y+2z=1, 2y-3z=1, 3x-2y+4z=2.
• 4)
An amount of Rs. 5000 is put into three investment at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is Rs. 358. If the combined income from the first two investment is Rs. 70 more than the income from the third, find the amount of each investment by matrix method.
• 5)
Resolve into partial fractions for the following:
$\frac { 1 }{ ({ x }^{ 2 }+4)(x+1) }$
#### 11th Business Maths - Full Portion Five Marks Question Paper - by 8682895000 - View & Read
• 1)
If A = $\begin{bmatrix}3 & -1 & 1 \\ -15 & 6 & -5\\5 & -2 & 2 \end{bmatrix}$ then, find the Inverse of A.
• 2)
If$A=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4 \end{bmatrix}$then verify that A (adj A) = |A| I and also find A-1.
• 3)
Let a, b and c denote the sides BC, CA and AB repectively of $\Delta$ ABC. If $\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0$, then find the value of sin2 A+sin2B+sin2C.
• 4)
The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it, we get 46. By using matrix inversion method find the numbers.
• 5)
Show that the matrices $A=\left[ \begin{matrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{matrix} \right]$and $B=\left[ \begin{matrix} \frac { -4 }{ 35 } & \frac { 11 }{ 35 } & \frac { -5 }{ 35 } \\ \frac { -1 }{ 35 } & \frac { -6 }{ 35 } & \frac { 25 }{ 35 } \\ \frac { 6 }{ 35 } & \frac { 1 }{ 35 } & \frac { -10 }{ 35 } \end{matrix} \right]$are inverses of each other.
#### 11th Business Maths - Full Portion Three Marks Question Paper - by 8682895000 - View & Read
• 1)
Evaluate: $\begin{bmatrix} 3&-2&4\\2&0&1\\1&2&3 \end{bmatrix}$
• 2)
Without actual expansion show that the value of the determinant $\begin{vmatrix}5 &5^2 &5^3 \\5^2 & 5^3 & 5^4\\5^4&5^5&5^6 \end{vmatrix}$is zero.
• 3)
Prove that $\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right|$ is independent of $\theta$
• 4)
Using co-factors of elements of second column evaluate $\left| \begin{matrix} 6 & -1 & 5 \\ 3 & 0 & 4 \\ -2 & 7 & -3 \end{matrix} \right|$
• 5)
The technology matrix of an economic system of two industries is $\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right]$ Test whether the system is viable as per Hawkins – Simon conditions.
#### 11th Business Maths - Full Portion Two Marks Question Paper - by 8682895000 - View & Read
• 1)
The technology matrix of an economic system of two industries is$\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}$. Test whether the system is viable as per Hawkins Simon conditions.
• 2)
If A $=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}$ and B = [-1 2 1], verify that (AB)T = BT. AT.
• 3)
Find the values of x if $\begin{vmatrix} 2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4\\6 & x \end{vmatrix}.$
• 4)
Evaluate:$\left| \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} \right|$
• 5)
Resolve into partial fractions for the following:
$\frac { 4x+1 }{ (x-2)(x+1) }$
#### 11th Business Maths - Revision Model Question Paper 2 - by Banumathi - Nilgiris - View & Read
• 1)
The inventor of input-output analysis is
• 2)
If $\begin{vmatrix} x & 2 \\ 8 &5 \end{vmatrix}=0$ then the value of x is
• 3)
If nPr = 720 (nCr), then r is equal to
• 4)
The term containing x3 in the expansion of (x - 2y)7 is
• 5)
The slope of the line 7x + 5y - 8 = 0 is
#### 11th Business Maths - Public Exam Model Question Paper 2019 - 2020 - by Banumathi - Nilgiris - View & Read
• 1)
If A is 3 x 3 matrix and |A|= 4, then |A-1| is equal to
• 2)
If $\triangle=\begin{vmatrix} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33} \end{vmatrix}$ and Aij is cofactor of aij, then value of $\triangle$ is given by
• 3)
If nC3 = nC2, then the value of nC4 is
• 4)
If nPr = 720 (nCr), then r is equal to
• 5)
The slope of the line 7x + 5y - 8 = 0 is
#### 11th Business Maths - Operations Research Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
In a network while numbering the events which one of the following statement is false?
• 2)
A solution which maximizes or minimizes the given LPP is called
• 3)
The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤20, x + 2y ≤ 20,x > 0 and y > 0 is
• 4)
In critical path analysis, the word CPM mean
• 5)
Given an L.P.P maximize Z=2x1+3x2 subject to the constrains x1+x2≤1, 5x1+5x2≥0 and x1≥0, x2≥0 using graphical method, we observe
#### 11th Business Maths - Correlation and Regression Analysis Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The correlation coefficient is
• 2)
The variable which influences the values or is used for prediction is called
• 3)
Scatter diagram of the variate values (X,Y) give the idea about
• 4)
If regression co-efficient of Y on X is 2, then the regression co-efficient of X on Y is
• 5)
Calculate the correlation coefficient from the following data
N=9, ΣX=45, ΣY=108, ΣX2=285, ΣY2=1356, ΣXY=597
#### 11th Business Maths - Descriptive Statistics and Probability Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
Which of the following is positional measure?
• 2)
When calculating the average growth of economy, the correct mean to use is?
• 3)
Harmonic mean is the reciprocal of
• 4)
Median is same as
• 5)
Harmonic mean is better than other means if the data are for
#### 11th Business Maths - Financial Mathematics Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
A person brought a 9% stock of face value Rs 100, for 100 shares at a discount of 10%, then the stock purchased is
• 2)
The Income on 7 % stock at 80 is
• 3)
A invested some money in 10% stock at 96. If B wants to invest in an equally good 12% stock, he must purchase a stock worth of
• 4)
The present value of the perpetual annuity of Rs 2000 paid monthly at 10 % compound interest is
• 5)
Example of contingent annuity is
#### 11th Business Maths - Applications of Differentiation Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The elasticity of demand for the demand function x = $\frac { 1 }{ p }$ os
• 2)
Relationship among MR, AR and ηd is
• 3)
if u = 4x2 + 4xy + y2 + 32 + 16 , then $\frac { \partial ^{ 2 }u }{ \partial y\partial x }$ is equal to
• 4)
If u = x3 + 3xy2 + y3 then $\frac { \partial ^{ 2 }u }{ \partial y\partial x }$
• 5)
if q = 1000 + 8p1 - p2 then, $\frac { \partial q }{ \partial { p }_{ 1 } }$ is
#### 11th Business Maths - Differential Calculus Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
Let $f\left( x \right) =\begin{cases} { x }^{ 2 }-4x\quad ifx\ge 2 \\ x+2\quad ifx<2 \end{cases}$, then f(5) is
• 2)
The graph of y = ex intersect the y-axis at
• 3)
Which one of the following functions has the property f (x) = $f\left( \frac { 1 }{ x } \right)$
• 4)
The graph of f(x) = ex is identical to that of
• 5)
$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =$
#### 11th Business Maths - Trigonometry Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The radian measure of 37o30' is
• 2)
The value of $\sin(-420^o)$ is
• 3)
The value of cos245o-sin245o is
• 4)
The value of $\frac{3\tan10^o-\tan^310}{1-3\tan^210}$ is
• 5)
If $\alpha$ and $\beta$ be between 0 and $\frac{\pi}{2}$ and if $\cos(\alpha+\beta)=\frac{12}{13}$ and $\sin(\alpha-\beta)=\frac{3}{5}$ then $\sin2\alpha$ is
#### 11th Business Maths - Analytical Geometry Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
If m1 and m2 are the slopes of the pair of lines given by ax2+ 2hxy + by2 = 0, then the value of m1 + m2 is
• 2)
If kx2 + 3xy - 2y2 = 0 represent a pair of lines which are perpendicular then k is equal to
• 3)
Length of the latus rectum of the parabola y2 = - 25x is
• 4)
ax2 + 4xy + 2y2 = 0 represents a pair of parallel lines then 'a' is
• 5)
The equation of the circle with centre (3,-4) and touches the x - axis
#### 11th Business Maths - Matrices And Determinants Important Questions - by Banumathi - Nilgiris - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 3)
If A is a square matrix of order 3, then |kA| is
• 4)
Which of the following matrix has no inverse
• 5)
If A = $\begin{vmatrix}cos \theta & sin \theta \\ -sin \theta&cons\theta \end{vmatrix}$ then |2A| is equal to
#### 11th Business Maths - Algebra Important Questions - by Banumathi - Nilgiris - View & Read
• 1)
If nC3 = nC2, then the value of nC4 is
• 2)
The middle term in the expansion of ${ \left( x+\frac { 1 }{ x } \right) }^{ 10 }$
• 3)
There are 10 true or false questions in an examination. Then these questions can be answered in
• 4)
The value of (5Co + 5C1) + (5C1 + 5C2) + (5C2 + 5C3) + (5C3 + 5C4) + (5C4 + 5C5 ) is
• 5)
Number of words with or without meaning that can be formed using letters of the word "EQUATION" , with no repetition of letters is
#### 11th Business Maths - Half Yearly Model Question Paper 2019 - by Banumathi - Nilgiris - View & Read
• 1)
The co-factor of -7 in the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ is
• 2)
Which of the following matrix has no inverse
• 3)
If A = $\begin{vmatrix}cos \theta & sin \theta \\ -sin \theta&cons\theta \end{vmatrix}$ then |2A| is equal to
• 4)
The possible out comes when a coin is tossed five times
• 5)
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for n $\in$ N is
#### 12th Business Maths - Term II Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
If A is a square matrix of order 3, then |kA| is
• 3)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
• 4)
The value of n, when nP2 = 20 is
• 5)
The term containing x3 in the expansion of (x - 2y)7 is
#### 11th Business Maths - Operations Research Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Solve the following LPP graphically.
Maximize $Z=6{ x }_{ 1 }+5{ x }_{ 2 }$ Subject to the constraints $3{ x }_{ 1 }+5{ x }_{ 2 }\le 15,\quad 5{ x }_{ 1 }+2{ x }_{ 2 }\le 10\quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0$
• 2)
Solve the following LPP graphically.$Maximize\quad Z=-{ x }_{ 1 }+2{ x }_{ 2 }$
Subject to the constraints $-{ x }_{ 1 }+3{ x }_{ 2 }\le 10,\quad { x }_{ 1 }+{ x }_{ 2 }\le 6,\quad { x }_{ 1 }{ -x }_{ 2 }\le 2\quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0$
• 3)
Solve the following LPP graphically. Minimize $Z={ x }_{ 1 }-5{ x }_{ 2 }+20$
Subject to the constraints ${ x }_{ 1 }-{ x }_{ 2 }\ge 0,\quad { -x }_{ 1 }+2{ x }_{ 2 }\ge 2,\quad { x }_{ 1 }\ge 3,{ x }_{ 2 }\le 4\quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0$
• 4)
Solve the following LPP graphically. $\\ \therefore \quad Maximize\quad Z=3{ x }_{ 1 }+4{ x }_{ 2 }$
subject to the constraints ${ x }_{ 1 }+{ x }_{ 2 }\le 4\quad \quad and\quad { x }_{ 1 },{ x }_{ 2 }\ge 0$
• 5)
Solve the following LPP graphically. Minimize$Z=-3{ x }_{ 1 }+4{ x }_{ 2 }$
Subject to the constraints ${ x }_{ 1 }+2{ x }_{ 2 }\le 8\quad ,{ 3x }_{ 1 }+{ 2x }_{ 2 }\le 12\quad and\quad \quad { x }_{ 1 }\ge 0,{ x }_{ 2 }\ge 2.$
#### 11th Business Maths - Correlation and Regression Analysis Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Calculate the correlation co-efficient for the following data.
X 5 10 5 11 12 4 3 2 7 1 Y 1 6 2 8 5 1 4 6 5 2
• 2)
Calculate the coefficient of correlation between X and Y series from the following data.
X Y Number of pairs of observation 15 15 Arithmetic mean 25 18 Standard deviation 3.01 3.03 Sum of squares of deviation from the arithmetic mean 136 138
Summation of product deviations of X and Y series from their respective arithmetic means is 122.
• 3)
A random sample of recent repair jobs was selected and estimated cost and actual cost were recorded.
Estimated cost 300 450 800 250 500 975 475 400 Actual cost 273 486 734 297 631 872 396 457
Calculate the value of spearman’s correlation coefficient.
• 4)
The rank of 10 students of same batch in two subjects A and B are given below. Calculate the rank correlation coefficient.
Rank of A 1 2 3 4 5 6 7 8 9 10 Rank of B 6 7 5 10 3 9 4 1 8 2
• 5)
The following table shows the sales and advertisement expenditure of a form
Coefficient of correlation r= 0.9. Estimate the likely sales for a proposed advertisement expenditure of Rs. 10 crores.
#### 11th Business Maths - Descriptive Statistics and Probability Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Compared to the previous year the overhead expenses went up by 32% in 1995, they increased by 40% in the next year and by 50% in the following year. Calculate the average rate of increase in overhead expenses over the three years.
• 2)
From the following data compute the value of Harmonic Mean.
Marks 10 20 30 40 50 No. of students 20 30 50 15 5
• 3)
Calculate the value of quartile deviation and its coefficient from the following data
Roll No. 1 2 3 4 5 6 7 Marks 20 28 40 12 30 15 50
• 4)
A factory has 3 machines A1, A2, A3 producing 1000, 2000, 3000 screws per day respectively. A1 produces 1% defectives, A2 produces 1.5% and A3 produces 2% defectives. A screw is chosen at random at the end of a day and found defective. What is the probability that it comes from machines A1?
• 5)
X speaks truth 4 out of 5 times. A die is thrown. He reports that there is a six. What is the chance that actually there was a six?
#### 11th Business Maths - Financial Mathematics Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Find the amount of an ordinary annuity of 12 monthly payments of Rs 1, 500 that earns interest at 12% per annum compounded monthly. [(1.01)12 = 1.1262 ]
• 2)
A bank pays 8% per annum interest compounded quarterly. Find the equal deposits to be made at the end of each quarter for 10 years to have Rs 30,200? [(1.02)40 = 2.2080]
• 3)
A man buys 400 of Rs 10 shares at a premium of Rs 2.50 on each share. If the rate of dividend is 12% find
(i) his investment
(ii) annual dividend received by him
(iii) rate of interest received by him on his money
• 4)
Sundar bought 4,500 of Rs 10 shares, paying 2% per annum. He sold them when the price rose to Rs 23 and invested the proceeds in Rs 25 shares paying 10% per annum at Rs 18. Find the change in his income.
• 5)
Find the amount of an annuity of Rs. 2000 payable at the end of every month for 5 years if money is worth 6% per annum compounded monthly.
#### 11th Business Maths - Applications of Differentiation Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Find the elasticity of demand in terms of x for the demand law $p={(a-bx)^{1\over 2}}.$ Also find the values of x when elasticity of demand is unity.
• 2)
Find the elasticity of supply for the supply law $x={p\over p+5}$ when p=20 and interpret your result.
• 3)
Revenue function ‘R’ and cost function ‘C’ are R=14x - x2 and C = x (x2 - 2). Find the
(i) average cost, (ii) marginal cost, (iii) average revenue and (iv) marginal revenue.
• 4)
Find the stationary value and the stationary points f(x)=x2+2x–5.
• 5)
For the production function P = $4L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }$ verify Euler’s theorem
#### 11th Standard Business Maths - Operations Research Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
The critical path of the following network is
• 2)
A solution which maximizes or minimizes the given LPP is called
• 3)
The maximum value of the objective function Z = 3x + 5y subject to the constraints x > 0 , y > 0 and 2x + 5y ≤10 is
• 4)
Which of the following is not correct?
• 5)
Network problems have advantage in terms of project
#### 11th Standard Business Maths - Differential Calculus Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
If $y=\sqrt { x } +\frac { 1 }{ \sqrt { x } }$ show that $2x\frac { dy }{ dx } +y=2\sqrt { x }$.
• 2)
Show that the function f(x) = [x] where [x] denotes the greatest integer function is discontinuous at all integral points
• 3)
Is the function defined by f(x) = x2 -sin x + 5 is continuous at x =$\pi$?
• 4)
Differentiate: $\sin ^{ -1 }{ \left( \sqrt { \cos { x } } \right) }$
• 5)
Differentiate: sin2 x + cos2 y = 1.
#### 11th Standard Business Maths - Trigonometry Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Find all other trigonometrical ratios if $\sin x=\frac{-2\sqrt6}{5}$ and x lies in III quadrant?
• 2)
Prove that $\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac12$
• 3)
Prove that$\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}=2\cot x$
• 4)
Write $\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) } ,\left| x \right| >1$ in the simplest form.
• 5)
Prove that $\frac{\sin(x+y)}{\sin(x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y}$
#### 11th Business Maths - Analytical Geometry Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
A point moves so that its distance from the point (-1, 0) is always three times its distance from the point (0, 2). Find its locus.
• 2)
Find the equation of a circle of radius 5 whose centre lies on X-axis and passes through the point (2, 3).
• 3)
Find the equation of a circle whose diameters are 2x-3y+12=0 and x+4y-5=0 and area is 154 square units.
• 4)
Find the condition that the straight lines y=m1x+C1, y=m2x+C2, and y=m3x+C3 may meet at a point?
• 5)
Find the locus of a point such that the sum of its distances from the points (0, 2) and (0, -2) is 6.
#### 11th Business Maths Algebra Three Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Resolve into partial factors:$\frac { x+4 }{ ({ x }^{ 2 }-4)(x+1) }$
• 2)
Solve : $\frac { (2x+1)! }{ (x+2)! } .\frac { (x-1)! }{ (2x-1)! } =\frac { 3 }{ 5 }$
• 3)
How may different numbers between 100 and 1000 can be formed using the digits 0, 1,2,3,4, 5, 6 assuming that in any number, the digits are not repeated.
• 4)
There are 6 gentlemen and 4 ladies to line at a round table. In how many ways can they seat themselves so that no two ladies together?
• 5)
In how many ways can n prizes be given to n boys, when a boy may receive any number of prizes?
#### 11th Business Maths - Matrices And Determinants Three Marks - by Banumathi - Nilgiris - View & Read
• 1)
Using matrix method, solve x+2y+z=7, x+3z = 11 and 2x-3y =1.
• 2)
if A=$\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right]$ is such that AT = A-1, find $\alpha$
• 3)
Write the minors and co-factors of the elements of $\begin{vmatrix}5 & 3 \\-6 & 2\end{vmatrix}$
• 4)
Verify that A(adj A) = (adj A) A = IAI·I for the matrix A = $\begin{bmatrix}2 & 3 \\-1 & 4\end{bmatrix}$
• 5)
Find the inverse of $\begin{bmatrix}-1 & 5 \\-3 & 2 \end{bmatrix}.$
#### 11th Business Maths - Correlation and Regression Analysis Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
Example for positive correlation is
• 2)
Correlation co-efficient lies between
• 3)
The correlation coefficient from the following data N=25, ΣX=125, ΣY=100, ΣX2=650, ΣY2=436, ΣXY=520
• 4)
From the following data, N=11, ΣX=117, ΣY=260, ΣX2=1313, ΣY2=6580,ΣXY=2827 the correlation coefficient is
• 5)
The variable which influences the values or is used for prediction is called
#### 11th Business Maths - Descriptive Statistics and Probability Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
Which of the following is positional measure?
• 2)
The best measure of central tendency is
• 3)
The geometric mean of two numbers 8 and 18 shall be
• 4)
The median of 10,14,11,9,8,12,6 is
• 5)
The first quartile is also known as
#### 11th Business Maths - Financial Mathematics Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
The dividend received on 200 shares of face value Rs.100 at 8% dividend value is
• 2)
A man purchases a stock of Rs 20,000 of face value 100 at a premium of 20%, then investment is
• 3)
A person brought a 9% stock of face value Rs 100, for 100 shares at a discount of 10%, then the stock purchased is
• 4)
An annuity in which payments are made at the beginning of each payment period is called
• 5)
Example of contingent annuity is
#### 11th Business Maths - Applications of Differentiation Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
Average fixed cost of the cost function C(x) = 2x3 +5x2 - 14x +21 is
• 2)
If demand and the cost function of a firm are p= 2–x and c = 2x2 +2x +7 then its profit function is
• 3)
For the cost function C =$\frac { 1 }{ 25 } { e }^{ 25 }$, the marginal cost is
• 4)
If u = x3 + 3xy2 + y3 then $\frac { \partial ^{ 2 }u }{ \partial y\partial x }$
• 5)
if q = 1000 + 8p1 - p2 then, $\frac { \partial q }{ \partial { p }_{ 1 } }$ is
#### 11th Standard Business Maths - Operations Research Two Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Draw the logic network for the following:
Activities C and D both follow A, activity E follows C, activity F follows D, activity E and F precedes B.
• 2)
Draw a network diagram for the project whose activities and their predecessor relationships are given below:
Activity: A B C D E F G H I J K Predecessor activity: - - - A B B C D F H,I F,G
• 3)
Construct a network diagram for the following situation:
A
• 4)
Construct the network for the projects consisting of various activities and their precedence relationships are as given below:
A, B, C can start simultaneously A<F, E; B<D, C; E, D<G
• 5)
Construct the network for each the projects consisting of various activities and their precedence relationships are as given below:
Activity A B C D E F G H I J K Immediate Predecessors - - - A B B C D E H,I F,G
#### 11th Standard Business Maths - Descriptive Statistics and Probability Two Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
Find the first quartile and third quartile for the given observations
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
• 2)
An aeroplane flies, along the four sides of a square at speeds of 100,200,300 and 400 kilometres per hour respectively. What is the average speed of the plane in its flight around the square.
• 3)
A die is thrown twice and the sum of the number appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
• 4)
Suppose one person is selected at random from a group of 100 persons are given in the following
Psychologist Socialist Democrate Total Men 15 25 10 50 Women 20 15 15 50 Total 35 40 25 100
What is the probability that the man selected is a Psychologist?
• 5)
A die is thrown. Find the probability of getting
(i) a prime number
(ii) a number greater than or equal to 3
#### 11th Standard Business Maths Unit 8 Financial Mathematics Two Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
The chairman of a society wishes to award a gold medal to a student getting highest marks in Business Mathematics. If this medal costs Rs 9,000 every year and the rate of compound interest is 15% what amount is to be deposited now.
• 2)
What is the amount of perpetual annuity of Rs 50 at 5% compound interest per year?
• 3)
Find the market value of 325 shares of amount Rs 100 at a premium of Rs 18.
• 4)
A man buys 500 shares of amount Rs 100 at Rs 14 below par. How much money does he pay?
• 5)
If the dividend received from 10% of Rs 25 shares is Rs 2000. Find the number of shares.
#### 11th Business Maths - Applications of Differentiation Two Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
The demand function for a commodity is $p={4\over x}$, where p is unit price. Find the instantaneous rate of change of demand with respect to price at p=4. Also interpret your result.
• 2)
The cost function of a firm is $C={1\over3}x^3-3x^2+9x$Find the level of output (x>0) when average cost is minimum
• 3)
Find the price elasticity of demand for the demand function x = 10 – p where x is the demand and p i the price. Examine whether the demand is elastic, inelastic or unit elastic at p = 6.
• 4)
Find the equilibrium price and equilibrium quantity for the following functions. Demand: x =100 – 2p and supply: x = 3p –50
• 5)
If f (x, y) = 3x2 + 4y3 + 6xy - x3y3 + 7 then show that fxy (1,1) = 18.
#### 11th Business Maths - Differential Calculus Two Marks Questions - by Banumathi - Nilgiris - View & Read
• 1)
If $f(x)={ x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } }$ then show that $f(x)+f\left( \frac { 1 }{ x } \right) =0$
• 2)
If $f(x)=\frac { x+1 }{ x-1 }$ ,then prove that f(f(x))=x
• 3)
Find the derivative of the following functions from first principles log (x+1)
• 4)
If $f\left( x \right) =\frac { 1 }{ 2x+1 } ,x\neq -\frac { 1 }{ 2 }$ then show that $f\left( f\left( x \right) \right) =\frac { 2x+1 }{ 2x+3 }$ provided that $x\neq -\frac { 3 }{ 2 }$
• 5)
Show that the function f(x) = 5x -3 is continous at x = +3
#### 11th Business Maths - Term 1 Model Question Paper - by Kalaivani - Palani - View & Read
• 1)
The inventor of input-output analysis is
• 2)
The possible out comes when a coin is tossed five times
• 3)
The slope of the line 7x + 5y - 8 = 0 is
• 4)
The degree measure of $\frac{\pi}{8}$ is
• 5)
The graph of y = 2x2 is passing through
#### 11th Business Maths - Trigonometry Two Marks Question - by Banumathi - Nilgiris - View & Read
• 1)
Convert the following degree measure into radian measure 240o
• 2)
Convert the following degree measure into radian measure -320o
• 3)
Find the degree measure corresponding to the following radian measure. -3
• 4)
Find the degree measure corresponding to the following radian measure. $\frac { 11\pi }{ 18 }$
• 5)
Evaluate $\cot\left(\frac{-15\pi}{4}\right)$
#### 11th Business Maths - Analytical Geometry Two Marks Question - by Banumathi - Nilgiris - View & Read
• 1)
Find the equation of the following circles having the center ( 0,0) and radius 2 units
• 2)
Find the centre and radius of the circle x2 + y2 = 16
• 3)
Find the center and radius of the circle (x + 2) ( x - 5) + (y -2 ) ( y -1) = 0
• 4)
Find the equation of the circle whose centre is (-3, -2) and having circumferences 16$\pi$
• 5)
Find the equation of the circle on the line joining the points (1,0), (0,1) and having its centre on the line x + y= 1
#### 11th Business Maths - Algebra Two Marks Question - by Banumathi - Nilgiris - View & Read
• 1)
Verify that 8C4+8C3=9C4
• 2)
Evaluate the following expression.$\frac { 7! }{ 6! }$
• 3)
If four dice are rolled, find the number of possible outcomes in which atleast one die shows 2.
• 4)
ResoIve into partial fractions :$\frac { 12x-17 }{ (x-2)(x-1) }$
• 5)
Show that 10P3 = 9 P3 + 3. 9P2
#### 11th Business Maths - Matrices And Determinants Two Marks Question - by Banumathi - Nilgiris - View & Read
• 1)
The technology matrix of an economic system of two industries is$\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}$ .Test whether the system is viable as per Hawkins-Simon conditions.
• 2)
Find the minors and cofactors of all the elements of the following determinants
$\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}$
• 3)
Solve: $\begin{vmatrix}2& x&3\\4&1&6\\1&2&7 \end{vmatrix}=0$
• 4)
Find |AB| if $A=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}and \begin{bmatrix} 3&0\\1&-2 \end{bmatrix}$
• 5)
If A $=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}$ and B = [-1 2 1], verify that (AB)T = BT. AT.
#### 11th Business Maths - Term 1 Five Mark Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
If A = $\begin{bmatrix}3 & -1 & 1 \\ -15 & 6 & -5\\5 & -2 & 2 \end{bmatrix}$ then, find the Inverse of A.
• 2)
Solve by matrix inversion method: x - y + 2z = 3; 2x +Z = 1; 3x + 2y + z = 4.
• 3)
Find adjoint of $A=\left[ \begin{matrix} 1 & -2 & -3 \\ 0 & 1 & 0 \\ -4 & 1 & 0 \end{matrix} \right]$
• 4)
Resolve into partial fractions for the following:
$\frac { x+2 }{ (x-1)(x+3)^{ 2 } }$
• 5)
By the principle of mathematical induction, prove the following.
an-bn is divisible by a-b, for all $n\in N$ .
#### 11th Business Maths Quarterly Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 3)
The number of Hawkins-Simon conditions for the viability of an input - output analysis is
• 4)
If A and B are non-singular matrices then, which of the following is incorrect?
• 5)
If A = $\begin{vmatrix}cos \theta & sin \theta \\ -sin \theta&cons\theta \end{vmatrix}$ then |2A| is equal to
#### 11th Business Maths Unit 10 Operations Research Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
Maximize: z=3x1+4x2 subject to 2x1+x2≤40, 2x1+5x2≤180, x1,x2≥0 in the LPP, which one of the following is feasible corner point?
• 2)
A solution which maximizes or minimizes the given LPP is called
• 3)
The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤20, x + 2y ≤ 20,x > 0 and y > 0 is
• 4)
In the context of network, which of the following is not correct
• 5)
The objective of network analysis is to
#### 11th Business Maths Unit 9 Correlation and Regression Analysis Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
Example for positive correlation is
• 2)
If the values of two variables move in opposite direction then the correlation is said to be
• 3)
Correlation co-efficient lies between
• 4)
The variable whose value is influenced or is to be predicted is called
• 5)
The variable which influences the values or is used for prediction is called
#### 11th Business Maths Chapter 8 Descriptive Statistics and Probability Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
When calculating the average growth of economy, the correct mean to use is?
• 2)
When an observation in the data is zero, then its geometric mean is
• 3)
The best measure of central tendency is
• 4)
Median is same as
• 5)
The median of 10,14,11,9,8,12,6 is
#### 11th Standard Business Maths - Financial Mathematics Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
The dividend received on 200 shares of face value Rs.100 at 8% dividend value is
• 2)
What is the amount related is selling 8% stacking 200 shares of face value 100 at 50.
• 3)
The Income on 7 % stock at 80 is
• 4)
If ‘a’ is the annual payment, ‘n’ is the number of periods and ‘i’ is compound interest for Rs 1 then future amount of the annuity is
• 5)
An annuity in which payments are made at the beginning of each payment period is called
#### 11th Standard Business Maths Chapter 6 Applications of Differentiation Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
Average fixed cost of the cost function C(x) = 2x3 +5x2 - 14x +21 is
• 2)
Marginal revenue of the demand function p= 20–3x is
• 3)
For the cost function C =$\frac { 1 }{ 25 } { e }^{ 25 }$, the marginal cost is
• 4)
Instantaneous rate of change of y = 2x2 + 5x with respect to x at x = 2 is
• 5)
If the average revenue of a certain firm is Rs 50 and its elasticity of demand is 2, then their marginal revenue is
#### 11th Standard Business Maths - Differential Calculus Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
If f(x) = x2 - x + 1, then f (x + 1) is
• 2)
The graph of the line y = 3 is
• 3)
Which one of the following functions has the property f (x) = $f\left( \frac { 1 }{ x } \right)$
• 4)
The range of f(x) = |x|, for all $x\epsilon R$, is
• 5)
A function f(x) is continuous at x = a if $\lim _{ x\rightarrow a }{ f\left( x \right) }$ is equal to
#### 11th Standard Business Maths - Analytical Geometry - by Banumathi - Nilgiris - View & Read
• 1)
If m1 and m2 are the slopes of the pair of lines given by ax2+ 2hxy + by2 = 0, then the value of m1 + m2 is
• 2)
If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0 are the diameters of a circle, then its centre is
• 3)
The locus of the point P which moves such that P is always at equidistance from the line x + 2y+ 7 = 0 is
• 4)
(1, - 2) is the centre of the circle x2 + y2 + ax + by - 4 = 0 , then its radius
• 5)
The length of the tangent from (4,5) to the circle x2 +y2 = 16 is
#### 11th Standard Business Maths - Trigonometry - by Banumathi - Nilgiris - View & Read
• 1)
The degree measure of $\frac{\pi}{8}$ is
• 2)
The value of $\sin(-420^o)$ is
• 3)
The value of sec A sin(270o+A) is
• 4)
The value of cos245o-sin245o is
• 5)
The value of $\frac{2\tan30^o}{1+tan^230}$ is
#### 11th Standard Business Maths Unit 2 Algebra Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
The value of n, when nP2 = 20 is
• 2)
The number of diagonals in a polygon of n seates is equal to
• 3)
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for n $\in$ N is
• 4)
For all n > 0, nC1 + nC2 + nC3 + ... +nCn is equal to
• 5)
The middle term in the expansion of ${ \left( x+\frac { 1 }{ x } \right) }^{ 10 }$
#### 11th Standard Business Maths Unit 1 Matrices And Determinants Book Back Questions - by Banumathi - Nilgiris - View & Read
• 1)
If A = $\begin{vmatrix}cos \theta & sin \theta \\ -sin \theta&cons\theta \end{vmatrix}$ then |2A| is equal to
• 2)
If $\triangle=\begin{vmatrix} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33} \end{vmatrix}$ and Aij is cofactor of aij, then value of $\triangle$ is given by
• 3)
If $\begin{vmatrix} x & 2 \\ 8 &5 \end{vmatrix}=0$ then the value of x is
• 4)
If $\begin{vmatrix} 4 & 3 \\ 3 & 1 \end{vmatrix}=-5$ then value of $\begin{vmatrix} 20 & 15 \\ 15 & 5 \end{vmatrix}$ is
• 5)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
#### 11th Standard Business Maths - Descriptive Statistics and Probability One Mark Question and Answer - by Banumathi - Nilgiris - View & Read
• 1)
Which of the following is positional measure?
• 2)
When calculating the average growth of economy, the correct mean to use is?
• 3)
When an observation in the data is zero, then its geometric mean is
• 4)
The correct relationship among A.M.,G.M.and H.M.is:
• 5)
Harmonic mean is the reciprocal of
#### 11th Standard Business Maths - Financial Mathematics One Mark Question with Answer Key - by Banumathi - Nilgiris - View & Read
• 1)
The dividend received on 200 shares of face value Rs.100 at 8% dividend value is
• 2)
What is the amount related is selling 8% stacking 200 shares of face value 100 at 50.
• 3)
A man purchases a stock of Rs 20,000 of face value 100 at a premium of 20%, then investment is
• 4)
A person brought a 9% stock of face value Rs 100, for 100 shares at a discount of 10%, then the stock purchased is
• 5)
The Income on 7 % stock at 80 is
#### 11th Business Maths - Applications of Differentiation One Mark Question with Answer - by Banumathi - Nilgiris - View & Read
• 1)
Average fixed cost of the cost function C(x) = 2x3 +5x2 - 14x +21 is
• 2)
Marginal revenue of the demand function p= 20–3x is
• 3)
If demand and the cost function of a firm are p= 2–x and c = 2x2 +2x +7 then its profit function is
• 4)
Relationship among MR, AR and ηd is
• 5)
For the cost function C =$\frac { 1 }{ 25 } { e }^{ 25 }$, the marginal cost is
#### 11th Standard Chapter 5 Differential Calculus One Mark Question and Answer - by Banumathi - Nilgiris - View & Read
• 1)
If f(x) = x2 - x + 1, then f (x + 1) is
• 2)
Let $f\left( x \right) =\begin{cases} { x }^{ 2 }-4x\quad ifx\ge 2 \\ x+2\quad ifx<2 \end{cases}$, then f(5) is
• 3)
For $f\left( x \right) =\begin{cases} { x }^{ 2 }-4x\quad ifx\ge 2 \\ x+2\quad ifx<2 \end{cases}$ then f(0) is
• 4)
If f(x) = $\frac{1-x}{1+x}$ then f(-x) is equal to
• 5)
The graph of the line y = 3 is
#### 11th Standard Chapter 4 Trigonometry - One Mark Questions and Answer - by Banumathi - Nilgiris - View & Read
• 1)
The degree measure of $\frac{\pi}{8}$ is
• 2)
The radian measure of 37o30' is
• 3)
If $\tan\theta=\frac{1}{\sqrt5}$ and $\theta$ lies in the first quadrant, then $\cos\theta$ is
• 4)
The value of $\sin15^o$ is
• 5)
The value of $\sin(-420^o)$ is
#### 11th Business Maths Chapter 3 Analytical Geometry One Mark Question Paper with Answer - by Banumathi - Nilgiris - View & Read
• 1)
If m1 and m2 are the slopes of the pair of lines given by ax2+ 2hxy + by2 = 0, then the value of m1 + m2 is
• 2)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
• 3)
If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0 are the diameters of a circle, then its centre is
• 4)
The x - intercept of the straight line 3x + 2y - 1 = 0 is
• 5)
The slope of the line 7x + 5y - 8 = 0 is
#### 11th Business Maths Algebra - One Mark Question with Answer Key - by Banumathi - Nilgiris - View & Read
• 1)
If nC3 = nC2, then the value of nC4 is
• 2)
The number of ways selecting 4 players out of 5 is
• 3)
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for n $\in$ N is
• 4)
If n is a positive integer, then the number of terms in the expansion (x + a)n is
• 5)
For all n > 0, nC1 + nC2 + nC3 + ... +nCn is equal to
#### 11th Mathematics Matrices And Determinants One Mark Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
The value of $\begin{vmatrix} 2x+y & x & y \\ 2y+z & y & z \\ 2z+x & z & x \end{vmatrix}$ is
• 3)
The co-factor of -7 in the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ is
• 4)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 5)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
#### 11th Business Maths Differential Calculus Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The graph of y = 2x2 is passing through
• 2)
Which one of the following functions has the property f (x) = $f\left( \frac { 1 }{ x } \right)$
• 3)
The range of f(x) = |x|, for all $x\epsilon R$, is
• 4)
A function f(x) is continuous at x = a if $\lim _{ x\rightarrow a }{ f\left( x \right) }$ is equal to
• 5)
If y = log x then y2 =
#### 11th Standard Business Maths First Mid Term Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
If A is a square matrix of order 3, then |kA| is
• 3)
If $\triangle=\begin{vmatrix} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33} \end{vmatrix}$ and Aij is cofactor of aij, then value of $\triangle$ is given by
• 4)
The number of permutation of n different things taken r at a time, when the repetition is allowed is
• 5)
The distance between directrix and focus of a parabola y2 = 4ax is
#### 11th Standard Business Maths Chapter 4 Trigonometry Sample Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The degree measure of $\frac{\pi}{8}$ is
• 2)
If $\tan\theta=\frac{1}{\sqrt5}$ and $\theta$ lies in the first quadrant, then $\cos\theta$ is
• 3)
The value of sin 15o cos 15o is
• 4)
The value of sec A sin(270o+A) is
• 5)
If sin A+ cos A=1, then sin 2A is equal to
#### 11th Standard Business Maths Chapter 3 Analytical Geometry Important Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
If m1 and m2 are the slopes of the pair of lines given by ax2+ 2hxy + by2 = 0, then the value of m1 + m2 is
• 2)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
• 3)
If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0 are the diameters of a circle, then its centre is
• 4)
The distance between directrix and focus of a parabola y2 = 4ax is
• 5)
The equation of directrix of the parabola y2 = - x is
#### 11th Standard Business Maths Unit 2 Algebra Important Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
If nC3 = nC2, then the value of nC4 is
• 2)
The value of n, when nP2 = 20 is
• 3)
The number of ways selecting 4 players out of 5 is
• 4)
If nPr = 720 (nCr), then r is equal to
• 5)
The possible out comes when a coin is tossed five times
#### 11th Business Maths - Unit 1 Model Question Paper - by Banumathi - Nilgiris - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 3)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 4)
Which of the following matrix has no inverse
• 5)
The Inverse of matrix of$\begin{pmatrix} 3 & 1 \\ 5 & 2\end{pmatrix}$ is
#### 11th Standard Business Maths Public Exam March 2019 Important Creative 3 Mark Questions and Answers - by Basky - View & Read
• 1)
Prove that $\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right|$ is independent of $\theta$
• 2)
Using matrix method, solve x+2y+z=7, x+3z = 11 and 2x-3y =1.
• 3)
if A=$\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right]$ is such that AT = A-1, find $\alpha$
• 4)
Verify that A(adj A) = (adj A) A = IAI·I for the matrix A = $\begin{bmatrix}2 & 3 \\-1 & 4\end{bmatrix}$
• 5)
Solve: 2x+ 5y = 1 and 3x + 2y = 7 using matrix method.
#### 11th Standard Business Maths Public Exam March 2019 Important One Mark Test - by Basky - View & Read
• 1)
The value of $\begin{vmatrix} 2x+y & x & y \\ 2y+z & y & z \\ 2z+x & z & x \end{vmatrix}$ is
• 2)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 3)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 4)
• 5)
The inverse matrix of $\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}$ is
#### Plus One Business Maths Public Exam March 2019 One Mark Question Paper - by Basky - View & Read
• 1)
The co-factor of -7 in the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ is
• 2)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 3)
• 4)
If A and B are non-singular matrices then, which of the following is incorrect?
• 5)
If A is a square matrix of order 3 and IAI = 3 then | adj A| is equal to
#### 11th Standard Business Maths Public Exam March 2019 Important One Mark Questions - by Basky - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 3)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 4)
If A is a square matrix of order 3, then |kA| is
• 5)
• 1)
The technology matrix of an economic system of two industries is$\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}$. Test whether the system is viable as per Hawkins Simon conditions.
• 2)
Find the minors and cofactors of all the elements of the following determinants.
$\begin{bmatrix} 1&-3&2\\4&-1&2\\3&5&2 \end{bmatrix}$
• 3)
Find |AB| if $A=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}and \begin{bmatrix} 3&0\\1&-2 \end{bmatrix}$
• 4)
If A $=\begin{bmatrix} 1 \\ -4\\3 \end{bmatrix}$ and B = [-1 2 1], verify that (AB)T = BT. AT.
• 5)
Using the property of determinant, evaluate $\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}.$
#### 11th Standard Business Maths Public Exam March 2019 Important 5 Marks Questions - by Basky - View & Read
• 1)
Evaluate:$\begin{vmatrix} 1&a&a^2-bc\\1&b&b^2-ca\\1&c&c^2-ab \end{vmatrix}$
• 2)
If A = $\begin{bmatrix}1 & 1 & 1 \\ 3 & 4 & 7\\1 & -1 & 1 \end{bmatrix}$ verify that A ( adj A ) = ( adj A ) A = |A| I3.
• 3)
Solve by using matrix inversion method: x - y + z = 2; 2x- y = 0 , 2y - z = 1.
• 4)
Without expanding show that $\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42 & 40 & 2 \end{matrix} \right| =0$
• 5)
If $A=\left[ \begin{matrix} 1 & tan\quad x \\ -tan\quad x & \quad \quad \quad 1 \end{matrix} \right]$, then show that ATA-1=$\left[ \begin{matrix} cos\quad 2x & -sin2x \\ sin\quad 2x & cos2x \end{matrix} \right] .$
#### Plus One Business Maths Public Exam Official Model Question Paper 2019 - by Basky - View & Read
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
• 3)
The value of n, when nP2 = 20 is
• 4)
13 guests have participated in a dinner. The number of handshakes happened in the dinner is
• 5)
The eccentricity of the parabola is
#### 11th Standard Business Maths Public Exam March 2019 Model Test Question Paper - by Basky - View & Read
• 1)
If A and B are non-singular matrices then, which of the following is incorrect?
• 2)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
• 3)
The term containing x3 in the expansion of (x - 2y)7 is
• 4)
If $\frac { kx }{ (x+4)(2x-1) } =\frac { 4 }{ x+4 } +\frac { 1 }{ 2x-1 }$ then k is equal to
• 5)
The focus of the parabola x2 = 16y is
#### Plus One Business Maths Matrices And Determinants Important Questions - by Basky - View & Read
• 1)
The value of $\begin{vmatrix} 2x+y & x & y \\ 2y+z & y & z \\ 2z+x & z & x \end{vmatrix}$ is
• 2)
The inverse matrix of $\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}$ is
• 3)
The value of $\begin{vmatrix} 5 & 5 & 5 \\ 4x & 4y & 4z \\ -3x & -3y & -3z \end{vmatrix}$is
• 4)
If A is 3 x 3 matrix and |A|= 4, then |A-1| is equal to
• 5)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
#### 11th Standard Business Maths 3rd Revision Test Question Paper 2019 - by Basky - View & Read
• 1)
The number of Hawkins-Simon conditions for the viability of an input - output analysis is
• 2)
The inventor of input-output analysis is
• 3)
The possible out comes when a coin is tossed five times
• 4)
Number of words with or without meaning that can be formed using letters of the word "EQUATION" , with no repetition of letters is
• 5)
The double ordinate passing through the focus is
• 1)
The number of Hawkins-Simon conditions for the viability of an input - output analysis is
• 2)
Which of the following matrix has no inverse
• 3)
If nC3 = nC2, then the value of nC4 is
• 4)
Number of words with or without meaning that can be formed using letters of the word "EQUATION" , with no repetition of letters is
• 5)
The slope of the line 7x + 5y - 8 = 0 is
#### 11th Standard Business Maths Model Revision Test Question Paper 2019 - by Basky - View & Read
• 1)
• 2)
If A is an invertible matrix of order 2, then det (A-1) be equal to
• 3)
The value of n, when nP2 = 20 is
• 4)
The last term in the expansion of (3 +$\sqrt{2}$ )8 is
• 5)
If kx2 + 3xy - 2y2 = 0 represent a pair of lines which are perpendicular then k is equal to
#### 11th Stateboard Maths Analytical Geometry Important Questions - by Basky - View & Read
• 1)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
• 2)
The slope of the line 7x + 5y - 8 = 0 is
• 3)
If kx2 + 3xy - 2y2 = 0 represent a pair of lines which are perpendicular then k is equal to
• 4)
The length of the tangent from (4,5) to the circle x2 +y2 = 16 is
• 5)
The focus of the parabola x2 = 16y is
• 1)
The inverse matrix of $\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}$ is
• 2)
Which of the following matrix has no inverse
• 3)
The term containing x3 in the expansion of (x - 2y)7 is
• 4)
The number of permutation of n different things taken r at a time, when the repetition is allowed is
• 5)
If kx2 + 3xy - 2y2 = 0 represent a pair of lines which are perpendicular then k is equal to
#### 11th Standard Business Maths Half Yearly Model Question Paper - by Basky - View & Read
• 1)
If A $=\begin{pmatrix} -1 & 2 \\ 1 & -4 \end{pmatrix}$ then A (adj A) is
• 2)
The value of $\begin{vmatrix} x & x^2 & -yz & 1 \\ y & y^2 & -zx & 1 \\ z & z^2 & -xy &1 \end{vmatrix}$ is
• 3)
The number of ways selecting 4 players out of 5 is
• 4)
The value of (5Co + 5C1) + (5C1 + 5C2) + (5C2 + 5C3) + (5C3 + 5C4) + (5C4 + 5C5 ) is
• 5)
(1, - 2) is the centre of the circle x2 + y2 + ax + by - 4 = 0 , then its radius
#### 11th Business Maths Correlation and Regression Analysis Important Questions - by Basky - View & Read
• 1)
Correlation co-efficient lies between
• 2)
The variable whose value is influenced or is to be predicted is called
• 3)
When one regression coefficient is negative, the other would be
• 4)
The lines of regression intersect at the point
• 5)
Cov(x,y)=–16.5, ${ \sigma }_{ x }^{ 2 }=2.89,{ \sigma }_{ y }^{ 2 }$=100. Find correlation coefficient
• 1)
The critical path of the following network is
• 2)
In a network while numbering the events which one of the following statement is false?
• 3)
In the given graph the coordinates of M1 are
• 4)
The maximum value of the objective function Z = 3x + 5y subject to the constraints x > 0 , y > 0 and 2x + 5y ≤10 is
• 5)
The objective of network analysis is to
• 1)
Suppose the inter-industry flow of the product of two industries are given as under.
Production sector Consumption sector Domestic demand Total output X Y X 30 40 50 120 Y 20 10 30 60
Determine the technology matrix and test Hawkin's -Simon conditions for the viability of the system. If the domestic demand changes to 80 and 40 units respectively, what should be the gross output of each sector in order to meet the new demands.
• 2)
An amount of Rs. 5000 is put into three investment at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is Rs. 358. If the combined income from the first two investment is Rs. 70 more than the income from the third, find the amount of each investment by matrix method.
• 3)
Resolve into partial fractions for the following:
$\frac { x-2 }{ (x+2)(x-1)^{ 2 } }$
• 4)
Using binomial theorem, find the value of ${ \left( \sqrt { 2 } +1 \right) }^{ 5 }+{ \left( \sqrt { 2 } -1 \right) }^{ 5 }$
• 5)
Show by the principle of mathematical induction that 23n–1 is a divisible by 7, for all n∈N.
#### 11th Business Maths Half Yearly Model Question Paper 1 - by Basky - View & Read
• 1)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 2)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
• 3)
The number of ways to arrange the letters of the word "CHEESE" is
• 4)
The number of permutation of n different things taken r at a time, when the repetition is allowed is
• 5)
The x - intercept of the straight line 3x + 2y - 1 = 0 is
#### 12th Standard Business Maths Important Question Paper - by S.B.O.A. Matric and Hr Sec School - View & Read
• 1)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 2)
The inverse matrix of $\begin{pmatrix} \frac { 1 }{ 5 } & \frac { 5 }{ 25 } \\ \frac { 2 }{ 5 } & \frac { 1 }{ 2 } \end{pmatrix}$ is
• 3)
The value of $\begin{vmatrix} 5 & 5 & 5 \\ 4x & 4y & 4z \\ -3x & -3y & -3z \end{vmatrix}$is
• 4)
If any three-rows or columns of a determinant are identical, then the value of the determinant is
• 5)
The number of ways selecting 4 players out of 5 is
• 1)
The value of x, if $\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0$ is
• 2)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 3)
• 4)
The number of Hawkins-Simon conditions for the viability of an input - output analysis is
• 5)
The Inverse of matrix of$\begin{pmatrix} 3 & 1 \\ 5 & 2\end{pmatrix}$ is
• 1)
The co-factor of -7 in the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ is
• 2)
If $\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ then $\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$ is
• 3)
The value of the determinant ${\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}$is
• 4)
If $\triangle=\begin{vmatrix} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33} \end{vmatrix}$ and Aij is cofactor of aij, then value of $\triangle$ is given by
• 5)
If nC3 = nC2, then the value of nC4 is
• 1)
If m1 and m2 are the slopes of the pair of lines given by ax2+ 2hxy + by2 = 0, then the value of m1 + m2 is
• 2)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
• 3)
If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0 are the diameters of a circle, then its centre is
• 4)
The locus of the point P which moves such that P is at equidistance from their coordinate axes is
• 5)
The locus of the point P which moves such that P is always at equidistance from the line x + 2y+ 7 = 0 is
• 1)
The value of n, when nP2 = 20 is
• 2)
The number of ways selecting 4 players out of 5 is
• 3)
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for n $\in$ N is
• 4)
If n is a positive integer, then the number of terms in the expansion (x + a)n is
• 5)
For all n > 0, nC1 + nC2 + nC3 + ... +nCn is equal to
• 1)
If $\tan\theta=\frac{1}{\sqrt5}$ and $\theta$ lies in the first quadrant, then $\cos\theta$ is
• 2)
The value of 1-2sin245o is
• 3)
The value 4cos340o-3cos40o is
• 4)
The value of $\frac{2\tan30^o}{1+tan^230}$ is
• 5)
If $\sin A=\frac{1}{2}$ then $4\cos^3A-3\cos A$ is
#### Analytical Geometry Important Question Paper In 11th Business Mathematics - by Basky - View & Read
• 1)
The angle between the pair of straight lines x2 - 7xy + 4y2 = 0
• 2)
If the lines 2x - 3y - 5 = 0 and 3x - 4y - 7 = 0 are the diameters of a circle, then its centre is
• 3)
The x - intercept of the straight line 3x + 2y - 1 = 0 is
• 4)
If the centre of the circle is (-a, -b) and radius $\sqrt{a^2-b^2}$then the equation of circle is
• 5)
Combined equation of co-ordinate axes is
### TN Stateboard Updated Class 11th Business Maths Syllabus
#### Matrices and Determinants
Determinants-Inverse of a Matrix-Input-Output Analysis
#### Algebra
Partial Fractions-Permutations-Combinations-Mathematical Induction-Binomial Theorem-Analytical Geometry-Locus-System of Straight Lines-Pair of Straight Lines-Circles-Conics
#### Analytical Geometry
Locus-System of Straight Lines-Pair of Straight Lines-Circles-Conics
#### Trigonometry
Trigonometric Ratios-Trigonometric Ratios of Compound Angles-Transformation Formulae-Inverse Trigonometric Functions
#### Differential Calculus
Functions and their Graphs-Limits and Derivatives-Differentiation Techniques
#### TN StateboardStudy Material - Sample Question Papers with Solutions for Class 11 Session 2019 - 2020
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Hello Students, if you like our sample question papers & study materials , please share these with your friends and classmates. | 21,252 | 61,147 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-24 | latest | en | 0.646941 |
https://www.coursehero.com/file/216343/fnt9/ | 1,498,289,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320227.27/warc/CC-MAIN-20170624064634-20170624084634-00497.warc.gz | 872,309,015 | 80,688 | # fnt9 - DL 9 FNTs 4 Look at a plot of the atom-atom...
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DL 9 FNTs 4) Look at a plot of the atom-atom potential. If the two atoms are momentarily at rest with a center to center separation of 1.4 sigma, what is the total energy of the system? If the sytem is at rest, it means one of three things. It could mean that this system isn't moving at all, that it is sitting at equilibrium, at rest. Since we're saying its at rest at 1.4 sigma, this isn't the case. The other two cases are when two masses are moving relative to one another. One case is when the bond between the atoms is compressed, and the atoms are momentarily at rest before moving away from each other. This occurs at values of r less than the equilibrium length, less than 1.12 sigma, so we know that's not the case. The third case is when two atoms are as far apart as they can get before they're pulled back together by their bond, when PE is equal to total E, and KE is zero. This occurs at some r max greater than 1.12 sigma. This must be the case! The total energy doesn't change, because this is a closed system, and
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## This note was uploaded on 07/31/2008 for the course PHY 7A taught by Professor Pardini during the Winter '08 term at UC Davis.
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fnt9 - DL 9 FNTs 4 Look at a plot of the atom-atom...
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Ask a homework question - tutors are online | 425 | 1,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-26 | longest | en | 0.95195 |
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=111962 | 1,576,462,827,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00060.warc.gz | 284,842,035 | 10,690 | # Difference between revisions of "2019 AMC 8 Problems/Problem 12"
## Problem
The faces of a cube are painted in six different colors: red (R), white (W), green (G), brown (B), aqua (A), and purple (P). Three views of the cube are shown below. What is the color of the face opposite the aqua face?
$\textbf{(A) }Red\qquad\textbf{(B) }White\qquad\textbf{(C) }Green\qquad\textbf{(D) }Brown\qquad\textbf{(E) }Purple$
## Solution 1
$B$ is on the top, and $R$ is on the side, and $G$ is on the right side. That means that (image 2)$W$ is on the left side. From the third image, you know that $P$ must be on the bottom since $G$ is sideways. That leaves us with the back, so the back must be $A$. The front is opposite of the back, so the answer is $\boxed{\textbf{(A)}\ R}$.~heeeeeeeheeeee
## Solution 2
2019 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions | 372 | 1,051 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | longest | en | 0.728435 |
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