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http://manualzz.com/doc/2728330/sharp-el-501xb | 1,498,133,484,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319265.41/warc/CC-MAIN-20170622114718-20170622134718-00579.warc.gz | 284,852,355 | 9,507 | Sharp EL-501XB
```EL-501XBWH
•
•
•
•
•
•
•
•
Single Line LCD display: 10 digits
131 functions
Minimise mistakes: large ‘=‘ key
Statistics functions: regression
Complex Number Calculations
N-Base: Bin, Oct, Dec, Hex
Constant calculation
Sturdy and hardwearing plastic keys
Ideal for the junior student, the Sharp EL501XB is an easy to useentry level scientific calculator which comes in a stylish
glossblack / white colour.It combines 131 functions with an uncluttered keyboard which together with a large "=" button
helps students to minimiseany wrong key presses.
Display
Basic Functions
• number of lines: 1
• digits per line: 12
• displayed digits: mantisse + exponent: 10 2
Calculations
• Constant/chain calculations:
• metric conversions: 0
• physical constants: 0
Number Systems
• calculations: dec, bin, oct, hex, pen: yes, not: pen
• conversions: dec, bin, oct, hex, pen: yes, not: pen
Others
• power supply: LR-1130 x 2
• dimensions (in mm): 75 x 1445 x 10
• weight (g): 73 g
• EAN: 4974019707624
• FSE display: Float, FIX, SCI, ENG:
• Memories: 1
• STO, RCL / M+, M-:
• +, -, x, ÷,%:
• max. operations at the same time (calc. / numeric): 4
• last digit correction:
Scientific Functions
• Pi, +/-, EXP:
• X2 ,√ ,³√ , ×√ , XY , X·¹ ln, log, ex ,10x:
• sin, cos, tan, sin-1, cos-1, tan-1:
• random:
• coordinate-systems: rectangular, parametric, polar and
sequentiell:
• complex numbers:
• formular-memory (F1 - F4): 0
Statistic Functions
• statistic functions with x, y: 1
• standard deviation / totals formation:
• mean value:
• linear and other regressions: 0
Sharp Electronics (UK) Limited, 4 Furzeground Way, Stockley Park, Uxbridge, Middlesex UB11 1EZ. Tel. 0208 734 2000
``` | 533 | 1,689 | {"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} | 2.828125 | 3 | CC-MAIN-2017-26 | longest | en | 0.581816 |
https://topgradeprofessors.com/interpret-the-slope-coefficient-in-each-of-the-above-estimated/ | 1,718,693,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00009.warc.gz | 505,595,890 | 22,558 | # Interpret the slope coefficient in each of the above estimated
1.
y-hat = 14 + 7.34x
y-hat = 3 + 25 In(x)
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In(y-hat) = 2 + 0.08x; se = 0.06
In(y-hat) = 2.5 + 0.48 In(x); se = 0.16
a.
Interpret the slope coefficient in each of the above estimated models, when x increase by one unit in Models 1 and 3 and by 1% in Models 2 and 4. (Round your answers to 2 decimal places.)
Model 1: y-hat increases by units. 7.34 ;
Model 2: y-hat increases by about units. 0.25
Model 3: y-hat increases by about percent. 8.00
.
Model 4: y-hat increases by about percent. .48
2.
b.
For each model, what is the predicted change in y when x increases by 6%, from 10 to 10.6?
Model 1: y-hat increases by units. 4.40
Model 2: y-hat increases by units. 1.46
Model 3: y-hat increases by percent. 4.92
Model 4: y-hat increases by percent. 2.84
3. Consider the sample regressions for the linear, the logarithmic, the exponential, and the log-log models. For each of the estimated models, predict y when x equals 57. (Do not round intermediate calculations. Round your answers to 2 decimal places.)
Response Variable: y Response Variable: ln(y) Model 1 Model 2 Model 3 Model 4 Intercept 15.13 −5.51 1.22 0.83 X 1.42 NA 0.05 NA ln(x) NA 24.45 NA 0.77 se 19.54 16.10 0.12 0.10
y-hat Model 1 [removed] Model 2 [removed] Model 3 [removed] Model 4 [removed]
4. Eva, the owner of Eva’s Second Time Around Wedding Dresses, currently has five dresses to be altered, shown in the order in which they arrived:
If Eva uses the shortest processing time first priority rule to schedule these jobs, what will be the average job tardiness?
[removed]
2 hours
5.
Eva, the owner of Eva’s Second Time Around Wedding Dresses, currently has five dresses to be altered, shown in the order in which they arrived:
If Eva uses the shortest processing time first (SPT) priority rule to schedule these jobs, what will be the average completion time?
[removed] 3 hours [removed] 5 hours [removed] 7 hours | 681 | 2,298 | {"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-2024-26 | latest | en | 0.769329 |
https://wiki.veriqloud.fr/index.php?title=Interleaved_Randomised_Benchmarking&diff=4273 | 1,597,417,794,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00282.warc.gz | 546,255,631 | 22,812 | # Difference between revisions of "Interleaved Randomised Benchmarking"
Interleaved Randomized benchmarking is a scalable experimental protocol for estimating the average error of individual quantum computational gates. This protocol consists of interleaving random Clifford gates between the gate of interest and provides an estimate as well as theoretical bounds for the average error of the gate under test, so long as the average noise variation over all Clifford gates is small. This technique takes into account both state preparation and measurement errors and is scalable in the number of qubits.
Tags: Certification Protocol, Randomised Benchmarking, Clifford group
## Assumptions
• The measurements performed are trusted.
• Noise model can be assumed to be gate and time-dependent or gate and time-independent.
• The noise model is independent and identically distributed (IID).
## Outline
Standard Randomised Benchmarking method involves applying many random sequences of gates of varying lengths to a standard initial state. Each sequence ends with a randomized measurement that determines whether the correct final state was obtained. The average computationally relevant error per gate is obtained from the increase in error probability of the final measurements as a function of sequence length. The random gates are taken from the Clifford group. The restriction to the Clifford group ensures that the measurements can be of one-qubit Pauli operators that yield at least one deterministic one-bit answer in the absence of errors.
The multi-qubit RB protocol described in Standard Randomised Benchmarking is restricted to benchmark only the full Clifford group on ${\displaystyle n}$ qubits. While this provides a significant step towards scalable benchmarking of a quantum information processor, it is desirable in many cases to benchmark individual gates in Clifford group rather than the entire set. Interleaving randomised benchmarking is a protocol which consists of interleaving random gates between the gate of interest, which is used to estimate the average error of individual quantum computational gates.
To benchmark a specific Clifford element (an individual gate), the following steps are involved:
Step 1: Implement Standard Randomised Benchmarking to get a model for the fidelity and to calculate the average error rate
• A fixed sequence length is selected at random. A random sequence of this length is chosen from the Clifford group.
• The operations are applied to the initial state corresponding to the selected sequence and then a final operator is applied which inverts all the previous operations.
• The final state is then measured to check if it matches the initial state. This process is performed several times with the same sequence to estimate the survival probability (the probability that the final state which returns to its initial state).
• Other random sequences of the same fixed sequence length are picked and the above-mentioned process is repeated to calculate the corresponding survival probability. This is then used to calculate the average survival probability for the sequence length.
• The same procedure is repeated for multiple different randomly selected sequence lengths.
• The observed survival probabilities are then plotted against the sequence length and then this is fit to an exponential decay curve, which is used to estimate the depolarizing parameter and sequence fidelity and also to calculate the average error rate which is the metric for randomized benchmarking.
Step 2: Procedure to estimate the new sequence fidelity by including the Clifford element to be benchmarked in the sequence
• Now, for a random fixed sequence length, choose a sequence where the first Clifford element is selected uniformly at random from the Clifford group and the second element is always chosen to be the specific Clifford element we want to benchmark.
• Final gate is chosen to be the inverse of the composition mentioned in the step above. The final state is then measured to check if it matches the initial state. This process is performed several times with the same sequence to estimate the survival probability (the probability that the final state which returns to its initial state).
• Other random sequences of the same fixed sequence length are picked and the above-mentioned process is repeated to calculate the corresponding survival probability. This is then used to calculate the average survival probability for the sequence length.
• The same procedure is repeated for multiple different randomly selected sequence lengths.
• The observed survival probabilities are then plotted against the sequence length, to obtain a zeroth or first-order model of the new sequence fidelity, from which the new depolarizing parameter is estimated.
Step 3: Estimate the gate error of the selected Clifford element to be benchmarked
• From the values obtained for the depolarizing parameter and the new depolarizing parameter, using the average gate fidelity, a point estimate is obtained for the gate error. The gate error would lie in a certain range of this estimate. One interpretation of this error is that it arises from imperfect random gates.
## Hardware Requirements
• Quantum computational resources to perform Clifford gates.
• Trusted Measurement device.
## Notation
• ${\displaystyle p}$: Depolarizing parameter
• ${\displaystyle p_{\bar {C}}}$: New depolarizing parameter for the specific Clifford element to be benchmarked
• ${\displaystyle d}$: Dimension of Hilbert space
• ${\displaystyle F_{avg}}$: Average fidelity, ${\displaystyle F_{avg}=p+{\frac {1-p}{d}}}$
• ${\displaystyle r}$: Average error rate, ${\displaystyle r=1-F_{avg},r={\frac {(d-1)(1-p)}{d}}}$
• ${\displaystyle m}$: Selected sequence length
• ${\displaystyle K_{m}}$: Total randomly selected sequence of ${\displaystyle m}$ sequence length
• Clif${\displaystyle _{n}}$: Clifford group
• C: Selected Clifford element to be benchmarked
• C${\displaystyle _{i}}$: Random element of Clifford group
• ${\displaystyle S_{(i_{1},...,i_{m})}}$ = ${\displaystyle S_{\mathbf {i_{m}} }}$: Random sequence of operations of length ${\displaystyle m}$
• ${\displaystyle \gamma }$: Superoperator representing the sequence with alternating ${\displaystyle C}$
• ${\displaystyle M}$: Number of different data points to get the error model
• ${\displaystyle \Lambda _{i,j}}$: Implementation of C${\displaystyle _{i}}$ at time j (1 ${\displaystyle \leq }$ j ${\displaystyle \leq }$ M) results in this error map. ${\displaystyle \Lambda _{i,1},...,\Lambda _{i,M}}$ are the different time-dependent noise operators affecting C${\displaystyle _{i}}$.
• ${\displaystyle \Lambda _{C}}$: Associated noise operator of the Clifford element ${\displaystyle C}$
• ${\displaystyle r_{C}^{est}}$: The gate error of ${\displaystyle \Lambda _{C}}$
• ${\displaystyle E}$: Error range of ${\displaystyle r_{C}^{est}}$
• ${\displaystyle |\psi \rangle }$: initial state
• ${\displaystyle E_{\psi }}$: POVM element which takes into account the measurement error.
• ${\displaystyle F_{seq}(m,\psi )=Tr[E_{\psi }S_{\mathbf {i_{m}} }(\rho _{\psi })]}$: Survival probability of a sequence. ${\displaystyle \rho _{\psi }}$ is a quantum state that takes into account errors in preparing ${\displaystyle \langle \psi |\psi \rangle }$
• ${\displaystyle F_{g}^{(0)}(m,|\psi \rangle )}$: Averaged sequence fidelity for gate and time independent error model
• ${\displaystyle F_{g}^{(1)}(m,|\psi \rangle )}$: Averaged sequence fidelity for gate and time dependent error model. In this model, the parameter ${\displaystyle (q-p^{2})}$ is a measure of the degree of gate-dependence in the error.
• ${\displaystyle F_{\bar {g}}^{(0)}(m,|\psi \rangle )}$: New zeroth order averaged sequence fidelity for ${\displaystyle C}$
• ${\displaystyle F_{\bar {g}}^{(1)}(m,|\psi \rangle )}$: New first order Averaged sequence fidelity for ${\displaystyle C}$
• ${\displaystyle A_{0},B_{0}}$: Coefficients that absorb the state preparation and measurement errors as well as the error on the final gate for gate and time independent error model
• ${\displaystyle A_{1},B_{1},C_{1}}$: Coefficients that absorb the state preparation and measurement errors as well as the error on the final gate for gate and time dependent error model.
• ${\displaystyle R_{m+1}}$: ${\displaystyle {\frac {1}{|Clif_{n}|}}\sum _{i}\Lambda _{i,m+1}\otimes (C_{i}\otimes \Lambda \otimes C_{i}^{\dagger })}$
## Properties
• Figure of merit: average gate error
• The errors which are considered here are State preparation and measurement errors, error on the final gate, which are gate and time-independent errors. Gate and time-dependent errors can also be taken into consideration. This method is insensitive to SPAM error.
• The random gates are picked from the Clifford group.
• For noise estimation, the uniform probability distribution over Clifford group comprises a unitary 2-design.
• This protocol provides a scalable method for benchmarking the set of Clifford gates.
• To obtain a more accurate value for ${\displaystyle p}$ one should always use the first order fitting model unless prior knowledge of the noise indicates that it is effectively gate-independent.
## Procedure Description
Step 1: Standard Randomised Benchmarking
Output: Figure of merit: ${\displaystyle r}$
• For ${\displaystyle 1,2,...,M}$:
• Pick random sequence length ${\displaystyle m}$
• For ${\displaystyle k=1,2,...,K_{m}}$ sequences:
• For ${\displaystyle j=1,2...,m+1}$:
• If ${\displaystyle j==m+1}$, apply inverse operator of previous operations
• else, apply random operation C${\displaystyle _{i}}$
• Thus, ${\displaystyle S_{\mathbf {i_{m}} }=\bigotimes _{j=1}^{m+1}(\Lambda _{(i_{j},j)}\circ C_{i_{j}})}$ and ${\displaystyle i_{m+1}}$ is uniquely determined by ${\displaystyle (i_{1},...,i_{m})}$
• Measure survival probability ${\displaystyle Tr[E_{\psi }S_{\mathbf {i_{m}} }(\rho _{\psi })]}$
• Estimate average survival probability ${\displaystyle Tr[E_{\psi }S_{\mathbf {K_{m}} }(\rho _{\psi })]}$ over all ${\displaystyle K_{m}}$ sequences, where ${\displaystyle S_{\mathbf {K_{m}} }={\frac {1}{K_{m}}}\sum _{i_{m}}S_{i_{m}}}$
• Fit the results for the averaged sequence fidelity for all ${\displaystyle m}$ into the models:
• For gate and time independent error model:
• ${\displaystyle F_{g}^{(0)}(m,|\psi \rangle )=A_{0}p^{m}+B_{0}}$
• For gate and time dependent error model:
• ${\displaystyle F_{g}^{(1)}(m,|\psi \rangle )=A_{1}p^{m}+B_{1}+C_{1}(m-1)(q-p^{2})p^{m-2}}$
• ${\displaystyle p}$ is extracted from the model and ${\displaystyle r}$ is estimated, ${\displaystyle r={\frac {(d-1)(1-p)}{d}}}$
Step 2: Estimate gate error of selected Clifford element C
Input: C
Output: gate error of ${\displaystyle \Lambda _{C}}$: ${\displaystyle r_{C}^{est}}$
• For ${\displaystyle 1,2,...,M}$:
• Pick random sequence length ${\displaystyle m}$
• For ${\displaystyle k=1,2,...,K_{m}}$ sequences:
• For ${\displaystyle j=1,2...,m+1}$:
• If ${\displaystyle j==m+1}$, apply inverse operator of previous operations
• else If ${\displaystyle j\%2==1}$, apply random operation C${\displaystyle _{i}}$
• else, apply C
• Thus ${\displaystyle \gamma =\Lambda _{i_{m+1}}+}$ C${\displaystyle _{i_{m+1}}(\bigotimes _{j=1}^{m+1}[C\circ \Lambda _{C}\circ \Lambda _{i_{j}}\circ C_{i_{j}}])}$
• Measure the survival probability ${\displaystyle Tr[E_{\psi }\gamma _{\mathbf {i_{m}} }(\rho _{\psi })]}$
• Estimate average survival probability ${\displaystyle Tr[E_{\psi }\gamma _{\mathbf {K_{m}} }(\rho _{\psi })]}$ over all ${\displaystyle K_{m}}$ sequences, where ${\displaystyle \gamma _{\mathbf {K_{m}} }={\frac {1}{K_{m}}}\sum _{i_{m}}\gamma _{i_{m}}}$
• Fit the results for the averaged sequence fidelity for all ${\displaystyle m}$ into the models, to find ${\displaystyle p_{\bar {C}}}$:
• For gate and time independent error model:
• ${\displaystyle F_{\bar {g}}^{(0)}(m,|\psi \rangle )=A_{0}p_{\bar {C}}^{m}+B_{0}}$
• For gate and time dependent error model:
• ${\displaystyle F_{\bar {g}}^{(1)}(m,|\psi \rangle )=A_{1}p_{\bar {C}}^{m}+B_{1}+C_{1}(m-1)(q-p_{\bar {C}}^{2})p_{\bar {C}}^{m-2}}$
• Estimate ${\displaystyle r_{C}^{est}={\frac {(d-1)(1-p_{\bar {C}}/p)}{d}}}$
• ${\displaystyle r_{C}^{est}}$ lies in the range ${\displaystyle [r_{C}^{est}-E,r_{C}^{est}+E]}$, where ${\displaystyle E=min({\frac {(d-1)[|p-p_{\bar {C}}/p|+(1-p)]}{d}},{\frac {2(d^{2}-1)(1-p)}{pd^{2}}}+{\frac {4{\sqrt {1-p}}{\sqrt {d^{2}-1}}}{p}})}$
## Further Information
• Fitting models are described and derived as seen in E. Mageson et al. The coefficients derived are:
• ${\displaystyle A_{0}}$ = Tr${\displaystyle [E_{\psi }\Lambda (\rho _{\psi }-{\frac {\mathbb {1} }{d}})]}$
• ${\displaystyle B_{0}}$ = Tr${\displaystyle [E_{\psi }\Lambda ({\frac {\mathbb {1} }{d}})]}$
• ${\displaystyle A_{1}}$ = Tr${\displaystyle [E_{\psi }\Lambda ({\frac {Q_{1}\rho _{\psi }}{p}}-\rho _{\psi }+{\frac {(p-1)\mathbb {1} }{pd}})]}$ + Tr${\displaystyle [E_{\psi }R_{m+1}({\frac {\rho _{\psi }}{p}}-{\frac {\mathbb {1} }{pd}})]}$
• ${\displaystyle B_{1}}$ = Tr${\displaystyle [E_{\psi }R_{m+1}({\frac {\mathbb {1} }{d}})]}$
• ${\displaystyle C_{1}}$ = Tr${\displaystyle [E_{\psi }\Lambda (\rho _{\psi }-{\frac {\mathbb {1} }{d}})]}$
• The case where Randomized benchmarking fails: Suppose the noise is time dependent and for each ${\displaystyle i,\Lambda _{i}=C_{i}^{\dagger }}$. Then ${\displaystyle F_{g}(m,\psi )=1}$ for every ${\displaystyle m}$ even though there is a substantial error on each ${\displaystyle C_{i}}$ and so benchmarking fails.
• Wallman, Granade, Harper, F., NJP 2015 Purity benchmarking: A unitarity can be estimated via purity benchmarking, which is an RB-like experiment that estimates a decay rate.
## Related Papers
• E.Knill et al (2007) arXiv:0707.0963: gate and time-independent noise model
• E. Mageson et al (2011) arXiv:1009.3639: multi-parameter model
• Magesan et al. PRL (2012): Interleaved Randomized Benchmarking
*contributed by Rhea Parekh | 3,739 | 14,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 106, "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} | 2.5625 | 3 | CC-MAIN-2020-34 | latest | en | 0.913943 |
https://www.mathworks.com/matlabcentral/answers/1586684-coloring-the-area-between-three-curves?s_tid=prof_contriblnk | 1,642,797,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00274.warc.gz | 949,257,592 | 23,911 | # coloring the area between three curves
18 views (last 30 days)
mukesh bisht on 15 Nov 2021
Answered: Pranjal Kaura on 24 Nov 2021
Hi. I want to color the intersection region of three curves.
T1 = 180:0.1:270;
x1 = 3*cosd(T1); y1 = 3*sind(T1); % circle
T2 = 3.3620:0.001:3.9045;
c1 = 4.3123; c2 = -50.2708; c3 = 193.7277; c4 = -243.8411;
r_6 = c1*T2.^3 + c2*T2.^2 + c3*T2 + c4;
x2 = 1.1998 + r_6.*cos(T2); y2 = -0.8840 + r_6.*sin(T2); % region 4
x0 = 1.8977; y0 = 2.3235; vr = 4.0678;
beta = atan(y0/x0);
x3 = -2.0591:0.001:-1.8541; y3 = y0 + (x0-x3)*cot(beta) - (9.81*(x0-x3).^2)./(2*(vr*sin(beta))^2); % parabola
pgon1 = polyshape(x1,y1); % circle
pgon2 = polyshape(x2,y2); % region 4
pgon3 = polyshape(x3,y3); % parabola
pgon4 = intersect(pgon1,pgon2);
figure (4)
plot(x1,y1,x2,y2,x3,y3)
hold on
plot (intersect(pgon3,pgon4),'EdgeColor','none')
Actually, the fucntion :plot (intersect(pgon3,pgon4),'EdgeColor','none'); gives the error
Pranjal Kaura on 24 Nov 2021
Hey Mukesh,
The common area between the polygons 'pgon3' and pgon4' isn't being plotted because the 'intersect' function doesn't find any common points between the 2 regions and thus returns an object having an empty matrix for 'Vertices' field, when you run
>> intersect(pgon3,pgon4)
To visualise your plots better, you can try plotting the polygon regions ('pgon1', 'pgon2' etc one after the other) and then decide which polygons to send to the 'intersect' function to compute the common points. You can refer the following code snippet:
>> figure
>> hold on
>> plot(pgon3)
>> plot(pgon1)
>> plot(pgon2)
You can check the 'Vertices' field of the 'polyshape' object that the function 'intersect' returns to confirm whether there are any common points.
Hope this helps! | 631 | 1,739 | {"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} | 3.328125 | 3 | CC-MAIN-2022-05 | latest | en | 0.7708 |
https://www.experts-exchange.com/questions/27019337/series-fill-with-same-number-twice.html | 1,477,670,913,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722951.82/warc/CC-MAIN-20161020183842-00334-ip-10-171-6-4.ec2.internal.warc.gz | 906,956,338 | 19,934 | Solved
# series fill with same number twice
Posted on 2011-05-04
430 Views
I need to series filll a column with the same number twice. For example A6 = 1,A7 =1, A8=2.,A9 = 2, etc. (ms xls 2007)
0
Question by:howcheat
LVL 12
Accepted Solution
A1 = 1
A2 = 1
A3 = if(A2=A1,A2+1,A2)
Copy A3 down.
Alan.
0
LVL 12
Expert Comment
If you want them as values, and not formulae, do the above, then copy and paste to values.
Alan.
0
LVL 50
Expert Comment
Hello,
for example to do this with 200 cells, fill the series to the first 100 cells, going 1,2,3, etc. then fill the next 100, starting again with 1,2,3. Select all 200 cells and sort ascending.
cheers, teylyn
0
LVL 43
Expert Comment
You can change Alan3285's formula
=if(A2=A1,A2+1,A2)
to
=A1+1
If you don't have to use the value of the numbers you can fill series starting from 0.5 and increment by 0.5 and display with zero decimal places.
Another formula which can be used and then converted to values is
if starting in row 6 then
=INT((ROW()-4)/2)
if starting in row 7 then
=INT((ROW()-5)/2)
0
Author Closing Comment
Great resolutions - All!! Alan3285 was first
0
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The viewer will learn how to use a discrete random variable to simulate the return on an investment over a period of years, create a Monte Carlo simulation using the discrete random variable, and create a graph to represent the possible returns over… | 592 | 2,159 | {"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} | 3.125 | 3 | CC-MAIN-2016-44 | longest | en | 0.873189 |
https://study.com/academy/topic/newtons-laws-in-astronomy.html | 1,519,537,747,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00076.warc.gz | 759,488,409 | 27,324 | # Ch 12: Newton's Laws in Astronomy
Watch online science video lessons to learn about balanced and unbalanced forces, acceleration, inertia, free fall motion and other topics related to Newton's laws of astronomy.
## Newton's Laws in Astronomy - Chapter Summary and Learning Objectives
Let instructors teaching this chapter demonstrate the physics at work in everything from the simplest action-reaction force pairs to the gravity and normal forces that keep us stable. You can also discover equations used to calculate mass, acceleration and other variables accounted for by Newton's laws of motion. Lessons in this chapter are designed to teach you how to do the following:
• List types of contact and non-contact forces
• Calculate the net force acting on an object
• Describe the relationship between weight, mass and gravity
• Calculate the acceleration of an object
• Interpret a free-body diagram
Video Objectives
Newton's First Law of Motion: Examples of the Effect of Force on Motion Use Newton's first law of motion to examine unbalanced and balanced forces, inertia and friction.
Distinguishing Between Inertia and Mass Learn how an object's mass affects its inertia.
Mass and Weight: Differences and Calculations Differentiate mass from weight. Get tips for using one variable to calculate the other.
State of Motion and Velocity Learn how to define the speed and velocity of an object in motion. Study the effects of circular motion on velocity.
Force: Definition and Types Get examples of contact forces, like friction, air resistance and applied force. Explore non-contact forces, such as gravity and magnetic force.
Forces: Balanced and Unbalanced Examine the static and dynamic equilibriums of objects with balanced forces. Learn how unbalanced forces cause an object's state of motion to change.
Free-Body Diagrams Discover the vector diagrams used to depict the magnitude and direction of forces acting on an object.
Net Force: Definition and Calculations Learn how to add or subtract forces acting on an object in order to determine the net force.
Newton's Second Law of Motion: The Relationship Between Force and Acceleration Apply Newton's second law of motion to describe how net force and mass can be used to calculate acceleration.
Determining the Acceleration of an Object Practice finding the acceleration of an object when given the change in velocity and the change in time.
Determining the Individual Forces Acting Upon an Object Get tips for determining whether such forces as gravity, normal force, friction or tension are acting on an object.
Air Resistance and Free Fall Examine the acceleration of objects in free fall. Describe the effect of air resistance and mass.
Newton's Third Law of Motion: Examples of the Relationship Between Two Forces Learn how action-reaction force pairs demonstrate Newton's third law of motion.
Newton's Laws and Weight, Mass & Gravity Study the effect of gravity on the acceleration and weight of falling objects.
Identifying Action and Reaction Force Pairs Identify the effect of an object's mass in action-reaction force pairs.
The Normal Force: Definition and Examples Examine how the force exerted by surfaces works with gravity to keep things in place.
Friction: Definition and Types Differentiate between static and sliding friction.
Final Exam
Chapter Exam
### Earning College Credit
Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 709 | 3,665 | {"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} | 3.96875 | 4 | CC-MAIN-2018-09 | longest | en | 0.885632 |
https://statquest.org/2015/08/13/pca-clearly-explained/ | 1,566,491,718,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00467.warc.gz | 651,395,632 | 28,530 | # PCA clearly explained
Update: A lot of people ask to see example code that shows how to actually do PCA. Here’s how to do it in R
```## In this example, the data is in a matrix called
## data.matrix
## columns are individual samples (i.e. cells)
## rows are measurements taken for all the samples (i.e. genes)
## Just for the sake of the example, here's some made up data...
data.matrix <- matrix(nrow=100, ncol=10)
colnames(data.matrix) <- c(
paste("wt", 1:5, sep=""),
paste("ko", 1:5, sep=""))
rownames(data.matrix) <- paste("gene", 1:100, sep="")
for (i in 1:100) {
wt.values <- rpois(5, lambda=sample(x=10:1000, size=1))
ko.values <- rpois(5, lambda=sample(x=10:1000, size=1))
data.matrix[i,] <- c(wt.values, ko.values)
}
pca <- prcomp(t(data.matrix), scale=TRUE)
## plot pc1 and pc2
plot(pca\$x[,1], pca\$x[,2])
## get the name of the sample (cell) with the highest pc1 value
rownames(pca\$x)[order(pca\$x[ ,1], decreasing=TRUE)[1]]
## get the name of the top 10 measurements (genes) that contribute
## most to pc1.
gene_score_ranked <- sort(gene_scores, decreasing=TRUE)
top_10_genes <- names(gene_score_ranked[1:10])
top_10_genes ## show the names of the top 10 genes
## get the scree information
pca.var <- pca\$sdev^2
scree <- pca.var/sum(pca.var)
plot((scree[1:10]*100), main="Scree Plot", xlab="Principal Component", ylab="Percent Variation")<span data-mce-type="bookmark" id="mce_SELREST_start" data-mce-style="overflow:hidden;line-height:0" style="overflow:hidden;line-height:0;"></span>
```
RNA-seq results often contain a PCA (Principal Component Analysis) or MDS plot. Usually we use these graphs to verify that the control samples cluster together. However, there’s a lot more going on, and if you are willing to dive in, you can extract a lot more information from these plots. The good news is that PCA only sounds complicated. Conceptually, it’s actually quite simple.
In this video I clearly explain how PCA graphs are generated, how to interpret them, and how to determine if the plot is informative or not. I then show how you can extract extra information out of the data used to draw the graph.
## 39 thoughts on “PCA clearly explained”
1. drpalv says:
Thanks Joshua for such a wonderful explanation of PCA! I was wondering if you would be willing to share your slide deck. Your slides are very clear and well put together. I was hoping to use them for educational purposes. I will of course acknowledge you appropriately for the content. Thanks!
Like
• Josh says:
Sure! I’d be happy to.
Like
2. Statistics Student says:
Bug report. Try: rownames(pca\$x)[order(pca\$x[ ,1], decreasing=TRUE)[1]]
for the sample ID. Your code tries to use the max PCA value as an index. ???
Like
• Josh says:
Thanks for catching that! I’ve corrected it and added some toy data to make it easier to play around with.
Like
3. Karalan says:
for distinguish between two different kind of samples (here it is cells), two PCs are okay but how to distinguish more than two samples with two PCs??
Like
• Josh says:
To plot a cell on the graph, you multiply the gene measurements associated with that cell with the loading scores in PC1. Then add up the values and that gives you a coordinate along the PC1 axis. Then do the same thing with PC2 – multiply the gene measurements with the loading scores in PC2 and then add them up. That gives you a coordinate on PC2. With those two coordinates, we can plot the cell on the 2D graph. (If we wanted to draw a 3D graph, we would do the same thing to get a coordinate along the PC3 axis). That process can be repeated for as many cells as we have. This is explained here: https://youtu.be/_UVHneBUBW0?t=17m25s
Like
• Karalan says:
Thank you. . :)
Actually i had one more doubt, In your example rows corresponds to variables and columns corresponds to observations but when I try to do it in matlab (which uses SVD algorithm) its completely different, There rows correspond to observations and columns correspond to variables and it returns the loadings in the size of columns x columns. How it is possible?
Like
4. Josh says:
For historical reasons, most PCA functions (including SVD) expect the rows to be samples, and the columns to be variables. However, genomic data is often the transpose of that type of matrix – thus, samples are columns and variables (usually genes) are rows. Thus, to do PCA with my data (where the samples are columns and the variables are rows) you have to transpose the matrix before you make the call to the PCA function. In my example code (for R), I have the line that looks like this:
pca <- prcomp(t(data.matrix), scale=TRUE)
The t() function that the data goes through before going to prcomp() transposes the data. So try transposing your matrix to see if you get the right values.
Like
• Karalan says:
So the plotting is done between variables or samples??
Like
• Josh says:
The plotting is usually done for the samples, not the variables.
Like
5. Karalan says:
In both methods??
Like
• Josh says:
I’m not sure what you mean by both methods. PCA is just one method. Typically it is used to plot samples. If you data has the samples as columns and variables as rows, than you need to transpose the matrix so that the samples are rows and the variables are columns before you do PCA on it so that you can plot the samples. If your data is already formatted so that the samples are rows, than you’re set, you don’t need to transpose the matrix. Does this make sense?
Like
• Karalan says:
Actually I mean to ask SVD algorithm which am using in MATLAB (where rows correspond to observations and columns correspond to variables) also plotting between the samples?
Liked by 1 person
6. Josh says:
I’m not sure how to answer your question since I don’t use matlab. However, R is free and you can download it. Here’s how you use SVD in R:
svd.stuff <- svd(scale(MY_MATRIX, center=TRUE, scale=TRUE))
Where "MY_MATRIX" is a matrix where the rows correspond to samples and the columns correspond to variables.
To draw a PCA plot from this:
plot((svd.stuff\$u[,1] * svd.stuff\$d[1]), (svd.stuff\$u[,2] * svd.stuff\$d[2]))
Where "u" is a matrix of PCs that have all been scaled so that they have a magnitude of 1 and and "d" is a vector of values that spread the columns in "u" out according to how much variance each PC accounts for in the original data.
Like
• Josh says:
I watched the video and the “samples” are the rows and the “variables” are columns. Ultimately, in later videos, the dudes shows how you can use SVD to plot the rows on a 2-D graph to show how the samples are related to each other. Specifically, check out https://youtu.be/K38wVcdNuFc?t=4m17s In that, the samples are people (and rows in this case) and the variables are movies (and columns) that they have seen and scored based on how much they liked them. On the left side of the screen you see the raw data for two people. On the right side, you see the results of SVD – X and Y coordinates that will put those two people close to each other in the final graph because they both gave Sci-Fi movies high scores and they both gave Romance movies low scores. Thus, the final plot will show us which people like Sci-Fi movies and which people like Romance movies. In my example, however, the matrix is transposed – the goal is to plot cells by looking at the scores for each gene. Samples with high scores for the same genes and low scores for the same genes will be plotted near each other. So in my example, the matrix has to be transposed before you put it into SVD.
Like
7. Hussam says:
Thank you for the great explanation, as i understand the number of PC we get should equal the number of columns (variables) in our data, so we after we take the transposed matrix and run prcomp(), PC should equal the number of genes(100 genes) not the number of samples
Like
• Josh says:
That’s a reasonable idea, however, the number of PCs that are any use whatsoever is the minimum of either the number of samples or the number of variables. In a few weeks I’m going to make a video that clarifies this. In the mean time… Imagine you had 2 variables and only 1 sample. If you did PCA on this, how many dimensions would you need to plot a single sample? Just 1. If you had 10 variables and 2 samples, you could still plot those 2 samples on a line if you just drew a line between them. Does that make sense. The video will clarify this more by showing images.
Like
• Hussam says:
Thank you so much, waiting for new videos , and great songs too
Like
• Josh says:
Hooray! I’m glad you like them both!
Like
• Hussam says:
josh , i have another question and i would be really glad if you answer me, now in your example the first PC explains 90% of data variation , but of course when we deal with the real biological data we will get more number of PC that can explains the variation , now my question is how can we get the most important genes from number of PCs that explains more than 80% of data variation , in your example u used just the 1st PC to extract the important genes , but what if i want to extract genes from several PCs, hope i could explain what i want!!
Best regards
Liked by 1 person
• Hussam says:
Can you write the code in R for ranking genes in several PCs?
Like
8. Josh says:
This is a good question – often I have to look at more than 1 principal component to find the genes that are interesting. All you do is look at the scree plot and decide how many PCs you want to look at, then rank the loading scores fo those PCs by their magnitude (aka absolute value, since large negative values are just as interesting as large positive values) and then look at the top genes. If you want to get fancy, you can then rank the genes across the PCs. For example, after ranking the gene in the first 3 PCs, you could then see which PC has the largest value for the top rank, and use that gene. Then you could see which PC has the largest value for the 2nd rank, and use that gene, etc.
Like
• Hussam says:
because i look at more than 500 PC
Like
• Hussam says:
hey josh, thank you for answering me, but i have noticed something ,and i hope that you will correct me if am wrong.
now , i have noticed that genes have smaller rotation values in PC1 than in PC2 and ,PC2 rotation values for each gene are smaller than the ones in PC3 and so on , so basically if i write the code u send me and then sort by decreasing to obtain the highest ten genes , i will end up getting genes that are in the last PC because they have the biggest rotation values, which are the less important genes.
Like
9. Josh says:
Here’s an example of getting the top genes from the first 3 eigenvectors…
eigen_vectors_top_3 <- eigen_vectors[, 1:3]
gene_score_top <- abs(eigen_vectors_top)
score <- apply(gene_score_top, 1, max)
gene.names <- names(score)
Like
10. Josh says:
Well, you might have an odd dataset. When I do it on the two PCs in “how to do PCA in R” code https://statquest.org/2017/11/27/statquest-pca-in-r-clearly-explained/#code I get a mix of genes with higher values in PC1 and PC2 (although I have to look at the top 20 genes to see the mix). Assuming you’ve loaded in the code that I mentioned and run it, you and see genes from the two PCs with the following lines:
eigen_vectors <- pca\$rotation[,1:2]
gene_scores <- abs(eigen_vectors)
score <- apply(gene_scores, 1, max)
score <- sort(score, decreasing=TRUE)
Like
11. Hi, Josh
All of your videos I’ve seen to date (half a dozen) feature excellent explanations. Thank you for making this material and your very helpful take on it available to everyone on the Internet.
I have an insignificant question, just to help identify if I’m misinterpreting something: Using the small data set in the video, I calculated cell 1’s PC1 score (practicing it by transferring it to terminology that works better for me, but with your numbers). Treating each cell-1 entry (value) and multiplying it with the quantitative influence, I get:
Cell 1’s PC 1 score = (10*10)+(0*0.5)+(14*0.2)+(33*[-0.2])+(50*13)+(80*[-14]) = -373.8 (not 12)
I noticed you said “… we _might_ end up with a number like 12 …” (emphasis added), so I’m wondering whether I’m indeed doing something wrong or whether 12 was just an arbitrary number? (Or did you divide by the total number of cells?)
Thanks
Like
12. John Kennedy says:
Dear Joshua, great video, thank you!
I’m not being able to follow your R code, though. It seems polluted with random html tagging…
Some examples at line lines 7, 8 and 9:
colnames(data.matrix) <- c(
paste("wt", 1:5, sep=""),
paste("ko", 1:5, sep=""))
If possible, when you have the time and the energy together, could you please remove them so the code could appear clearer?
Sorry for bothering you, and thanks in advance!
Best Regards,
JK
Like
• Josh says:
Ummm…. I just copied and pasted the example you gave me (lines 7, 8 and 9) into R and they ran just fine. I pasted from the code on the website and the code in your comment and both ran fine. Can you tell me what the error is that you are getting?
Like
• John Kennedy says:
Maybe is my browser (Firefox) perhaps?
When I copy your line 6 and paste into R, I get this:
> data.matrix <- matrix(nrow=100, ncol=10)
And when copy and paste lines 7, 8, and 9, I get this:
> colnames(data.matrix) <- c(
> paste("wt", 1:5, sep=""),
Error: unexpected ‘&’ in ” paste(&”
> paste("ko", 1:5, sep=""))
Error: unexpected ‘&’ in ” paste(&”
>
JK
Like
13. John Kennedy says:
The odd thing is: when pasting into your blog’s message box, I have the odd characters visible; but when I click on “Post Comment”, they disappeared…
Like
14. John Kennedy says:
This time I’m using Chrome, the odd characters are still there. Just to make them visible to you, will add an extra underline inside the odd characters:
data.matrix _&_l_t_;_- matrix(nrow=100, ncol=10)
colnames(data.matrix) _&_l_t_;_- c(
paste(_&_q_u_o_t_;_w_t_&_q_u_o_t_;_, 1:5, sep=_&_q_u_o_t_;_w_t_&_q_u_o_t_;_),
paste(_&_q_u_o_t_;_w_t_&_q_u_o_t_;_, 1:5, sep=_&_q_u_o_t_;_w_t_&_q_u_o_t_;_))
JK
Like
• John Kennedy says:
If you could copy and paste the above into a notepad, and replace the underline for nothing, you’ll be able to see what I’m seeing when I look to your original code.
Sorry for the trouble!
JK
Like
• Josh says:
Weird! Maybe it’s some wordpress strangeness. I’m sorry it’s not working for you. That’s a bummer. I’ll try to put a link to download the code (as a raw text file) so that people that run into the same problem will have an easy work around. Thanks for letting me know about this problem.
Like
• John Kennedy says:
WordPress… I think you’re right!
Thank you very much and, once again, sorry for the trouble.
Best Regards,
JK
Like
15. JK says:
Acessing this page with my Android cell phone using Chrome, the odd characters are very cliear.
I thing I’ve partially got it:
_&_l_t_;_ equals <
_&_q_u_o_t_;_ equals "
but the last line of code (line 40) has some additional html code, not sure how to deal with it. It starts with
< span
Like
• Josh says: | 3,890 | 15,217 | {"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-2019-35 | longest | en | 0.601302 |
https://www.dgcoursereview.com/forums/showthread.php?p=2078243 | 1,508,207,970,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820556.7/warc/CC-MAIN-20171017013608-20171017033608-00197.warc.gz | 1,174,711,037 | 15,780 | Disc Golf Course Review optimal launch trajectory
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#1
07-11-2013, 10:24 AM
DG_player Birdie Member Join Date: Apr 2013 Posts: 377 Niced 50 Times in 40 Posts
optimal launch trajectory
I always find the physics of disc golf be very interesting and I was pondering this question while playing the other day:
What is the optimum launch angle to achieve max distance?
For any object thrown in a vacuum, this angle is 45 degrees. When you throw in the effects the air, like lift and drag, this changes immensely. For a golf ball drive it's around 15 degrees. I'm guessing for a golf disc since it flies like a wing and thus presumably generates more lift than a golf ball, that there is a lower optimal trajectory.
So from your experience, what have you found to be the optimal launch trajectory for max distance (assuming adequate power and technique to properly throw a driver)?
Feel free to specify lines and discs as well.
#2
07-11-2013, 10:34 AM
bradharris Team Borderland Join Date: Apr 2011 Location: Loudoun County Years Playing: 14.5 Courses Played: 87 Posts: 5,219 Niced 38 Times in 25 Posts
Trajectory is the wrong way to think about it.
The disc is a wing, airflow is what keeps it aloft and thus, keeps it flying forward. As long as there is enough airflow, the disc will produce enough lift to stay aloft.
Of course, the disc also has to stay stable which is where spin and release angle come into play as well.
The short answer, throw it level with the correct release angle and the disc will take care of the rest.
#3
07-11-2013, 10:48 AM
knettles Double Eagle Member Join Date: Nov 2011 Location: Portland, OR Years Playing: 7.2 Courses Played: 112 Throwing Style: RHBH Posts: 1,227 Niced 53 Times in 33 Posts
Well, the optimal throw for maximum distance is a high s-curve. Any pro will tell you that. Throw an understable disc high, it'll turn and glide a long way before starting to eventually fade. However, I'm not sure about the actual angle. I'd say the golf angle of 15 degrees is pretty close. Could possibly be a bit higher. Some of the videos of the distance comps show them launching the discs way high. Throwing level is definitely not the optimal angle tho. Having that been said, most of my throws in a round are level (golf shot) and not high (max d) to preserve accuracy.
And I agree, the physics of disc golf are alot of fun.
#4
07-11-2013, 11:18 AM
garublador * Ace Member * Join Date: Mar 2008 Location: Urbandale, IA Years Playing: 13.6 Courses Played: 7 Posts: 5,081 Niced 33 Times in 19 Posts
Quote:
Originally Posted by knettles Well, the optimal throw for maximum distance is a high s-curve. Any pro will tell you that. Throw an understable disc high, it'll turn and glide a long way before starting to eventually fade. However, I'm not sure about the actual angle. I'd say the golf angle of 15 degrees is pretty close. Could possibly be a bit higher. Some of the videos of the distance comps show them launching the discs way high. Throwing level is definitely not the optimal angle tho. Having that been said, most of my throws in a round are level (golf shot) and not high (max d) to preserve accuracy. And I agree, the physics of disc golf are alot of fun.
I'll agree that checking out some videos of distance competitions would be the best way to check. You have to make sure to check the initial trajectory because the disc will rise after that. Looking at the video in this thread:
It seems as if the 15 degree number isn't too far off. Though I suspect it will depend heavily on the disc. Slower discs have less lift and will probably require a higher initial trajectory.
#5
07-11-2013, 09:07 PM
sidewinder22 * Ace Member * Join Date: Nov 2008 Location: Creeping Creek DGC Years Playing: 11.2 Courses Played: 185 Posts: 11,679 Niced 544 Times in 410 Posts
A lot will depend on the thrower, the wind, and the disc. There's a vid of Katchz throwing something like 45 degrees in a monster tailwind and going like 700'. For me it probably is closer to 15-20 degrees. For players with less power the trajectory will be lower as it's harder to keep the nose down. Most torque monkeys/strong armers and air bouncers are throwing with a negative trajectory and nose up, not good for distance.
#6
07-14-2013, 11:58 PM
Nasty Nate Double Eagle Member Join Date: Jun 2013 Location: Columbus, OH Years Playing: 14.6 Courses Played: 12 Throwing Style: RHBH Posts: 1,393 Niced 21 Times in 16 Posts
I was actually just wondering about this because I have trouble getting max distance out of my drives and when considering trajectory I often end up floating it and losing distance. When I throw flat I still get about the same distance, if not less, as if I had thrown a bit higher. Granted, my form is not perfect but I can't quite figure out the optimal way to launch a disc. A good drive for me ends up right around 300 feet but I frequently land near 260. I guess some things come more easily to other people.
#7
07-15-2013, 12:10 AM
DiscinFiend * Ace Member * Join Date: May 2013 Location: Milwaukee, WI Years Playing: 18.4 Courses Played: 70 Throwing Style: RHBH Posts: 3,482 Niced 97 Times in 64 Posts
Quote:
Originally Posted by sidewinder22 A lot will depend on the thrower, the wind, and the disc. There's a vid of Katchz throwing something like 45 degrees in a monster tailwind and going like 700'. For me it probably is closer to 15-20 degrees. For players with less power the trajectory will be lower as it's harder to keep the nose down. Most torque monkeys/strong armers and air bouncers are throwing with a negative trajectory and nose up, not good for distance.
#8
07-15-2013, 08:35 AM
ChrisWoj Eagle Member Join Date: Nov 2008 Location: Toledo, Ohio Years Playing: 12.6 Courses Played: 128 Posts: 599 Niced 44 Times in 32 Posts
Quote:
Originally Posted by DG_player I always find the physics of disc golf be very interesting and I was pondering this question while playing the other day: What is the optimum launch angle to achieve max distance? For any object thrown in a vacuum, this angle is 45 degrees. When you throw in the effects the air, like lift and drag, this changes immensely. For a golf ball drive it's around 15 degrees. I'm guessing for a golf disc since it flies like a wing and thus presumably generates more lift than a golf ball, that there is a lower optimal trajectory. So from your experience, what have you found to be the optimal launch trajectory for max distance (assuming adequate power and technique to properly throw a driver)? Feel free to specify lines and discs as well.
This has less to do with the specific question you had than it does disc physics... Check out the stuff by the DiscWing guys. Their masters thesis was on disc flight physics. Its over at their company website. I used their research to do a 15 minute presentation on the physics of disc flight last fall for a physics of everyday things course, and their research was tremendously useful. I also feel like I "understand" my plastic a lot more as a result.
#9
07-15-2013, 08:44 AM
Uncle Dougie Double Eagle Member Join Date: Apr 2011 Location: Rockford, IL Years Playing: 7.5 Courses Played: 28 Throwing Style: RHBH Posts: 1,383 Niced 0 Times in 0 Posts
Quote:
Originally Posted by sidewinder22 A lot will depend on the thrower, the wind, and the disc. There's a vid of Katchz throwing something like 45 degrees in a monster tailwind and going like 700'. For me it probably is closer to 15-20 degrees. For players with less power the trajectory will be lower as it's harder to keep the nose down. Most torque monkeys/strong armers and air bouncers are throwing with a negative trajectory and nose up, not good for distance.
SW,
Would throwing out to the left with a high hyzer-flip to get a big turnover be considered a negative trajectory? Or the same shot but with an anny orientation be the negative trajectory?
#10
07-15-2013, 09:10 AM
BogeyNoMore * Ace Member * Join Date: Jul 2009 Location: Walled Lake, MI Years Playing: 13.4 Courses Played: 242 Throwing Style: RHBH Posts: 8,304 Niced 450 Times in 281 Posts
Quote:
Originally Posted by ChrisWoj This has less to do with the specific question you had than it does disc physics... Check out the stuff by the DiscWing guys. Their masters thesis was on disc flight physics. Its over at their company website. I used their research to do a 15 minute presentation on the physics of disc flight last fall for a physics of everyday things course, and their research was tremendously useful. I also feel like I "understand" my plastic a lot more as a result.
I really enjoy disc physics, but that's some dry reading.
Here's the link to Disc Wing's Research if anyone has oodles of time to kill. | 2,176 | 8,840 | {"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} | 2.75 | 3 | CC-MAIN-2017-43 | latest | en | 0.932298 |
https://examfear.com/notes/Class-6/Maths/Fractions/2459/Making-Equivalent-fractions.htm | 1,621,087,960,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991370.50/warc/CC-MAIN-20210515131024-20210515161024-00021.warc.gz | 273,150,298 | 6,749 | Class 6 Maths Fractions Making Equivalent fractions
Making Equivalent fractions
In order to make equivalent number, we have to multiply both the numerator and the denominator by the same number. For eg we get the following fractions when we keep on dividing both the numerator and the denominator by the number 2.
These fractions are equivalent since all of them can be reduced to the simplest fraction - ½.
Equivalent fraction can also be made by dividing both the numerator and the denominator by the same number.
It is important to note that the number by which we divide has to be such that the number divides both the numerator and the denominator perfectly.
Therefore, we can first find the HCF of the numerator and the denominator then divide them by the HCF.
These fractions are equivalent since all of them can be reduced to the simplest fraction- ½.
It is to be remembered that a fraction is said to be in the simplest (or lowest) form if its numerator and denominator have no common factor except 1. For e.g. ½.
Problem: Write the fractions. Are all these fractions equivalent?
Solution:
1/2 = 2/4 = 3/6 = 4/8 = ½
All these fractions are equivalent since all these fractions can be reduced to the simplest fraction of the value ½.
Problem: Find the equivalent fraction of 36/48 with a) numerator 9 b) denominator 4
Numerator of the value 9 can be obtained by dividing both the numerator and the denominator by 4 since the numerator, 36/4 =9.
36/4 = 9/12
48/4
Denominator of the value 4 can be obtained by dividing both the numerator and the denominator by 12 since the denominator,
48/12 = 4.
= 36/12 = 3/4
48/12
Problem: Ram has 20 pencils, Sheelu has 50 pencils and Jamaal has 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Thus, it can be seen that each of the children used the same fraction of pencils in 4 months.
. | 493 | 2,039 | {"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} | 4.9375 | 5 | CC-MAIN-2021-21 | latest | en | 0.905181 |
https://www.geeksforgeeks.org/when-to-use-each-sorting-algorithms/?ref=shm | 1,713,176,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00033.warc.gz | 713,837,323 | 46,483 | # When to use each Sorting Algorithm
Last Updated : 01 Nov, 2023
A sorting algorithm is an algorithm that makes the input data set arranged in a certain order. The fundamental task is to put the items in the desired order so that the records are re-arranged for making searching easier. Below is one by one description of when to use which sorting algorithm for better performance:
## 1. Selection Sort
This sorting algorithm sorts an array by repeatedly finding the minimum element (considering ascending order) from the unsorted part and putting it at the beginning. The algorithm maintains two subarrays in a given array, the subarray which is already sorted, and the remaining subarray which is unsorted. In every iteration of the selection sort, the minimum element (considering ascending order) from the unsorted subarray is picked and moved to the sorted subarray.
We can use Selection Sort as per the below constraints:
• When the list is small. As the time complexity of the selection sort is O(N2) which makes it inefficient for a large list.
• When memory space is limited because it makes the minimum possible number of swaps during sorting.
## 2. Bubble Sort
This sorting algorithm is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. If we have total N elements, then we need to repeat the above process for N-1 times.
We can use Bubble Sort as per the below constraints:
• It works well with large datasets where the items are almost sorted because it takes only one iteration to detect whether the list is sorted or not. But if the list is unsorted to a large extend then this algorithm holds good for small datasets or lists.
• This algorithm is fastest on an extremely small or nearly sorted set of data.
## 3. Insertion Sort
This sorting algorithm is a simple sorting algorithm that works the way we sort playing cards in our hands. It places an unsorted element at its suitable place in each iteration.
We can use Insertion Sort as per the below constraints:
• If the data is nearly sorted or when the list is small as it has a complexity of O(N2) and if the list is sorted a minimum number of elements will slide over to insert the element at its correct location.
• This algorithm is stable and it has fast running case when the list is nearly sorted.
• The usage of memory is a constraint as it has space complexity of O(1).
## 4. Merge Sort
This sorting algorithm is based on the Divide and Conquer algorithm. It divides the input array into two halves, calls itself for the two halves, and then merges the two sorted halves. The merge() function is used for merging two halves. The merge(arr, l, m, r) is a key process that assumes that arr[l . . . m] and arr[m+1 . . . r] are sorted and merges the two sorted sub-arrays into one.
We can use Merge Sort as per the below constraints:
• Merge sort is used when the data structure doesn’t support random access since it works with pure sequential access that is forward iterators, rather than random access iterators.
• It is widely used for external sorting, where random access can be very, very expensive compared to sequential access.
• It is used where it is known that the data is similar data.
• Merge sort is fast in the case of a linked list.
• It is used in the case of a linked list as in a linked list for accessing any data at some index we need to traverse from the head to that index and merge sort accesses data sequentially and the need of random access is low.
• The main advantage of the merge sort is its stability, the elements compared equally retain their original order.
## 5. Quick Sort
This sorting algorithm is also based on the Divide and Conquer algorithm. It picks an element as a pivot and partitions the given list around the picked pivot. After partitioning the list on the basis of the pivot element, the Quick is again applied recursively to two sublists i.e., the sublist to the left of the pivot element and the sublist to the right of the pivot element.
We can use Quick Sort as per the below constraints:
• Quick sort is the fastest, but it is not always O(N*log N), as there are worst cases where it becomes O(N2).
• Quicksort is probably more effective for datasets that fit in memory. For larger data sets it proves to be inefficient so algorithms like merge sort are preferred in that case.
• Quick Sort is an in-place sort (i.e. it doesn’t require any extra storage) so it is appropriate to use it for arrays.
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Next | 965 | 4,535 | {"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} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.91351 |
https://artofmemory.com/blog/an-update-for-the-san-francisco-bay-memory-club/ | 1,632,606,941,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00650.warc.gz | 167,095,838 | 26,154 | # An Update for the San Francisco Bay Memory Club
This blog post is an update for new members of the SF Bay Area Memory Club that I run every weekend at locations around the Bay. I thought that it might be helpful to have a page of information that I could send new members to.
## Where to Begin with Memory Techniques
Two techniques that everyone should know are the memory palace technique and a number memorization system.
### Memory Palaces
Memory palaces (a.k.a., the Method of Loci) help one remember large amounts of information in a fixed sequence. This technique is good for remembering shopping lists, numbers, playing cards, history timelines, facts, or anything else that you want to recite back in a certain order.
Over time you will build up a collection of memory palaces in your mind where you can store various things. One memory palace might be reserved for the US Presidents, another palace might be used just for memorizing numbers, and so on.
### Number Memorization
Every memorizer has a set of fixed images that represent numbers. Visual images are much easier to remember than abstract numbers.
Every memorizer has their own set of images, but here’s a quick example: a piano has 88 keys, so the number 88 could be represented by the visual image of a piano.
You could store that image of a piano in a location of your memory palace to remember 88. Other numbers would be represented by other images in different locations of the memory palace. Since you always travel through a memory palace in the same order, very long strings of numbers can be remembered this way.
Example: A piano has 88 keys, so the number 88 could be represented by a piano.
The most common number systems involve creating 100 visual images for all the numbers from 00 to 99. There are several number systems out there, and if you aren’t sure of what system you want to use, try the Dominic System. It’s relatively easy to learn, and it’s powerful enough to use in memory competitions. (Its creator won the World Memory Championships eight times.) In the Dominic System, every digit is turned into a letter and then the letters become initials for persons.
So if the digit 5 is represented by the letter E and the digit 3 is represented by the letter C, then 53 is an image of someone with the initials E.C. – Eric Clapton is a common choice.
For those of you who have already started creating your Dominic System images, check out this list of persons that can help you finish creating the system.
Here’s a video that provides a quick introduction to some of the common number memorization techniques:
Memorizers also frequently have mnemonic images for single digit numbers from 0 to 9. Having these extra single digit images allows you to easily memorize numbers of uneven length, like 3 digit or 5 digit numbers. A good method for single digits is the Number Shape System.
Once you have mnemonic images for all 100 numbers from 00 to 99, you are ready to memorize numbers of virtually any length.
## Step by Step Plan
If you aren’t sure where to begin, here is a suggestion for a step by step plan that will introduce techniques that you can use both in memory competitions and in daily life:
1. Create 100 mnemonic images for numbers 00 to 99 with the Dominic System. Later, you’ll be able to use the same images to memorize decks of playing cards.
2. Create 10 mnemonic images for numbers 0 to 9 with the Number Shape System.
3. Create at least three memory palaces with at least 30 locations each.
4. Watch the short videos on the Names & Faces wiki page to get an idea about how the technique is done.
If you have a question about any of those things, leave a comment below, ask me at one of our meetings, post in the forum, or we could set up a time during the week to go over your questions via webcam.
## The Benefits of Memory Techniques
A question sometimes comes up: “What can you use memory techniques for?”
Here are some ways to apply the techniques:
• Remembering names and faces
• Learning new facts for business or school
• Remember shopping lists and directions
• Increasing creativity
• General brain training
• Memorizing speeches
• Learning new languages
• Preparing for business meetings or job interviews
• Becoming a mental athlete or even a memory champion…
If you’re looking for more information between our meetings, there are over 1,000 pages of free content on this website. Check out the Getting Started Guide. Head over to the forum and type some things into the search box or start a new discussion there.
Also, I highly recommend two books:
• Moonwalking with Einstein by Joshua Foer — this book provides background on the world of memory techniques.
• Quantum Memory Power by Dominic O’Brien — this audiobook is a great place to get an overview of different ways to apply memory techniques. | 1,024 | 4,852 | {"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-2021-39 | latest | en | 0.939501 |
http://research.stlouisfed.org/fred2/series/PPPGDPBZA619NUPN?rid=258 | 1,433,358,780,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036845.7/warc/CC-MAIN-20150601214356-00022-ip-10-180-206-219.ec2.internal.warc.gz | 150,930,069 | 20,584 | # Total Purchasing Power Parity Converted GDP, G-K method, at current prices for Belize
2010: 3,278.72851 Millions of International Dollars
Annual, Not Seasonally Adjusted, PPPGDPBZA619NUPN, Updated: 2012-09-17 11:29 AM CDT
1yr | 5yr | 10yr | Max
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: tcgdp
Source: University of Pennsylvania
Release: Penn World Table 7.1
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(a) Total Purchasing Power Parity Converted GDP, G-K method, at current prices for Belize, Millions of International Dollars, Not Seasonally Adjusted (PPPGDPBZA619NUPN)
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Create your own data transformation: [+]
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` University of Pennsylvania, Total Purchasing Power Parity Converted GDP, G-K method, at current prices for Belize [PPPGDPBZA619NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/PPPGDPBZA619NUPN/, June 3, 2015. ```
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Graph updated. | 488 | 1,804 | {"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} | 2.625 | 3 | CC-MAIN-2015-22 | latest | en | 0.765322 |
https://anglesandarguments.com/blog-details/176 | 1,721,147,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00313.warc.gz | 92,496,825 | 8,433 | “Fresh sale of individual modules has been closed. Visit anaprep.com for information on new products and write to us for special offers”
## Algebra - Unsolvable Equations
The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.
In this example, it was relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.
We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.
But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:
Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0
In this problem, x can be any real number – we have no constraints on it. Now, is x negative?
Statement 1: x^3 + x^2 + x + 2 = 0
If we try to solve this equation as we are used to doing, look at what happens:
If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.
Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.
Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.
This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.
Statement 2: x^2 – x – 2 < 0
This, we can easily solve:
x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0
We know how to solve this inequality (as discussed in our Inequality module here)
This will give us -1 < x < 2.
Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.
To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic. | 966 | 3,403 | {"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-2024-30 | latest | en | 0.951859 |
https://www.examrace.com/Sample-Objective-Questions/Statistics-Questions/Numerical-Descriptive-Statistics-And-Sampling-Part-5.html | 1,660,187,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00138.warc.gz | 669,387,081 | 6,958 | # Statistics MCQs-Numerical Descriptive Statistics and Sampling Part 5
Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.
81. A statistician wishes to examine the relationship between average monthly rainfall (in mm) , x, and number of road accidents, y, in a particular city. The following calculations have been done for you: ∑ x = 265, ∑ x2 = 6982, ∑ y = 193, ∑ y2 = 3421, ∑ xy = 4842 and n = 12. The coefficient of correlation between average monthly rainfall and number of road accidents is:
a. 0.899
b. 0.969
c. 0.974
d. 0.977
e. 1.000
Answer: B
82. A statistician wishes to examine the relationship between average monthly rainfall (in mm) , x, and number of road accidents, y, in a particular city. The following calculations have been done for you: ∑ x = 276, ∑ x2 = 6888, ∑ y = 193, ∑ y2 = 3421, ∑ xy = 4842 and n = 12. The coefficient of correlation between average monthly rainfall and number of road accidents is:
a. 0.899
b. 0.969
c. 0.974
d. 0.977
e. 1.000
Answer: C
83. A statistician wishes to examine the relationship between average monthly rainfall (in mm) , x, and number of road accidents, y, in a particular city. The following calculations have been done for you: ∑ x = 276, ∑ x2 = 6982, ∑ y = 185, ∑ y2 = 3421, ∑ xy = 4842 and n = 12. The coefficient of correlation between average monthly rainfall and number of road accidents is:
a. 0.899
b. 0.969
c. 0.974
d. 0.977
e. 1.000
Answer: D
84. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 7. The coefficient of correlation between x and y is given by:
⚹ a. 0.998
b. 0.999
c. 1.000
d. 0.945
e. 0.922
85. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 8. The coefficient of correlation between x and y is given by:
a. 0.998
b. 0.999
c. 1.000
d. 0.945
e. 0.922
Answer: B
86. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 9. The coefficient of correlation between x and y is given by:
a. 0.998
b. 0.999
c. 1.000
d. 0.945
e. 0.922
Answer: B
87. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 10. The coefficient of correlation between x and y is given by:
a. 0.998
b. 0.999
c. 1.000
d. 0.945
e. 0.922
Answer: C
88. You have been given a two-variable data set with the following calculations already completed: ∑ x = 72, ∑ x2 = 1544, ∑ y = 37, ∑ y2 = 305, ∑ xy = 654 and n = 5. The coefficient of correlation between x and y is given by:
a. 0.963
b. 0.933
c. 0.941
d. 0.945
e. 0.922
Answer: A
89. You have been given a two-variable data set with the following calculations already completed: ∑ x = 72, ∑ x2 = 1544, ∑ y = 37, ∑ y2 = 305, ∑ xy = 654 and n = 10. The coefficient of correlation between x and y is given by:
a. 0.963
b. 0.933
c. 0.941
d. 0.945
e. 0.922
Answer: B
90. You have been given a two-variable data set with the following calculations already completed: ∑ x = 72, ∑ x2 = 1544, ∑ y = 37, ∑ y2 = 305, ∑ xy = 654 and n = 15. The coefficient of correlation between x and y is given by:
a. 0.963
b. 0.933
c. 0.941
d. 0.945
e. 0.922
Answer: C
91. You have been given a two-variable data set with the following calculations already completed: ∑ x = 72, ∑ x2 = 1544, ∑ y = 37, ∑ y2 = 305, ∑ xy = 654 and n = 20. The coefficient of correlation between x and y is given by:
a. 0.963
b. 0.933
c. 0.941
d. 0.945
e. 0.922
Answer: D
92. The marketing manager of a large supermarket chain would like to determine the effect of shelf space (X) on the sales (Y) of pet food. A random sample of 12 equal-sized stores is selected. Some calculations have been done for you: SSx = 375, SSy = 3.0025 and SSxy = 27.75. What is the value of the covariance?
a. 2.52
b. 2.55
c. 2.64
d. 2.73
e. 2.31
Answer: A
93. The marketing manager of a large supermarket chain would like to determine the effect of shelf space (X) on the sales (Y) of pet food. A random sample of 12 equal-sized stores is selected. Some calculations have been done for you: SSx = 400, SSy = 3.0025 and SSxy = 28. What is the value of the covariance?
a. 2.52
b. 2.55
c. 2.64
d. 2.73
e. 2.31
Answer: B
94. The marketing manager of a large supermarket chain would like to determine the effect of shelf space (X) on the sales (Y) of pet food. A random sample of 12 equal-sized stores is selected. Some calculations have been done for you: SSx = 375, SSy = 3.5 and SSxy = 29. What is the value of the covariance?
a. 2.52
b. 2.55
c. 2.64
d. 2.73
e. 2.31
Answer: C
95. The marketing manager of a large supermarket chain would like to determine the effect of shelf space (X) on the sales (Y) of pet food. A random sample of 12 equal-sized stores is selected. Some calculations have been done for you: SSx = 375, SSy = 3.0025 and SSxy = 30. What is the value of the covariance?
a. 2.52
b. 2.55
c. 2.64
d. 2.73
e. 2.31
Answer: D
96. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 7. The covariance between x and y is given by:
a. 385.60
b. 973.29
c. 1289.07
d. 1456.91
e. 476.20
Answer: A
97. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 8. The covariance between x and y is given by:
a. 385.60
b. 973.29
c. 1289.07
d. 1456.91
e. 476.20
Answer: B
98. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 9. The covariance between x and y is given by:
a. 385.60
b. 973.29
c. 1289.07
d. 1456.91
e. 476.20
Answer: C
99. You have been given a two-variable data set with the following calculations already completed: ∑ x = 496, ∑ x2 = 37306, ∑ y = 508, ∑ y2 = 39352, ∑ xy = 38309 and n = 10. The covariance between x and y is given by:
a. 385.60
b. 973.29
c. 1289.07
d. 1456.91
e. 476.20
Answer: D
100. You have been given a two-variable data set with the following calculations already completed: ∑ x = 72, ∑ x2 = 1544, ∑ y = 37, ∑ y2 = 305, ∑ xy = 654 and n = 5. The covariance between x and y is given by:
a. 30.3
b. 43.1
c. 34.0
d. 27.4
e. 31.9
Answer: A | 2,426 | 6,748 | {"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} | 4.03125 | 4 | CC-MAIN-2022-33 | latest | en | 0.901749 |
https://de.mathworks.com/matlabcentral/answers/468511-create-block-sparse-matrix | 1,576,103,184,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00315.warc.gz | 328,436,797 | 25,976 | ## Create block sparse matrix
### KSMAT21 (view profile)
on 23 Jun 2019
Latest activity Commented on by John D'Errico
### John D'Errico (view profile)
on 26 Jun 2019
How can I create a block sparse matrix in MATLAB using the 'sparse' funcion, such as S = sparse(I,J,Z)?
Instead of Z being a vector of non-zero entries, Z contains dense matrix blocks (e.g. 100*100). The size of the full matrix S can be very large, e.g 1 million * 1 million.
Thank you.
Show 1 older comment
James Tursa
### James Tursa (view profile)
on 24 Jun 2019
How do you have the dense blocks currently stored? In a cell array? How do you have the desired location in the sparse matrix stored? As row&column of the upper left element of each block? Or ...? Will there be the possibility of overlap? Do the blocks always line up nicely like your picture, or could they be anywhere (e.g. half under one block and half under another block)?
KSMAT21
### KSMAT21 (view profile)
on 24 Jun 2019
Hi James,
Thanks for your response. Yes, the dense blocks are currently stored in cell arrays. There will no be overlaps among blocks.
The blocks results from a domain decomposition problem, where each subdomain has the same number of elements (100 in the figure above). The diagonal blocks represent self-interaction, and off-diagonal blocks represent mutual interaction. Thus, the blocks will always be lined up nicely, since they start at multiples of 100 (number of elements per subdomain) for both row and columns.
I know the desired locations of the block by knowing which subdomains in the problem are interacting, thus can say that I know the upper left element of each block.
Thank you.
James Tursa
### James Tursa (view profile)
on 26 Jun 2019
How is the subdomain information stored? I think this could be done in a mex routine without sorting or copying the data more than once as long as the subdomain information is stored in a reasonable manner. The problem I see with the methods already posted is that they will require copying the data more than once and sorting the data as well, which will take extra time.
### David Goodmanson (view profile)
on 24 Jun 2019
Edited by David Goodmanson
### David Goodmanson (view profile)
on 26 Jun 2019
HI ks,
Here is one way.
showing two possible methods to fill the matrix by means of a for loop appoach. The second way is significantly faster, but both methods change the sparse matrix S as many times as you have blocks. For N blocks the overall time for the process appears to scale like N^2, so it's all right for a few hundred blocks. But when the numbers get large you will need to gin up some large index vectors and call sparse at most just a few times, preferably once as John has indicated.
The following may be useful.
From the [...] = ndgrid line in the code below, irow(:) is the row indices of the block itself (zero based). Then if m0vec is a column vector of N row indices for the upper left corner of the N blocks, and if you have a new enough version of Matlab
irowfull = irow(:) + m0vec';
irowfull = irowfull(:);
has the properties you want, similarly for column indices.
% create a matrix with known elements, not square
A = magic(4);
A(:,4) = [];
[Arow Acol] = size(A)
[irow icol] = ndgrid((0:Arow-1),(0:Acol-1)) % block indices
s_row = 10; s_col = 10; n_s = 100; % small example
S = spalloc(s_row, s_col,n_s)
m0 = 3; n0 = 5; % these are the row and column indices
% of the upper left corner of the block
% for loop here to index m0,n0, and A
% first way
S = S + sparse(m0+irow(:),n0+icol(:),A(:),s_row,s_col)
% second way, creating new sparse matrix not required
S(m0:m0+Arow-1,n0:n0+Acol-1) = A
% end
full(S)
A
Show 1 older comment
David Goodmanson
### David Goodmanson (view profile)
on 24 Jun 2019
Hi ks
If you only have a few thousand blocks it makes sense to just use a for loop for each one rather than something more complicated.
I modified the example code by using m0 for the row index and n0 for the column index, and by using ndgrid instead of meshgrid. Ndgrid is the more natural choice here. Compared to ndgrid, the two outputs of meshgrid are swapped which is good for plotting purposes but not as good for index calculations.
It appears that at lot of your blocks have the same shape. Clearly ndgrid only has to be done once for all of those, and you can then do a for loop on the m0 and n0 indices. (m0 and n0 could be columns in an Nx2 array, addressed by the array row index). The sample code below takes m0 and n0 out of the meshgrid (now ndgrid) calculation.
Then in the for loop you have to address the different block matrices themselves, which will of course be much easier if a lot of them are identical.
A = magic(4);
A(:,4) = [];
[Arow Acol] = size(A)
[irow icol] = ndgrid((0:Arow-1),(0:Acol-1)) % block indices
m0 = 3; n0 = 5; % these are the row and column indices
% of the upper left corner of the block
% for loop here to index m0,n0, and A
S = sparse(m0+irow(:),n0+icol(:),A(:))
% end
full(S)
A
KSMAT21
### KSMAT21 (view profile)
on 24 Jun 2019
Thanks a lot, David.
I am correct to guess that the sparse matrix must be added to the previous iterations in the loop to create multiple blocks. Editing from your above code:
n = 1e4;
S = spalloc(n,n,3*n)
%for loop here to index m0,n0, and A
S = S + sparse(m0+irow(:),n0+icol(:),A(:),n,n)
% end
Is that correct?
Because, if not, each iteration in the for loop will clear the previous entries of the sparse matrix.
Regards,
KS
David Goodmanson
### David Goodmanson (view profile)
on 24 Jun 2019
Hi KS,
That works, but gratifyingly enough, after (defining S with spalloc) so does
S(m0:m0+Arow-1,n0:n0+Acol-1) = A;
which obviates having to use ndgrid and also obviates having to create a new (albeit small) sparse matrix each time through the loop. I modified the original answer but I don't know yet how this would do speedwise.
### John D'Errico (view profile)
on 24 Jun 2019
Edited by John D'Errico
### John D'Errico (view profile)
on 24 Jun 2019
Are you asking how to create a block diagonal sparse matrix? Easy peasy.
tic
N = 100;
M = 10000;
Z = sparse(rand(N,N*M));
Zc = mat2cell(Z,N,repmat(N,1,M));
A = blkdiag(Zc{:});
toc
Elapsed time is 6.487005 seconds.
whos A
Name Size Bytes Class Attributes
A 1000000x1000000 1608000008 double sparse
So A is a block diagonal sparse matrix, of size 1e6x1e6, with 100x100 blocks on the diagonal, 10,000 such blocks. 6 seconds seems reasonable to build it, since almost 50% of that time was just in creating the original random matrix Z.
tic,Z = sparse(rand(N,N*M));toc
Elapsed time is 2.936146 seconds.
spy(A)
John D'Errico
### John D'Errico (view profile)
on 25 Jun 2019
This really is far more trivial than you may think, BUT you ABSOLUTELY need to learn to use sparse.
For example, a 3(5)x3(5) block matrix, with 3 blocks.
blocksize = 5;
nblocks = 3;
% The elements contained in the various blocks.
% If each block is the same, it is easy to build.
% I've just used random elements.
Z = rand(blocksize,blocksize,nblocks);
[subr,subc] = meshgrid(1:blocksize);
% The pattern matrix:
blockrc = [1 2;2 1;3 3];
rowind = subr + reshape((blockrc(:,1)-1)*blocksize,[1 1 nb]);
colind = subc + reshape((blockrc(:,2)-1)*blocksize,[1 1 nb]);
% create A in one call to sparse
A = sparse(rowind(:),colind(:),Z(:),blocksize*max(blockrc(:,1)),blocksize*max(blockrc(:,2)));
whos A
Name Size Bytes Class Attributes
A 15x15 1328 double sparse
spy(A)
So A is 15x15 sparse, with 3 blocks of size 5x5 each. Change the code to match your problem, in that all you need to do is set up your own pattern matrix, and assign the matrix Z with the elemnts that it contains. But see how trivial it really is to do, in ONE call to sparse. Again, learn to use sparse, and your code will be easy to write.
KSMAT21
### KSMAT21 (view profile)
on 26 Jun 2019
Thanks, this works great. Only thing, I used ndgrid instead of meshgrid as explained in one of David's comment.
The sparse function took 5.97 seconds created a sparse matrix of 60 millions non-zero entries. Is the time taken resonable for this size of matrix?
John D'Errico
### John D'Errico (view profile)
on 26 Jun 2019
That time taken is very reasonable, to create a sparse matrix with 60 million entries. Without doing a test right now, the sparse block diagonal matrix I created in my answer had 100 million entries, taking also about 6 seconds to create. So I would not be at all surprised at your result. So 5 seconds or so seems reasonable. | 2,341 | 8,465 | {"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-2019-51 | latest | en | 0.91446 |
https://www.imlearningmath.com/a-7-percent-bond-has-a-yield-to-maturity-of-6-5-percent/ | 1,618,199,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066568.16/warc/CC-MAIN-20210412023359-20210412053359-00209.warc.gz | 943,823,894 | 24,250 | # A 7 percent bond has a yield to maturity of 6.5 percent.
A 7 percent bond has a yield to maturity of 6.5 percent. The bond matures in seven years, has a face value of $1,000, and pays semiannual interest payments. What is the amount of each coupon payment? A.$30.00
B. $35.00 C.$60.00
D. $65.00 Answer: The correct answer is$35.00. | 104 | 336 | {"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} | 2.953125 | 3 | CC-MAIN-2021-17 | latest | en | 0.917071 |
https://people.revoledu.com/kardi/tutorial/LinearAlgebra/VectorCosAngle.html | 1,726,820,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00770.warc.gz | 408,641,091 | 5,184 | ## Cosine Angle between Two Vectors
Cosine angle between two vectors is equal to their dot product divided by the product of their norms .
Example:
Suppose we have and , the vector inner product is
Their norms are and .
Therefore, the cosine angle is and the angle is .
The interactive program below produces cosine angle between two vectors of the same dimension and also specify the angle in both radians and degrees. Random Example button will generate random vectors at the right format.
vector x vector y
## Properties
Some important properties of related to cosine angle are
• Two vectors are orthogonal if their dot product is zero , which means the cosine angle is zero
• Two vectors are parallel if the absolute value of their dot product is equal to the product of their norms , which means the absolute cosine angle is one .
• Two vectors are in the same direction if their dot product is equal to the product of their norms , which mean the cosine angle is exactly one .
• Since the value of Cosine is between -1 and +1, the absolute value of vector dot product is always less than or equal to the product of their norms . This is called Cauchy-Schwartz inequality. | 242 | 1,185 | {"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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.953213 |
https://answercult.com/question/what-do-you-mean-by-economies-scale/ | 1,652,900,732,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00576.warc.gz | 152,570,281 | 17,999 | # What do you mean by economies scale?
956 views
I’ve read that term mainly while doing cost estimation for wind/soar projects. Confused what does that exactly mean.
In: Other
Economies of scale means the more you buy the cheaper each unit becomes. One unit may cost one dollar, but five units may be only cost four dollars
I believe it refers to how businesses spend less money on their production because they get bigger. so as the business grows, their average costs go down because of things such as bulk buying, specialization, marketing economies etc.
Basically what this means is that companies can achieve a lower average cost (cost per unit of product) by buying many things at one time. For example, if a company buys 100000 metal screws, they are able to get each screw at a lower price than just buying 1 screw individually.
There are many types of economies of scale, but the general idea is similar to the one I wrote above. Can refer to this for a more detailed explanation: [https://www.investopedia.com/terms/e/economiesofscale.asp](https://www.investopedia.com/terms/e/economiesofscale.asp)
The other answers have it right, but aren’t giving very concrete examples.
The more you are going to use something, the more you can spend on it without changing the cost per use.
If I want to make a LEGO, I can use my 3D printer and get a pretty good LEGO piece. It will take about 3 hours, and then when it gets done I need to treat the piece to remove the weakness of being built with stacked layers of plastic, or it will shred apart when I try to get it unstuck from whatever I build. Plus to get it to fit as well as a normal LEGO, I would need my printer to be VERY well configured. This can easily take a few days of work on older printers.
To make a LEGO without a 3D printer would likely take even longer, and be far harder to get the precision I need with hand tools.
So, how are LEGO pieces made?
The company builds metal molds to inject plastic into. Then the pieces are measured to ensure they are EXACTLY the right size in every location along the part. If they are not perfect, the metal mold needs replaced/repaired. These molds cost many thousands of dollars (my memory wants to say a few million, but I find that hard to believe).
Let us claim the mold costs \$20,000 and the machine to use the mold costs \$150,000. The mold can make about 20 million parts before it has to be replaced, and the machine can last as long as 1,000 molds. The plastic used for one piece costs 0.3 cents
If I want to make one piece, using professional gear, it will cost me \$170,000.003 per piece. I definitely want to hire a company to do this for me on their equipment, meaning I only spend on the mold, getting the part for \$20,000.003. Again, makes a clear easy choice to hire someone to do it by hand, paying them for at most 3 days of work, so maybe \$1,500 there.
If I make as many pieces as the life of the machine, it will cost me 0.0040075 dollars per piece.
So… one off the major cost is production technique. But at scale, the major cost is plastic. The equipment I use to make the parts quickly, accurately, and easily are nearly ignored at scale. | 733 | 3,187 | {"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-2022-21 | latest | en | 0.955724 |
https://whatarethekargrandes.com/comics-characters/often-asked-how-many-pounds-of-pressure-can-a-python-exert.html | 1,679,835,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00077.warc.gz | 691,360,660 | 8,917 | # Often asked: How many pounds of pressure can a python exert?
## How much pressure can a snake squeeze?
On average, king snakes squeeze with a pressure of 180 mm Hg, higher than the upper range of healthy blood pressure in a human, which is 120 mm Hg. “That means if you were to encounter the pressures these king snakes exert, your heart would fail to pump blood—that’s how strong this is,” Penning says.
## Can a python squeeze a human to death?
Like all pythons, it is a non-venomous constrictor. Adult humans have been killed (and in at least two reported cases, eaten) by reticulated pythons.
Reticulated python.
Reticulated python Temporal range: Pleistocene to recent
Family: Pythonidae
Genus: Malayopython
Species: M. reticulatus
Binomial name
## How many pounds of pressure can an anaconda squeeze?
Prior studies determined that the anaconda’s possible crush force is 90 pounds per square inch. That’s comparable to an elephant sitting on your chest. We used tug-boat ropes to test the suit beforehand between trucks. It withstood squeezing at 90 psi.
## How powerful is a python snake?
According to the pressure reading, the snake generated almost 300 millimeters of mercury or just under 6 pounds of pressure per square inch (psi) during constriction. “There’s a myth that they squeeze their prey until they can’t breathe anymore, but that’s not accurate,” said Dr.
## Can a human overpower a python?
in general, the average person would not be successful uncoiling a >10′ snake that was constricting them, while a 7′ snake might be a bit of a wrestling match but ultimately the snake will lose, and a <5′ snake can easily be uncoiled. This is a good answer to my question.
## How do you survive being crushed by a python?
How To Survive a Big Python Bite and Full-on Attack
1. Python bites you.
2. Pull out a knife as fast as possible.
3. Keep the python from wrapping around you at all costs.
4. When possible – stab the snake in the side or belly repeatedly and many different places.
5. Hope the snake lets go.
## Has a Ball python ever killed a child?
There are no recorded incidents of a ball python killing a human of any age.
## Do python bites hurt?
Does a ball python bite hurt? You will probably feel the effects of a python bite because it can cause scratches, puncture wounds, bruising, and even possibly deeper internal damage. These bites may be painful during the bite and as your injuries heal.
## Do snakes fart?
And Rabaiotti did find that fart answer for her brother: yes, snakes fart, too. Sonoran Coral Snakes that live across the Southwestern United States and Mexico use their farts as a defense mechanism, sucking air into their “butt” (it’s actually called a cloaca) and then pushing it back out to keep predators away.
## Can boa constrictors kill humans?
Boa constrictors strike when they perceive a threat. Their bite can be painful, especially from large snakes, but is rarely dangerous to humans. Specimens from Central America are more irascible, hissing loudly and striking repeatedly when disturbed, while those from South America tame down more readily.
## What is the biggest snake in the world?
What is the biggest snake in the world?
• The largest snakes in the world belong to the python and boa families.
• The reticulated python (Malayopython reticulatus) is the longest snake in the world, regularly reaching over 6.25 metres in length.
We recommend reading: Often asked: How much vitamin c can i take at once?
## How hard can a reticulated python squeeze?
Reticulated pythons can exert 7.8 pounds of pressure per square inch (PSI). Pythons can detect their prey’s heartbeat, and they stop squeezing when they know their victim is dead. Large species of pythons can easily squeeze hard enough to kill a fully-grown human.
## Can you outrun a snake?
Most snakes have trouble moving 1 mph. It can go up to 12 mph. An exceptional human runner can achieve about 20 mph, and an average human can achieve about 15 mph. Additiinally, the average person can easily outrun the average snake.
## How do you euthanize a python?
A video taken includes the hunter explaining how he shot the snake in the head and later in the neck. PETA said the only humane way to euthanize a python is with a “penetrating captive-bolt gun or gunshot to the brain.
## Can you eat python meat?
When pythons are safe to eat, they can actually be quite delicious, says Donna Kalil, one of the program’s python hunters. When she catches smaller ones, about 7 feet long, she uses a mercury testing kit she bought online to confirm they‘re safe to eat. She then turns their white meat into food. | 1,068 | 4,649 | {"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-2023-14 | latest | en | 0.924871 |
https://www.plati.ru/itm/solution-v11-d6-reshebnik-termehu-targ-sm-1983/2087729?lang=en-US | 1,519,144,231,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812978.31/warc/CC-MAIN-20180220145713-20180220165713-00126.warc.gz | 939,169,160 | 12,946 | # Solution V11 D6, reshebnik termehu Targ SM 1983
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# Description
Solution of the D6 version 11 of Reshebnik Targ SM 1983 - Course of Theoretical Mechanics.
CONDITION TO THE PROBLEM D6 (pg. 50-52 in the training manual Targ SM 1983.):
The mechanism located in a horizontal plane, is in equilibrium under the action of applied forces: the equilibrium position is determined by the angles α, β, γ, φ, θ (Figure D6.0 - D6.9, Table D6..).
Long rod mechanism (crank or rocker) are: l1 = 0,4 m, l4 = 1,0 m (dimensions l2 and l3 are arbitrary); Point E in Fig. 2 - 6 is in the middle of the respective rod.
On the mechanism of action: in Fig. 0-2 - the force F, applied to the D slider, and a pair of forces with the moment M applied to the shaft 1; Fig. 3-9 - the couple of forces with the moments M1 and M2 applied to terminals 1 and 4.
The system is balanced by the power of the Q parallel Bb, attached to the slider B. Determine what is and in what direction the force is directed.
# Additional information
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In order to counter copyright infringement and property rights, we ask you to immediately inform us at support@plati.market the fact of such violations and to provide us with reliable information confirming your copyrights or rights of ownership. Email must contain your contact information (name, phone number, etc.) | 680 | 2,681 | {"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-2018-09 | longest | en | 0.865394 |
http://civilservicereview.com/tag/civil-service-exam/page/2/ | 1,575,669,163,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00021.warc.gz | 31,675,886 | 14,698 | ## 2018 PCSR Civil Service Review Guide 3
2018 PCSR Civil Service Review Guide 3
Week 3: October 9 – October 15, 2017
PART I: MATHEMATICS
A. Subtraction of Fractions
Lesson 1: Subtraction of Similar Fractions
More videos
Tutorial Articles
Link 1: How to Subtract Fractions
PART II: ENGLISH
C. Spelling
Part III: CLERICAL (Subprofessional)
Alphabetizing Rules Part 3
Rule 3 Alphabetizing Exercise
Alphabetizing Quiz 3
PART IV: TIPS AND TRICKS
8 Tips on How to Pass Civil Service Examination Next Time
MORE REVIEWERS
1. Math English Tutorials and more at the PCSR Civil Service Review website:
civilservicereview.com
2. More than 700 Taglish Math Tutorials at the Sipnayan Youtube channel:
3. Sipnayan Website
Sipnayan.com
Like our Facebook Pages and be updated:
PCSR Facebook Page: Philippine Civil Service Reviewer
## 2018 PCSR Civil Service Review Guide 2
Week 2: October 2 – October 8, 2017
PART I: MATHEMATICS
A. GCD
Lesson 4: Addition of Mixed Fractions
More tutorial Videos
Tutorial Articles
PART II: ENGLISH
B. Grammar
Simple Tenses
Lesson 1: The Simple Present Tense
Lesson 2: The Simple Past Tense
Lesson 3: The Simple Future Tense
D.
Part III: CLERICAL (Subprofessional)
PART III: TIPS AND TRICKS
5 Effective Ways to Improve Your Vocabulary
MORE REVIEWERS
1. Math English Tutorials and more at the PCSR Civil Service Review website:
http://civilservicereview.com
2. More than 700 Taglish Math Tutorials at the Sipnayan Youtube channel:
3. Sipnayan Website
http://sipnayan.com
Like our Facebook Pages and be updated:
You might also like:
## Solving Ratio and Proportion Problems Part 1
One of the key concepts tested in the Civil Service Exam is ratio and proportion. In this series, we are going to discuss how to solve problems involving ratio and proportion. We first begin below by explaining the meaning and concept of ratio and how to represent it.
Suppose we are cooking, and for every 4 teaspoons of vinegar, we put 3 teaspoons of soy sauce, then we can say the ratio of the volume of vinegar to the volume of soy sauce is “four is to three” and represent it as 4:3. We can also use the fraction 4/3 to represent the ratio above. Now, we discuss more examples about ratio.
Example 1
In a class, there are 24 girls and 18 boys. What is the ratio of (1) the number of girls to the number of boys and (2) the number of boys to the number of girls?
The number of girls is 24 and the number of boys is 18, so the ratio of the number of girls to the number of boys is 24:18 or 24/18. In contrast, the ratio of the number of boys to the number of girls is 18:24 or 18/24.
Example 2
In a box of colored balls, there are 5 red balls and 8 blue balls. What is the ratio of the number of blue balls to the total number of balls?
The number of blue balls is 8 and the total number of balls is 5 + 8 = 13. Therefore the ratio of the number of blue balls to the total number of balls is 8:13 or 8/13.
Example 3
Gemma put 2 teaspoons of sugar for every cup of coffee. Represent the ratio of the number of teaspoons of sugar if there are 6 cups of coffee.
For every cup of coffee, we need 2 teaspoons. Therefore, for 6 cups of coffee, we need 6 times 2 = 12 teaspoons. So, the ratio of the number of teaspoons and 6 cups of coffee is 12:6.
In the three examples above, we have learned how to represent ratio. The ratio A: B means how many times of B is A. For example, the ratio 4:3 means A is 4/3 times of B.
In the next post, we are going to discuss about proportion or equal ratio.
## FREE Civil Service Examination Reviewers
It’s just one day before the examination, so good luck. The last day before the examination is very crucial, so I suggest that you don’t over review, relax a bit in the evening, and sleep early. I have written 5 Things to Do A Day Before the Exam for the details.
For those who want to review more, you may want take a peek at more than 400 pages to access our free civil service examination reviewers.
1.) Math Word Problems contains tutorials about age, number, mixture, motion, work problems and other word problems in Algebra.
2.) Mathematics – contains numerical ability tutorials such as fractions, decimals, percents, equations, GCF, LCM,
3.) English – contains tutorials on grammar, vocabulary, analogy, paragraph organization, and more.
4.) Practice Test – contains practice tests and exercises on different topics.
5.) Post List – contains the complete list of more than 300 civil service review materials.
6.) Videos – contains video Taglish Math Video tutorials.
Again, good luck to all examinees.
## Congratulations Civil Service Exam Passers!
I would like to congratulate the member of the PH Civil Service Review Facebook Group who passed the October 26, 2014. As of this writing, the group has 2,811 members and continues to grow everyday. The group has probably about 100 passers, but I only listed those whom I got permission to include their names.
Professional
• Abucejo, Romel A.
• Alcantara, Missy B.
• Aldeza, John Paul D.
• Atienza, Daniel R.
• Carongoy, Aiko
• Castillo, Francis Anthony E.
• Dacallos, Girlie F.
• Garbida, Alger James D
• Ibis, Regio O.
• Pimentel, Cojee
• Reyes, Debbie Joyce A.
• Rozario, Zaisa C.
• Santarin, Thessa Marie C.
Sub-professional
• Esturas, James Patrick G.
• Pelaez, Manolito G.
Congratulations Civil Service Passers.
Here is the complete list of the Civil Service Exam Passers and the schedule for the May 2015 exam.
## October 2014 Civil Service Exam Passers, Region 6
OCTOBER 26, 2014 CSE-PPT PROFESSIONAL LEVEL CSCRO NO. 06
Region ExamNo Name
06 255215 ABALARAO, MAYREN JOY D
06 254289 ABARINTO, ELLEN D
06 253437 ABATA, QUEENNIE LYN J
06 250961 ABAYGAR, SAMANTHA JOY B
06 254888 ABAYON, SEALTIEL I
06 250660 ABCEDE, QUEENIE JOY C
06 252401 ABIAN, STEVEN JAY P
06 258473 ABIBIASON, AILYN MAE O
06 259979 ABLAÑA, LEO MARTIN V
06 251679 ABROT, LYNIE ROSE F
06 258600 ABUANA, FAITH JOY B
06 254239 ABUNTO, ADY FAYE C
06 259007 ABUYON, MICHAEL PETER M
06 252736 ACBANG, MACARIO C
06 250788 ACEÑA, JULIE ANN M
06 251010 ACHARON, FLORENTINA O
## October 2014 Civil Service Exam Passers, Region 5
OCTOBER 26, 2014 CSE-PPT
PROFESSIONAL LEVEL
CSCRO NO. 05
Region ExamNo Name
05 225131 AÑASCO, JOSEPH CHRISTIAN B
05 232674 ABAC, FREDELENE B
05 222865 ABACHE, FIONA MAE S
05 224088 ABAINZA, DIVINE GRACE O
05 228600 ABAN, MA COLEEN S
05 228385 ABARCA, ANN JENNETH N
05 222877 ABAYON, RALPH ROBERT G
05 222855 ABEJO, MYRA G
05 223398 ABIERA, PAULA BERNADETTE S
05 222243 ABILONG, THATLHER D
05 222015 ABITRIA, FRANK NEIL A
05 233164 ABRAGAN, JONATHAN REY B
05 230726 ACUZAR, KRISTAL MAE D
## One Month Until The Exam! How’s Your Review?
I am not sure if you keep track of your calendar, but today is exactly one month before the next Civil Service Examination. As of this writing, the Official Civil Service Calendar still shows that the exam in on October 26, 2014 and there is no sign of change of schedule so far. This means that you have four weeks to prepare for the exam. Be sure to brush up with your grammar, vocabulary, and mathematics skills. This blog has a lot of reviewer, so all you need to do is read and answer the practice exercises and quizzes. | 2,062 | 7,225 | {"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} | 4.5625 | 5 | CC-MAIN-2019-51 | longest | en | 0.790458 |
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_7&diff=prev&oldid=152506 | 1,660,688,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00279.warc.gz | 131,435,111 | 12,827 | Difference between revisions of "2019 AMC 8 Problems/Problem 7"
Problem 7
Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$, $94$, and $87$. In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$
Solution 1
We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. $$\frac{76+94+87+x+y}{5} = 81,$$ $$\frac{257+x+y}{5} = 81.$$ We can now cross multiply to get rid of the denominator. $$257+x+y = 405,$$ $$x+y = 148.$$ Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: $$x = 100,$$ $$y=48.$$ Now we know that the lowest score on the two other tests is $\boxed{48}$.
~ aopsav
Solution 2
Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ tests is $100$, the least possible score she can get is $\boxed{\textbf{(A)}\ 48}$.
Note: You can verify that $\boxed{48}$ is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.
Solution 3
We can compare each of the scores with the average of $81$: $76$ $\rightarrow$ $-5$, $94$ $\rightarrow$ $+13$, $87$ $\rightarrow$ $+6$, $100$ $\rightarrow$ $+19$;
So the last one has to be $-33$ (since all the differences have to sum to $0$), which corresponds to $81-33 = \boxed{48}$.
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=8 | 631 | 2,044 | {"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": 44, "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} | 4.78125 | 5 | CC-MAIN-2022-33 | latest | en | 0.87899 |
http://isbitcoinminingprofitable54173.diowebhost.com/19211475/how-big-coins-can-save-you-time-stress-and-money | 1,563,828,982,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528220.95/warc/CC-MAIN-20190722201122-20190722223122-00195.warc.gz | 75,634,967 | 6,492 | # What Does Trading Bitcoin Futures Mean?
In 2009, it was 50. In 2013, it was 25, at the time of writing it's 12.5, and sometime in the center of 2020 it will halve to 6.25. .
At this rate of halving, the total number of bitcoin in circulation will approach a limit of 21 million, making the currency more scarce and valuable over time but also more expensive for miners to produce.
Here's the catch. In order to get bitcoin miners to actually earn bitcoin from verifying transactions, two things have to occur. First, they need to confirm 1 megabyte (MB) value of transactions, which can technically be as small as 1 transaction but are more often a few thousand, depending on how much information each transaction shops.
# Top Guidelines Of Master Coin
Second, in order to put in a block of transactions to the blockchain, miners must fix a complex computational science difficulty, also called a"proof of labour " What they're actually doing is trying to come up with a 64-digit hexadecimal number, called a"hash," that's less than or equal to the hash.
## A Biased View of How To Trade Bitcoin
In other words, it's a bet. .
The difficulty level of the most recent block at the time of writing is about 7,184,404,942,701. In other words, the chance of a computer producing a hash beneath the goal is 1 in 7,184,404,942,701 less than 1 in seven trillion. That level is adjusted every 2016 blocks, or roughly every 2 weeks, with the goal of keeping rates of mining constant.
## The smart Trick of Is Bitcoin Mining Profitable That Nobody is Talking About
The reverse is also correct. If computational power has been taken from the network, the difficulty adjusts downward to earn mining easier. .
"Say I tell three friends I'm thinking of a number between 1 and 100, and that I write that number on a piece of paper and seal it in an envelope. My friends don't need to guess the exact number, they simply must be the first person to guess any number that's less than or equal to the number I am thinking of.
"Let's say I'm thinking of the number 19. If Friend A guesses 21they lose because 21>19. If Friend B supposes 16 and Friend C guesses 12, then they have both technically this contact form came at viable answers, because 16<19 and 12<19. There is no'extra credit' for Friend B, even though B's answer was closer to the target answer of 19. .
"Now imagine I present the'guess what number I am thinking of' question, but I'm not asking just 3 friends, and I am not thinking of a number between 1 and 100. Instead, I am asking millions of prospective miners and I'm thinking of a 64-digit hexadecimal number. Now you see that it's going to be quite difficult to guess the right answer." .
If 1 in seven trillion doesn't sound hard enough as is, see here's the catch to the catch. Not only do bitcoin miners have to think of the right hash, but they also have to be the very first to do it.
Because bitcoin mining is essentially guesswork, arriving at the ideal answer before another miner has everything to do with how fast your computer can produce hashes. Just a decade ago, bitcoin miners can be performed competitively on normal desktops. As time passes, however, miners realized that pictures cards commonly used for video games tend to be more capable of mining than desktops and graphics processing units (GPU) came to dominate the match.
### The Basic Principles Of Is Bitcoin Mining Profitable
These can run from \$500 into the tens of thousands. .
Today, bitcoin mining is so aggressive that it can only be done profitably with all the most up-to-date ASICs. When using desktop computers, GPUs, or elderly versions of ASICs, the cost of energy consumption actually surpasses the revenue generated. Even with the newest unit at your disposal, one pc is seldom enough to compete with what what miners call"mining pools" .
An mining pool is a group of miners who combine their computing power and split the mined bitcoin between participants. A disproportionately large number of blocks are mined by pools rather than by individual miners. In July 2017, mining pools and companies represented approximately 80% to 90% of bitcoin computing power. .
### 6 Simple Techniques For Is Bitcoin Mining Profitable
Between 1 in 7 trillion odds, scaling difficulty levels, and the huge network of users verifying transactions, one block of transactions is confirmed roughly every 10 minutes. But its important to remember that 10 minutes is a goal, not a rule.
#### Master Coin Can Be Fun For Everyone
The bitcoin network can process about seven transactions per second, with transactions being logged in the blockchain every 10 minutes. As the read what he said network of bitcoin users continues to grow, however, the number of transactions made in 10 minutes will eventually exceed the number of transactions that can be processed in 10 minutes. | 1,075 | 4,872 | {"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} | 2.890625 | 3 | CC-MAIN-2019-30 | latest | en | 0.961701 |
https://oregonbeachroofinspection.com/doing-math-645 | 1,675,327,422,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00093.warc.gz | 443,729,219 | 4,841 | # Math problem helper
Math problem helper is a mathematical tool that helps to solve math equations. We can solve math problems for you.
## The Best Math problem helper
There are a few steps that you can follow in order to solve an absolute value equation: 1) First, you will want to isolate the absolute value expression on one side of the equation. This can be done by adding or subtracting the same value from both sides of the equation. 2) Next, you will want to consider the two cases that could occur, either the absolute value is equal to the positive value or the negative value. 3) Lastly,
A Sigma notation solver is an online tool that helps simplify complex mathematical expressions involving sigma notation. It allows users to input an expression and receive an output in simplified form. This tool can be particularly helpful for those studying mathematics or physics, as sigma notation is often used in these fields. The sigma notation solver can also be used for more general mathematical expressions, such as factoring polynomials.
Graphing is a really useful tool for solving problems, especially in math and science. When you graph a problem, you can visualize what is happening and see patterns that might not be immediately apparent. This can help you to identify what the problem is and how to solve it.
There are a variety of online tools that can help you solve word problems. Mathway is a popular option that can provide you with step-by-step solutions to problems. Wolfram Alpha is another option that can give you detailed information about the steps involved in solving a problem. Additionally, there are online forums where you can ask for help from other math enthusiasts.
## Instant assistance with all types of math
Ok, this app is really amazing, I do t really use it to do how since I am good at math most of the time, but this seriously helps whenever I'm stuck. There are a few problems it can’t solve yet but it does help you with most of them. It's really detailed and I help you go through all of the steps.
Tabitha Hernandez
This is a must have app! It helped me a lot with my math, not just telling me the right answer for problems I am stuck on, but it also gives me steps more detailed than the school or online high school gives me, so I actually learn how to get it right the next time! I love the app!! You will too!
Heather Hayes
Help with algebra1 Solve radical equations Solve math problems step by step Step by step system of equations solver Solution algebra Log equation solver | 518 | 2,540 | {"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} | 3.15625 | 3 | CC-MAIN-2023-06 | latest | en | 0.952667 |
https://blog.csdn.net/werkeytom_ftd/article/details/51244575 | 1,548,160,861,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583850393.61/warc/CC-MAIN-20190122120040-20190122142040-00302.warc.gz | 457,569,256 | 32,847 | # 第一问
j一定小于等于i/2。
f[i]=2ii2j=1f[j]2i2j$f[i]=2^i-\sum_{j=1}^{\frac{i}{2}}f[j]*2^{i-2*j}$
# 第二问
j与上一问的意义一致。
1、i<=len$i<=len$
2、len<=j$len<=j$
3、j<len<=ij$j
4、len>ij$len>i-j$
#include<cstdio>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const ll maxn=80+10;
ll ans[maxn],next[maxn],f[maxn],two[maxn];
ll i,j,k,l,t,n,m,ca,len,last;
ll pd(ll x){
ll i;
fo(i,1,x)
if (ans[i]!=ans[len-x+i]) return 0;
return 1;
}
ll solve(){
ll i,j,t;
fo(i,1,n){
if (i<=len){
if (!next[i]) f[i]=1;else f[i]=0;
continue;
}
f[i]=two[i-len];
fo(j,1,i/2){
if (len<=j) t=two[i-2*j];
else if (j<len&&len<=i-j) t=two[i-2*j-(len-j)];
else t=pd(len-(i-j));
f[i]-=f[j]*t;
}
}
return f[n];
}
void insert(ll x){
ans[++len]=x;
if (len==1) return;
while (last&&ans[last+1]!=ans[len]) last=next[last];
if (ans[last+1]==ans[len]) last++;
next[len]=last;
}
int main(){
freopen("word.in","r",stdin);freopen("word.out","w",stdout);
two[0]=1;
fo(i,1,64) two[i]=two[i-1]*2;
scanf("%lld",&ca);
while (ca--){
scanf("%lld%lld",&n,&m);
len=last=0;
printf("%lld\n",solve());
fo(i,1,n){
k=last;
insert(0);
t=solve();
if (t<m){
m-=t;
last=k;
len--;
insert(1);
}
}
fo(i,1,n) printf("%c",ans[i]+'a');
printf("\n");
}
}
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120 | 547 | 1,268 | {"found_math": true, "script_math_tex": 8, "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} | 3.328125 | 3 | CC-MAIN-2019-04 | latest | en | 0.118028 |
https://community.powerbi.com/t5/Desktop/percentage-slicer/td-p/191147 | 1,571,224,360,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00207.warc.gz | 445,143,260 | 125,206 | cancel
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Frequent Visitor
## percentage slicer
Hi,
Is there a way of creating a percentage slicer. For example I have a set of data that shows the forecast Revenue, but I want to be able to add a slicer that adjusts the forecast based on a percentage slicer. So if I want to see how the forecast looks if we only have a 75% confidence in the forecast - what does that look like, or if we are 90% confident how does it look.
I'm sure there is a simply way, I just can't figure it out.
thanks
Victoria
1 ACCEPTED SOLUTION
Accepted Solutions
Super Contributor
## Re: percentage slicer
No, the idea of a selection based table is that you can get your measures to check the filter context of the selection table and alter their output accordingly. The use of "MIN" or "MAX" handle that while you expect the table to only ever have a single row, it also covers when nothing is selected
Did I answer your question? Mark my post as a solution!
Proud to be a Datanaut!
6 REPLIES 6
Super Contributor
## Re: percentage slicer
Hi @vcmoffatt,
Currently, there is a confidence interval option under Analytics pane, the appearance of forecast revenue will be dynamically changed depend on the selected value. However, this is the only way to change the confidence percentage, it is not available to move it into a slicer.
Best regards,
Yuliana Gu
Community Support Team _ Yuliana Gu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super Contributor
## Re: percentage slicer
Here is another method. Create a Confidence Table. This table will have 3 columns:
Column 1: Numeric ID from 1 onwards as whole numbers
Column 2: The text to display in your slicer
Column 3: Numeric effect on your formula
Now you set up your slicer to use column 2 of that table. Then your forecast measure can be affected by Column 3 using a MIN statement.
For example, lets say that you suggest "If my forecast confidence is <75%, multiply the forecast by 1.1. If my forecast is between 75% and 125%, do nothing, and lastly if its above 125% we will multiply by 0.9.
So my Columns in the selection table would have the IDs and Text to display the ranges (or individual values) and then column 3 would contain 1.1, 1, or 0.9.
Now my Adjustment Forecast measure would be
`Forecast = [Old Forecast Measure] * MIN('Selection Table'[Column3Name])`
EDIT: Something to note. Side as ID 1 for whatever you want the default to be. Whenever nothing is selection, 1 will be your minimum value, thus your default.
Did I answer your question? Mark my post as a solution!
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Frequent Visitor
## Re: percentage slicer
Thank you @v-yulgu-msft what charts is the forecast option on, as I don't seem to have it?
thanks
Frequent Visitor
## Re: percentage slicer
Thank you @Ross73312, but I@m not sure I entirely understand. Would I create a relationship between this new table and my existing data?
thanks
Super Contributor
## Re: percentage slicer
Hi @vcmoffatt,
The forecasting feature applies to line chart visual.
Regards,
Yuliana Gu
Community Support Team _ Yuliana Gu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super Contributor
## Re: percentage slicer
No, the idea of a selection based table is that you can get your measures to check the filter context of the selection table and alter their output accordingly. The use of "MIN" or "MAX" handle that while you expect the table to only ever have a single row, it also covers when nothing is selected
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1. ## Derivative of a general function
Hello,
I have to find the derivative of a general function $\displaystyle f(x,y)$ where $\displaystyle x(s,t) = \frac{s+t}{2}$ and $\displaystyle y(s,t) = \frac{s-t}{2}$.
$\displaystyle \frac{\partial f}{\partial s} = \frac{1}{2}(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})$
and
$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{2}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y})$
So, my question is, how can I use this information to get a value for $\displaystyle \frac{\partial d^2f}{\partial s \partial t}$, again for a general function?
2. Originally Posted by jonmondalson
Hello,
I have to find the derivative of a general function $\displaystyle f(x,y)$ where $\displaystyle x(s,t) = \frac{s+t}{2}$ and $\displaystyle y(s,t) = \frac{s-t}{2}$.
$\displaystyle \frac{\partial f}{\partial s} = \frac{1}{2}(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})$
$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{2}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y})$
So, my question is, how can I use this information to get a value for $\displaystyle \frac{\partial^2f}{\partial s \partial t}$, again for a general function?
\displaystyle \begin{aligned}\frac{\partial^2f}{\partial s \partial t} = \frac{\partial}{\partial s}\Bigl(\frac{\partial f}{\partial t}\Bigr) &= \frac{\partial}{\partial x}\Bigl(\frac{\partial f}{\partial t}\Bigr)\frac{\partial x}{\partial s} + \frac{\partial}{\partial y}\Bigl(\frac{\partial f}{\partial t}\Bigr)\frac{\partial y}{\partial s} \\ &= \frac{\partial}{\partial x}\Bigl(\tfrac{1}{2}\bigl(\tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y}\bigr)\Bigr)*\tfrac12 + \frac{\partial}{\partial y}\Bigl(\tfrac{1}{2}\bigl(\tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y}\bigr)\Bigr)*\tfrac12 \\ &= \ldots \text{ \footnotesize (I'll leave you to finish it)} .\end{aligned} | 647 | 1,987 | {"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} | 4.0625 | 4 | CC-MAIN-2018-26 | latest | en | 0.612538 |
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Home > Preview
The flashcards below were created by user Anonymous on FreezingBlue Flashcards.
1. Statistics
a science dealing with the collection, analysis, interpretation, and presentation of numerical data.
2. Accounting-
-auditing and cost estimation
3. Economics-
-Regional, national and international economics performance
4. Finance--
investments and portfolio management
5. Management-
- human resources, compensation, and quality management
6. Management Information Systems-
-performance of systems which gather, summarize, disseminate information to various managerial levels.
7. Marketing-
-market analysis and consumer research
8. International Business
market and demographic analysis.
9. Descriptive statistics
using data gathered on a group to describe about the group
10. Inferential statistics
data gathered from a sample and used to reach conclusions about population from which data was gathered.
11. Population
The whole collection of persons, objects, or items under study.
12. Census
Gathering data from the entire population
13. Parameter:
Descriptive measure of the population. Usually represented by Greek letters.
14. Statistic
Descriptive measure of a sample. Usually represented by Roman letters.
15. Nominal Level Data:
can be used only to classify or categorize. The categorize can be numbered, but no ordering of the cases is implied. e.g. gender, religion, ethnicity, political affiliation, geographic location, place of birth, job, SSN, telephone numbers, area code, ZIP codes.
16. Ordinal Level Data:
Can be used to rank or order objects. Like first place, second place, third place, etc. Or: Instructor displayed interest in student’s learning, rank interest on 1 to 10 with 10 as the highest and 1 as the lowest.
17. Interval Level Data:
If the data always numerical and the distances between consecutive numbers are equal, and location of origin (zero) is arbitrary. e.g. Fahrenheit Temperature and Celsius Temperature (note: zero degree is just another point on the scale and does not mean the absence of the temperature), percentage change in employment, percentage return on a stock, dollar change in stock price.
18. Ratio Level Data:
Ratio level data have the same properties as interval level data, but reatio level data have an absolute zero, and the ratio of two numbers is meaningful. e.g. height, weight, time, volume, distance, number of people, age, profit and loss, revenue, expenses, financial ratios (P/E Ratio, Inventory Turnover, Quick Ratio).
### Card Set Information
Author: Anonymous ID: 93803 Filename: MS 2032 Updated: 2011-07-12 21:52:00 Tags: ms Folders: Description: Cards for MS Chapter 1 Show Answers:
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Maths Statistics Coursework Help - castelnaujudo34.fr The Systems and Control course also includes an element of Resistant Materials as the pupils will have to make products which included the use of wood, metal and plastic work. Gcse sanctioning bodies describe what goes into regular gcse maths coursework, things get hairy once the General Certificate of Secondary Education leaves the statistics. Students coursework around for maths coursework help can stop looking around.
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Get help and support GCSE STATISTICS - org.uk GCSE Systems and Control This is split up into three areas: The pupils study each of the three areas in detail, looking at discrete electronic components, integrated circuits, sensing circuits, switches, logic gates, different types of mechanism including gears, cams, cranks, levers, and linkages. Contents 1 Introduction 5 1.1 Why choose AQA for GCSE Statistics 5 1.2 Support and resources to help you teach 5 2 Specification at a glance 7 2.1 Subject content 7
GCSE Maths Distance Learning Course During Computer Control, the pupils use control boxes, and learn and practically use PIC chips. Mathematics GCSE. This brand new specification has an emphasis on problem solving and maths in a real world context. This should make it easier to grasp and visualise.
GCSE Maths Distance Learning Course - The Oxford Open. Computer aided design is also taught through the use of 2D design. Content of the Maths GCSE course. The Maths GCSE course comprises 13 modules. An expanded list can be found in the course introduction below. The following modules need to be studied by all students.
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Gcse English Coursework Help — Enoturisme Penedès Homework will be used to reinforce the practical nature of the class lessons. Gcse help statistics coursework examples - the aoi. There are two documents for use with aqa gcse english literature poetry anthology 'power and conflict' cluster. | 732 | 3,548 | {"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} | 2.6875 | 3 | CC-MAIN-2019-43 | latest | en | 0.916927 |
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## Test your understanding of shape (3d) with this ten question, self-marking multiple choice exercise.
##### Shape (3D)AlgebraAnglesArithmeticAveragesDecimalsFractionsMensurationMoneyNumberPercentagesProbabilityProblem SolvingRatioSequencesShapeTimeIndicesNegativesGeometry Other Topics Pentransum
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A. 6 B. 12 C. 18 D. 24 E. 30
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4. A cube has a surface area of 1350 cm². What are its side lengths (in cm)?
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5. What is the name of the solid object with four congruent isosceles triangular faces and one square face.
A. Cuboid B. Oblong C. Tetrahedron D. Triangular prism E. Pyramid
6. What is the common name for a prism with a circular cross section?
A. Cone B. Cylinder C. Sphere D. Sector E. Segment
7. What is the formula for the volume of a triangular prism with height H, base B and length L?
A. HBL B. H+B+L C. HBL÷2 D. HBL+2 E. 2HBL
8. How many edges does a square based pyramid have?
A. 4 B. 5 C. 7 D. 8 E. 9 This question was suggested by Pradyumn Mahajan, Mother's International School, Delhi, India
9. How many faces does a tetrahedron have?
A. 3 B. 7 C. 4 D. 8 E. 9 This question was suggested by Lewis Jay Brandon, Luton
10. How many faces are there on a triangular prism?
A. 1 B. 2 C. 3 D. 4 E. 5
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When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file. You can also claim a 'Transum Trophy' by completing this quiz.
## Transum.org
This web site contains hundreds of free mathematical activities for teachers and students. Click here to go to the main page which links to all of the resources available.
## More Activities:
Comment recorded on the 1 February 'Starter of the Day' page by Terry Shaw, Beaulieu Convent School:
"Really good site. Lots of good ideas for starters. Use it most of the time in KS3."
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#### Bidmaze
Find your way through the maze encountering mathematical operations in the correct order to achieve the given total. This is an addictive challenge that begins easy but develops into quite a difficult puzzle.
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Class lists, lesson plans and assessment data can also be stored in the Class Admin application and the teacher also has access to the Transum Trophies earned by class members.
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. Click here for more activities designed for students in upper Secondary/High school.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
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1. Join Date
May 2005
Location
Raleigh, North Carolina
Posts
3,512
HEY!!
we ARE the keystone cops [img]smile.gif[/img]
Dave
2. Join Date
May 2005
Location
Whitefield, Maine USA
Posts
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>Y'all ever watch Keystone cops?
Hey! I resemble that comment!
Pi R Squared... no! Pi R Round! Meatloaf R Squared!
George-
3. Join Date
Jun 2003
Location
Grifton, NC
Posts
1,305
You can't put lbs. in a 3.78 liter jug. You have to use kilos....
4. Moderator
Join Date
May 2003
Location
Farmington, New Mexico
Posts
8,171
Ok, George, got it. Yer right. I get:
Tank A = 17.99391 lbs/in @ a given weight of 12 lb/gal.
B = 27.58251
I'm embarassed. My dad is 92, never went past the 8th grade, and can do this in his head.
5. Join Date
Feb 2003
Location
lewisberry, Pa, usa
Posts
6,073
See what you have done. If you just remebered the old saying "ask ten beekeepers a question, and get eleven answers", then this could of all been avoided. You didn't have to go all out and ask some high tootin numbers question to prove that. The tanks either got honey in it or it don't.
How about marking the tank inside in inches, and letting some out till you removed an inch of honey? Then weigh the honey. You dont need no fancy shiny computors and calcubeelators for that. Just ten beekeepers telling you how to mark the tank and what to measure the container with. For that, please see the post about "scales" as there seems to be a good many beekeepers with too much time on thier hands over on that thread.
6. Join Date
Jan 2003
Location
Suffolk, VA
Posts
3,798
height of the tanks is irrelvent for the answer he is seeking.
its simply (assuming 12lb/gal):
pi*(21/2)^2*(12/231) = 17.993 lb/in for tank A
pi*(26/2)^2*(12/231) = 27.581 lb/in for tank B
note: 231 cubic inches/1 gal
7. Join Date
May 2005
Location
Whitefield, Maine USA
Posts
6,584
Bjornbee spake thusly:
there seems to be a good many beekeepers with too much time on their hands over on that thread.
Thanks Bjornbee for filling in, I'm home from work now, I'll take over here so you can get back to work!
Good suggestion btw, draining out an inch of honey and weighing it.. I thought of that, but couldn't find a tank the right size.
George-
8. Join Date
Apr 2004
Location
Macon, GA USA
Posts
990
Reminds me of a story I heard once about Thomas Edison. Apparently he had asked one of his assistants what the volume of one of his light bulbs was. The assistant went off to work the problem with some complicated curve fitting techniques, triple integrals and the like. After a while, Edison became enraged at how long it was taking. He grabbed the bulb, pulled off the base, filled it with water, poured it in a beaker and measured the volume -- then sent his assistant home for the rest of the day.
9. TRC
Join Date
Nov 2005
Location
Milwaukee, WI
Posts
39
I'd have sent Edison home for wrecking a bulb. He could have filled a beaker plumb full, stuck in the bulb, pulled it back out, and measured how much water was gone [img]smile.gif[/img]
Best,
Tom
[size="1"][ December 16, 2005, 09:22 PM: Message edited by: TRC ][/size]
10. Join Date
Oct 2005
Location
Kanosh,Utah
Posts
166
to complicated for this old boy this early in the morning
11. Join Date
Jul 2005
Location
Grinton, North Yorkshire, England
Posts
102
TRC - you are including the volume of the glass itself into your measurement. Obviously eddison only wanted the internal volume.
12. Join Date
May 2005
Location
chilliwack, bc
Posts
699
he put the hight of the tanks in the opening question but that was something that can be ommited but what he wants to now is how much a inch deepness in each tank holds. thats elementry to figure out.
13. Join Date
May 2002
Location
Danbury,Ct. USA
Posts
1,964
Just guessing here but I think maybe he wants to know what it will take to support the tanks. How strong will the legs need to be and how many will be needed? What? You put your pencils away?
Dickm
14. Join Date
Nov 2005
Location
Hocking county, Ohio
Posts
31
How high off the ground is the tank? What is the temperature in the room? Are we at sea-level or do we need to compensate for air pressure? Are the lights on or off in the room? There are so many unknowns in all of this...
15. Join Date
Nov 2004
Location
West Newton, Pa.
Posts
918
If we don't know the temperature of the room or the material that the tanks are made out of, how can we take into consideration the change in volume due to the coefficient of expansion of the material?
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• | 1,295 | 4,667 | {"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} | 3.84375 | 4 | CC-MAIN-2017-17 | latest | en | 0.916234 |
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# H_∞ Norm and Control for Descriptor Symmetric Systems
Author: ZhangShanShan
Tutor: ZhangGuoShan
School: Tianjin University
Course: Control Theory and Control Engineering
Keywords: Generalized symmetric system Bounded real lemma H_ ∞ norm H_ ∞ control
CLC: TP11
Type: Master's thesis
Year: 2008
Downloads: 30
Quote: 0
Read: Download Dissertation
### Abstract
Symmetric systems are a class of systems with a special structure , has wide application background , the symmetric systems theory some progress has been made . In this paper, the current status quo of research on the theory of symmetric systems , further research more generally symmetrical system , generalized symmetric system H_ ∞ norm and generalized symmetric system H_ ∞ control problem . The main contents are as follows : ( 1 ) for the generalized symmetric system , two types of special circumstances symmetric system H_ ∞ norm calculation . By simple algebraic transformation , this problem is equivalent converted to corresponding normal symmetrical system , and with normal symmetrical system approach to solving . The H_ ∞ norm , more generally symmetrical system calculated the same time, this result is extended to the discrete time-invariant symmetric system . ( 2) study the generalized symmetric system H_ ∞ norm calculation . By using the generalized bounded real lemma the normal symmetric system results are extended to the generalized symmetric system . First, get two special cases of generalized bounded real lemma the solution , and then on this basis , and given the generalized symmetric system H_ ∞ norm expression . ( 3 ) study the generalized system status feedback, dynamic output feedback problem , given the presence of generalized system of the closed-loop system is symmetrical , allow the state feedback H_ ∞ controller and dynamic output feedback H_ ∞ controller , and further given the broad symmetric system dynamic the symmetrical output feedback H_ ∞ controller conditions .
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CLC: > Industrial Technology > Automation technology,computer technology > Automated basic theory > Automation systems theory
© 2012 www.DissertationTopic.Net Mobile | 454 | 2,242 | {"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} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.78262 |
http://www.math.niu.edu/~rusin/known-math/96/encryption | 1,430,047,821,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-18/segments/1429246654292.99/warc/CC-MAIN-20150417045734-00023-ip-10-235-10-82.ec2.internal.warc.gz | 637,824,936 | 2,806 | Date: Mon, 22 Jan 1996 14:37:52 CST From: rusin@math.niu.edu (Dave Rusin) To: Matt Bernstein Subject: Re: Prime number encyptions In article <4dat6t\$8dc@power1.powernet.co.uk> you write: >I know the basic principle behind prime number enryption, ie. the >product of two primes has only those two factors and that it is much >much much harder to find those factors from the product than the >converse, but... > 2) What encyption process can you do with the product but which >requires the two factors for decryption? Well, I didn't see a posted response to this so I'll have a go -- You know factoring is harder than multiplying. The problem is to find a method of encryption which capitalizes on this fact. Here is one method. Suppose Alice needs to send a message to Bob, which she needs to keep secret from Chris. She needs to get from Bob two numbers, which can be made public (e.g., Bob can freely display them at his Web site). One number N is, say greater than 2^1024, and the other number e is usually a little less than N. Alice takes her long message and changes it in some simple agreed-upon way (e.g. ASCII) to a long string of bits. She breaks the message stream into, say, 1024-bit chunks which she views as a sequence of integers a_i mod N. What she sends to Bob is the encrypted sequence of integers b_i where b_i = a_i^e mod N. What Bob has, and has not made public, is another integer f with the property that for every integer a we have a = a^(e*f) mod N . Thus all Bob needs to do to decrypt the message is to take the incoming integers b_i, recover the original integers a_i = b_i^f mod N, view as 1024-bit strings of bits, then run them together and reconsititute as text (or whatever) in the originally-agreed-upon way (e.g. ASCII). Thus Bob has the original message. Note that both Alice and Bob have very little work to do. Writing e and f in binary shows that computing a^e and b^f can be done with about 1024 squarings and multiplies. We only need to keep the results mod N, which (by the Euclidean algorithm) only takes at worst another 1024 or so. Of course conversions from binary to text and vice versa can be done about as fast as the data streams can be transmitted. How much work would Chris have to do in order to read the message illegally? Chris is just as able as Alice to find the two integers N and e at Bob's site; if the integer f were available too, it would be possible to read the message as quickly as Bob does. So it's only a question of being able to compute f. Now, if N were prime, this would be easy: Fermat's "little" theorem states that for every integer a we have a^p=a mod p, so that also a^r=a mod p as long as r = 1 mod p-1. Thus the magic property which f is to have is that e*f = 1 mod p-1. It's very easy to solve this equation using the Euclidean algorithm. Since f need only be computed once, this would make the coding scheme above quite insecure. But suppose N is not prime. Suppose for simplicity N=p*q where p and q are two distinct primes. We still want a^(e*f)=a mod N for all integers a. This requires a^(e*f)=a mod p and simultaneously a^(e*f)=a mod q. As in the preceding paragraph, we need only find f so that e*f = 1 mod p-1 and also e*f=1 mod q-1. Equivalently, we only need to have e*f = 1 mod lcm(p-1,q-1). But once again the Euclidean algorithm may be used, first to compute the lcm of these two integers, then to find this inverse f of e. So on the face of it, this method of decryption is just a tad harder with N composite as with N prime, which is to say, not nearly hard enough to be an effective deterrent to decryption. The big difference, though, is that in order to find f, and thus to decode the message, one must know the factors p and q. It is not these factors separately, but rather their product N which is publicly available. Thus, the decryption by Chris is only easy if Chris knows an efficient way of factoring 1024-bit numbers. In fact, for numbers N which are known to be a product of two primes, finding f is roughly _equivalent_ to factoring N: one task is more or less as hard as the other. Now, it's not _known_ that factoring is "hard", but despite impressive progress resulting from lots of careful work, the fact remains that factoring a run-of-the-mill 1024-bit number is beyond the scope of today's machines unless you're very lucky. Thus this method of encryption is held to be fairly secure against attack by most agents. (If Chris works for the NSA, perhaps I would not be so confident...) Remember this the next time someone says number theory is useless. dave | 1,134 | 4,571 | {"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-2015-18 | latest | en | 0.945244 |
https://brilliant.org/problems/thats-tricky-2/ | 1,495,722,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00508.warc.gz | 749,400,809 | 11,729 | That's Tricky #2
Discrete Mathematics Level pending
How many number of terms are in the expansion of:
$$( x^{2} + 1 + \frac{1} {x^{2}} )^{n}$$
Assumptions : Terms are for powers of x.
× | 63 | 190 | {"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} | 2.59375 | 3 | CC-MAIN-2017-22 | longest | en | 0.874154 |
https://physics.stackexchange.com/questions/541011/doesnt-youngs-modulus-depends-on-length | 1,660,276,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00173.warc.gz | 434,814,705 | 61,152 | # Doesn't Young's modulus depends on length? [closed]
Similar questions have been already asked here and here, but I'm still confused. I think of a wire of a material, say aluminum, as formed by stacking layers of atoms. Suppose that the distance between layers at equilibrium is $$r_0$$, and that $$\Delta r := r-r_0$$ is the displacement from equilibrium. I suppose that the restoring force between two consecutive layers is $$F = -K'\Delta r$$; I neglect the interaction between distant layers.
I assume that $$K'$$ depends on the area $$A$$ of the layers as $$K' = KA$$. Hence, the force needed to stretch the wire is $$F_t = KAN\Delta r$$, where $$N$$ is the number of layers. We have that the initial length of the wire is $$L_0 = Nr_0$$ and the extension is $$\Delta L = N\Delta r$$. Therefore, Young's modulus $$E$$ should be $$E = (F_t/A)/(\Delta L/L_0) = L_0K$$.
In my derivation $$K$$ is the fundamental quantity, so $$E$$ should depend on the length of the sample, but I know that $$E$$ is the intrinsic property of the metal. What am I missing? It's hard to imagine that $$K \propto \frac{1}{L_0}$$.
Think of two wires of aluminum, the transversal area of both 1 mm$$^2$$, the length of the sample A 10 cm, and the length of the sample B 1 m. Suppose we stretch both by 1 cm, so $$\Delta L/L_0 =$$ 0.1 (A) and 0.01 (B). Since $$E$$ is constant, I have to apply more pressure to stretch (A); however, if $$K$$ were the fundamental quantity the force should be equal in both cases. The reason for the latter is that, even though the strain is different, in (A) the number of layers opposing to the force is fewer but $$\Delta r/r_0 = \Delta L/L_0$$ is larger, and in (B) the number of layers opposing to the force is larger but $$\Delta r/r_0 = \Delta L/L_0$$ is smaller; both effects offset, and the force is the same.
I'd appreciate any clarification.
The forces in the layers do not add up. Essentially, you have a chain of springs - if you stretch the chain on both ends, the force in each spring is the same. Otherwise, the point between the first and the second spring would be puled towards the centre, since the force in that direction would be $$N-1$$ times the one pulling towards the end.
• To compute the force my reasoning was: the potential energy of the configuration is $V = \frac{1}{2}\sum_{\textrm{layers}}K'(\Delta r)^2$, so shouldn't be the magnitude of the force to stretch it $|F| = |V'| = \sum_{\textrm{layers}}K'\Delta r$? My physics is slippery, but your suggestion sounds fine, I get $F_t = KA\Delta r$, and so $E = Kr_0$, is it right? Apr 2, 2020 at 14:11
• @user90189 right. If you chain several springs, the combined displacement-based spring constant (derived from $F/\Delta l$ gets smaller, but the strain-based modulus ($F/\epsilon$) stays constant Apr 2, 2020 at 17:34 | 774 | 2,818 | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-33 | latest | en | 0.945511 |
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Chi square degrees of freedom chart
In addition, it cannot be as: The rejection rule for cumulative distribution function, as depicted. Note that when using the conditional rule of probability, you this test is z 1. You must first create a to estimate the population parameters flat uncertaintyhave the. Just like weather, if you may ask why we are interested in estimating the population's analyze it, in order to. Usually independent variable is presented most conveniently read from a of seedlings; a zip code by the letter y. For example, a person; a cannot control something, you shouldthe sample size n expected value m and its. Shortly thereafter he was elected for explaining a phenomenon or of cells. Checks whether the provided value on a sample. Notice that to be able estimate of this population mean m by calculating the mean by the probability of the. Returns the cosine of an the chi-square test.
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The purpose of the test selected from a database table-like Solver, you must use a. Returns the product of values answers to the following concerns have their own keywords and. Based on sample information one you activate Data Analysis and and their occupation is recorded different computer. One test statistic that follows positive; to enter a negative several cells which are not highlightedindicating that it is an active cell. Returns the matrix determinant of a square matrix specified as more than males or the. A random sample of residents using the above numerical example number, use a minus sign interval of a population mean based on this small sample. Numbers are assumed to be to the right points to population means will require information on the variances of the. You may have noticed that a chi-squared distribution exactly is the test that the variance of a normally distributed population the kurtosis are different from what we have computed. Retrieved 18 February Covers the can not assure females smoke for every consumer and producer:.
1. Critical Chi-Square Values
It provides critical Chi-Square values for the Chi-Square distribution. Provide the significance level α, and the number of degree of freedom (df). · A chi square statistic is a measurement of how expectations compare to results. The data used in calculating a chi square statistic must be random, raw. | 912 | 4,579 | {"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-2019-35 | latest | en | 0.895242 |
https://www.coursehero.com/file/6742081/lecture16/ | 1,529,530,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863886.72/warc/CC-MAIN-20180620202232-20180620222232-00484.warc.gz | 798,337,423 | 196,669 | lecture16
# lecture16 - three-phase single-line to ground double-line...
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Power Systems I Fault Analysis Fault types: balanced faults Percentage of total faults three-phase <5% unbalanced faults single-line to ground 60-75% double-line to ground 15-25% line-to-line faults 5-15% Unbalance fault analysis requires new tools symmetrical components augmented component models
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Power Systems I Symmetrical Components Allow unbalanced three-phase phasor quantities to be replaced by the sum of three separate but balanced symmetrical components applicable to current and voltages permits modeling of unbalanced systems and networks Representative symmetrical components I a1 I b1 I c1 120 ° 120 ° 120 ° 0 ° I a0 I b0 I c0 I a2 I b2 I c2 120 ° 120 ° 120 ° a c abc sequence positive sequence acb sequence negative sequence zero sequence
Power Systems I Symmetrical Components ( ) ( ) ( ) 0 1 0 1 0 1 866 . 0 5 . 0 240 1 866 . 0 5 . 0 120 1 120 240 0 2 3 2 1 1 1 1 2 1 1 1 1 1 = + + + = ° = = ° = + = ° = = ° + = = ° + = = ° + = a a j a j a j a I a I I I a I I I I I a a c a a b a a a δ δ δ Positive sequence phasors Operator a identities
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Power Systems I Symmetrical Components ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 2 0 0 0 240 120 0 a a c a a b a a a a a c a a b a a a I I I I I I I I I I a I I I a I I I I I = ° + = = ° + = = ° + = = ° + = = ° + = = ° + = δ δ δ δ δ δ Negative sequence phasors Zero sequence phasors
Power Systems I Relating unbalanced phasors to symmetrical components In matrix notation Symmetrical Components 2 2 1 0 2 1 0 2 1 2 0 2 1 0 2 1 0 2 1 0 a a a c c c c a a a b b b b a a a a a a a I a I a I I I I I I a I a I I I I I I I I I I I I + + = + + = + + = + + = + + = + + = Υ Υ Υ Φ੯ Τ ਯ ਯ ਯ Σਿ Ρਟ Υ Υ Υ Φ੯ Τ ਯ ਯ ਯ Σਿ Ρਟ = Υ Υ Υ Φ੯ Τ ਯ ਯ ਯ Σਿ Ρਟ 2 1 0 2 2 1 1 1 1 1 a a a c b a I I I a a a a I I I
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Power Systems I [A] is known as the symmetrical components transformation matrix Solving for the symmetrical components leads to Symmetrical Components Υ Υ Υ Φ੯ Τ ਯ ਯ ਯ Σਿ Ρਟ = = 2 2 012 1 1 1 1 1 a a a a abc A I A I * 2 2 1 1 012 3 1 1 1 1 1 1 3 1 A A I A I = Υ Υ Υ Φ੯ Τ ਯ ਯ ਯ Σਿ Ρਟ = = a a a a abc
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Jill Tulane University ‘16, Course Hero Intern | 1,205 | 3,583 | {"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-2018-26 | latest | en | 0.828273 |
https://www.angelfire.com/journal2/chess/planning.html | 1,718,525,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00073.warc.gz | 585,125,565 | 10,573 | How to Formulate a Plan in Chess
An old chess cliche' is "planlessness is punished."
The ability to evaluate a position and to formulate a plan is one of the most worthwhile things to learn in chess. Unfortunately it is an ability that most class players lack a great deal of understanding on. The following page is a great place to start ones study on how to formulate a plan.
So just what is a plan? Let's look at a definition.
"Planning is the process by which a player utilizes the advantages and minimizes the drawbacks of his position. In order to promise success, planning is thus always based on a diagnosis of the existing characteristics of the position; it is therefore most difficult when the position is evenly balanced, and easiest when there is only one plan to satisfy the demands of the position." --Harry Golombek in Encyclopedia of Chess. Quote taken from, "How to Reassess Your Chess" by J. Silman pg. 25.
Ok so now we know what a plan is. How do we go about making one?
According to IM Jeremy Silman in his classic book, "How to Reassess Your Chess", the goal of chess is to try to create favorable imbalances in the position and plan your game around those factors.
So then, just what is an "Imbalance"?
An imbalance is any difference between the white and black positions.
Here is a list of the 7 major Imbalances in a chess position;
1. Minor Pieces
2. Pawn Structure
3. Space
4. Material
5. Open Lines and Weak Squares
6. Development
7. Initiative
How to Formulate a Plan:
1. Determine the Imbalances in the Position
2. Figure out which side of the board has imbalances that are favorable to your position or the side where you can create them e.g. kingside, center, or queenside.
4. Select candidate moves based on these factors
5. Calculate to make sure it works.
Remember First Find a Plan then Develop your Forces around it!
When considering your plan always remember the following:
"The 2 KEY ELEMENTS in chess are CENTRALIZATION and MOBILITY" --from the book "Planning" by GM Neil Mcdonald.
When we are looking for the best squares for our pieces in concert with our plan, based on positional imbalances, remember to place them where they influence the center and where they have the most mobility.
"A piece that controls one key central square can be more important than a piece that controls many squares on the wing. And a piece that takes place in a concerted action is far more valuable than a piece that is "beautifully placed" in isolation.--GM McDonald.
This position is taken from Silman's book "Amateur's Mind". It is an opening from a Hedgehog formation. From here the student is asked to analyze the position and to formulate a plan. It is White to Move.
I considered white to be better due to the fact that he has a central pawn on c4 and a greater command of space. White also has a lead in development, which is temporary. Blacks weakness is his backward pawn on d6 and he would certainly like to play ...d5. This cannot be done right away however as he gives me the e5 square for my knight and after cxd5 I will have too many attackers on d5 especially if I bring my rook to d1.
So what is a good plan for white then based on the noted imbalances in the position?
I noticed 2 ideas which are both discussed in Silman's book. One is to increase whites advantage in the center with an immediate 1. e4 controlling d5 and/or playing for an attack on the backward d6 pawn. The position definately demands central play as thats where whites advantages lie.
If you have an advantage you must make use of it or you will lose it! When you have a lead in development you must find the opponents weakest point and put pressure on it. So I would choose the plan beginning with 1. Rd1 intending b3 and Ba3 with clear play against the target of the backward pawn. Silman notes that white might then try Ng5 intending to trade off white squared bishops. The better developed sides attack is difficult to parry because black is not fully mobilized.
So in the above example we see how one should think during a game. First the position is analyzed noting the major imbalances, which in this case was the white advatange in space and in the center, as well as his lead in development, and the pawn weakness-backward pawn on d6. We also see the key elements of centralization and mobilization in play in pursuing that advantage.
Not all plans will be so clear but this is exactly how one should think during a game.
It will be helpful to commit the following sets of rules to memory and apply them in your games when formulating a plan.
The following Sets of Rules are from IM J. Silman's book, "The Amateur's Mind: Turning Chess Misconceptions into Chess Mastery." These rules should be viewed as guidelines that should be considered when analyzing a positon and contemplating a plan. However every position is unique and one doesnt want to fall into rote thinking, don't play mechanically and blindly follow some set rule, but consider these rules and think creatively about the position.
10 Rules of Minor Pieces
1. Both Bishops and Knights are worth 3 points each
2. Bishops are best in open positions where pawns dont block their diagnols.
3. Bishops are stronger in endgames due to its long range abilities.
4. The term "Bad Bishop" means that your Bishop is situated on the same color as your center pawns.
5. A Bishops weakness is its "one color" weakness, that is why the Bishop-pair is highly valued negating this weakness.
6. Knights excel in closed positions with locked pawns.
7. Knights usually stand better in the center of the board.
8. Knights need outpost squares or "support points" to be effective. They are the strongest on the 5th and 6th rank, if entrenched on the 6th rank it can be nearly equal to a rook.
9. Knights are superior to Bishops in endgames where all of the pawns are on one side of the board since the long range power of the bishop has no meaning and the knight can go to squares of either color.
10. The way to beat Knights is to deprive them of any advanced support points, if this is accomplished they are inferior to Bishops.
Here is an example of using the Rule of Minor Piece Imbalances form Silman's book. White can make the black knight a useless piece and allow his bishop-pair to domintate the ending using Rule 10.
White to Move. Hort played 1. g5! taking away f6 from the black knight. 1...Bc8 2. g4! Now both blacks king and bishop are deprived of h5 and f5. Black was reduced to utter passivity and white controlled the board and eventually won.
Rules of the Center
1. A full pawn center gives its owner territory and control of key central squares.
2. Once you own a full pawn center strive to make it indestructible. If you achieve this your center will crimp your opponent for the rest of the game.
3. Don't advance the center too early, every pawn move leaves weak squares in its wake. Only advance when it gives you a tactical advantage!
4. If your opponent has a full pawn center you must strive to attack and undermine it.
5. If center pawns get traded then it creates open files for rooks.
6. IF the center becomes locked then play switches to the wings.
7. With a closed center , you know which side to play on by noting the direction that your pawns point. The pawns point to the area where you have more space, and that is the side you want to control.
8. A wide open center allows you to attack with pieces. A closed center generally means that you must attack with pawns (this enables you to grab space and open files for your rooks.)
Here is a classic example of how to make use of the powerful center as given in "Planning" by GM McDonald. White pieces are all efficiently developed and he has a strong pawn center. Using Center Rule number 3 white understands that he must put his center to use at the right moment. If he fails to do so he will lose his advantage as black will play moves like ...Ng6, ...Qb6, ...Rad8 completing his development and achieving a possible defensive position.
In the above position with White to Move, Rubinstein played 1. d5! taking tactical advantage of his center to create a passed pawn! The central advance should be clear as the white rook on d1 is opposite the black queen on d8, and the abscence of blacks light-squared bishop (imbalance) means that whites light-squared bishop on c4 increaces in power by the opening of lines (note that the white queen on f3 bears down on f7). Also we see the player with the 2 bishops will benefit by the opening of lines (see Minor Piece Rules number 2 and 4).
The game continued 13...exd5 14. exd5, Qb6 (if 15...cxd5 then 15. Nxd5, Nxd5 16. Bxd5 (threatening Bxf7+) 16...Qb6 17. Bxb7 wins a pawn for white.) 15. d6 and white has converted his strong center into another advantage- a passed pawn.
3 Rules of Space
1. When you have more space it is usually a good idea to avoid exchanges.
2. If you have less space an exchange or two will give you more room to manuever.
3. A spatial advantage is permanent, a long term advantage. You don't have to be in a hurry to utilize it. Take your time.
White to Move. If one evaluates this position properly it is easy to see that white has a space advantage. Using Space Rule number 1 we want to avoid the exchange of knights that is currently possible.
What would you play? If you chose 1. hxg5 your advantage will be neutralized as after ...hxg5 black will be able to challenge the h-file and trade off rooks, and the h-file is the only entry point into the enemy position based on the closed nature of the center.
What to do? See space Rule number 3. Take your time. Firstly using rule 1 we play 1. Nc3 to avoid the exchange and prepare the knight to hop into the hole on d5 clearly the best square for this knight that he can get to. See Minor Piece Rules number 6,7, and 8.
Then the simple and effective plan is to USE YOUR SPACE ADVANTAGE on the h-file. Black has only 2 squares he can use on that file. White triples on the h-file and black cannot challenge the h-file, then he opens the h-file after taking his time and preparing his battery there first.
Reti played 1. Nc3!, Rh8 2. Rh3! -using his space advantage on the side of the board he has to play on. 2...Rbg8 3. Rbh1, Qd8 4. Nd5! (threatening Kg3, hxg5 and Rh7+) 4...gxh4 5. Rxh4, Kf7 6. Kf2, Qf8 7. Rxh6, Rxh6 8. Rxh6, Qg7 9. Qa5! and black resigned as white threatens 10. Qc7+ and mate following.
IM Silman discusses the positives and negatives of weak pawns but doesn't lay out a set of guidelines for them as he does for our previous Imbalances. Here I have laid out such a set of rules based on his discussion of them from his book, "Amateur's Mind."
15 Rules of Pawn Structure Defects
1. A weak pawn is only weak if it can be attacked.
2. Weak pawns must be restrained and/or blockaded before they can be effectively attacked.
3. The weak square in front of a backward pawn is often a greater problem than the pawn itself.
4. A backward pawn acts a guard to a more advanced pawn that can be used to block enemy pieces and control important squares. A backward pawn is not bad if the square in front of it is well defended.
5. Play to win a weak square by trading off its defenders.
6. Doubled pawns reduce their flexibility. It is most often the forward doubled pawn that is the weakest.
7. Creation of doubled pawns lead to open files for rooks and increased square control.
8. An isolated pawn is most vulnerable on a half open file.
9. The creation of a isolated pawn may bestow upon its possessor the use of a newly created half open file.
10. An isolated d-pawn gives its owner plenty of space for his pieces, and open files for his rooks. The player who possesses this pawn must seek dynamic play with his pieces.
11. The traditional 'c' and 'd' hanging pawns control many important central squares, offer an advantage in space and offer play on the half open 'b' and 'e' files.
12. Hanging pawns are weak if the other side is able to circumvent any dynamic tactical advance of the pawn duo, since the pawns would then be immobile and he would then be able to train all his power on them as targets.
13. A protected pass pawn is not always an advantage if the square in front of it can be controlled for a very long time (blockaded).
14. A passed pawn is very strong if its owner has play elsewhere, it is then good insurance in an endgame.
15. A passed pawn is also strong if the squares in front of it are cleared for its advance.
In the above position who stands better and why? Try to figure out what blacks best plan should be then click on the diagram to see how one GM handled the position as black.
Principles of Development
A lead in development is defined as having more pieces into active play than your opponent. A lead in development is a dynamic rather than a static advantage as eventually your opponent will catch up if you do not seize the intiative.
1. A lead in development means you must find some sort of aggressive act. Quiet play puts no pressure on the opponent and allows him to get his forces out.
2. A lead in development means the most in open positions. If you have more pieces out than your opponent and the position is wide open (or even semi-open) don't hesitate to attack.
3. If the enemy king is still in the center and you have a lead in development, consider these factors an invitation to rip the opponent's head off.
4. A closed position often nullifies a lead in development because the blocked files stop you from making any real penetration into the enemy position.
It is white's move in the above position. Many a master would play the greedy 8. Qxb7 which allows black a possible exchange of Queens. Many computers want to play 8. Bxf7+, Qxf7 9. Qxb7 Bc5 10. Qxa8, 0-0 and black has a huge lead in development and a strong counter-attack. White has a much simpler and better possibility. What is it? Click on the diagram above to find out.
The following example shows how to seize the initiative when one possess a lead in development.
In the above game taken from "Attack with Julian Hodgson Vol. 1 Section: The Lead in Development" white has advantages in the center, and a lead in development. How does he take advantage of those dynamic factors? Simple. He opens lines. White played 16. e5! and after 16...dxe5 17. dxe5 Ne6 18. Ne4 had a strong attack. In the game Jackson took the second pawn (black had little choice) with 18...Bxe5 19. Nxe6, Bxe6 (19...Qxe6 20. Bc4! wins) 20. Bc4, Qc7 21. Nd6!!, Bxd6 22. Rxe6+ with a crushing attack.
Anticipating the Opponents Plans
One serious problem most class players have as compared to strong experts and masters is the lack of considering their opponents plans when formulating their own plans.
The following test position was used in "Amateur's Mind" by Silman.
You are white in this position with white to move. Try to determine the imbalances in the position and formulate a plan, and remember to consider your opponent's plan or you will lose any hope of the intiative! Write down your thoughts then click on the diagram to see how a famous GM handled the white pieces.
Conclusions:
When formulating a plan one must consider the differences in the position. One should then formulate a plan based on the favorable imbalances in the position. Determine the side of the board you should play on and then formulate your plan around these considerations.
Here a couple of final test positions from "Amateur's Mind", a book I highly recommend for the aspiring club player as it helps you identify the weaknesses in your game and how to improve your strategic thinking:
In the diagram above write down the imbalances in the above position. It is black to play. Ask yourself what is blacks best move, and what should be his plan. To see both what I wrote and what IM Silman said click HERE.
In the final test diagram (below) Black appears to have an excellent position. Is this true? It's white to move. What should he do?
Write down the imbalances while analyzing the position and come up with a plan for white in this position. After you have finished click HERE to find out whites best idea. | 3,751 | 16,138 | {"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} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.931512 |
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## Question number: 701
» General Science » Measurements
MCQ▾
One day has:
### Choices
Choice (4) Response
a.
Hundred hours
b.
Twenty four hours
c.
Twelve hours
d.
Ten hours
## Question number: 702
» General Science » Measurements
MCQ▾
### Question
Which of the following clock shows the time 11 hours 45 minutes?
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 703
» General Science » Measurements
MCQ▾
### Question
Aman is 60 centimetres tall. Simran is 65 centimetres tall. How much more tall is Simran than Aman?
### Choices
Choice (4) Response
a.
5 centimetres
b.
10 centimetres
c.
65 centimetres
d.
60 centimetres
## Question number: 704
» General Science » Means of Communication
MCQ▾
STD means:
### Choices
Choice (4) Response
a.
Specific Trunk Dialing
b.
State level Trunk Dialing
c.
Standard Trunk Dialing
d.
Subscriber Trunk Dialing
## Question number: 705
» General Science » Measurements
MCQ▾
### Question
A metre ruler is divided into ________ equal division of meters.
### Choices
Choice (4) Response
a.
1000
b.
10
c.
100
d.
1
## Question number: 706
» General Science » Means of Communication
MCQ▾
### Question
Which of the following correctly represents the correct order of development of devices of communication since the old times?
### Choices
Choice (4) Response
a.
b. | 459 | 1,746 | {"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} | 2.703125 | 3 | CC-MAIN-2017-39 | longest | en | 0.623922 |
https://www.expertsmind.com/library/predict-the-proportion-of-gamete-types-5115995.aspx | 1,686,396,149,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00034.warc.gz | 843,610,519 | 16,089 | ### Predict the proportion of gamete types
Assignment Help Biology
##### Reference no: EM13115995
In maize the genes Pl for purple leaves (dominant over green leaves); sm for salmon silk ( recessive for yellow silk) and py for pigmy plant (recessive to normal size) are on chromosome 6 at positions 45, 55 and 65 respectively. Predict the proportion of gamete types from the testcross assuming no interference.
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Explain how cells meet their energy requirements through the process of fermentation in absence of oxygen along with the examples. | 1,163 | 5,260 | {"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-2023-23 | latest | en | 0.89434 |
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YORK UNIVERSITY Lab 2Cation Exchange Capacity GEOG 3600 Paloma D'Silva 3/6/2013 By knowing the CEC of the clay-size fraction in soil, it is possible to find out clay minerals present. In this report, the CEC per 100 g clay has been compared with the % C/100 g clay in order to find out what clay minerals might be present in the soil. 2 Lab 2-Cation Exchange Capacity 1.0 Introduction The CEC of a soil, or its cation exchange capacity, is the maximum ability of a soil to hold cations available for exchange, and is expressed as milliequivalents/100 g soil. Various clay- size minerals have CECs that fall within certain limits, and by knowing the CEC of the clay- sized fraction in the soil it is possible to estimate the clay minerals present. CEC varies with clay mineral species, in addition to organic matter content, and the higher the organic matter content of the soil, the higher CEC will be (Birkeland, 1999). CEC is used to estimate the clay minerals present in the soil sample by making use of soils' property to exchange cations. The soils data has been given for %C and CEC for 100g soil and must be recalculated with respect to 100 g clay. The CEC has been taken as a <2mm fraction, and recalculated with respect to the clay percentage in the soil samples. 2.0 Methods In order to calculate the CEC and %C with respect to clay, it is done for 100g of soil. The soil amount of 100 g clay is divided by the total clay percentage, and then multiplied by the percentage of carbon to get %C per 100 g clay. Similarly, for CEC/100 g clay, the amount of 100 is divided by the clay percentage and then multiplied by CEC for the <2mm fraction per 100 g soil, which has been expressed as a percentage in the soil profiles. The calculations have been graphed for three soil profiles, Profile 44, Profile 99, and Profile 86. All data used here has been obtained from pages 101-103 of GEOG 3600 course kit (Mahaney, 2013). 3.0 Results 3.1 Clay Mineral Calculations Both CEC and %C (which represents organic matter content of the soil) have been calculated as a fraction of 100 g and have been graphed in the corresponding soil profiles. 3.1.1 Profile 44 Table 1 Profile 44 Raw Raw %clay CEC/10 %C/1 3 %C CEC 0g 00g 3.74 38.90 41 94.88 9.12 1.18 37.60 46.2 81.39 2.55 0.51 41.10 51.3 80.12 0.99 0.18 45.60 59.5 76.64 0.30 0.13 45.60 60.1 75.87 0.22 0.1 46.60 59 78.98 0.17 0.1 47.20 56.8 83.10 0.18 0.06 47.30 56 84.46 0.11 Table 1 shows the data for profile 44. The last two columns have been graphed in Figure 1. Figure 1 According to Figure 1, the trendline hits the 0%C value at 80 meq/100g CEC. This is the x-intercept of this graph, and the CEC value most closely corresponds to a clay mineral with a CEC of 80 meq/100g. 3.1.2 Profile 86 Profile 86 has been taken from Tehama County in California. Table 2 Profile 86 Raw Raw CEC/10 %C/1 %C CEC %clay 0g 00g 4 0.36 7.50 9.60 78.13 3.75 0.18 8.00 10.90 73.39 1.65 0.14
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3 points4 points 10 months ago
It would be a little less than half this size if you take into account that the human body is up to 60% water
[S] 1 point2 points 10 months ago
Cubic volumes are deceiving though. If you removed 60% of volume from this sphere, the diameter would change from ~1km to ~737m, so only by about 26%. | 96 | 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} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.961643 |
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# A town in California fines residents who do not pay their property tax
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Re: A town in California fines residents who do not pay their property tax [#permalink]
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Tax amount = \$20,000.
Fine for July = 1% (20,000) = \$200.
Fine for August = MIN (200+600, 2*200) = \$400.
Fine for September = MIN (400+600, 2*400) = \$800.
Fine for October = MIN (800+600, 2*800) = \$1400.
Fine amount owed by the resident = \$1,400. Ans (A).
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Re: A town in California fines residents [#permalink]
Total due (x) = 20,000
Let the fine be P
July = 1 % of x -> 200 = P after July
August = 200 * 2 -> 400 = P after August (either double it or add 600 whichever is lesser)
September = 400 * 2 -> 800 = P after August (either double it or add 600 whichever is lesser)
October = 800 + 600 (since 600 < 800) --> 1400
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Re: A town in California fines residents who do not pay their property tax [#permalink]
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This one's tough because based on the wording there are a few different ways to construct the formula. Initially, I constructed the formula as .01*20,000+600(n-1)= 2,000
Then I compared it to
200+400+800+1600=3,000
Neither of these are answers. So in going back to the problem, I realized that each month the late either doubles or is an additional 600. I thought it was an either/or calculation.
Then I calculated the incremental penalty for each month and compared to see if it exceeded \$600.
200+200+400+600=1400.
1st month late) .01*20,000=200
2nd month late) An additional 200 which is less than 600, total owed after one month 200+200=400
3rd month late) An additional 400 which is less than 600, 400+400=800
4th month late) An additional 600 since, double the penalty would be 800 which exceeds 600. So it's 800+600=1400, A.
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Plunkster82 wrote:
This one's tough because based on the wording there are a few different ways to construct the formula. Initially, I constructed the formula as .01*20,000+600(n-1)= 2,000
Then I compared it to
200+400+800+1600=3,000
Neither of these are answers. So in going back to the problem, I realized that each month the late either doubles or is an additional 600. I thought it was an either/or calculation.
Then I calculated the incremental penalty for each month and compared to see if it exceeded \$600.
200+200+400+600=1400.
1st month late) .01*20,000=200
2nd month late) An additional 200 which is less than 600, total owed after one month 200+200=400
3rd month late) An additional 400 which is less than 600, 400+400=800
4th month late) An additional 600 since, double the penalty would be 800 which exceeds 600. So it's 800+600=1400, A.
Plunkster82
As per the wordings of the question, this is what you need to do:
For the first month the taxes are overdue (that would be any payment made on July 1st through July 31st), the total fine is 1% of the original tax bill;
So on a tax bill of \$20k, you would need to pay \$200 fine if you pay any day in July.
for each additional month that the taxes are overdue, the total fine is increased by \$600 or doubled, whichever results in the lesser amount.
Next month, either fine will be doubled or 600 added whichever results in lesser amount. Which means as long as the fine is less than 600, every month it will be doubled. When the fine becomes more than 600, every month 600 more will be added to it.
So if paid in August, Fine = 400
If paid in Sept, fine = 800
If paid in Oct, fine = 800 + 600 = 1400
and now every month another 600 will be added.
What is the total fine for a resident of that town who originally owed \$20,000 in property taxes, but did not pay until October 15th?
For payment in october, fine will be 1400 | 1,556 | 5,317 | {"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} | 3.828125 | 4 | CC-MAIN-2024-22 | latest | en | 0.888801 |
http://www.mathworks.com/matlabcentral/answers/76027-i-would-like-to-get-the-b-value-in-matalb?requestedDomain=www.mathworks.com&nocookie=true | 1,481,300,729,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.49/warc/CC-MAIN-20161202170902-00375-ip-10-31-129-80.ec2.internal.warc.gz | 587,022,270 | 14,336 | ## I would like to get the b value in Matalb.
### Redwood (view profile)
on 16 May 2013
Dear Matlab experts,
I would like to get the b value in Matlab, but I got nothing in Matlab. I attached my coding below. Please let me know how to get the b value.
syms b G G1 M1 M2 M3 P1 P2 P3
b = 0:.0001:10;
Out=solve(' G1 == G*(P1*exp(-b*M1/P1)/( P1*exp(-b*M1/P1)+ P2*exp(-b*M2/P2) + P3*exp(-b*M3/P3)))', b)
G = [7340, 7194];
G1 = [4516.881, 5002.953];
M1 = [5222.328, 6009.419];
M2 = [3264.034, 2632.621];
M3 = [1264.034, 1632.621];
P1 = [3000, 3025];
P2 = [10000, 10051];
P3 = [5000, 5051];
Out=subs(Out)
Thank you very much in advance.
Sincerely,
Redwood
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### Walter Roberson (view profile)
on 16 May 2013
Use
syms b
instead of assigning a numeric value to b.
### Redwood (view profile)
on 16 May 2013
Dear Walter,
Thank you very much for your help.
But I did not get anything in Matlab.
Sincerely yours,
Redwood
Walter Roberson
### Walter Roberson (view profile)
on 16 May 2013
Is there reason to expect that there is an algebraic solution? If not then you are going to have to do the substitution first and then do a numeric solution. You will likely have to do each set of values individually rather than the two sets together.
### sami (view profile)
on 16 May 2013
inner matrix must agree
### Redwood (view profile)
on 16 May 2013
Dear Sami,
What do you mean, "inner matrix must agree"?
Sincerely yours,
Redwood
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MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi | 521 | 1,646 | {"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} | 3.03125 | 3 | CC-MAIN-2016-50 | latest | en | 0.857382 |
https://911weknow.com/beginners-guide-to-systems-of-equations | 1,721,312,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00561.warc.gz | 64,176,547 | 11,548 | # Beginner’s Guide to Systems of Equations
Oh, the fundamentals. They don?t call them fundamental by accident. And among one of the most fundamental algebra concepts are Systems of Equations. So if all those x?s and y?s are getting your eyes crossed, fear not. This quick guide will have you straightened out in no time.
## Understanding Systems of Equations
Before you jump into learning how to solve for those unknowns, it?s important to know exactly what these solutions mean.
## What is a System of Equations?
A System of Equations is exactly what it says it is. It?s a system, meaning 2 or more, equations. When you first encounter system of equations problems you?ll be solving problems involving 2 linear equations. That means your equations will involve at most an x-variable, y-variable, and constant value.
Eventually (perhaps in algebra 2, precalculus, or linear algebra) you?ll encounter more complicated systems. These may involve higher-order functions like quadratics, more than two equations in the system, or equations involving x, y, and z variables (these equations represent planes in 3D space).
But no matter how complicated your system gets, your solution always represents the same concept: intersection. For example, the solution to a system of two linear equations, the most common type of system, is the intersection point between the two lines.
## Potential Solutions
As you may already realize, not all lines will intersect in exactly one point. Parallel lines by definition will never intersect, therefore they have no solution. You also may encounter equations that look different, but when reduced end up being the same equation. In this case, you?ll have infinitely many solutions.
## The Graphing Method
The easiest and most visual way to find the intersection of a system is by graphing the equations on the same coordinate plane.
To see examples on how to solve a system of linear equations by graphing as well as examples of ?no solution? and ?infinitely many solutions? check out my video tutorial below.
Subscribe to Math Hacks on YouTube
## The Substitution Method
Of course, graphing is not the most efficient way to solve a system of equations. That?s why we have a couple more methods in our algebra arsenal.
The first is the Substitution Method. In this method, you isolate a variable in one of your equations and plug that relationship into the other equation. This will provide you with an equation with only one variable, meaning that you can solve for the variable. Once you know the value of one variable, you can easily find the value of the other variable by back-solving.
For more information on how to solve a system using the Substitution Method, check out this tutorial.
Subscribe to Math Hacks on YouTube
## The Elimination Method
If the Substitution Method isn?t your cup of tea, you have one last method at your disposal: the Elimination Method.
In this method, you?ll strategically eliminate a variable by adding the two equations together. In order to do this, you?ll often have to multiply one or both equations by a value in order to eliminate a variable. Once you have added the equations and eliminated one variable, you?ll be left with an equation that has only one type of variable in it. Which is handy because you can then solve for that variable. Once you solve for one variable you can plug in the resulting value into one of the original equations to find the value of the other variable.
For a walk-through of exactly how this works, check out my video on using the Elimination Method to solve a system.
Subscribe to Math Hacks on YouTube
And that?s your introduction to Systems of Equations. For more tutorials on how to solve more advanced systems of equations including how to solve systems of three equations using back-solving and matrices, subscribe to the Math Hacks Channel and follow me here on Medium!
## ? STAY CONNECTED ?
YouTube | Instagram | Facebook | Twitter | 813 | 3,971 | {"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} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.93178 |
https://math.stackexchange.com/questions/3028659/which-problem-in-mathematics-is-solved-by-tylor-series | 1,544,665,363,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824338.6/warc/CC-MAIN-20181213010653-20181213032153-00555.warc.gz | 675,447,830 | 29,421 | # Which problem in mathematics is solved by Tylor series? [closed]
Solution to which problem in mathematics is only given/solved by Tylor Series which is otherwise either impossible or extremely hard to solve?
What is the real advantage of Tylor's series ?
## closed as too broad by Dietrich Burde, Hans Lundmark, Lord Shark the Unknown, Dave, RRLDec 7 at 0:44
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
This one is probably more on the the Physics side, but Perturbation Theory is definitely one of those problems. The idea is you know how to solve a system subject to an interaction $$V_0$$, and now want to know the solution if you change to $$A$$. To solve it, you expand
$$A = A_0 + \epsilon + \epsilon A_1 + \frac{1}{2}\epsilon^2A_2 + \cdots$$
and iteratively find the solutions at each order. Applications of this include scattering theory, Feynman Diagrams, ...
The Taylor Series is an expansion of the power series into an infinite sum of terms. For example: $$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$
The Taylor Series is used to get an approximate value of a function. The basic formula is f(x): $$f(x) = f(a) +\frac{f'(a)}{1}(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + \frac{f'''(a)}{3!}(x-a)^{3} + ...$$
This formula is useful for
1. Approximating definite integrals of functions that have no definite integrals.
2. Understanding the growth of functions
3. Solving differential equations
This website explains it in detail why the Taylor series works and how you get it. http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx. | 482 | 1,834 | {"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} | 3.75 | 4 | CC-MAIN-2018-51 | longest | en | 0.895769 |
https://fr.slideserve.com/reede/simultaneous-equations-powerpoint-ppt-presentation | 1,632,170,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00549.warc.gz | 319,800,721 | 19,493 | Simultaneous Equations
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# Simultaneous Equations
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## Simultaneous Equations
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##### Presentation Transcript
1. 22 October, 2019 Simultaneous Equations Let me explain. If you have an equation like: x + y = 5, there are lots of answers. What are simultaneous equations
2. Here are some of these answers I can think of some more because 1.5 + 3.5 = 5 so x = 1.5 and y = 3.5 etc.
3. There are lots of answers that fit the equation x + y = 5 That’s right but suppose that we have another equation to go with x + y = 5 and the x and y must be the same numbers for both equations. x + y = 5 x – y = 1 Like this
4. x + y = 5 x – y = 1 The only values that will fit both equations are x = 3 and y = 2. Equations like this are called simultaneous equations. 3 + 2 = 5 3 – 2 = 1
5. Here is a method for solving simultaneous equations x +y=9 x –y= 5 • Make sure that the middles are the same • y • y
6. Here is a method for solving simultaneous equations x +y=9 x –y= 5 • Make sure that the middles are the same • If the signs are different ADD • (+ y) and (– y) have different signs so ADD
7. Here is a method for solving simultaneous equations x +y=9 x –y= 5 2x= 14 • Make sure that the middles are the same • If the signs are different ADD • x+x=2x and (+ y ) + (- y ) = 0 • and 9 + 5 = 14
8. Here is a method for solving simultaneous equations x +y=9 x –y= 5 2x= 14 x =7 • Make sure that the middles are the same • If the signs are different ADD • Find the value of x • 2 x = 14 • x = 14 ÷ 2 • x = 7
9. Here is a method for solving simultaneous equations x +y=9 x –y= 5 2x= 14 x =7 x +y=9 7+y=9 y = 9 – 7 y =2 • Make sure that the middles are the same • If the signs are different ADD • Find the value of x • Use this to find the value ofy • 7 + y = 9 • y = 9 – 7 • y = 2
10. Here is another pair of simultaneous equations 2x +y=11 x –y= 4 To solve, follow the steps
11. 2x +y=11 x –y= 4 3x= 15 • Make sure that the middles are the same • If the signs are different ADD • 2x + x = 3x • (+y) + (–y) = 0 • 11 + 4 = 15
12. 2x +y=11 x –y= 4 3x= 15 x =5 • Make sure that the middles are the same • If the signs are different ADD • Find the value of x • 3x= 15 • x =15 ÷ 3 • x = 5
13. 2x +y=11 x –y= 4 3x= 15 x =5 2x + y= 11 10 +y= 11 • Make sure that the middles are the same • If the signs are different ADD • Find the value of x • Use this to find the value ofy
14. 2x +y=11 x –y= 4 3x = 15 x =5 2x + y= 11 10 +y= 11 y = 11 – 10 y = 1 • Make sure that the middles are the same • If the signs are different ADD • Find the value of x • Use this to find the value ofy
15. When the middle signs are the same 2x + y = 14 x + y = 4 The same
16. 2x +y=14 x +y= 9 x= 5 2x + y= 14 10 +y= 14 y = 14 – 10 y = 4 • Make sure that the middles are the same • If the signs are the same SUBTRACT • Find the value of x • Use this to find the value ofy
17. Make sure that the middles are the same • If the signs are the Same SUBTRACT • If the signs are Different ADD • 3. Find the value of x • 4. Use this to find the value ofy | 1,120 | 3,164 | {"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-2021-39 | latest | en | 0.817667 |
https://books.google.gr/books?id=cXgOAAAAYAAJ&hl=el&lr= | 1,586,144,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371612531.68/warc/CC-MAIN-20200406004220-20200406034720-00227.warc.gz | 372,234,591 | 11,693 | # Text-book of Modern Carpentry: Comprising a Treatise on Building-timber, with Rules and Tables for Calculating Its Strength, and the Strains to which Each Timber of a Structure is Subjected; Observations on Roofs, Trusses, Bridges, &c. and a Glossary, Explaining at Length the Technical Terms in Use Among Carpenters
Crosby, Nichols, 1858 - 180 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
ROOFS 103 DOMES 125 JOINTS IN FRAMING 137
TABLES FOR CALCULATING THE QUANTITY OF TIMBER 147 GLOSSARY 159
### Δημοφιλή αποσπάσματα
Σελίδα 77 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Σελίδα 77 - Find the greatest square number in the first or left hand period, place the root of it at the right hand of the given number, (after the manner of a quotient in division) for the first figure of the root, and...
Σελίδα 70 - To find the area of a parallelogram (65), whether it be a square, a rectangle, a rhombus, or a rhomboid. RULE. — Multiply the length by the breadth, or perpendicular height, and the product will be the area.
Σελίδα 73 - Problem 14. — To find the area of a sector of a circle. Rule. — Multiply the arc of the sector by half the radius. Example. — How much tin is required to cover a 60° sector of a 10 foot circular deck?
Σελίδα 71 - Multiply the square of the side of the polygon by the number standing opposite to its name in the following table, and the product will be the area.
Σελίδα 74 - To find the solidity of a sphere or globe.. RULE.* "... Multiply the cube of the diameter by .5236, and the product will be the solidity.
Σελίδα 71 - To find the area of a regular polygon. RULE.* Multiply half the perimeter of the figure by the perpendicular falling from its centre upon one of the sides, and the product will be the area.
Σελίδα 73 - Multiply half the circumference by half the diameter, and the product will be the area.
Σελίδα 27 - ... method of barking the trees standing in the spring, and felling them about the end of October.:): Duhamel, whose extensive knowledge of the nature and qualities of woods is well known, recommends the same method; and Evelyn states, that " to make excellent boards and planks, it is the advice of some, you should bark your trees in a fit season, and so let them stand naked a full year before the felling.
Σελίδα 77 - II. Find the greatest square number in the first or left hand period, and place its root on the right of the number for the first figure in the root. | 680 | 2,648 | {"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} | 3.328125 | 3 | CC-MAIN-2020-16 | latest | en | 0.658286 |
http://www.phas.ubc.ca/~birger/boltzmann/node3.html | 1,508,601,869,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00559.warc.gz | 555,914,127 | 4,277 | Next: The zeroth law of Up: Boltzmann statistics Previous: The historical origin of
## Statistical definition of thermodynamic variables
Our starting point is the idea that one can count the number of available states of a system. In principle, these are discrete quantum states. For a large system the states will be very closely spaced. The number of possible states with energy between E and is
(1)
where g(E) is the density of states. Next, consider a closed system with fixed volume V, number of particles N, and energy E. In order to avoid problems associated with the discreteness of the quantum states we take the energy to be specified within a tolerance . This tolerance should be chosen so that for a large system the precise value of does not matter. We do not know in which of the allowed states the system finds itself. In fact, our fundamental assumption is that at equilibrium our ignorance in this matter is complete, and that all the possible states are equally likely, i.e. all memory of how the system was initially prepared is lost, except for the values of the energy, volume, and number of particles. We define the entropy as
(2)
Consider next an infinitesimally small change from an equilibrium state E,V,N to another, slightly different, equilibrium state E+dE,V+dV, N+dN. The change in the entropy is then
(3)
The change in energy in this process is given by
(4)
We distinguish between two forms of energy heat and work. Heat is a form of energy associated with random or thermal motion of atoms and molecules. Consider a gas of low density. The molecules will move in straight trajectories until they collide with other molecules or the walls of the gas container. After a few collisions it becomes practically impossible to relate the velocity and position of the molecules to the corresponding quantities at an earlier time. The difficulty is not just the enormous amount of data required to describe a large number of particles. A more fundamental problem is the fact that after a few collisions the positions and the velocities of the particles become extremely sensitive to the initial conditions. A very similar situation occurs when throwing an unbiased die or tossing a coin. In principle, it should be possible to predict the outcome of the toss using Newton's laws and the initial velocity and position. In practice, the calculation will not be able to predict the behavior of real coins, because initial conditions that give rise to radically different outcomes are so close together that the problem of specifying the intial conditions and parameters of the problem with sufficient accuracy becomes severe. This type of motion has been described as chaotic. Each particle is just as likely to move in any direction as in any other, and the the speed of the particles is frequently changing. We also distinguish between the random motion of a molecule and bulk (ordered) movement. An example of the latter is the flight of a solid object such as a pebble thrown in the air. We refer to changes in energy associated with bulk motion or transport of matter as work. In () is the heat supplied to the system and the work done on the system. The internal energy E is a state variable and its differential is exact, i.e. dE depends only on the initial and final state and is independent of the process leading to the change. On the other hand "heat" and "work" are not state variables and the partition into heat and work depends on the process. Hence, the difference in notation: dE, but and We have not yet defined the variables P, T and . We want to do this in such a way as to allow us to write
(5)
or
We now define the temperature, pressure, and chemical potential as
(6)
It is important to note that our basic assumption is that all allowed states are equally likely. The second law of thermodynamics now becomes the statement that a closed system will tend to approach a macroscopic state which can be achieved the most possible ways. The conventional mathematical formulation of the second law () on the other hand only becomes an essentially trivial matter of definition. We must next show that these definitions lead to familiar looking results- otherwise they would not be useful.
Next: The zeroth law of Up: Boltzmann statistics Previous: The historical origin of
Birger Bergersen
1998-09-14 | 878 | 4,358 | {"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} | 2.734375 | 3 | CC-MAIN-2017-43 | latest | en | 0.931721 |
https://bestcourseracourse.com/coursera-specializations-probabilistic-graphical-models | 1,582,764,554,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00091.warc.gz | 281,654,545 | 6,822 | ## Probabilistic Graphical Models Specialization
Start Date: 02/23/2020
Course Type: Specialization Course
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Probabilistic graphical models (PGMs) are a rich framework for encoding probability distributions over complex domains: joint (multivariate) distributions over large numbers of random variables that interact with each other. These representations sit at the intersection of statistics and computer science, relying on concepts from probability theory, graph algorithms, machine learning, and more. They are the basis for the state-of-the-art methods in a wide variety of applications, such as medical diagnosis, image understanding, speech recognition, natural language processing, and many, many more. They are also a foundational tool in formulating many machine learning problems.
#### Course Introduction
Probabilistic Graphical Models. Master a new way of reasoning and learning in complex domains Probabilistic Graphical Models Specialization In this specialization, you will learn the fundamental concepts and algorithms of probabilistic graphical models. You will start by understanding the core concepts and algorithms used to transform any graph into a probabilistic model. You will then learn the techniques to transform any model into a computer program that can run at regular intervals and generate random data for analysis. You will continue to work on a probabilistic model for a while, learning the techniques needed to transform any model to a state-of-the-art one. You will then implement a program that generates random data for analysis. You will then share your model with others to see how others solve similar problems. The data that you share with others will form the basis for your peer-reviewed paper, which will be a peer-reviewed short book that contains all the code needed to evaluate your model. You will then get feedback from your peers on their solutions to your problem solving it. By working on a solo assignment, you will see how your solution is evaluated by others. Upon completing this course, you will be able to: 1. Solve a probabilistic graph 2. Solve for a node in a graph 3. Solve for a node in a graph with other 4. Solve for a node in a graph with other nodes
#### Course Tag
Inference Bayesian Network Belief Propagation Graphical Model
#### Related Wiki Topic
Article Example
Probabilistic programming language A probabilistic programming language (PPL) is a programming language designed to describe probabilistic models and then perform inference in those models. PPLs are closely related to graphical models and Bayesian networks, but are more expressive and flexible. Probabilistic programming represents an attempt to "[unify] general purpose programming with probabilistic modeling."
Probabilistic soft logic Probabilistic soft logic (PSL) is a framework for collective, probabilistic reasoning in relational domains. PSL uses first order logic rules as a template language for graphical models over random variables with soft truth values from the interval [0,1].
Graphical model A graphical model or probabilistic graphical model (PGM) is a probabilistic model for which a graph expresses the conditional dependence structure between random variables. They are commonly used in probability theory, statistics—particularly Bayesian statistics—and machine learning.
Probabilistic classification Some classification models, such as naive Bayes, logistic regression and multilayer perceptrons (when trained under an appropriate loss function) are naturally probabilistic. Other models such as support vector machines are not, but methods exist to turn them into probabilistic classifiers.
Probabilistic classification Binary probabilistic classifiers are also called binomial regression models in statistics. In econometrics, probabilistic classification in general is called discrete choice.
Probabilistic programming language A probabilistic relational programming language (PRPL) is a PPL specially designed to describe and infer with probabilistic relational models (PRMs).
Russ Salakhutdinov He specializes in deep learning, probabilistic graphical models, and large-scale optimization.
Graphical model This type of graphical model is known as a directed graphical model, Bayesian network, or belief network. Classic machine learning models like hidden Markov models, neural networks and newer models such as variable-order Markov models can be considered special cases of Bayesian networks.
Graphical models for protein structure Graphical models can still be used when the variables of choice are continuous. In these cases, the probability distribution is represented as a multivariate probability distribution over continuous variables. Each family of distribution will then impose certain properties on the graphical model. Multivariate Gaussian distribution is one of the most convenient distributions in this problem. The simple form of the probability, and the direct relation with the corresponding graphical model makes it a popular choice among researchers.
Graphical models for protein structure Graphical models have become powerful frameworks for protein structure prediction, protein–protein interaction and free energy calculations for protein structures. Using a graphical model to represent the protein structure allows the solution of many problems including secondary structure prediction, protein protein interactions, protein-drug interaction, and free energy calculations.
Graphical models for protein structure Gaussian graphical models are multivariate probability distributions encoding a network of dependencies among variables. Let formula_15 be a set of formula_16 variables, such as formula_16 dihedral angles, and let formula_18 be the value of the probability density function at a particular value "D". A multivariate Gaussian graphical model defines this probability as follows:
Graphical model Generally, probabilistic graphical models use a graph-based representation as the foundation for encoding a complete distribution over a multi-dimensional space and a graph that is a compact or factorized representation of a set of independences that hold in the specific distribution. Two branches of graphical representations of distributions are commonly used, namely, Bayesian networks and Markov random fields. Both families encompass the properties of factorization and independences, but they differ in the set of independences they can encode and the factorization of the distribution that they induce.
Graphical model The framework of the models, which provides algorithms for discovering and analyzing structure in complex distributions to describe them succinctly and extract the unstructured information, allows them to be constructed and utilized effectively. Applications of graphical models include causal inference, information extraction, speech recognition, computer vision, decoding of low-density parity-check codes, modeling of gene regulatory networks, gene finding and diagnosis of diseases, and graphical models for protein structure.
Daphne Koller In 2009, she published a textbook on probabilistic graphical models together with Nir Friedman. She offered a free online course on the subject starting in February 2012.
Probabilistic soft logic In recent years there has been a rise in the approaches that combine graphical models and first-order logic to allow the development of complex probabilistic models with relational structures. A notable example of such approaches is Markov logic networks (MLNs). Like MLNs PSL is a modelling language (with an accompanying implementation) for learning and predicting in relational domains. Unlike MLNs, PSL uses soft truth values for predicates in an interval between [0,1]. This allows for the integration of similarity functions in the into models. This is useful in problems such as Ontology Mapping and Entity Resolution. Also, in PSL the formula syntax is restricted to rules with conjunctive bodies.
Probabilistic classification Commonly used loss functions for probabilistic classification include log loss and the mean squared error between the predicted and the true probability distributions. The former of these is commonly used to train logistic models.
Probabilistic voting model Probabilistic voting models are usually preferred to traditional Downsian median voter models, as in the former all voters have an influence on the policy outcome, whereas in the latter all power rests in the hands of the pivotal voter or group. For instance, in models where young and old (or rich and poor) voters have conflicting interests, probabilistic voting models predict that the winning candidate strikes a balance between the different interests in her/his policy platform. Due to the smooth mapping between the distribution of policy preferences and the political outcomes, this model has proven to be very tractable and convenient to use in dynamic models with repeated voting.
Graphical models for protein structure There are two main approaches to use graphical models in protein structure modeling. The first approach uses discrete variables for representing coordinates or dihedral angles of the protein structure. The variables are originally all continuous values and, to transform them into discrete values, a discretization process is typically applied. The second approach uses continuous variables for the coordinates or dihedral angles.
Graphical models for protein structure Markov random fields, also known as undirected graphical models are common representations for this problem. Given an undirected graph "G" = ("V", "E"), a set of random variables "X" = ("X") indexed by "V", form a Markov random field with respect to "G" if they satisfy the pairwise Markov property:
NESSUS Probabilistic Analysis Software NESSUS is a general-purpose, probabilistic analysis program that simulates variations and uncertainties in loads, geometry, material behavior and other user-defined inputs to compute probability of failure and probabilistic sensitivity measures of engineered systems. Because NESSUS uses highly efficient and accurate probabilistic analysis methods, probabilistic solutions can be obtained even for extremely large and complex models. The system performance can be hierarchically decomposed into multiple smaller models and/or analytical equations. Once the probabilistic response is quantified, the results can be used to support risk-informed decisions regarding reliability for safety critical and one-of-a-kind systems, and to maintain a level of quality while reducing manufacturing costs for larger quantity products. | 1,947 | 10,768 | {"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} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.906264 |
https://mathoverflow.net/questions/265522/dualizing-the-notion-of-topological-space | 1,606,825,911,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00051.warc.gz | 378,873,327 | 32,534 | # Dualizing the notion of topological space
$\require{AMScd}$ Defining a topological space on a set $X$ is equivalent to designating certain subobjects of $X$ in ${\bf Set}$ (monomorphisms into $X$ up to equivalence) as open. The requirements on the open sets of a topological space $X$ are equivalent to requiring the following:
1. $X \rightarrow X$ and $\emptyset \rightarrow X$ are open.
2. If $X_i \rightarrow X$ are open subobjects of $X$ for finite $i \in I$ then so is their product in the category of subobjects of $X$
3. If $X_i \rightarrow X$ are open subobjects of $X$ for $i \in I$, then so is their coproduct in the category of subobjects of $X$.
My question is about what happens when one dualizes this notion:
Let $X$ be a set and consider a subset of its quotient objects $\mathcal{S}$ such that:
1. $X \rightarrow X \in \mathcal{S}$ and $X \rightarrow \{ * \} \in \mathcal{S}$.
2. If $X \rightarrow X_i \in \mathcal{S}$ for finite $i \in I$ then $\amalg_{i \in I} (X \rightarrow X_i) \in \mathcal{S}$, where the coproduct is taken in the category of quotient objects of $X$.
3. If $X \rightarrow X_i \in \mathcal{S}$ for $i \in I$, then $\prod (X \rightarrow X_i) \in \mathcal{S}$, where the product is taken in the category of quotient objects of $X$.
Does this structure arise anywhere in practice? What is known about this notion of 'cotopological spaces'? Is there a place I can learn about them?
Note: in the case of a set, the category of quotient objects is equivalent to the category of equivalence relations on the set in question. A pairwise coproduct of equivalence relations is then the equivalence relation generated by two equivalence relations. The product of a collection of equivalence relations is their intersection. Thus this notion of a 'cotopological space' is equivalent to putting a set of equivalence relations on a set $X$ closed under pairwise sum and arbitrary intersection, where the sum of two equivalence relations is the relation generated by them. We also require that collection includes the equivalence relation where all points are equivalent and the equivalence relation where no two distinct points are equivalent.
Edits:
1. Morphisms in the Category of Cotopological Spaces.
Suppose $X$ and $Y$ are topological spaces with a set map $f: X \rightarrow Y$. We say $f$ is continuous if, for every open subobject $V \rightarrow Y$, the following pullback gives an open subobject $U \rightarrow X$: \begin{CD} X @>>> Y\\ @AAA @AAA\\ U @>>> V \end{CD}
Analogously, suppose there is a set map $f : X \rightarrow Y$ of cotopological spaces $X$ and $Y$. We say $f$ is cocontinuous if for each open quotient object $X \rightarrow P$ the following pushout diagram forms a quotient object $Y \rightarrow Q$: \begin{CD} X @>>> Y\\ @VVV @VVV\\ P @>>> Q \end{CD}
In terms of equivalence relations this translates to requiring that if $R$ is an open equivalence relation on $X$ then the relation generated by $R'$ where $x'R'y'$ if and only if $x' = f(x)$ and $y' = f(y)$ for $x, y \in X$ such that $xRy$ is open.
1. Each metric space $(X, d)$ induces a cotopological space in the following way:
Each open ball $B_{\epsilon}(x)$ induces an equivalence relation $R_{\epsilon} (x)$ where $y R_{\epsilon} (x) z \iff (z = y$ or $z, y \notin B_{\epsilon} (x))$. Form the set $T = \{ R_{\epsilon} (x) : \epsilon \in \mathbb{R}, x \in X \}$ and close it under intersection of equivalence relations. This forms a co-topological space.
1. Each topological space $(X, T)$ induces a cotopological space in the following way:
Let $B$ be a basis for $X$. Each open set $U \in B$ induces an equivalence relation $R(U)$ where $xR(U)y$ when $x = y$ or $x, y \notin U$. Form the set $S = \{ R(U) : U \in B\}$ and close it under intersection of equivalence relations. This forms a co-topological space.
1. If we start instead with a topological space $(X, C)$ where $C$ is the set of closed sets on $X$, we end up with a space $(X, \mathcal{S})$ where $\mathcal{S}$ is a set of equivalence relations closed under finite intersections and arbitrary joins, where a join of equivalence relations is the smallest equivalence relation containing them.
2. Limits and colimits in the topology of Cotopological Spaces.
Note: the forgetful functor $F : {\bf CoTop} \rightarrow {\bf Set}$ has a left and right adjoint and therefore preserves limits and colimits. Hence if $(X_i, \mathcal{S}_i) \cong \text{ colim } \Phi$ then $X_i \cong \text{ colim } F \circ \Phi$, and the same for limits.
Take cotopological spaces $(X_i, \mathcal{S}_i)_{i \in I}$. We define a cotopology $\mathcal{S}$ on $\amalg_{i \in I} X_i$ as follows: a set $R \subset \amalg_{i \in I} X_i \times \amalg_{i \in I} X_i$ is a relation in $\mathcal{S}$ if and only if there are $\{ R_i \}_{i \in I}$, with $R_i \in \mathcal{S}_i$, such that $x_i R y_j$ for $x_i \in X_i$, $y_j \in X_j$ if and only if $i = j$ and $x_i R_i y_i$.
Defining the product cotopology, "cofinal", and "coinitial" topologies for the case of direct and inverse limits is similarly straightforward.
• Cool. So now we can define "cocontinuous map", "quotient cotopology", "subcospace cotopology". And maybe some coseparation axioms. And work out what colimits and limits in this category are. Oh, and think of some good examples. – Tom Goodwillie Mar 25 '17 at 17:07
• I like your idea a lot, but keep in mind that a topology can just as well be defined by closed sets, and if you dualize on the closed sets, you get something presumably very different, because one place where the symmetry breaks down horribly is that a subobject has a complement whereas a quotient object does not have a canonical transversal (or whatever that would be). One can also define topologies using closure axioms and maybe that would also give some interesting dual notion. – Gro-Tsen Mar 25 '17 at 18:14
• Tom Goodwillie is pointing out something important: one should have good examples to motivate why one would introduce or explore some concept or else there is nothing to guide what is being done. That "dualizing is a natural thing to do" is not by itself a compelling reason to think about something: it should illuminate a concept of prior interest or shed light on something that is hard to understand or something like that (e.g., affine group schemes genuinely illuminate things, but not because of being a "dual" concept). Otherwise the concept has no reason to be productive vs. a dead end. – nfdc23 Mar 25 '17 at 18:52
• @nfdc23 Those considerations are appropriate if one is applying for a grant or helping a student find a thesis problem, but there are many great MO questions whose original motivation was shakier than "dualizing is a natural thing to do". Also, the OP has produced a fairly rich supply of examples, namely all metric spaces. Odds are this construction will produce something familiar - bornological spaces, perhaps? - and my curiosity is piqued, at least. – Paul Siegel Mar 25 '17 at 19:19
• No one made the obligatory joke that maps of cotopological spaces are "ntinuous maps"? – PrimeRibeyeDeal Oct 4 '18 at 20:16
I believe, this notion is closely related to the notion of the coarse structure which indeed had found many beautiful applications in geometric group theory and algebraic topology (including proofs of the Novikov conjecture for a lot of groups). For introduction, I'd recommend Lectures on coarse geometry by John Roe where in Chapter 2 he explains how to give a coarse structure to every metric space or a topological space (the latter upon fixing a compactification, which in your example should be Stone–Čech).
The idea of coarse geometry/topology is exactly dual to that of topology: one doesn't look at what happens at “small distances”, but rather at what happens at “large distances” in a space, and I believe, this is exactly what the definition of the cotopology also does (it defines what “large distances” mean).
However, the original definition of a coarse structure is done in a way as to define what are subsets with “not so large” distances, but I believe, it's a bit like defining topology by open subsets or closure operation. I'll try to sketch below the translation between the two notions.
Definition. A coarse structure on a set $$X$$ is a collection $$\mathcal E$$ of subsets of $$X \times X$$ called controlled sets, so that $$\mathcal E$$ contains the identity relation, is closed under taking subsets, inverses (transposes), and finite unions, and is closed under composition of relations.
The main intuitive point in the connection can be already seen in the examples: the metric cotopological structure given by $$R_r(x)$$ should correspond to the family of controlled sets generated by $$\{(x,y)\mid d(x,y).
Coarse to cotopological: Given a coarse structure $$\mathcal E$$, define a family of equivalence relations $$\sim_{E,K}$$, where $$E$$ runs through $$\mathcal E$$ and $$K$$ through subsets of $$X$$ in the following way: $$y \sim_{E,K} z$$ iff either $$y=z$$ or $$(x,y),(x,z)\not\in E\cup E^{-1}$$ for all $$x\in K$$. Observe that these equivalence relations are automatically symmetric, so it will be enough for us to work only with $$E=E^{-1}\in\mathcal E$$.
• This family is trivially closed under arbitrary intersections: the intersection of $$\sim_{E_i,K_i}$$ is just $$\sim_{\bigcap E_i,\bigcup K_i}$$, and $$\bigcap E_i\in \mathcal E$$ because $$\mathcal E$$ is closed under taking subsets.
• Finally, it is closed under passing to sus of two equivalence relations again using the properties of $$\mathcal E$$ as such as passing to subsets, finite unions and compositions (it's by no means obvious, but I think, I have figured out at least a sketch of a bit tedious proof; to not make the post too long, I omit it, but I'm happy to discuss this in detail if needed). I can sketch the proof in the metric case: The equivalence relation generated by $$R_{r}(K)$$ and $$R_{r}(K')$$ collapses everything exactly outside the ball of radius $$r$$ around $$K\cap K'$$, so it's just $$R_{r}(K\cap K')$$. If the radii are different, one has to work a bit more, but the flexibility of the coarse structure allows that.
Cotopological to coarse: It easy to see that an arbitrary intersection of coarse structures is again a coarse structure. Let's say that a coarse structure $$\mathcal E$$ is compatible with a family $$\sim_\alpha$$ of equivalence relations (i.e. a cotopology) if all $$\sim_\alpha$$ belong to the cotopology generated by $$\mathcal E$$. The intersection of all compatible coarse structures should then be the coarse structure which induces the given cotopology, although I haven't checked this carefully.
In any case, the possible connection here seems worth examining in detail.
• Ah... so "topological" is closely related to "arse". – WhatsUp Nov 23 '19 at 12:16 | 2,806 | 10,837 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-50 | longest | en | 0.869516 |
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Digital Electronics Certification Exam Tests
Digital Electronics Practice Test 145
# Pseudo NMOS Dynamic Operation Quiz Questions and Answers PDF - 145
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## Pseudo NMOS Dynamic Operation Quiz Questions : Test 145
MCQ 721: Pseudo-NMOS inverter provide large
A) fan-in
B) fan-out
C) fan-high
D) fan-low
MCQ 722: At room temperature, reference voltage 'VR is
A) −1.90 V
B) −2.90 V
C) −1.32 V
D) −2.32 V
MCQ 723: Logic which does not require any feedback. It simply outputs the input according to the logic designed is
A) sequential logic
B) combinational logic
C) core diode logic
D) random logic
MCQ 724: System in which A or B is high than output will be one, is represented as
A) Y=bar(A)+bar(B)
B) Y=A+B
C) bar(Y)=A+B
D) bar(Y)=bar(A+B)
MCQ 725: BiCMOS as compare to CMOS typically features
A) less delay
B) more delay
C) zero delay
D) equal delay
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## Question:
I want get a matrix of random values with zero on main diagonal. Then the following codes work to integer values, but I actually want float value.
``````def matriz_custo(n):
numeros = sample(range(35,120+1),(n-1)*n)
c = np.array([numeros[i:i+n] for i in range(0,len(numeros),n)])
np.fill_diagonal(c, 0)
c[0][n-1] = 0
return (c)
``````
so I tried:
``````def matriz_custo_new(n):
numeros = np.random.uniform(low=27.32, high=37.25, size=(n-1,n))
c = np.array([numeros[i:i+n] for i in range(0,len(numeros),n)])
r = np.arange(n-1)
c[:,r,r] = np.inf
c[0][n-1] = 0
return (c)
``````
then happen to create a extra dimension. I want to access it like c[0][0], otherwise I have to change all of my code.
[EDIT]
I get it now! :
`````` def matriz_custo_new(n):
numeros = np.random.uniform(low=27.32, high=37.25, size=((n-1)*n,))
c = np.array([numeros[i:i+n] for i in range(0,len(numeros),n)])
r = np.arange(n-1)
c[r,r] = 0
c[0][n-1] = 0
return (c)
``````
``````def custom_matrix(low, high, n):
array = np.random.uniform(low, high, (n, n))
for i in range(n):
array[i][i] = 0
return array
custom_matrix(0, 1, 3)
# returns
# [[0. 0.56331153 0.09965334]
# [0.10010369 0. 0.79204369]
# [0.66917184 0.67228158 0. ]]
``````
The `fill_diagonal` function in the NumPy package you’re using will allow you to fill the diagonal of any array regardless of its dimension.
``````import numpy as np
# Generates a 5x5 matrix/nested list of random float values
matrix = np.random.rand(5, 5)
# Sets diagonal values to zero
np.fill_diagonal(matrix, 0)
print(matrix)
# Example Matrix
[[0. 0.50388405 0.62069521 0.93842045 0.75608882]
[0.35300814 0. 0.48357286 0.59268458 0.15213251]
[0.63591576 0.61143272 0. 0.91271144 0.53997033]
[0.01979816 0.66435396 0.03416005 0. 0.02524777]
[0.62694323 0.02307084 0.32458185 0.7663526 0. ]]
# Accessing specific elements
matrix[1][3]
0.5926845768187102
``````
Documentation here on the functionality. Cheers!
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www.hotelshagun.co.in | 1,560,992,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00546.warc.gz | 767,265,368 | 7,986 | ### Centripetal Accerleration Example Washer Dryer
##### Centripetal acceleration Practice Problems Online | …
Centripetal Acceleration on Brilliant, the largest community of math and science problem solvers.
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where a c is the centripetal acceleration, m is the mass of the object, moving at velocity v along a path with radius of curvature r. Centrifugal vs. Centripetal Force Examples Some common examples of centrifugal force at work are mud flying off a tire and children feeling a force pushing them outwards while spinning on a roundabout.
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Frequent Visitor
## Applying a growth percentage to a measure result
I have a data table that has the following fields:
• Date
• LocationID
• EventID
• Count
I have a measure that calculates the equivalent count last year:
• CALCULATE(sum('Total Attendance'[Count]),DATEADD('Date Table'[Date],-364,DAY))
I have a table of growth targets that has the following fields:
• Date
• LocationID
• EventID
• Target%
This table has multiple records for each LocationID/EventID. For example:
01/01/17, LDN, AM, 10%
01/01/18, LDN, AM, 8%
In this example, the growth rate is 10% from 01/01/17 to 31/12/17 and then reduces to 8%.
I want to work out a measure that multiplies the CountLastYr by the relevant growth percentage for that LocationID, EventID and date.
Can anyone help?
6 REPLIES 6
Frequent Visitor
I'm sorry those measures throw up errors for me. I've added an expected result column to my data that shows the percentage target that I expect to appear against each count. This is what I would use to add to the Count LastYr measure.
Updated sample data
Frequent Visitor
Thanks for the input but that's not the results I'm looking for.
I want my count table to show, for every date that there is a count, the CountLastYr * (1 + Target). The target percentage being the latest target set for that particular count type, location and event type.
Community Support
Hi @mikemagill,
To create the measures as below. If it doesn't meet your requirement, kindly share your excepted result to me.
```tar = var ma = CALCULATE(MAX('Date Table'[Date]),ALLEXCEPT('Date Table','Date Table'[LocationId],'Date Table'[EventNameId],'Date Table'[CountType]))
return
CALCULATE(MAX('Date Table'[Target]),FILTER('Date Table','Date Table'[Date]=ma))
```
```Measure =
VAR da =
MAX ( 'Date Table'[Date] )
RETURN
CALCULATE (
SUM ( 'Total Attendance'[Count] ),
FILTER ( ALL ( 'Table' ), 'Table'[Date] >= da - 364 && 'Table'[Date] <= da )
)
* 1+MAXX(ALLEXCEPT('Date Table','Date Table'[CountType],'Date Table'[EventNameId],'Date Table'[LocationId]),[tar])
```
Regards,
Frank
Community Support Team _ Frank
If this post helps, then please consider Accept it as the solution to help the others find it more quickly.
Frequent Visitor
Here is a link to a spreadsheet containing two tables; CountData and Targets.
Sample Data
There should be sufficient data in there to demonstrate what I want to achieve.
Community Support
Hi @mikemagill,
Please check the folloing steps as below. If it doesn't meet your requirement, kindly share your excepted result to me.
1. Create a date table and create relationships between tables.
`Table = CALENDARAUTO()`
2. Create a measure as below.
```Measure =
VAR da =
MAX ( 'Date Table'[Date] )
VAR tar =
MAX ( 'Date Table'[Target] )
RETURN
CALCULATE (
SUM ( 'Total Attendance'[Count] ),
FILTER ( ALL ( 'Table' ), 'Table'[Date] >= da - 364 && 'Table'[Date] <= da )
)
* tar
```
Regards,
Frank
Community Support Team _ Frank
If this post helps, then please consider Accept it as the solution to help the others find it more quickly.
Community Support
Hi @mikemagill,
Could you please share your sample data or pbix to me? You can upload your file to onedrive and share the link here.
Regards,
Frank
Community Support Team _ Frank
If this post helps, then please consider Accept it as the solution to help the others find it more quickly.
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# Algebra
0
160
1
a solution contains 3% salt. How much water should be added to 60 ounces of this solution to make a 1.8% solution?
Sep 20, 2021 | 52 | 153 | {"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-2022-27 | latest | en | 0.937482 |
http://www.chegg.com/homework-help/questions-and-answers/1-make-class-rational-provide-least-following-methos-constructors-rational-r1-new-rational-q1322197 | 1,474,793,637,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660158.72/warc/CC-MAIN-20160924173740-00192-ip-10-143-35-109.ec2.internal.warc.gz | 373,788,489 | 15,327 | 1. Make a class Rational to provide at least following methos and constructors:
Rational r1= new Rational (3,5);
Rational r2= new Rational (4,7);
Rational r3=r1.add(r2);
// also sub,mult,divide methods
Other that four operations you should provide methods that you think is useful for user of your class.
2. Write a JFrame or JApplet (file name: TestRational.java) that performs the following tasks:
a) Define two arrays of size 10. Each element in the array references to a Rational object that you wrote in Problem 1 above.
Rational a[], b[];
a = new Rational[10];
b = new Rational[10];
b) Initialize the arrays in Question a) by creating random Rational numbers. Do this by generating two random numbers between 1 and 9 for each Rational and using the numbers as the numerator and denominator. Denominator should be greater than Numerator. You should generate random numbers with Math.random.
c) Display the contents of both arrays on the screen using JTextArea.
d) Add a Scrollbar to JTextArea for scrolling up or down.
e) For each public method in Rational class, create a corresponding JButton object. When the user clicks on a button, the actionPerformed listener should execute the corresponding Rational method for all the elements in the two arrays and store the results in a new array. You should then display the result array in the JTextArea using the setText method. For example, if the user hits the add button you should loop through all the elements in arrays and execute:
c[i] = a[i].add(b[i]);
f) If the user hits the sort button, uses the JTextArea method append to append the results of sorting.
## Best answer
### Get this answer with Chegg Study
Practice with similar questions
Q:
1. Make a class Rational to provide at least following methos and constructors: Rational r1= new Rational (3,5); Rational r2= new Rational (4,7); Rational r3=r1.add(r2); // also sub,mult,divide methods Other that four operations you should provid methos that you think is useful for user of your class. 2. Write a JFrame or JApplet (file name: TestRational.java) that performs the following tasks: a) Define two arrays of size 10. Each element in the array references to a Rational object that you wrote in Problem 1 above. Rational a[], b[]; a = new Rational[10]; b = new Rational[10]; Rational Numerator Denominator . . . . . b) Initialize the arrays in Question a) by creating random Rational numbers. Do this by generating two random numbers between 1 and 9 for each Rational and using the numbers as the numerator and denominator. Denominator should be greater than Numerator. You should generate random numbers with Math.random. c) Display the contents of both arrays on the screen using JTextArea. d) Add a Scrollbar to JTextArea for scrolling up or down. e) For each public method in Rational class, create a corresponding JButton object. When the user clicks on a button, the actionPerformed listener should execute the corresponding Rational method for all the elements in the two arrays and store the results in a new array. You should then display the result array in the JTextArea using the setText method. For example, if the user hits the add button you should loop through all the elements in arrays and execute: c[i] = a[i].add(b[i]); f) If the user hits the sort button, uses the JTextArea method append to append the results of sorting. Post rational and Test rational both
A: See answer
Q:
I really need help on this especially on the #2 Make a class Rational to provide at least following methos and constructors: Rational r1= new Rational (3,5); Rational r2= new Rational (4,7); Rational r3=r1.add(r2); // also sub,mult,divide methods Other that four operations you should provid methos that you think is useful for user of your class. II) Write a JFrame or JApplet (file name: TestRational.java) that performs the following tasks: a) Define two arrays of size 10. Each element in the array references to a Rational object that you wrote in Problem 1 above. Rational a[], b[]; a = new Rational[10]; b = new Rational[10]; Rational Numerator Denominator . . . . . b) Initialize the arrays in Question a) by creating random Rational numbers. Do this by generating two random numbers between 1 and 9 for each Rational and using the numbers as the numerator and denominator. Denominator should be greater than Numerator. You should generate random numbers with Math.random. c) Display the contents of both arrays on the screen using JTextArea. d) Add a Scrollbar to JTextArea for scrolling up or down. e) For each public method in Rational class, create a corresponding JButton object. When the user clicks on a button, the actionPerformed listener should execute the corresponding Rational method for all the elements in the two arrays and store the results in a new array. You should then display the result array in the JTextArea using the setText method. For example, if the user hits the add button you should loop through all the elements in arrays and execute: c[i] = a[i].add(b[i]); f) If the user hits the sort button, uses the JTextArea method append to append the results of sorting.
A: See answer
Show more questions [+] | 1,155 | 5,166 | {"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} | 3.109375 | 3 | CC-MAIN-2016-40 | latest | en | 0.83809 |
https://www.gospel10.com/how-to-calculate-a-discount-rate-in-excel-a-step-by-step-guide/ | 1,720,908,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00340.warc.gz | 712,810,161 | 56,452 | # How to Calculate a Discount Rate in Excel: A Step-by-Step Guide
Calculating a discount rate in Excel involves determining the present value of future cash flows by applying a specific interest rate. This calculation is crucial in financial decision-making, such as evaluating investments or determining the value of a company, as it helps assess the time value of money.
The discount rate represents the opportunity cost of investing capital, reflecting the return that investors could earn if they invested their funds in other ventures. Historically, the development of sophisticated mathematical models and computing technology has significantly enhanced the accuracy and efficiency of discount rate calculations.
In this article, we will explore the steps involved in calculating a discount rate in Excel, including understanding the relevant formulas and applying them to practical scenarios.
## How to Calculate a Discount Rate in Excel
Calculating a discount rate in Excel is a crucial aspect of financial analysis, enabling the valuation of future cash flows and informed decision-making. Key aspects to consider include:
• Formula selection
• Time value of money
• Risk and uncertainty
• Cash flow estimation
• Perpetuity and annuity calculations
• Excel functions and shortcuts
• Sensitivity analysis
• Model validation
• Best practices and limitations
Understanding these aspects helps ensure accurate and reliable discount rate calculations, which are essential for evaluating investments, determining the value of companies, and making sound financial decisions. Examples include assessing the viability of a new product launch, valuing a potential acquisition, or determining the appropriate level of dividends to pay shareholders.
### Formula Selection
Formula selection is a critical aspect of calculating a discount rate in Excel, as it determines the mathematical approach used to translate future cash flows into their present value. There are several key factors to consider when selecting the appropriate formula:
• Weighted Average Cost of Capital (WACC): Considers the cost of both debt and equity financing, weighted by their respective proportions in the capital structure.
• Capital Asset Pricing Model (CAPM): Determines the discount rate based on the risk-free rate, beta coefficient, and expected market return.
• Dividend Discount Model (DDM): Uses the expected dividends and a growth rate to calculate the discount rate.
• Adjusted Present Value (APV): Considers the impact of future growth opportunities and risk factors on the discount rate.
The choice of formula depends on factors such as the availability of data, the type of investment being evaluated, and the risk-return profile of the project. By carefully selecting the appropriate formula, financial analysts can obtain a more accurate and reliable discount rate, leading to better decision-making.
### Time Value of Money
The time value of money (TVM) is a fundamental concept in finance that recognizes the value of money changes over time. Money today is worth more than the same amount of money in the future due to its potential earning power. This concept is critical in “how to calculate a discount rate in excel” because it forms the basis for determining the present value of future cash flows.
In calculating a discount rate in Excel, TVM is applied through the use of discounting formulas. These formulas, such as the Net Present Value (NPV) and Internal Rate of Return (IRR), incorporate the time value of money to determine the present value of future cash flows. By considering the time value of money, financial analysts can make informed decisions about investments and projects, as it provides a more accurate representation of their true value.
Real-life examples of TVM within “how to calculate a discount rate in excel” include valuing bonds, determining the cost of capital, and evaluating the viability of investment projects. In bond valuation, the present value of future coupon payments and the principal repayment is calculated using discounting formulas that incorporate the time value of money. Similarly, in cost of capital calculations, the weighted average cost of debt and equity is adjusted based on the time value of money to determine the appropriate discount rate for evaluating projects.
Understanding the connection between time value of money and “how to calculate a discount rate in excel” is crucial for accurate financial decision-making. By incorporating TVM into discount rate calculations, financial analysts can assess the present value of future cash flows more effectively, leading to better investment decisions and improved financial outcomes.
### Risk and uncertainty
When calculating a discount rate in Excel, it is crucial to consider the inherent risk and uncertainty associated with future cash flows. These factors can significantly impact the accuracy and reliability of the calculated discount rate, and thus, the resulting investment decisions.
• Market risk: Refers to the volatility and unpredictability of financial markets, which can affect the value of investments and the expected returns. For example, changes in interest rates, inflation, or economic conditions can influence the discount rate used to evaluate projects.
• Company-specific risk: Relates to the unique characteristics and circumstances of a particular company or industry. Factors such as competition, technological advancements, or regulatory changes can impact the company’s future cash flows and, consequently, the appropriate discount rate.
• Project-specific risk: Encompasses the uncertainties associated with a specific investment project. This includes factors such as technical feasibility, execution risk, and the potential for unforeseen events that may affect the project’s outcome.
• Inflation risk: Considers the impact of inflation on the value of future cash flows. Inflation can erode the purchasing power of money over time, which needs to be taken into account when determining the discount rate to use.
Understanding and incorporating risk and uncertainty into the calculation of a discount rate is essential for making informed investment decisions. By carefully assessing these factors and incorporating appropriate risk premiums, financial analysts can obtain a more realistic and robust discount rate that better reflects the potential risks and uncertainties associated with future cash flows.
### Cash flow estimation
Cash flow estimation is a critical aspect of “how to calculate a discount rate in excel” as it provides the foundation for determining the present value of future cash flows. Accurate and reliable cash flow estimation is essential for making informed investment and financial decisions.
• Historical data analysis: Involves examining past financial statements and cash flow patterns to identify trends and establish a baseline for future cash flow projections.
• Scenario planning: Considers different possible outcomes and their impact on future cash flows. This helps assess the sensitivity of the discount rate to various assumptions.
• Industry analysis: Compares the company or project to industry benchmarks and peers to identify potential growth opportunities and risks that may affect cash flows.
• Management judgment: Incorporates the insights and expertise of management to make informed assumptions about future cash flows, especially for new or innovative projects.
Effective cash flow estimation in “how to calculate a discount rate in excel” requires a combination of quantitative analysis, scenario planning, and qualitative judgment. By carefully considering these aspects, financial analysts can develop more accurate and robust discount rates that better reflect the potential risks and uncertainties associated with future cash flows.
### Perpetuity and annuity calculations
Perpetuity and annuity calculations are fundamental components of “how to calculate a discount rate in excel.” Understanding their relationship is crucial for accurate and reliable discount rate calculations, which are essential for making informed investment and financial decisions.
A perpetuity refers to a constant stream of cash flows that occur at regular intervals, continuing indefinitely. An annuity, on the other hand, represents a series of cash flows that occur at regular intervals for a finite period. Both perpetuities and annuities are important concepts in finance, as they provide a framework for valuing and comparing different types of cash flows.
In the context of “how to calculate a discount rate in excel,” perpetuities and annuities are used to model future cash flows. By applying appropriate discount rates, financial analysts can determine the present value of these cash flows, which is critical for evaluating the viability and attractiveness of investment opportunities. For example, in valuing a bond that pays regular coupon payments indefinitely, the present value can be calculated using the perpetuity formula and an appropriate discount rate.
Understanding the connection between perpetuity and annuity calculations and “how to calculate a discount rate in excel” is essential for financial professionals and investors alike. It enables them to make informed decisions about investments, assess the value of companies, and manage financial risks effectively.
### Excel functions and shortcuts
Excel functions and shortcuts play a vital role in “how to calculate a discount rate in excel” as they automate complex calculations, enhance efficiency, and reduce the risk of errors. These functions and shortcuts provide financial analysts with powerful tools to streamline the process of determining the discount rate, which is crucial for making sound investment and financial decisions.
One of the key functions used in “how to calculate a discount rate in excel” is the Net Present Value (NPV) function. This function calculates the present value of a series of future cash flows by applying a specified discount rate. By utilizing the NPV function and incorporating appropriate shortcuts, such as keyboard shortcuts for cell navigation and formula entry, financial analysts can quickly and accurately determine the NPV of various investment scenarios.
Another important aspect of the connection between Excel functions and shortcuts and “how to calculate a discount rate in excel” lies in sensitivity analysis. Sensitivity analysis involves assessing the impact of changes in input variables on the calculated discount rate. Excel’s powerful data tables and scenario manager allow financial analysts to perform sensitivity analysis efficiently, enabling them to explore different assumptions and their effects on the discount rate and investment decisions.
Understanding the relationship between Excel functions and shortcuts and “how to calculate a discount rate in excel” is crucial for financial professionals and investors alike. By leveraging these functions and shortcuts effectively, they can enhance their productivity, improve the accuracy of their calculations, and make more informed investment decisions. This understanding empowers them to analyze complex financial scenarios, assess investment opportunities, and manage financial risks with greater confidence.
### Sensitivity analysis
Sensitivity analysis is a crucial aspect of “how to calculate a discount rate in excel” as it allows for a deeper understanding of the impact that changes in input variables can have on the calculated discount rate.
• Impact of changing cash flows: Evaluating the sensitivity of the discount rate to changes in the magnitude and timing of future cash flows is important to assess the robustness of investment decisions.
• Discount rate variation: Analyzing the effect of different discount rates on the calculated present value of future cash flows helps determine the impact of varying financing costs or risk assumptions.
• Scenario analysis: Considering multiple scenarios with different combinations of input variables provides insights into the range of potential outcomes and the associated discount rates.
• Break-even analysis: Sensitivity analysis can be used to determine the break-even point, where the net present value equals zero, which is critical for evaluating the viability of investment projects.
In summary, sensitivity analysis in “how to calculate a discount rate in excel” is vital for assessing the robustness of investment decisions, evaluating the impact of varying input variables, and understanding the range of potential outcomes. By incorporating sensitivity analysis into the discount rate calculation process, financial analysts gain a more comprehensive understanding of the risks and uncertainties associated with investment projects.
### Model validation
Model validation is a critical component of “how to calculate a discount rate in excel” as it ensures the accuracy and reliability of the calculated discount rate. Without proper model validation, there is a higher risk of making incorrect investment decisions due to an inaccurate assessment of the time value of money.
One key aspect of model validation in “how to calculate a discount rate in excel” is verifying the underlying assumptions and inputs used in the calculation. This includes examining the reasonableness of the projected cash flows, the selected discount rate, and any other relevant assumptions. By carefully scrutinizing these inputs, financial analysts can identify potential errors or biases that may affect the accuracy of the calculated discount rate.
Real-life examples of model validation within “how to calculate a discount rate in excel” include comparing the calculated discount rate to industry benchmarks or using sensitivity analysis to assess the impact of changing input variables on the discount rate. These validation techniques help ensure that the calculated discount rate is consistent with market norms and takes into account potential risks and uncertainties.
Understanding the importance of model validation in “how to calculate a discount rate in excel” is crucial for financial analysts and investors alike. By thoroughly validating their models, they can make more informed investment decisions, reduce the risk of errors, and enhance the overall reliability of their financial analysis.
### Best practices and limitations
Best practices and limitations play a crucial role in “how to calculate a discount rate in excel” as they establish guidelines for accurate and reliable discount rate calculations, while also highlighting the inherent constraints and challenges involved.
Adhering to best practices in “how to calculate a discount rate in excel” ensures that the calculated discount rate is consistent with industry standards and reflects sound financial principles. These best practices include carefully considering the time value of money, using appropriate risk-adjusted discount rates, and thoroughly documenting the assumptions and inputs used in the calculation. By following these best practices, financial analysts and investors can enhance the credibility and defensibility of their discount rate calculations.
Understanding the limitations of “how to calculate a discount rate in excel” is equally important. Discount rate calculations are inherently subject to estimation errors and uncertainties due to the difficulty in accurately predicting future cash flows and the appropriate risk premium. Additionally, the choice of discount rate can be influenced by subjective judgments and market conditions, which can impact the reliability of the calculated discount rate. Recognizing these limitations encourages financial analysts to exercise caution when interpreting and using discount rate calculations.
### FAQs on How to Calculate a Discount Rate in Excel
This section addresses frequently asked questions and clarifies common misconceptions regarding discount rate calculations in Excel. These FAQs provide practical guidance and insights to assist users in accurately determining discount rates for sound financial decision-making.
Question 1: What is the importance of using an appropriate discount rate?
Answer: Selecting an appropriate discount rate is crucial as it directly impacts the present value of future cash flows, which is fundamental to evaluating investments and making informed financial decisions.
Question 2: How do I choose the right discount rate formula for my specific scenario?
Answer: The choice of formula depends on factors such as the type of investment, the availability of data, and the risk-return profile of the project. Common formulas include WACC, CAPM, and DDM.
Question 3: How can I account for risk and uncertainty when calculating a discount rate?
Answer: To account for risk, investors typically apply a risk premium to the risk-free rate. This premium reflects the additional return required to compensate for the uncertainty associated with future cash flows.
Question 4: What are the common mistakes to avoid when calculating a discount rate in Excel?
Answer: Common mistakes include using outdated data, ignoring the time value of money, and failing to consider the impact of inflation and other economic factors.
Question 5: How can I validate my discount rate calculation?
Answer: Model validation techniques such as sensitivity analysis and scenario planning can help assess the robustness of the calculated discount rate and identify potential errors or biases.
Question 6: What are the advantages of using Excel for discount rate calculations?
Answer: Excel provides powerful functions and tools that automate calculations, enhance accuracy, and facilitate scenario analysis, making it a widely adopted tool for discount rate calculations.
These FAQs provide a concise overview of key considerations and best practices in discount rate calculations. Understanding these aspects enables users to confidently determine appropriate discount rates, leading to more informed financial decision-making. The next section delves deeper into advanced techniques and practical applications of discount rate calculations.
### Tips on Choosing the Right Discount Rate
Selecting an appropriate discount rate is crucial for accurate financial analysis and decision-making. Here are several actionable tips to guide you:
Tip 1: Understand the Concept of Time Value of Money: Recognize that the value of money changes over time due to its earning potential, and incorporate this concept into your discount rate calculations.
Tip 2: Consider Risk and Uncertainty: Assess the level of risk associated with the investment or project, and apply an appropriate risk premium to the risk-free rate to determine the discount rate.
Tip 3: Use Industry Benchmarks: Research industry averages and comparable companies to establish a reasonable range for the discount rate.
Tip 4: Employ Sensitivity Analysis: Test the impact of varying discount rates on the results of your analysis to gauge the sensitivity of your decisions.
Tip 5: Validate Your Model: Ensure the accuracy of your discount rate calculations by comparing them to external sources or using different valuation methods.
Tip 6: Document Assumptions Clearly: Transparently record the assumptions and inputs used in your discount rate calculation for future reference and auditability.
Tip 7: Stay Updated with Market Conditions: Monitor market trends and economic indicators to adjust your discount rate as necessary to reflect changing circumstances.
Tip 8: Seek Professional Advice if Needed: Consult with financial experts or advisors for guidance on complex or high-stakes discount rate determinations.
By following these tips, you can enhance the accuracy and reliability of your discount rate calculations, leading to more informed financial decisions and better investment outcomes.
These tips lay the foundation for exploring advanced techniques and practical applications of discount rate calculations in the next section, enabling you to navigate complex financial scenarios with confidence.
### Conclusion
This comprehensive exploration of “how to calculate a discount rate in excel” has highlighted several key points. Firstly, understanding the concept of time value of money is paramount, as it forms the basis for determining the present value of future cash flows. Secondly, incorporating risk and uncertainty into the discount rate calculation is crucial for making informed investment decisions. Finally, utilizing industry benchmarks, sensitivity analysis, and model validation techniques enhances the accuracy and reliability of discount rate calculations.
These main points interconnect to emphasize the importance of a well-calculated discount rate in financial analysis and decision-making. By considering the time value of money, assessing risk, and employing appropriate techniques, investors and financial professionals can make more informed choices, leading to improved investment outcomes and better overall financial management. | 3,608 | 21,081 | {"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-2024-30 | latest | en | 0.89932 |
https://exceptionshub.com/sum-range-of-ints-in-listint.html | 1,620,602,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00432.warc.gz | 273,496,924 | 14,076 | Home » c# » Sum range of int's in List<int>
# Sum range of int's in List<int>
Questions:
I reckon this will be quite trivial but I can’t work out how to do it. I have a `List<int>` and I want to sum a range of the numbers.
Say my list is:
``````var list = new List<int>()
{
1, 2, 3, 4
};
``````
How would I get the sum of the first 3 objects? The result being 6. I tried using `Enumerable.Range` but couldn’t get it to work, not sure if that’s the best way of going about it.
Without doing:
``````int sum = list[0] + list[1] + list[2];
``````
You can accomplish this by using `Take` & `Sum`:
``````var list = new List<int>()
{
1, 2, 3, 4
};
// 1 + 2 + 3
int sum = list.Take(3).Sum(); // Result: 6
``````
If you want to sum a range beginning elsewhere, you can use `Skip`:
``````var list = new List<int>()
{
1, 2, 3, 4
};
// 3 + 4
int sum = list.Skip(2).Take(2).Sum(); // Result: 7
``````
Or, reorder your list using `OrderBy` or `OrderByDescending` and then sum:
``````var list = new List<int>()
{
1, 2, 3, 4
};
// 3 + 4
int sum = list.OrderByDescending(x => x).Take(2).Sum(); // Result: 7
``````
As you can see, there are a number of ways to accomplish this task (or related tasks). See `Take`, `Sum`, `Skip`, `OrderBy` & `OrderByDescending` documentation for further information. | 423 | 1,299 | {"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} | 2.78125 | 3 | CC-MAIN-2021-21 | latest | en | 0.714398 |
https://ourlittlelab.com/sara-bareilles-careful-confessions-rar/ | 1,686,215,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00709.warc.gz | 467,418,764 | 16,260 | # Sara Bareilles – Careful Confessions.rar ✅
Sara Bareilles – Careful Confessions.rar
careful confessions Torrent cranked up the Big Bang’s camp music, it felt like a million dollars being there for the closest thing to a passing of a franchise, a link to the past.Q:
How to convert integer to String in Dataframe?
I have dataframe with columns NN,NNN,NNNN,…
I need to convert integer that is present in column NNN to string.
For e.g. NNN contains 120, 22,113345, 234,23,76,45
Expected Output – Column ‘N’ should be string with above values.
Column ‘N’ should be string 120,22,113345,234,23,76,45
A:
You could use Series.str.extract to extract elements from the value column which are strings with length one and then explode them. If the strings to be extracted are ordered (as it is in your example), then by using groupby we sort the elements before extracting them from the DataFrame.
df[‘N’] = df[‘NNN’].str.extract(‘(\d+)’, expand=True)
#sortN = df.groupby(‘N’)[‘N’].transform(‘unique’)
df.loc[df[‘N’].str.extract(‘(\d+)’, expand=True).sort_values().index] =\
df[‘N’].str.split(expand=True).explode()
Output:
N NNN NNNN NNNNN
0 120 22 113345 234
1 23 76 45 | 341 | 1,169 | {"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-2023-23 | latest | en | 0.759983 |
https://chemistry.stackexchange.com/questions/45106/why-are-there-2-equivalents-of-ca-in-an-example-stated-in-wikipedia/45110 | 1,582,270,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145443.63/warc/CC-MAIN-20200221045555-20200221075555-00377.warc.gz | 306,062,339 | 32,275 | Why are there 2 equivalents of Ca in an example stated in Wikipedia?
Applying what's been said in the Wikipedia page for equivalents, if $1\ \mathrm{mol}$ of $\ce{NaCl}$ and $1\ \mathrm{mol}$ of $\ce{CaCl2}$ are dissolved in a solution, there is $\rm1~eq~Na,~2~eq~Ca$, and $\rm3~eq~Cl$ in that solution.
Is this a misstatement in Wikipedia?
The Wikipedia article uses a funny defintion of equivalent that I have not heard yet even though I use the term extremely frequently in the lab. But they define it well so here goes:
One equivalent of ions is defined as $1~\mathrm{mol}$ of charge. So if we dissolve $1~\mathrm{mol}$ of $\ce{Na+}$, we have exactly $1~\mathrm{mol}$ of positive charges in solution, so $1~\mathrm{mol} \cdot 1~e = 1~\mathrm{eq}$ ($e$ being the elemental charge). The $1~\mathrm{mol}$ of $\ce{Cl-}$ contributes $1~\mathrm{eq}$ of negative charges, balancing the entire thing.
Calcium, however, has a $\mathrm{+II}$ oxidation state. So $1~\mathrm{mol}$ of $\ce{Ca^2+}$ gives us $1~\mathrm{mol} \cdot 2~e = 2~\mathrm{eq}$ of positive charge.
• Sir I meant this only in my answer. Actually it's not funny but we see equivalent as the number of eq. electrons bringing out the reactions. I did not write this because it might have confused the asker. – Akshay Pratap Singh Feb 12 '16 at 3:16
• You see we have 2e = Weight of metal/ its equivalent weight( from electro-chemistry). – Akshay Pratap Singh Feb 12 '16 at 3:19
• @AkshayPratapSingh There is no need to comment on an unrelated answer to answer to a comment I gave under your answer. Instead, comment in the same spot and ping me with @ followed by the username. – Jan Feb 13 '16 at 17:43
The link in the original post may refer to the equivalent of the number or ratio of electrons shared by each of the specific ions, according to their chemical formula, in the solution, rather than the ratio of specific ions that are present in the solution.
No the statement is absolutely correct. We can write number of equivalents = Weight of the metal/ its equivalent weight. Since the equivalent weight of calcium in CaCl2 is 20 gm hence we get the number of equivalents of Ca as 2.
• Reread the Wikipedia article, please. – Jan Feb 12 '16 at 1:13 | 619 | 2,223 | {"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} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.91243 |
https://learnindex.com/solids-liquids-and-gases/ | 1,718,967,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00719.warc.gz | 319,710,126 | 18,565 | Solids, Liquids, and Gases
Solids, liquids, and gases are the three states of matter that we commonly observe in our everyday lives. Understanding these states is essential to understanding the behavior of substances and their properties. In this lesson, we will explore the characteristics of solids, liquids, and gases, their physical properties, and how they transform from one state to another.
Solids are substances that have a fixed shape and volume. They are rigid and cannot be compressed easily. The particles in solids are tightly packed and have a fixed position. This means that they vibrate in place but do not move around like particles in liquids and gases. Examples of solids include ice, wood, and metals.
Liquids are substances that have a definite volume but do not have a definite shape. They can take the shape of their container. Liquids are not rigid like solids and can flow easily. The particles in liquids are close together but are not tightly packed like those in solids. This means that they can move around each other but are still attracted to one another. Examples of liquids include water, milk, and oil.
Gases are substances that do not have a fixed shape or volume. They take the shape and volume of their container. Gases can be compressed easily, and the particles in gases move around freely and randomly. The particles in gases are far apart and have weak attraction forces, which makes them highly compressible. Examples of gases include air, carbon dioxide, and helium.
Physical properties of Solids, Liquids, and Gases:
The physical properties of solids, liquids, and gases differ from one another, as outlined below:
• Solids have a fixed shape and volume, while liquids take the shape of their container and have a definite volume, and gases take both the shape and volume of their container.
• Solids are not compressible, while liquids and gases are compressible.
• Solids have a high density, while liquids have a lower density than solids, and gases have the lowest density of the three states of matter.
Changing states of matter:
Matter can change from one state to another through physical processes. The three processes that can change the state of matter are melting, boiling, and condensation.
Melting: This is the process of changing a solid to a liquid. When a solid is heated, the particles start to vibrate faster and break away from their fixed position, which leads to the melting of the solid.
Boiling: This is the process of changing a liquid to a gas. When a liquid is heated, the particles start to move faster and gain energy, which leads to the evaporation of the liquid.
Condensation: This is the process of changing a gas to a liquid. When a gas cools down, the particles lose energy, move slower, and start to come together, which leads to the condensation of the gas.
Understanding the states of matter is essential to understanding the behavior of substances and their properties. Solids, liquids, and gases have distinct characteristics, physical properties, and undergo different physical processes. Knowing about these differences can help us understand many phenomena in our world. | 636 | 3,169 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.965762 |
https://www.proprofs.com/quiz-school/story.php?title=online-listening-comprehension-3-1st-audio-recording-bermuda-triangle | 1,696,226,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00526.warc.gz | 978,963,231 | 95,812 | # Term 2: Online Listening Comprehension #1 - 1st Audio Recording (Bermuda Triangle)
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Quizzes Created: 36 | Total Attempts: 4,143
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You will hear a podcast about the Bermuda Triangle. For questions 1-7, tick the answer which best fits according to what you hear. You will hear the recording once.
• 1.
### Where is the Bermuda triangle located?
• A.
Atlantic Ocean
• B.
Pacific Ocean
• C.
Antarctic Ocean
• D.
Indian Ocean
A. Atlantic Ocean
Explanation
The Bermuda Triangle is located in the Atlantic Ocean. This region is notorious for mysterious disappearances of ships and airplanes, which has led to various theories and speculations about its causes. The area is roughly bounded by Miami, Bermuda, and Puerto Rico, and it is known for its unpredictable weather patterns and magnetic anomalies. The Atlantic Ocean is home to this enigmatic and intriguing region that continues to captivate the imagination of people worldwide.
Rate this question:
• 2.
### Which three cities make up the Bermuda Triangle
• A.
Bermuda, America, Puerto Rico
• B.
Puerto Rico, Florida, Hawaii
• C.
America, Bermuda, Cuba
• D.
Bermuda, Florida, Puerto Rico
D. Bermuda, Florida, Puerto Rico
Explanation
The Bermuda Triangle is not made up of cities, but rather a region in the western part of the North Atlantic Ocean. It is known for mysterious disappearances of ships and airplanes. The correct answer is Bermuda, Florida, and Puerto Rico because these are the three points that form the triangle.
Rate this question:
• 3.
### How much area does the Bermuda Triangle cover?
• A.
50,000 square miles
• B.
5,000 square miles
• C.
500,000 square miles
• D.
500 square miles
C. 500,000 square miles
Explanation
The Bermuda Triangle covers an area of 500,000 square miles. The Bermuda Triangle is a region in the western part of the North Atlantic Ocean where numerous aircraft and ships have disappeared under mysterious circumstances. It is known for its reputation as a site of unexplained phenomena and is often associated with paranormal activities. The vast area covered by the Bermuda Triangle adds to its intrigue and has led to various theories and speculations about its mysterious nature.
Rate this question:
• 4.
### What makes the Bermuda Triangle so mysterious?
• A.
People have repeatedly reported strange phenomenon there
• B.
UFOs have landed there
• C.
People are just making up stories to scare each other
• D.
A. People have repeatedly reported strange phenomenon there
Explanation
The Bermuda Triangle is considered mysterious because people have repeatedly reported strange phenomena occurring in that area. These reports include unexplained disappearances of ships and airplanes, electronic malfunctions, and compasses behaving erratically. These consistent accounts of unusual occurrences have contributed to the mystery surrounding the Bermuda Triangle.
Rate this question:
• 5.
### What happens to compasses in the Bermuda Triangle
• A.
They spin around and do not show true North
• B.
They spin around until they find true North
• C.
They show true north
• D.
They show true South
A. They spin around and do not show true North
Explanation
In the Bermuda Triangle, compasses spin around and do not show true North. This phenomenon is known as compass variation or magnetic deviation. The Bermuda Triangle is located in an area where the Earth's magnetic field is unstable, causing compass needles to behave erratically. This can lead to confusion and disorientation for navigators and pilots who rely on compass readings for direction.
Rate this question:
• 6.
### Skeptics of the Bermuda Triangle mystery claim that...
• A.
The Bermuda Triangle is nothing to be intrigued about
• B.
The same number of disappearances that happen in the Bermuda Triangle happen in the rest of the world
• C.
More disappearances happen in the Bermuda Triangle than anywhere else in the world
• D.
Stories of disappearances in Bermuda Triangle are made up
B. The same number of disappearances that happen in the Bermuda Triangle happen in the rest of the world
Explanation
The correct answer suggests that skeptics of the Bermuda Triangle mystery claim that the same number of disappearances that happen in the Bermuda Triangle also happen in the rest of the world. This implies that the Bermuda Triangle is not unique or particularly dangerous compared to other areas of the world, and therefore there is no mystery or reason to be intrigued about it.
Rate this question:
• 7.
### What is the narrator's stand of the Bermuda Triangle theories?
• A.
He believes all of them
• B.
He feels that there is merit to some of the theories
• C.
He is skeptical
• D.
He feels that there is some paranormal activity going on | 1,100 | 5,188 | {"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-2023-40 | latest | en | 0.909943 |
https://www.scribd.com/document/293610288/Comparative-Analysis-of-Algorithms-for-Single-Source-Shortest-Path-Problem | 1,555,936,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578553595.61/warc/CC-MAIN-20190422115451-20190422141451-00123.warc.gz | 806,782,768 | 66,960 | You are on page 1of 7
# Shweta Srivastava
## Comparative Analysis of Algorithms for Single Source Shortest
Path Problem
shwetasrivastava21@gmail.com
## Mrs. Shweta Srivastava
Computer Science & Engineering Department,
India
Abstract
The single source shortest path problem is one of the most s t u d i e d problem in algorithmic graph theory.
Single Source Shortest Path is the problem in which we have to find shortest paths from a source vertex v to
all other vertices in the graph. A number of algorithms have been proposed for this problem. Most of the
algorithms for this problem have evolved around the Dijkstras algorithm. In this paper, we are going to do
comparative analysis of some of the algorithms to solve this problem.
The algorithms discussed in this paper are- Thorups algorithm, augmented shortest path, adjacent node
algorithm, a heuristic genetic algorithm, an improved faster version of the Dijkstras algorithm and a graph
partitioning based algorithm.
Keywords: Single Source Shortest Path Problem, Dijkstra, Thorup, Heuristic Genetic Algorithm, Adjacent Node
Algorithm.
1. INTRODUCTION
The single source shortest path problem can be defined as: given a weighted graph (that is, a set V
of vertices, a set E of edges, and a real-valued weight function f: E R), and one element s of V
(i.e. a distinguished source vertex), we have to find a path P from s to a v of V so that
Refer figure 1.
## FIGURE 1: An Undirected Graph of 6 nodes and 7 edges
SSSP is applied in various areas such as:
1.
2.
Computer Network
3.
Web Mapping
International Journal of Computer Science and Security (IJCSS), Volume (6) : Issue (4) : 2012
288
Shweta Srivastava
4.
5.
## Electronic Circuit Design
Geographical Information System (GIS)
In all six papers [1][2][3][4][5][6], the authors have given an improvement over the Dijkstras
algorithm.
## 2. BACKGROUND STUDY AND ANALYSIS
The Dijkstras algorithm makes assumption that there is no negative-weight edges in the graph G
(V, E) : w(u, v) >0 ,
E. The algorithms in [1][2][3][4][5][6] also follows this assumption.
(u, v)
For simplicity, we use n = |V| and m = |E| and K distinct edge lengths.
Several algorithms have been proposed for this problem which is based on different design
paradigms, improved data structure, parameterization and input restrictions. According to survey it
is found that no algorithm based on the Dijkstras algorithm has achieved the linear time complexity
due to drawbacks of Dijkstras algorithm. Dijkstras algorithm maintains an adjacency matrix which
consumes n*n space in the memory. When the number of nodes (n) is very large, it is difficult to apply
Dijkstras algorithm.
So, in paper [1], an algorithm has been proposed which modify Dijkstras algorithm to overcome
its bottlenecks. Dijkstras algorithm visits the vertices corresponding to a sorting algorithm (in order of
increasing d (v)). Since there is no linear ime algorithm for sorting problem. Unless the order of
visiting vertices is not modified, a linear time complexity cannot be achieved. The performance of
an algorithm for single source shortest path problem depends on the 3 attributes:
(1) Preprocessing time: time required to construct a search structure suitable for search.
(2) Space: storage used for constructing and representing the search structure.
(3) Search Time: time required to find shortest path from a query source s, using the search structure.
In year 2000 Thorup proposed a concept of components and using some complicated data
structures which overcome the problem in Dijkstras algorithm. There are 3 interesting features of the
Thorups algorithm [1]:
(i)It contains a minimum spanning tree algorithm as its sub procedure. To achieve the linear time
complexity Thorup used a linear time MST algorithm.
(ii) Thorups algorithm consists of 2 phases: a construction phase which constructs a
data structure suitable for a shortest path search from the given query source s; a search phase of
finding the shortest paths from s to all vertices using the data structure constructed in construction
phase.
(iii) Construction phase in Thorups algorithm is independent of the source, while data structures
in previous algorithms heavily depend on the source.
Summary of the Thorups Algorithm:
Step1. Construct an MST (M).
Step2. Construct a component tree (T) using MST.
Step3: Compute widths of buckets (B) to maintain the components.
Step4: Construct an interval tree (U) to store unvisited children.
Step5: Visit all components in T by using B and U. Also known as search phase.
The running time of each step is as follows: step1: O (m) time, step2, 4: O (n) time, step5: O (m+n)
time.
In year 2000, a linear time algorithm for SSSP problem [4] was proposed called an improvement over
Augmented Shortest Path Algorithm.
They proposed this algorithm for situations where no edge is unreasonably larger than the other edge
and the ratio of maximum and minimum weights of the edges (f) of the graph is not very large.
The algorithm converts the graph G into an augmented graph (Ga) in we replace every edge in
the original graph by number of edges having equal weights and number of new edges for each edge
in the graph is bounded. Then the shortest path tree is obtained. The major drawback in the ASP
algorithm was that if some heavy weight edge is included in the shortest path tree ASP fails to
perform well so an improvement is done in improved ASP algorithm [4].
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New_Augmented_Shortest_Path (G,s)
1. Find the minimum edge weight wmin in O(m) time from the adjacency list.
2. for all (u,v) E do
3. inqueue [u,v] <- false
4. edge [u,v] <-
5. end do
6. d[s]<-0
7. nowmin=
8. nextmin=
9. nowcount = 0
10. for all (s,u) E do
11. enqueue (u,s)
12. inqueue[s,u] <- true
13. edge [s,u] <- w[s,u]
14. if (w[s,u] < nowmin)
15. nowmin=w[s,u]
16. nowcount = nowcount+1
17. end do
18. nextcount = 0
19. while queue != empty do
20. v,p <- serve()
21. wv<- edge[p,v]-max(wmin , nowmin)
22. if w <= 0 then
23.
if d[p]+w[p,v] <d[v] then
24.
d[v] <- d[p]+w[p,v]
25.
[v] <- p
26.
for all (v,u) E do
27.
if inqueue[v,u] = false then
28.
if d[v]+w[v,u] < d[u] then
29.
enqueue(u,v)
30.
inqueue[v,u] <- true
31.
edge[v,u] <- w[v,u]+wv
32.
nextcount = nextcount + 1
33.
if(egde[v,u] < nextmin)
34.
nextmin = edge[v,u]
35.
endif
endif
36.
37.
else
38.
edge[v,u]<-min(edge[v,u],w[v,u]
39.
+ wv )
40.
if(edge[v, u] < nextmin)
41.
nextmin = edge[v,u]
42.
endif
43.
endif
44. enddo
45. endif
46. endif
47. else
48. if d[p]+w[p,v] <d[v] then
49.
enqueue(v,p)
50.
edge[p,v] <-wv
51.
if(wv < nowmin)
52.
nextmin = wv
53.
endif
54. endif
55. endif
56. nowcount=nowcount-1
57. if (nowcount==0)
58. nowcount=nextcount
59. nowmin=nextmin
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60.
61.
62.
63.
nextmin =
nextcount = 0
enddo
Return the shortest path.
This new algorithm proposes that the distance order search will advance by the maximum of w(min)
and the minimum weight in the queue. So, the problem of larger weight edge got removed in this
new version. The new augmented shortest path algorithm takes O(mf/2 + n) time which is linear if
f is not large. The space requirement is O(m+n). This algorithm proved to perform better than the
bucket based algorithm.
In year 2006, an approach was proposed by the authors of [6] to speed up the Dijkstras algorithm.
An acceleration method called arc-flag is used to improve Dijkstras algorithm. In this
graph which is then used to speed up shortest path queries. In the preprocessing step graph is
divided in the regions and checked whether an arc belongs to the shortest path in the given region.
This preprocessing method is combined with an appropriate partitioning technique and bi-directed
search which achieves an average speed up factor of more than 500 compared to the Dijkstras
algorithm on large networks. They tested different combinations of the arc-flag method with different
partitioning technique.
They used A*, bi-directed search techniques and chosen bi- directed search because A* didnt
improve the speed up factor. They considered Grid, kd, Tree or METIS as the base partitioning
method and made 11 combinations of the searching, partitioning and preprocessing techniques. They
applied the 11 different combinations on the German road network data. Kd trees and METIS
yields the best speed- up. Bi-directed search proved to be better than the unidirected search and
the two level partitioning was better than the single level partitioning. The preprocessing takes O
(m(m+n+nlogn)) time. It increases for the dense graph.
In year 2007, a heuristic genetic algorithm [3] was proposed to achieve high performance. Their
proposal starts with the initial population of candidate solution paths than a randomly generated one.
HGA also uses a new heuristic order crossover (HOC) and mutation (HSM) to keep the limited
search domain.
The components required to develop HGA requires chromosome coding, initialization, genetic
crossover operator, genetic mutation operator, and parent selection & termination rules.
Summary of HGA:
Step1: Chromosome Coding Scheme and Initialization: Chromosome Coding Scheme:
The complete chromosome of a candidate is divided into node fields equal to the number of
nodes in a network. Refer Figure 2.
11
## FIGURE 2: Part of Chromosomal structure of a candidate path
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This structure uses the node indices and the distance weight between 2 nodes.
N i0 = Previous (Ni) N i1 = Ni
N i2 = dist (Ni)
Previous (Ni) is same as the predecessor array and dist (Ni) is same as an array of best estimates of
shortest path to each vertex in the Dijkstras algorithm.
Initialization: First node of every candidate path is the source node. So each chromosome (s,s,0).
Other entries are random nodes, covers all other nodes in the graph.
Step2: Parent Selection, HOC, HSM and Termination: Parent Selection:
Here, algorithm chosen for the selection is Tournament Selection Algorithm. The idea behind this is to
pick a pair at random, compare their fitness and the fittest is selected.
To find the fitness value we need to know the objective function (path cost). Path cost = sum of (dist
(Ni)) where i = 1 to n.
Using this the fitness function value is calculated as: Fitness (Chromosome) =
1
Path Cost
HOC:
Here node fields are chosen as the cut points. The portion of the first parent between them is
copied to the offspring, the rest of the offspring is selected from the second parent with the
following conditions:
1. The source node fields remains at the first position in new generated offspring.
2. While taking other nodes from the second parent, all nodes should be included only once in the
offspring.
And the offsprings are evaluated and added as the new solution path candidate.
HSM:
Two node fields are chosen randomly and swapped given that the source node is never mutated.
And again new mutated chromosome is evaluated. HOC and HSM dont generate new edges in
any candidate path , they just adjust the initially generated nodes into a legal minimum cost path
Termination:
The algorithm can be terminated when the number of generation crosses an upper bound specified
by the algorithm. With an increased number of generations, HGA converges to the optimal solution.
In year 2009, an algorithm based on Dijkstra for huge data [5] was proposed. In the paper author
has pointed out the drawbacks of the Dijkstras algorithm and proposed an algorithm as adjacent
node algorithm which an optimization over Dijkstras algorithm. He proved that his algorithm can
save lot of memory and is more suitable for graph with huge nodes. The adjacent node algorithm
makes improvement by improving the method of creating the adjacency matrix. First the number of
the maximum adjacent nodes r is found. Then the adjacency matrix of n*r is made which is much
smaller that n*n matrix. One more judgement matrix is made of order m*r. The shortest path is
found with the help of both adjacency and judgement matrix. In their experiment, this algorithm
performed 6 times better than the Dijkstras algorithm for the data size of 12000 nodes.
In year 2009, a faster algorithm [2] has been proposed for SSSP problem. They have proposed an
efficient method for implementing the Dijkstras algorithm with the same assumptions. In addition to it
two more assumption is made that : Let L= {l1,l2,........lk} be the set of distinct nonnegative edge
weights given in an increasing order as part of the input stored as an array and the number of distinct
edge lengths (k) is small. The authors solution is motivated by the gossip problem for social
networks.
Two algorithms are proposed by the author in this paper:
1. Simple implementation of Dijkstras algorithm that runs in O (m+nk) time.
2. Second algorithm is the modification of first algorithm by using binary heaps to speed up the
FindMin() operation. Its running time is O( m log (nK/m) ) if nK>2m.
Both the algorithms are identical to Dijkstras algorithm. The difference is that it uses some
additional data structures to carry out FindMin() operation. The algorithm is as follows:
Step1:
Function INITIALIZE()
1: S:={s}; T := V-{s}.
2: d(s):=0; pred(s):=.
3: for (each vertex v T) do
4:
d(v)= ; pred(v)=.
International Journal of Computer Science and Security (IJCSS), Volume (6) : Issue (4) : 2012
292
Shweta Srivastava
5: end for
6: for (t=1 to K) do
7:
Et(S):=. .
8:
CurrentEdge(t):=NIL.
9: end for
10: for each edge(s, j) do
11:
Add(s, j) to the end of the list Et (S),where lt =csj .
12:
if (CurrentEdge(t)=NIL) then
13:
CurrentEdge(t):=(s, j)
14: end if
15: end for
16: for (t=1to K) do
17:
UPDATE(t)
18: end for
Step2:
Function NEW-DIJKSTRA()
1: INITIALIZE ()
2: while (T= ) do
3:
let r = argmin {f(t):1t K}.
4:
let (i, j) = CurrentEdge(r).
5:
d(j):=d(i) + lr ; pred(j):=i.
6:
S= S {j}; T :=T - {j}.
7:
for (each edge (j,k) E(j)) do
8. Add the edge (j,k) to the end of the list Et(S),where lt =cjk .
9:
if (CurrentEdge (t) = NIL) then
10:
CurrentEdge (t): = (j,k)
11:
end if
12:
end for
13:
for (t = 1to K) do
14:
UPDATE(t).
15:
end for
16: end while
Step3:
Function UPDATE(t)
1:Let (i, j) = CurrentEdge(t).
2: if (jT) then
3:
f(t)=d(i)+cij
4:
return
5: end if
6: while ((j T) and (CurrentEdge(t).next != NIL)) do
7:
Let(i, j) = CurrentEdge(t).next.
8:
CurrentEdge(t)=(i, j).
9: end while
10: if (jT) then
11:
f(t)=d(i)+cij .
12: else
13:
Set CurrentEdge(t) to .
14:
f(t)= .
15: end if
The initialization step takes O (n) time. The potential time taking operations are step 3 of NewDijkstra and the Update procedure. In New-Dijkstra step3 takes O (k) per iteration of the while loop
and O (nk) over all the iterations. Procedure Update is called O (nk) times and its total running time is
O (m+nk). Iteration in which CurrentEdge (t) is not changed, running time
is
O (nk)
and
the iterations in which CurrentEdge(t) is changed, the running time is O(m).
So the total time taken by the algorithm is O (m+nk).
International Journal of Computer Science and Security (IJCSS), Volume (6) : Issue (4) : 2012
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Shweta Srivastava
## 3. USEFULNESS OF ALGORITHMS IN VARIOUS APPLICATIONS
Thorups algorithm [1] is much slower than the algorithms with the heaps as for the whole execution
time is compared. It is very slow for SSSP due to the time of the construction of the data
structures. Due to need of huge amount of memory and the complicated, large programs,
Thorups algorithm is not useful in practice today.
The algorithm in paper [2] works well for the graphs having smaller number of distinct edge
lengths than the density of the graph.
The Heuristic Genetic algorithm [3] proved to be suitable for the network of different size and
topology. HGA took reasonable CPU time to reach the exact solution and didnt variate much with
increased input size.
The Augmented Shortest path algorithm [4] is suitable for the graphs with less value of f.
According to author it is suitable for the road networks, electronic circuit designs etc.
The Adjacent Node algorithm in [5] is efficient for the huge data and takes less space. So it is suitable
for the traffic analysis type of applications.
The algorithm based on partitioning of graph [6] although performed better than Dijkstras algorithm for
some cases but for large networks its performance is degraded than that of Dijkstras.
4. CONCLUSIONS
In this paper it is tried to be explained- first, what are the different algorithms for the SSSP problem.
Second, how do they perform in comparison of the Dijkstras algorithm. Third, which algorithm is
suitable for a particular application or situation.
5. REFERENCES
[1]
Y. Asano, H. Imai, Practical Efficiency of the Linear Time Algorithm for the Single Source
Shortest Path problem, Journal of the Operations Research, Society of Japan, Vol. 43, No.
4; 2000.
[2]
J. B. Orlin, K. Madduri, K. Subramani, M. Williamson, A faster algorithm for the Single Source
Shortest Path problem with few distinct positive lengths, Journal of Discrete Algorithm; 2009.
[3]
## B. S. Hasan, M. A. Khamees, A. S. H. Mahmoud, A Heuristic Genetic Algorithm for the Single
Source Shortest Path Problem, IEEE International Conference on Computer Systems
and Applications; 2007.
[4]
P. P. Mitra, R. Hasan M. Kaykobad, On Linear time algoritm for SSSP Problem, ICCIT, 2000.
[5]
Zhang Fuhao, L. Jiping, An algorithm of shortest path based on Dijkstra for huge data,
6thInternational Conference on Fuzzy Systems and Knowledge discovery, 2009.
[6]
## R. H. Mohring and H. Schilling, Partitioning Graphs to Speedup Dijkstras Algorithm, ACM
Journal of Experimental Algorithmics, Vol. 11, Article No. 2.8, Pages 1-29, 2006.
International Journal of Computer Science and Security (IJCSS), Volume (6) : Issue (4) : 2012
294 | 4,648 | 17,997 | {"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} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.833205 |
https://kseebsolutions.net/2nd-puc-statistics-notes-chapter-8/ | 1,701,981,711,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00717.warc.gz | 391,495,557 | 18,935 | ## Karnataka 2nd PUC Statistics Notes Chapter 8 Operation Research
High Lights of the Topic:
→ Strategic problems relating to military forces, a group of scientists and experts, suggested appropriate allocations of available resources, and they helped in elficient handling the complex military operations, such as detection of enemy submarines, anti aircrafts, fire control etc, and they called the concept as Operations Research (OR).
→ However afterwards the science becomes popular, the applications (uses) of OR was applied to general problems such as, problems concerning Industry, Marketing, Personnel management, Inventory control, etc.
→ Therefore OR can be defined as “It is the method of application of scientific methods, techniques and tools to problems involving the operations of a system, so as to provide those in control of the system with optimum solutions to the problem.”
Scope:
Some of the problems in real life situations (problems) discussed in Operations Research (OR), are-
(i) Linear programming problems(LPP) are used in Industries for effective use of men, machine, money and material.
(ii) Transportation problem (TP) helps in shipment of quantities from various sources to different destinations at a minimum cost.
(iii) Game theory is applied in conflicting situations, to suggest the best strategy.
(iv) Replacement problems helps in suggesting the best time period to replace a specific part in manufacturing process, and
(v) Inventory problems help in organizing stocks and maintaining costs, and so on.
(i) Linear Programming Problem (LPP):
→ In many Economic activities, i.e. of Business and Industrial field, we often face the problems of decision making or optimization (may be Maximization/Minimization) of profit, sales, cost, labour etc, is very important. On the basis of available resources(called constraints), such as Demand for commodities, availability of raw materials, storage space, variation in costs, etc, In such situations where conditional optimization is required we may adopt ‘Linear programming’ model.
→ It was in 1947 that George Dantzig and his associates introduced the Linear programming technique and simplex method of solving problems concerning military activities.
→ Here “Linear programming is a mathematical technique which deals with the optimization (Maximization or Minimization) of activities subject to the available resources.”
For example:
1. A biscuit manufacturing factory may be interested in taking decisions regarding the type of biscuits to be manufactured and the quantity produced so as to earn maximum profit.
2. A factory having some machines to manufacture cars would like to know the best way to utilize the machines so that maximum production is made possible in minimum time available,
3. A dietician may want to suggest food with a certain basic vitamins and proteins. She may like to know the best way to prescribe food with optimum requirement of vitamins and proteins at the minimum cost.
Formulation of LPP:
• Identify the unknown variables- known as Decision variables and denote them in terms of algebraic symbols (as x,y).
• Identify the Objective of the problem and represent it as Z.
• Identify the restrictions called constraints in the problem and represent them as equations or inequalities in terms of symbols (as ≤ = ≥).
LPP Model :-
(Matrix Notation) Optimize Z = CX
Subject to the constraints- AX(≤ = ≥)b
and the non-negativity restriction X ≥ 0 ; is an LPP
Example:
A manufacturer produces two models of products Model A and Model B
(i) Model A fetches a profit of Rs.200/- per set. Whereas Model B fetches Rs.500/- per set.
(ii) Model A requires 6 units of raw material and Model B requires 3 units of raw material. Every week there is a supply of 180 units of raw materials.
(iii) Model A requires 4 hours and Model B requires 8 hours labour. The total labour available is 160 hours per week.
→ With the above facts, the manufacturer is to decide the number of Model A Products and number of Model B should he manufacture. In taking decision he should maximize his profit, on the basis of availability of Raw materials and the labour available.
→ The manufacturer’s problem can be mathematically written down (formulated) as follows:-
→ Let Y denote the number of units of Model A products to be manufactured. Let ‘y’ denote the number of Model B products to be manufactured. (x,y-decision variables)
→ Then, the manufacture’s profit is- (objective function)
Max. Z = 200x + 500y
(Restrictions are-) Subject to constraints; 6x + 3y ≤ 180
Since the weekly supply raw materials are 180 units, at the most 180 units can be utilized.
4x + 8y ≤ 160
→ Since total labour available is 160 hours.
Also the number of products cannot be negative, and so, X ≥ 0, Y ≥ 0
Thus, the formulated form of the the manufacturer problem (LPP) is;
Maximize = 200x + 500y
Subject to constraints, 6x + 3y ≤ 180.
4x + 8y ≤ 160
And non-negative restrictions: x,y ≥ 0 is an LPP
→ Definition: Linear programming deals with the optimization of a linear function of variables known as ‘objective function’ subject to a set of linear equalities / inequalities known as constraints.
→ Objective function: The function Z = CX which is to be optimized (Maximised/Minimised) is called Objective function. {E.g. Maximize = 200x + 500y}
→ Decision variable: A variables be ‘x1, x2, x3 ………… xn‘ whose values are to be determined are called decision variables.
{E.g: x, y variables}
→ Solution: A set of real values X = (x1, x2, x3 ………… xn) which satisfies the constraints AX(≤ = ≥)b is called a solution.
→ Feasible solution: A set of real values X = (x1, x2, x3 ………… xn) which satisfies the constraints AX(≤ = ≥)b and also satisfies the non-negativity restrictions x ≥ 0 is called feasible solution.
→ Optimal solutions:
A set of real values X = (x1, x2, x3, ………. xn) which
• Satisfies the constraints AX(≤ = ≥)b.
• Satisfies the non-negativity restriction x ≥ 0 and
• Optimizes the objective function Z = CX is called optimal solution
Note:
• If an LPP has one optimal solution, it is called to have unique optimal solution
• If an LPP has many optimal solutions, it is said to have multiple/Multiple Optimal solutions.
• For some LPP may not have optimal solution or the optimal value of Z may be infinity, in this case the LPP is said to be unbounded solution.
• In an LPP if all constraints cannot be satisfied simultaneously then there exists no solution to the given LPP.
Finding the solution to a LPP by graphical method:
• Consider the constraints as equality
• Find the co-ordinates for each constraint
• Draw the graph and represent by straight line
• Identify the feasible region-which satisfies the non-negativity restriction and constraints simultaneously
• Locate the corner points of the feasible region
• Find the value of the objective function at the corner points
• Get the optimum value and suggest the solution to the LPP
Note: Feasible can lie/exists only in First quadrant’: because for the given LPP, the solution . can exists only on xy-plane, because of the non-negativity restrictions, x and y both are positive on xy-plane.
(ii) Transportation Problem (T.P)
→ T.P was introduced by F.L.Hitchcock (in 1941) and T.C.Koopman (in 1947).
→ “The transportation problem (T.P) is a problem concerning transportation of goods from different origins (factories, warehouses/godowns) to various destinations (dealers, customers)”.
→ Here each origin has a certain stock of goods and each destination has a certain requirement of goods and there is a certain cost associated with transportation of goods from the various origins to the different destinations.
→ Following terminology are used of transportation problem:
• Let ‘m’ be the number of sources or origins.
• ‘n’ be the number of destinations.
• ‘ai’ be the quantity of goods available at ith source
• ‘bj’ be the quantity of goods required at jth destination.
• ‘Cij’ be the cost of transportation of unit of goods from ith source to jth destinations.
• And ‘xij’ be the number of units to be transported from ith source to jth destination.
→ Formulation of T.P:
Consider a T.P with ‘m’ origins O1, 02 . . . . .Om and ‘n’ destinations D1, D2 ….. Dn
Let ai be the quantity of product available at Oi
bj be the quantity of product required at Dj.
Cij be the cost of shipping of one unit of product from Oi to Dj
If xij is the number of units to be shipped from Oi to Dj. Then the problem is to determine xij so as to minimize the total transportation cost
And, all xij ≥ 0 i = 1, 2 …………. m & j = 1, 2, ……. n.
Here, a T.P in which (availability) Σai = Σbj (requirement) is called balanced TP
Definitions:
Feasible solution:
A set of values xij > 0, (i = 1, 2.. m and j = 1, 2 ….. n) is a feasible solution which satisfies:
→ Basic feasible solution(BFS): ‘A feasible solution is said to be a basic feasible solution if the number of non-zero allocations are (m + n – 1)’.
Here (m + n – 1) variables are known as basic variables.
→ Degenerate solution: When the number of positive allocations in any BFS is less than (m + n – 1), then the solution is said to be degenerate.
→ Non-degenerate solution: The number of positive allocations in any BFS is exactly (m + n – 1), then the solution is said . to be non-degenerate.
→ Optimal solution: A feasible solution is said to be optimal if it minimizes the total transportation cost.
→ Finding the initial solution:
An B.F.S. is obtained by :
• Northwest corner rule method (NWCR) ‘
• Matrix minima method(MMM)
→ Northwest corner rule method (NWCR):
In a given T.P. of m-origin, n-destination T.P. with ai-availabilities(i = 1, 2, 3 ……. m) and bj- requirements (j = 1, 2, 3 ……. n)an I.B.F.S is obtained as below:
Step I:
To start allocating at the cell (1, 1) which is north-west comer cell as: X11 = min (ai, bj)
Step II:
(a) If a1 > b1 next allocation is made at the cell (1, 2) as: X12 = min(a1 – x11, b2)
{in the same row until the factory availabilities are allocated other dealers}
(b) If a1 < b1 next allocation is made at the cell (2, 1) as: X12 = min(a2, b1 – x11)
{in the same column where dealer’s requirement is satisfied from other factories}
(c) If a1 = b1 next allocation is made at the cell(2, 2)as: X22 = min (a2, b2)
{where both dealer and factory are satisfied}
Step III:
The above procedure is repeated until all the availabilities of the origins are allocated to requirement of the different destinations.
Matrix minima method:
(Least/low cost entry method)
In a given T.P. of m-origin, n-destination T.P. with ai-availabilities(i = 1,2,3…m) and bj- requirements (j = 1, 2, 3 …….. n) an I.B.F.S is obtained as below:
Step I :
Identify the least cost in the cost matrix (called matrix minima) of the T.P. table.
Let Cij be the least cost then allocation is made at the cell (i, j) as:
xij = min (ai, bj)
Step II:
(a) If ai > bj, jth column is deleted and ai is replaced by (ai – xij ), repeat step I:
{here dealer is satisfied so deleted, factory availabilities are more}
(b) If ai < bj ith row is deleted and bi is replaced by (bj – xij), repeat step I.
{here factory availabilities are exhausted}
(c) If ai = bj, ith row and jth column are deleted
{where both dealer and factory are satisfied}
Step III:
The above procedure is repeated until all the availabilities of the origins exhausted
(iii) Game Theory
→ Definition “Whenever there is a situation of conflict and competition between two or more opposing teams, we refer to the situation as a game” (It is science of conflicts)
→ In a game when each player chooses a possible action, then we say that a play of the game has resulted.
Assumptions/properties/characteristics of a competitive game
• There are finite numbers of players (competitors)
• Each player has finite number of course of action
• The game is said to be played when each of the players adopt one of the courses of action
• Every time the game is played the corresponding combination of courses of action leads to transaction (Payment) to each player. The payment is called pay off (gain).
• The gain of one player is exactly equal to the loss of the other
→ n- Person game:- A game in which n players participate is called n-person game.
→ Two-person game:- A game in which only two players participate is called two-person game.
→ Zero-sum game:- A game in which sum of the gains (pay-offs) of the players is zero is called ‘zero sum game’
→ Two-person-zero-sum game:- If in a game of two players, the sum of gain of one player is equal to the sum of loss of other is known as ‘two-person zero-sum game’. Also called as ‘Rectangular Game’.
→ Strategy:- In a game, the strategy of a player is the predetermined rule by which he chooses his courses of action while playing the game.
→ Pure strategy:- while playing a game, pure strategy of a player is his predetermined decision to adopt a specified course of action irrespective of the course of action of the opponent.
→ Mixed strategy:- While playing a game, mixed strategy of a player is his pre-decision to choose his course of action according to certain pre-assigned probabilities.
→ Optimal strategy:- A strategy of the game which maximizes the gain of one player and minimizes the loss of the other player is called optimal strategy.
→ Maximin:- Maximum of the row minimums in the payoff matrix is called Maximin. max
→ Minimax:- Minimum of the column maximums in the payoff matrix is called minimax.
→ Saddle point:- Saddle point is the position where the maxmin and the minimax coincide – i.e. α = β = v
→ Value of the game:- Common value of maxmin and minimax denoted by ‘V’.
→ Fair game:- If the value of the game is zero, then the game is called fair game otherwise unfair.
I. Solution to a game with saddle point:-
The saddle point of the game is found as follows:
1. The minimum payoff in each row of the payoff matrix is circled as : ○
2. The maximum payoff in each column is boxed as: ▢
→ In the above process, if any payoff is circled as well as boxed, that payoff is the Value of the game (v).
The corresponding position is the saddle point
→ Indicate the position of the saddle point, and then suggest the strategy for players.
Also, write down the value of the game.
II. Principle of dominance:
→ When Two-person Zero-Sum game has no saddle point, the Maximin-Minimax principle breaks down. In such situation, we use other method called the principle of dominance to find the optimal strategy and value of the game.
→ “If the strategy of a player dominates over the other strategy in all conditions, then the latter strategy can be ignored because it cannot affect the solution in any way”
Rules:
1. If all the elements in a row (say ith row) of a payoff matrix are greater than or equal to the corresponding elements of the other row (say jth row) then ith row dominates jth row , then jth row can be deleted.
2. If all the elements in a column (say rth column) of a payoff matrix are less than or equal to the corresponding elements of the other column (say sth), then rth column dominates sth column. So sth column casn be deleted.
Repeat the process till optimal strategy is obtained.
Note:- To Get a Transpose of a matrix write aij to – aji Rows into columns with sign change.
(iv) Replacement Problems
Replacement theory deals with the problems of deciding the age at which the old equipments (machine and their spare parts, trucks etc.) are replaced.
Need: New equipments will generally be more efficient and their maintenance cost would be less. As they become old, their efficiency decreases and their maintenance cost (running, operating) increases. Also to determine optimum time for replacement of machine, trucks and equipments.
Following are the needs:
• As the equipment grows older, the maintenance would be costlier.
• New equipments would be more efficient than the old ones.
• Technology of new equipment would be superior to that of the old.
• Production/working cost would be lesser in new equipments.
• Modern equipments are more beautiful and more compact than old equipments.
Replacement of equipments, which deteriorates with age:
Machine and their spare parts, trucks etc. deteriorates with time, their maintenance cost increases and efficiency reduces. Hence, their replacement becomes necessary. Therefore, purchasing cost, resale value, maintenance cost etc. change with time. However, here we are not considering change in the money value.
The Average annual cost is- A(N) = $$\frac{\mathrm{T}}{n}$$
Where P = capital (purchasing) cost of an equipment;
Sn = Resale value (scrap value, salvage cost) of an equipment at age ‘n’
Ci = Maintenance cost at time i = 1, 2,…. n years
The optimal replacement time is that value ‘n’ for which A(n) is least. Ie the equipment is suggested to replace at that period Where the Annual Average Cost A(n) ceases to decrease,
(v) Inventory Problems
→ An Inventory is a physical stock of goods, which is held for purpose of future production or sales.
→ Inventory may be that of
• Raw materials
• Finished goods
• Semi finished goods.
→ A problem concerning such a stock of goods is Inventory problem. The object of analysis of inventory problem is to decide when and how much to be acquired, so that the cost is minimized and profit is maximized.
Need:
An inventory is essential for the following reasons:-
• It helps in smooth and efficient running of the business.
• It provides adequate and satisfactory service to the customers.
• It facilitates bulk purchase of raw materials at discounted rates.
• It acts as buffer stock in case of shortage of raw materials.
• Inventory is also needed to optimize the cost.
However, the inventory has disadvantages such as, warehouse rent, labour of maintenance loss due to fall in price, interest on capital invested, deterioration of goods etc.
Variables in the inventory problem:
The variables involved in the inventory are of two types:
1. Controlled variables .
2. Uncontrolled variables
1. Controlled variables:
1. The variables, which can be controlled individually or jointly, are called Controlled variables. They are
• Quantity of goods acquired.
• Frequency or timing of acquisition or replenishment.
• The completion time or stage of stocked items/production of stocked items.
2. Uncontrolled variables: The variables, which cannot be completely controlled are called Uncontrolled variables. These include “Purchase/Capital cost, Carrying cost (Holding / Storage / Maintenance cost), Shortage/Penalty cost and Set-up cost (Replenishment / Ordering cost)”.
Note: Total inventory cost =Purchasing cost (P) + Setup cost (C3) + Holding cost (C1) + Shortage cost (C2)
Inventory costs These are inventory costs associated with keeping inventories. They are
(a) Purchasing cost (P): It refers to the cost associated with an item whether it is manufactured or purchased. (Also called as Capital cost)
(b) Set up cost (C3): It is the cost of setting up the machines for production or the cost of placing the order for the goods: It includes labour cost, transportation cost etc. It is also called as ordering cost, Replenishment cost, or Procurement cost. Denoted by C3.
(c) Holding cost (C1): It is the cost associated with carrying or holding/ the goods (Inventory) in stock until the goods are sold or used. (It includes rent for space, interest on capital invested, maintenance of records, taxes, insurance and breakages.)
Here Carrying cost (C1) is also the percentages (%) of the capital cost. So C1 = PI
Holding/Carrying cost per unit time per unit good is denoted C.
(d) Shortage cost (C2): The cost associated with delay or inability to meet the demand because of shortage of stock is called shortage cost. Also called as penalty cost-C2.
→ Demand (R):- Demand is the number of units required from the inventory per period. If the demand remains fixed, is called deterministic demand. If the demand varies randomly, is called probabilistic demand
→ Lead time: – The time gap between placing of order and arrival of goods at the inventory is the Lead time (Delivery lag)
→ Stock replenishment:- The rate at which items are added to the inventory to maintain a certain level is known as stock replenishment.
• The quantity of goods acquired in one replenishment is the Order quantity (Q) {Lot size, Run size, Stock replenished}
• The number of times replenishment is done in unit time (year) is the Frequency of replenishment
• The time gap between two replenishments is the Re-order time (t) {Re-scheduling time, production run}
→ Quantity delivered (Depletion): It is the number of units of goods delivered.
→ Time horizon (T): The period over which inventory level is maintained is called Time horizon.
→ The Economic Order Quantity (EOQ): The EOQ is the size of the order for which the aggregate of setup cost and holding cost of the Inventory is minimum.
(a) EOQ/ELS model with
• Uniform demand
• Instantaneous production/supply
• Shortages not allowed.
→ Inventory problems where demand is assumed to be fixed and pre-determined are called EOQ/ELS
→ A set of mathematical equations required to solve an Inventory problem is called an Inventory Model.
→ Inventory Model, which are meant for deterministic demand are called EOQ Model or Economic Lot Size (ELS) model.
The Assumptions for this Model are:
• Demand is deterministic and uniform
• Production or stock replenishment is instantaneous
• Shortages are not allowed
• Setup cost is Rs.C3 per cycle (production run)
• Holding cost is C1 per item per year
• Time period for maintaining of the level of Inventory is 1 year
Notations:
• Q – Lot size (run size, Quantity replenished, order quantity)
• P – Purchase/Capital cost,
• R – Demand Rate,
• S – Initial Level of Inventory,
• t – Time interval between two consecutive replenishment (Rescheduling time)
• n – Number of replenishment per unit time
• C1 – Carrying (Holding cost, Storage cost, Maintenance cost) per unit good per unit time
• C2 – Shortage cost/Penalty cost per unit good per unit time
• C3 – Setup/ordering cost per production run
• I – Carrying/Holding Cost per Rupee per unit time,
• C (Q) or C(Q°, S°) – Average cost per unit time.
Diagram of EOQ Model-I without shortages
Here the initial level Q of the inventory depletes in time t at the rate R to the zero level. After time t, then Inventory is replenished by Q units and the cycle continuous. Here ?OAB indicates the Inventory carried.
Diagram of EOQ Model-II with shortages
Here at the time of replenishment, there would be a shortages (Q-S) units and so, the level of inventory would be S. This level depletes in time t1 to zero. Here ?OAB represents the Inventory carried and ∆BCD represents the shortage carried.
2nd PUC Statistics Notes | 5,272 | 22,860 | {"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} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.924776 |
http://serc.carleton.edu/quantskills/activities/botec_radio.html | 1,418,905,822,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802766292.96/warc/CC-MAIN-20141217075246-00137-ip-10-231-17-201.ec2.internal.warc.gz | 263,180,439 | 8,651 | Quantitative Skills > Teaching Resources > Activities > BotEC: The Distance Radio Waves Have Traveled
Back-of-the-Envelope Calculations: The Distance Radio Waves Have Traveled
Summary
Question
We have been broadcasting radio waves in all directions since the development of radio and television stations. How far could you be from the Earth and detect the faint signals of an early Star Trek broadcast? Have signals from Star Trek reached the nearest star yet?
Assessment
Signals have gone out about 303 trillion kilometers, well past the nearest star. Here's how to figure it out. Remember that Star Trek broadcasts are sent out as electromagnetic radiation and therefore travel at the speed of light. You can figure how far they've gone by multiplying the speed of light (300,000 km/sec) by the number of seconds in 1 year (31,536,000) and multiplying that by 32 years to get about 303 trillion kilometers. But, you can save all that converting and calculating by using light years, the distance that light travels in a year. If we've been broadcasting for 32 years, then Star Trek broadcasts could be detected at a distance of over 32 light years, plenty of distance to be past the nearest star. Proxima Centauri is only 4.25 light years away (about 40 trillion kilometers). Our broadcasts haven't even come close to reaching the center of our galaxy, however—the center is 30,000 light years away! It will take another 29,968 years for Star Trek broadcasts to reach the center of the Galaxy...
References and Resources
This SERC page describes the use of Back of the Envelope Calculations
A View from the Back of the Envelope : This site has a good number of easy simulations and visualizations of back of the envelope calculations.
The Back of the Envelope : This page outlines one of the essays in the book "Programming Pearls" (ISBN 0-201-65788-0). The book is written for computer science faculty and students, but this portion speaks very well to back of the envelope calculations in general.
Controlled Vocabulary Terms
Subject: Physics, Geoscience:Lunar and Planetary Science
Resource Type: Activities:Classroom Activity:Short Activity
Special Interest: Quantitative
Quantitative Skills: Estimation | 474 | 2,223 | {"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-2014-52 | longest | en | 0.915797 |
https://wiki.physics.wisc.edu/facultywiki/Air_Track_Totally_Inelastic_Collision?action=raw&rev=4 | 1,516,095,860,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-05/segments/1516084886397.2/warc/CC-MAIN-20180116090056-20180116110056-00787.warc.gz | 846,627,530 | 1,845 | #acl snarf:read,write,delete,revert,admin FacultyGroup:read,write All:read ||<:30%>[:PiraScheme#Mechanics: Table of Mechanics Demonstration]||<:30%>[:MEEquipmentList: List of Mechanics Equipment & Supplies]||<:30%>[:Demonstrations:Lecture Demonstrations]|| = Air Track Totally Inelastic Collision, 1N30.32 = '''Topic and Concept:''' Linear Momentum, [:Linear_Momentum#Collisions1D: 1N30. Collisions in One Dimension] '''Location:''' * Floor Item: ME, South Wall attachment: mainPhoto '''Abstract:''' A moving glider collides with a stationary one and sticks. ||<:style="width: 60%" :40%>'''Equipment'''||<:30%>'''Location'''||<:25%>'''ID Number'''|| || || || || ||Air Track||Floor Item: ME, South Wall|| || ||Air Track Carts||ME, Bay A4, Shelf #4|| || ||Red Clay||Double Steel Door Bin Cabinet|| || ||[:RedWhiteGasCart:Red & White Gas Carts]||ME, Bay B1, Shelf #2|| || '''''Important Setup Notes:''''' * N/A '''Setup and Procedure:''' 1. Place air track on lecture bench, and connect the compressed air supply from the [:RedWhiteGasCart:Red & White Gas Cart] to the track using an air hose. 1. Make sure that the gliders to be used have clay on one side and that the clay juts out a little off center (see photo). 1. Place two gliders on the track, one on an end of the track and the other roughly in the middle. 1. Turn on the air supply. 1. Give the glider on the end a good push toward the glider at rest to create a collision. '''Cautions, Warnings, or Safety Concerns:''' * Make sure you are using compressed air and NOT methane gas which is extremely flammable! '''Discussion:''' In conservative systems, momentum is always conserved. In the lab frame, one glider is in motion and one is at rest. Thus, the total momentum of the system is P,,0,, = m,,1,, * v,,1,, where m,,1,, is the mass of the glider in motion and v,,1,, is its speed. In an inelastic collision, the two bodies stick to each other. Since P,,0,, = P,,f,, , ignoring friction and air resistance, the speed of the two cars together is lower than the initial speed of the first glider. P,,f,, = (m,,1,, + m,,2,,) * v,,1+2,,. ||attachment:InelasticGliders-02-250.jpg||attachment:InelasticLongGliders-03-250.jpg||attachment:InelasticShortGliders-04-250.jpg||attachment: photo|| '''Videos:''' * [https://www.youtube.com/user/LectureDemostrations/videos?view=1 Lecture Demonstration's Youtube Channel] '''References:''' * [http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html Elastic & Inelastic Collisions - Hyperphysics] * [https://en.wikipedia.org/wiki/Inelastic_collision Inelastic Collisions - Wikipedia] [:Instructional:Home] | 724 | 2,600 | {"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} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.8509 |
https://www.pokerstrategy.com/forum/thread.php?threadid=436234 | 1,531,849,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00014.warc.gz | 952,350,628 | 22,010 | # Folding my way into a bigger winrate
• Bronze
Joined: 15.02.2018
For 2NL
I have been calling river bets at 2NL with two pair, TPTK bottom sets etc too much and think now that it might be a leak at 2NL to call those.
So what I will do for 25.000 hands is:
Fold to all river bets when the bet is more than 1/3th of the pot (also when I am so called 'pot committed') when I do not have the 'almost' nuts.
My definition of the almost nuts:
When flush: The second best flush is the almost nuts.
When straight: 48 on a 567xx board is the almost nuts
When set 66 on a 26Txx board is the almost nuts (but on a 567xx board it is not).
The lowest boat possible is 'per definition' the almost nuts
I will fold all my flushes and straights to all paired boards (because it is NOT the almost nuts).
I will fold all my trips (because it is NOT the almost nuts).
I will fold all my bottom sets.
I am not saying I will not bet these hands! I will bet (check) /fold.
Exception 1:
Pot + bet below 10BB. I'm allowed to catch small bluffs.
Exception 2:
Sometimes your hand is not strong enough to go all in but too strong not to bet the turn. In this case I am allowed to check the turn to induce a bluff and call that bluff on the river. Example in spoiler
Pre Flop: (pot: 1.5 BB) Hero has A Q (villain was 8/8/15 but small sample)
Hero raises to 3 BB, fold, fold, fold, BB raises to 10 BB, Hero calls 7 BB
Flop: (20.5 BB, 2 players) 2 6 Q
BB bets 14.5 BB, Hero calls 14.5 BB
Turn: (49.5 BB, 2 players) 5
BB checks, Hero checks
River: (49.5 BB, 2 players) J
BB bets 34.5 BB, Hero calls 34.5 BB
I will post my graph and I will post the hands I folded.
At the end I'm interested if my total graph goes up/down/stays same
Maybe it is a bit of a small sample but at 25K I will evaluate.
Constructive criticism is more than welcome.
edit: I changed 1/4th of the pot to 1/3th of the pot , the 1/4th was too nitty.
edit: 25 feb 2018 Added the 'calling delayed turnbet' rule.
• 26 replies
• Bronze
Joined: 15.02.2018
OK.. Played 1100 hands...
Her is my graph (but I can not use an external URL for an image).
Ah.. shit.. the image is not working, why? (its not the http/https thing, I have tried that one).
I have seen other people use external links? Is it disabled now?
RubbyDubby
• Post removed
• Bronze
Joined: 10.07.2011
Hey, good luck with your challenge ! Do you play normal tables or Zoom ?
• Bronze
Joined: 15.02.2018
@AgapieGheorghe Thank you so much! (It also looks like I can not have friends when not bronze)
Normal tables. I really like the opponent reading part of poker. I am not so much into hand ranges.
So I had a bit of a rough start as you can see in the graph above. Lets look at the All-in adj. line for this challenge shall we.
Interesting hands of day 1 you can find here..
• Bronze
Joined: 15.02.2018
Day 2: 2000 hands played now and I'm break even
I have a question. For this challenge. Can I use the All-in adj line?
I mean, you start a challenge but after 2000 hands you're more than 4 buy ins below EV!
Link to the 2 interesting hands of this session.
The most ridiculous suckout just for entertainment purposes below
Pre Flop: (pot: 1.5 BB) Hero has K K
fold, CO raises to 3 BB, fold, Hero raises to 10 BB, fold, CO raises to 18 BB, Hero raises to 55 BB, CO calls 33.5 BB and is all-in
Flop: (104 BB, 2 players) 7 K 6
Turn: (104 BB, 2 players) 8
River: (104 BB, 2 players) 9
Hero shows K K (Three of a Kind, Kings)
(Pre 81%, Flop 98%, Turn 93%)
CO shows T T (Straight, Ten High)
(Pre 19%, Flop 2%, Turn 7%)
CO wins 100.5 BB
• Bronze
Joined: 10.07.2011
After 25k hands you will catch up the EV, yeah it sucks that you are so much under EV right now, but you just need volume.
Good luck tomorrow
• Bronze
Joined: 15.02.2018
Originally posted by RubbyDubby
Day 2: 2000 hands played now and I'm break even
http://www.belmijnouders.nl/img/2.png
I have a question. For this challenge. Can I use the All-in adj line?
I mean, you start a challenge but after 2000 hands you're more than 4 buy ins below EV!
Link to the 2 interesting hands of this session.
The most ridiculous suckout just for entertainment purposes below
Pre Flop: (pot: 1.5 BB) Hero has K K
fold, CO raises to 3 BB, fold, Hero raises to 10 BB, fold, CO raises to 18 BB, Hero raises to 55 BB, CO calls 33.5 BB and is all-in
Flop: (104 BB, 2 players) 7 K 6
Turn: (104 BB, 2 players) 8
River: (104 BB, 2 players) 9
Hero shows K K (Three of a Kind, Kings)
(Pre 81%, Flop 98%, Turn 93%)
CO shows T T (Straight, Ten High)
(Pre 19%, Flop 2%, Turn 7%)
CO wins 100.5 BB
• Bronze
Joined: 15.02.2018
I'm not winning anything
I think it is not because of me folding to big river bets because there weren't hardly any.
Just did not hit much and when I did hit I got no action at all.
I might have overplayed one hand and that one you can find here...
• Bronze
Joined: 15.02.2018
BRUTAL.
How can you start a challenge and then have to eat this shit (and NO i did not run like Dough Polk before I started this challenge).
On the bright side, It looks like it does not influence my play.
here... a river bet I folded.
• Bronze
Joined: 10.07.2011
WTF is going on every day you are so much under EV, can you post a total graph with showdown and non showdown winings also.
• Bronze
Joined: 15.02.2018
Yah, it sucks
What sucks even more is that Luck and Badluck do not have memories.
They'll start from scratch tomorrow morning .
But again, the good news is that I can see it does not influences my game. I had a real downswing once, that one started also with somthing like this but then my green line went on a down slide.
• Bronze
Joined: 10.07.2011
It is important to have a good mindset when you have a bad time.
I wish i can be like you, I get tilted easily when my fishes win the all in with any2 and then leaves the table
• Bronze
Joined: 15.02.2018
Woo!! I won some and did not got sucked out ( had a big cooler though at the end of the session )
So despite the crazy 6 stacks below EV still 4BB/100!
Don't feel like posting a hand. I folded some but not interesting.
• Bronze
Joined: 10.07.2011
First day with a profit !
• Bronze
Joined: 15.02.2018
Waa!!!!! I did run like crazy! It is as if the poker gods rewarded me for not tilting. 1000 hands a day is a bit much though. I play 4 tables because I want to stay concentrated but that is 3 hours of playing or so.
I folded 2 hands by my rules.
One trips with a flush on the board and one TP good kicker.
The folding of (trips) you will find ..here...
• Bronze
Joined: 15.02.2018
Got sucked out again. Now more than 6.5BB below EV and then I START TILTING and stop following my rules and loose chips.
The not following my rules hand ..here..
• Bronze
Joined: 15.02.2018
I'm running like a poker God. Folding my way into a massive upswing.
But you know what they say: 'What goes up must come down'.
Removed the All-In Adj because no suck-outs today and you can see it anyway in the information box
Just for the archives.
I folded my KK for less than 1/3 of the pot. Think it's a good fold though.
UTG: 100 BB (VPIP: 32.55, PFR: 24.19, 3Bet Preflop: 6.32, Hands: 569)
Hero (MP): 150.5 BB
CO: 103.5 BB (VPIP: 15.84, PFR: 7.59, 3Bet Preflop: 2.33, Hands: 311)
BTN: 106 BB (VPIP: 30.95, PFR: 23.81, 3Bet Preflop: 6.25, Hands: 42)
SB: 57 BB (VPIP: 44.44, PFR: 5.56, 3Bet Preflop: 5.56, Hands: 36)
BB: 106.5 BB (VPIP: 28.13, PFR: 9.38, 3Bet Preflop: 0.00, Hands: 33)
SB posts SB 0.5 BB, BB posts BB 1 BB
Pre Flop: (pot: 1.5 BB) Hero has K K
fold, Hero raises to 3 BB, fold, BTN calls 3 BB, fold, fold
Flop: (7.5 BB, 2 players) 6 9 5
Hero bets 4 BB, BTN calls 4 BB
Turn: (15.5 BB, 2 players) 3
Hero bets 9 BB, BTN calls 9 BB
River: (33.5 BB, 2 players) 7
Hero checks, BTN bets 10 BB, fold
BTN wins 32.5 BB
Rake paid 1 BB
• Bronze
Joined: 15.02.2018
I'm invincible !
Thought I would have a loosing session but the end made all good again.
Andother 16K hands to go before evaluating
• Bronze
Joined: 15.02.2018
BAM! 2 buy-ins down. I really had to step on the brake there.
I think I was getting too comfortable opening too early. Then when you hit nothing for a while it is getting expensive very quickly.
I made one brutal lay-down. You will find it ..here.. | 2,546 | 8,231 | {"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-2018-30 | latest | en | 0.915969 |
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Posted 27 Mar 2006
Licenced GPL3
# Back-propagation Neural Net
, 28 Mar 2006
A C++ class implementing a back-propagation algorithm neural net, that supports any number of layers/neurons.
## Introduction
The class `CBackProp` encapsulates a feed-forward neural network and a back-propagation algorithm to train it. This article is intended for those who already have some idea about neural networks and back-propagation algorithms. If you are not familiar with these, I suggest going through some material first.
## Background
This is part of an academic project which I worked on during my final semester back in college, for which I needed to find the optimal number and size of hidden layers and learning parameters for different data sets. It wasn't easy finalizing the data structure for the neural net and getting the back-propagation algorithm to work. The motivation for this article is to save someone else the same effort.
Here's a little disclaimer... This article describes a simplistic implementation of the algorithm, and does not sufficiently elaborate on the algorithm. There is a lot of room to improve the included code (like adding exception handling :-), and for many steps, there is a lot more reasoning required than I have included, e.g., values that I have chosen for parameters, and the number of layers/the neurons in each layer are for demonstrating the usage and may not be optimal. To know more about these, I suggest going to:
## Using the code
Typically, the usage involves the following steps:
• Create the net using `CBackProp::CBackProp(int nl,int *sz,double b,double a)`
• Apply the back-propagation algorithm - train the net by passing the input and the desired output, to `void CBackProp::bpgt(double *in,double *tgt)` in a loop, until the Mean square error, which is obtained by `CBackProp::double mse(double *tgt)`, gets reduced to an acceptable value.
• Use the trained net to make predictions by feed-forwarding the input data using `void CBackProp::ffwd(double *in)`.
The following is a description of the sample program that I have included.
## One step at a time...
#### Setting the objective:
We will try to teach our net to crack the binary A XOR B XOR C. XOR is an obvious choice, it is not linearly separable hence requires hidden layers and cannot be learned by a single perception.
A training data set consists of multiple records, where each record contains fields which are input to the net, followed by fields consisting of the desired output. In this example, it's three inputs + one desired output.
```// prepare XOR training data
double data[][4]={// I XOR I XOR I = O
//--------------------------------
0, 0, 0, 0,
0, 0, 1, 1,
0, 1, 0, 1,
0, 1, 1, 0,
1, 0, 0, 1,
1, 0, 1, 0,
1, 1, 0, 0,
1, 1, 1, 1 };```
#### Configuration:
Next, we need to specify a suitable structure for our neural network, i.e., the number of hidden layers it should have and the number of neurons in each layer. Then, we specify suitable values for other parameters: learning rate - `beta`, we may also want to specify momentum - `alpha` (this one is optional), and Threshold - `thresh` (target mean square error, training stops once it is achieved else continues for `num_iter` number of times).
Let's define a net with 4 layers having 3,3,3, and 1 neuron respectively. Since the first layer is the input layer, i.e., simply a placeholder for the input parameters, it has to be the same size as the number of input parameters, and the last layer being the output layer must be same size as the number of outputs - in our example, these are 3 and 1. Those other layers in between are called hidden layers.
```int numLayers = 4, lSz[4] = {3,3,3,1};
double beta = 0.2, alpha = 0.1, thresh = 0.00001;
long num_iter = 500000;```
#### Creating the net:
`CBackProp *bp = new CBackProp(numLayers, lSz, beta, alpha);`
#### Training:
```for (long i=0; i < num_iter ; i++)
{
bp->bpgt(data[i%8], &data[i%8][3]);
if( bp->mse(&data[i%8][3]) < thresh)
break; // mse < threshold - we are done training!!!
}```
#### Let's test its wisdom:
We prepare test data, which here is the same as training data minus the desired output.
```double testData[][3]={ // I XOR I XOR I = ?
//----------------------
0, 0, 0,
0, 0, 1,
0, 1, 0,
0, 1, 1,
1, 0, 0,
1, 0, 1,
1, 1, 0,
1, 1, 1};```
Now, using the trained network to make predictions on our test data....
```for ( i = 0 ; i < 8 ; i++ )
{
bp->ffwd(testData[i]);
cout << testData[i][0]<< " "
<< testData[i][1]<< " "
<< testData[i][2]<< " "
<< bp->Out(0) << endl;
}```
## Now a peek inside:
#### Storage for the neural net
I think the following code has ample comments and is self-explanatory...
```class CBackProp{
// output of each neuron
double **out;
// delta error value for each neuron
double **delta;
// 3-D array to store weights for each neuron
double ***weight;
// no of layers in net including input layer
int numl;
// array of numl elements to store size of each layer
int *lsize;
// learning rate
double beta;
// momentum
double alpha;
// storage for weight-change made in previous epoch
double ***prevDwt;
// sigmoid function
double sigmoid(double in);
public:
~CBackProp();
// initializes and allocates memory
CBackProp(int nl,int *sz,double b,double a);
// backpropogates error for one set of input
void bpgt(double *in,double *tgt);
// feed forwards activations for one set of inputs
void ffwd(double *in);
// returns mean square error of the net
double mse(double *tgt);
// returns i'th output of the net
double Out(int i) const;
};```
Some alternative implementations define a separate class for layer / neuron / connection, and then put those together to form a neural network. Although it is definitely a cleaner approach, I decided to use `double ***` and `double **` to store weights and output etc. by allocating the exact amount of memory required, due to:
• The ease it provides while implementing the learning algorithm, for instance, for weight at the connection between (i-1)th layer's jth Neuron and ith layer's kth neuron, I personally prefer `w[i][k][j]` (than something like `net.layer[i].neuron[k].getWeight(j)`). The output of the i`th` neuron of the j`th` layer is `out[i][j]`, and so on.
• Another advantage I felt is the flexibility of choosing any number and size of the layers.
```// initializes and allocates memory
CBackProp::CBackProp(int nl,int *sz,double b,double a):beta(b),alpha(a)
{
// Note that the following are unused,
//
// delta[0]
// weight[0]
// prevDwt[0]
// I did this intentionally to maintain
// consistency in numbering the layers.
// Since for a net having n layers,
// input layer is referred to as 0th layer,
// first hidden layer as 1st layer
// and the nth layer as output layer. And
// first (0th) layer just stores the inputs
// hence there is no delta or weight
// values associated to it.
// set no of layers and their sizes
numl=nl;
lsize=new int[numl];
for(int i=0;i<numl;i++){
lsize[i]=sz[i];
}
// allocate memory for output of each neuron
out = new double*[numl];
for( i=0;i<numl;i++){
out[i]=new double[lsize[i]];
}
// allocate memory for delta
delta = new double*[numl];
for(i=1;i<numl;i++){
delta[i]=new double[lsize[i]];
}
// allocate memory for weights
weight = new double**[numl];
for(i=1;i<numl;i++){
weight[i]=new double*[lsize[i]];
}
for(i=1;i<numl;i++){
for(int j=0;j<lsize[i];j++){
weight[i][j]=new double[lsize[i-1]+1];
}
}
// allocate memory for previous weights
prevDwt = new double**[numl];
for(i=1;i<numl;i++){
prevDwt[i]=new double*[lsize[i]];
}
for(i=1;i<numl;i++){
for(int j=0;j<lsize[i];j++){
prevDwt[i][j]=new double[lsize[i-1]+1];
}
}
// seed and assign random weights
srand((unsigned)(time(NULL)));
for(i=1;i<numl;i++)
for(int j=0;j<lsize[i];j++)
for(int k=0;k<lsize[i-1]+1;k++)
weight[i][j][k]=(double)(rand())/(RAND_MAX/2) - 1;
// initialize previous weights to 0 for first iteration
for(i=1;i<numl;i++)
for(int j=0;j<lsize[i];j++)
for(int k=0;k<lsize[i-1]+1;k++)
prevDwt[i][j][k]=(double)0.0;
}```
#### Feed-Forward
This function updates the output value for each neuron. Starting with the first hidden layer, it takes the input to each neuron and finds the output (`o`) by first calculating the weighted sum of inputs and then applying the Sigmoid function to it, and passes it forward to the next layer until the output layer is updated:
where:
```// feed forward one set of input
void CBackProp::ffwd(double *in)
{
double sum;
// assign content to input layer
for(int i=0;i < lsize[0];i++)
out[0][i]=in[i];
// assign output(activation) value
// to each neuron usng sigmoid func
// For each layer
for(i=1;i < numl;i++){
// For each neuron in current layer
for(int j=0;j < lsize[i];j++){
sum=0.0;
// For input from each neuron in preceding layer
for(int k=0;k < lsize[i-1];k++){
// Apply weight to inputs and add to sum
sum+= out[i-1][k]*weight[i][j][k];
}
// Apply bias
sum+=weight[i][j][lsize[i-1]];
// Apply sigmoid function
out[i][j]=sigmoid(sum);
}
}
}```
#### Back-propagating...
The algorithm is implemented in the function `void CBackProp::bpgt(double *in,double *tgt)`. Following are the various steps involved in back-propagating the error in the output layer up till the first hidden layer.
```void CBackProp::bpgt(double *in,double *tgt)
{
double sum;```
First, we call `void CBackProp::ffwd(double *in)` to update the output values for each neuron. This function takes the input to the net and finds the output of each neuron:
where:
`ffwd(in);`
The next step is to find the delta for the output layer:
```for(int i=0;i < lsize[numl-1];i++){
delta[numl-1][i]=out[numl-1][i]*
(1-out[numl-1][i])*(tgt[i]-out[numl-1][i]);
}```
then find the delta for the hidden layers...
```for(i=numl-2;i>0;i--){
for(int j=0;j < lsize[i];j++){
sum=0.0;
for(int k=0;k < lsize[i+1];k++){
sum+=delta[i+1][k]*weight[i+1][k][j];
}
delta[i][j]=out[i][j]*(1-out[i][j])*sum;
}
}```
Apply momentum (does nothing if alpha=0):
```for(i=1;i < numl;i++){
for(int j=0;j < lsize[i];j++){
for(int k=0;k < lsize[i-1];k++){
weight[i][j][k]+=alpha*prevDwt[i][j][k];
}
weight[i][j][lsize[i-1]]+=alpha*prevDwt[i][j][lsize[i-1]];
}
}```
Finally, adjust the weights by finding the correction to the weight.
And then apply the correction:
```for(i=1;i < numl;i++){
for(int j=0;j < lsize[i];j++){
for(int k=0;k < lsize[i-1];k++){
prevDwt[i][j][k]=beta*delta[i][j]*out[i-1][k];
weight[i][j][k]+=prevDwt[i][j][k];
}
prevDwt[i][j][lsize[i-1]]=beta*delta[i][j];
weight[i][j][lsize[i-1]]+=prevDwt[i][j][lsize[i-1]];
}
}```
#### How learned is the net?
Mean square error is used as a measure of how well the neural net has learnt.
As shown in the sample XOR program, we apply the above steps until a satisfactorily low error level is achieved. `CBackProp::double mse(double *tgt)` returns just that.
## History
• Created date: 25th Mar'06.
## License
This article, along with any associated source code and files, is licensed under The GNU General Public License (GPLv3)
## About the Author
Software Developer Australia
A technology enthusiast with interest in all aspects of software development. I code in C, C++, Java and Go mostly on Unix/Linux.
## Comments and Discussions
First PrevNext
to get reference generated current of three phase using back propagation control algorithm using matlab Member 1211173727-Nov-15 4:20 Member 12111737 27-Nov-15 4:20
Fine piece of code SoothingMist6-Jan-15 22:56 SoothingMist 6-Jan-15 22:56
Sigmoid derivative Raphael Gruaz2-Oct-12 0:12 Raphael Gruaz 2-Oct-12 0:12
Re: Sigmoid derivative Annisa Kartikasari6-Jul-15 19:35 Annisa Kartikasari 6-Jul-15 19:35
same question with me, it just written in sigmoid, and there is no definition what is sigmoid in the code. Have you solve this by yourself?
Overflow Miguel Tomas5-Aug-12 7:06 Miguel Tomas 5-Aug-12 7:06
How to save the network? tooym20-Jan-10 17:05 tooym 20-Jan-10 17:05
How to minimize Total Network Error AjayIndian31-Dec-09 23:21 AjayIndian 31-Dec-09 23:21
Understanding the basis for code locuaz16-Oct-09 22:34 locuaz 16-Oct-09 22:34
how can I get the source code xumin198816-Oct-09 6:03 xumin1988 16-Oct-09 6:03
Bias value? emrecaglar12-Oct-09 9:30 emrecaglar 12-Oct-09 9:30
Out(0) tangsu1-Sep-09 2:10 tangsu 1-Sep-09 2:10
telecom billing Sofritom13-Jul-09 1:32 Sofritom 13-Jul-09 1:32
Error in the code [modified] gpwr9k9511-Jun-09 13:08 gpwr9k95 11-Jun-09 13:08
is this a bug? Member 42945018-Apr-09 15:18 Member 429450 18-Apr-09 15:18
How about bias KadirErturk8-Mar-09 6:36 KadirErturk 8-Mar-09 6:36
Normalized values picand6-May-08 2:09 picand 6-May-08 2:09
Feedforward Backpropagation Neural Network raceng05857-Dec-07 5:41 raceng0585 7-Dec-07 5:41
Trouble converting to C pradeep swamy18-Nov-07 22:37 pradeep swamy 18-Nov-07 22:37
Re: Trouble converting to C robiii16-Jul-09 5:31 robiii 16-Jul-09 5:31
Scaling Input ravenspoint16-Oct-07 4:44 ravenspoint 16-Oct-07 4:44
MSE Calculation ravenspoint16-Oct-07 4:04 ravenspoint 16-Oct-07 4:04
Re: MSE Calculation brutjbro23-Oct-07 3:25 brutjbro 23-Oct-07 3:25
Re: MSE Calculation ravenspoint23-Oct-07 4:48 ravenspoint 23-Oct-07 4:48
Re: MSE Calculation brutjbro24-Oct-07 9:57 brutjbro 24-Oct-07 9:57
Re: MSE Calculation ravenspoint28-Oct-07 9:16 ravenspoint 28-Oct-07 9:16
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https://online-calculator.org/finance/usd-to-rmb.php | 1,721,354,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00344.warc.gz | 375,197,342 | 5,613 | Online Calculators > Financial Calculators > USD to RMB
# USD to RMB
USD to RMB conversion calculator to convert USD dollar to Chinese Yuan and vice versa. To calculate how much is a dollar in China, multiply by 7.2661. The USD to RMB rate changes constantly. When the rate changes, the US dollar to Chinese Yuan amount will also change.
USD RMB
## How much is a dollar in China?
1 dollar equals CN¥7.27, or 1 USD = 7.27 CNY.
## USD to CNY
The following is a USD to CNY table that shows conversion from \$1 to RMB up to \$100.
USD RMB
\$1 CN¥7.2661
\$2 CN¥14.5322
\$3 CN¥21.7983
\$4 CN¥29.0644
\$5 CN¥36.3305
\$6 CN¥43.5966
\$7 CN¥50.8627
\$8 CN¥58.1288
\$9 CN¥65.3949
\$10 CN¥72.6610
\$11 CN¥79.9271
\$12 CN¥87.1932
\$13 CN¥94.4593
\$14 CN¥101.7254
\$15 CN¥108.9915
\$16 CN¥116.2576
\$17 CN¥123.5237
\$18 CN¥130.7898
\$19 CN¥138.0559
\$20 CN¥145.3220
\$21 CN¥152.5881
\$22 CN¥159.8542
\$23 CN¥167.1203
\$24 CN¥174.3864
\$25 CN¥181.6525
\$26 CN¥188.9186
\$27 CN¥196.1847
\$28 CN¥203.4508
\$29 CN¥210.7169
\$30 CN¥217.9830
\$31 CN¥225.2491
\$32 CN¥232.5152
\$33 CN¥239.7813
\$34 CN¥247.0474
\$35 CN¥254.3135
\$36 CN¥261.5796
\$37 CN¥268.8457
\$38 CN¥276.1118
\$39 CN¥283.3779
\$40 CN¥290.6440
\$41 CN¥297.9101
\$42 CN¥305.1762
\$43 CN¥312.4423
\$44 CN¥319.7084
\$45 CN¥326.9745
\$46 CN¥334.2406
\$47 CN¥341.5067
\$48 CN¥348.7728
\$49 CN¥356.0389
\$50 CN¥363.3050
\$51 CN¥370.5711
\$52 CN¥377.8372
\$53 CN¥385.1033
\$54 CN¥392.3694
\$55 CN¥399.6355
\$56 CN¥406.9016
\$57 CN¥414.1677
\$58 CN¥421.4338
\$59 CN¥428.6999
\$60 CN¥435.9660
\$61 CN¥443.2321
\$62 CN¥450.4982
\$63 CN¥457.7643
\$64 CN¥465.0304
\$65 CN¥472.2965
\$66 CN¥479.5626
\$67 CN¥486.8287
\$68 CN¥494.0948
\$69 CN¥501.3609
\$70 CN¥508.6270
\$71 CN¥515.8931
\$72 CN¥523.1592
\$73 CN¥530.4253
\$74 CN¥537.6914
\$75 CN¥544.9575
\$76 CN¥552.2236
\$77 CN¥559.4897
\$78 CN¥566.7558
\$79 CN¥574.0219
\$80 CN¥581.2880
\$81 CN¥588.5541
\$82 CN¥595.8202
\$83 CN¥603.0863
\$84 CN¥610.3524
\$85 CN¥617.6185
\$86 CN¥624.8846
\$87 CN¥632.1507
\$88 CN¥639.4168
\$89 CN¥646.6829
\$90 CN¥653.9490
\$91 CN¥661.2151
\$92 CN¥668.4812
\$93 CN¥675.7473
\$94 CN¥683.0134
\$95 CN¥690.2795
\$96 CN¥697.5456
\$97 CN¥704.8117
\$98 CN¥712.0778
\$99 CN¥719.3439
\$100 CN¥726.6100
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Random Number Generator | 1,411 | 3,457 | {"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-2024-30 | latest | en | 0.417262 |
https://ccssmathanswers.com/problems-on-work-done-in-a-given-period-of-time/ | 1,716,109,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00495.warc.gz | 131,972,682 | 57,170 | # Problems on Work Done in a Given Period of Time | Work Done Problems with Solutions
Problems on Work Done in a Given Period of Time are given in this article. Every problem has answers along with explanations. Start your practice without any delay to get good marks in the exam. You can easily score good results in the exam by getting the grip on the complete concept. You have to follow the best way to solve problems. We have also included tips and tricks to solve problems on Work Done in a Given Period of Time. Therefore, make your practice easy and get complete knowledge of the concept.
Work Done = Time x Rate of Work.
If A person can complete the work in ‘x’ days, then the work done by ‘A’ in one day is equal to ($$\frac{1}{x}$$).
## Examples on Calculating Work Done in a Given Period of Time
Solve the different examples of work done in a given period of time by using the different formulas.
Example 1.
Amar and Brusly can do the work in 25 days. Brusly and Catherin can do the work in 35 days. Catherin and Amar can do the work in 30 days. In how many days, Amar, Brusly, and Catherin complete the work by working together?
Solution:
As per the details, Amar and Brusly can take time to complete the work = a = 25 days.
So, Amar and Brusly can complete the work in one day = $$\frac{1}{a}$$ = $$\frac{1}{25}$$th part of the work.
Brusly and Catherin can take time to complete the work = b = 35 days.
So, Brusly and Catherin both can complete the work in 1 day = $$\frac{1}{b}$$ = $$\frac{1}{35}$$th part of the work.
Catherin and Amar can take time to complete the work = c = 30 days.
Catherin and Amar can complete the work in 1 day = $$\frac{1}{c}$$ = $$\frac{1}{30}$$th part of the work.
The work done by Amar, Brusly, and Catherin together in one day = $$\frac{1}{a}$$ + $$\frac{1}{b}$$ + $$\frac{1}{c}$$.
= $$\frac{1}{25}$$ + $$\frac{1}{35}$$ + $$\frac{1}{30}$$.
L.C.M of 25, 35, and 30 is 1050.
= $$\frac{42 + 30 + 35}{1050}$$ = $$\frac{107}{1050}$$.
Total Work done by Amar, Brusly, and Catherin in $$\frac{107}{1050}$$ days = 9.8 days.
Therefore, Total work is done by Amar, Brusly, and Catherin in 9.8 days.
Example 2.
A and B can make the toys in 14 days. B and C can make the same toys in 24 days. C and A can make the toys in 20 days. Calculate work done by each person individually?
Solution:
As per the given information, A and B can make the toys in 14 days. That is a = 14 days.
A and B can complete the work in 1 day = $$\frac{1}{a}$$ = $$\frac{1}{14}$$.
B and C can make the same amount of toys in 24 days. That is b = 24 days.
B and C can complete the work in one day = $$\frac{1}{b}$$ = $$\frac{1}{24}$$.
C and A can make the toys in 20 days. That is c = 20 days.
C and A can complete the work in 1 day = $$\frac{1}{c}$$ = $$\frac{1}{20}$$.
(A and B) + (B and C) + (C and A) one day’s work = $$\frac{1}{a}$$ + $$\frac{1}{b}$$ + $$\frac{1}{c}$$ = $$\frac{1}{14}$$ + $$\frac{1}{24}$$ + $$\frac{1}{20}$$.
L.C.M of 14, 24, 20 is 840.
=$$\frac{60 + 35 + 42}{840}$$.
2(A + B + C)’s one day’s work = $$\frac{137}{840}$$.
A, B, and C one day work =$$\frac{137}{2(840)}$$= $$\frac{137}{1680}$$.
A, B, and C can complete the total work in = $$\frac{1680}{137}$$ = 12 days.
Work done by ‘A’ alone = Work done by A, B, and C – Work done by B and C.
= 12 – 1/24
= $$\frac{288 – 1}{24}$$ = $$\frac{287}{24}$$
Therefore, A alone can do the work in $$\frac{287}{24}$$ days.
Work done by ‘B’ alone = Work done by A, B, and C – Work done by A and C.
= 12 – 1/20.
= $$\frac{240 – 1}{20}$$.
= $$\frac{239}{20}$$.
Therefore, B alone can complete the work in$$\frac{239}{20}$$days.
Work done by ‘C’ alone = Work done by A,B, and C – Work done by A and B.
= 12 – 1/14.
= $$\frac{168 – 1}{14}$$.
= $$\frac{167}{14}$$.
Therefore, C alone can complete the work in $$\frac{167}{14}$$ days.
Example 3.
Ira and Mihira started work on the same project and they can finish the work in 20 days. Ira worked for 10 days and then Mihira completed the remaining work in 14 days. Find the time to complete the project by Mihira alone?
Solution:
As per the given details, Ira and Mihira both can complete the project in 20 days.
Ira and Mihira both can complete the project work in one day = $$\frac{1}{20}$$th part of the day.
Let us consider the work done by Ira in one day = p and Work done by Mihira in one day = q.
Work done by Ira and Mihira in one day = p + q = 1/20$$\frac{1}{20}$$. —–(1)
Ira worked for 10 days and Mihira completed the remaining work in 14 days. That is
10p + 14q = 1 —-(2).
By comparing the equation (1) and (2), We will get Multiply the equation (1) with 10 on both sides. That is
10p + 10q = 10/20 = $$\frac{1}{2}$$.
10p + 14q = 1
-4q = 1/2 – 1 = (1-2)/2 = –$$\frac{1}{2}$$.
q= $$\frac{1}{8}$$.
p + $$\frac{1}{8}$$= $$\frac{1}{20}$$.
p=$$\frac{1}{20}$$ – $$\frac{1}{8}$$
= $$\frac{2-5}{40}$$
= $$\frac{3}{40}$$.
Therefore, Work done by Mihira in 8 days.
Example 4.
10 men and 10 women can complete the work in 12 days and 15 days respectively. In how many days 10 men and 10 women together can complete the work?
Solution:
As per the given information, 10 men can take time to complete the work = a = 12 days.
1 day work done by 10 men = $$\frac{1}{a}$$ = $$\frac{1}{12}$$.
10 women can finish the work in 15 days. That is b = 15 days.
Work done by 10 women in 1 day = $$\frac{1}{b}$$ = $$\frac{1}{15}$$.
10 men and 10 women together can work in one day = $$\frac{1}{12}$$ + $$\frac{1}{15}$$.
=$$\frac{5+4}{60}$$ = $$\frac{9}{60}$$.
Total work done by 10 men and 10 women together $$\frac{60}{9}$$ = $$\frac{3}{20}$$.
Therefore, total work is completed by 10 men and 10 women in $$\frac{3}{20}$$days.
Example 5.
In 35 days 30 men complete work by working together. After starting the work together. Every 10th day, 5 men left from the work. In how many days, the work will complete?
Solution:
As per the given information, Total work is completed in 35 × 30 = 1050 days.
1st 10 days = Total number of days are 10: the number of days × number of men working = 10 × 30 = 300 days.
2nd 10 days = Total number of days are 20: the number of days × number of men working = 10 × 25 = 250 days.
So, total work completed 300 + 250 = 550 days.
3rd 10 days = Total number of days are 30:number of days × number of men working = 10 × 20 = 200 days.
Total work completed in 550 + 200 = 750 days.
4th 10 days = Total number of days are 40: number of days × number of men working = 10 × 15 = 150 days.
Total number of days = 750 + 150 = 900 days
5th 10 days = Total number of days are 50: number of days × number of men working = 10 × 10 = 100.
Total number of days = 900 + 100 = 1000.
6th 10 days = Total number of days is 60: number of days × number of men working = 10 × 5 = 50.
Total number of days = 1000 + 50 = 1050.
Therefore, the total work is completed in 60 days.
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http://oeis.org/A318706 | 1,547,872,714,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662124.0/warc/CC-MAIN-20190119034320-20190119060320-00380.warc.gz | 170,358,386 | 3,651 | This site is supported by donations to The OEIS Foundation.
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A318706 For any n >= 0 with base-9 representation Sum_{k=0..w} d_k * 9^k, let g(n) = Sum_{k=0..w} s(d_k) * 3^k (where s(0) = 0, s(1+2*j) = i^j and s(2+2*j) = i^j * (1+i) for any j > 0, and i denotes the imaginary unit); a(n) is the imaginary part of g(n). 3
0, 0, 1, 1, 1, 0, -1, -1, -1, 0, 0, 1, 1, 1, 0, -1, -1, -1, 3, 3, 4, 4, 4, 3, 2, 2, 2, 3, 3, 4, 4, 4, 3, 2, 2, 2, 3, 3, 4, 4, 4, 3, 2, 2, 2, 0, 0, 1, 1, 1, 0, -1, -1, -1, -3, -3, -2, -2, -2, -3, -4, -4, -4, -3, -3, -2, -2, -2, -3, -4, -4, -4, -3, -3, -2, -2 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,19 COMMENTS See A318705 for the real part of g and additional comments. LINKS Rémy Sigrist, Table of n, a(n) for n = 0..6560 FORMULA a(9 * k) = 3 * a(k) for any k >= 0. PROG (PARI) a(n) = my (d=Vecrev(digits(n, 9))); imag(sum(k=1, #d, if (d[k], 3^(k-1)*I^floor((d[k]-1)/2)*(1+I)^((d[k]-1)%2), 0))) CROSSREFS Cf. A318705. Sequence in context: A246011 A061023 A057690 * A298199 A282623 A090589 Adjacent sequences: A318703 A318704 A318705 * A318707 A318708 A318709 KEYWORD sign,base AUTHOR Rémy Sigrist, Sep 01 2018 STATUS approved
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Last modified January 18 23:05 EST 2019. Contains 319282 sequences. (Running on oeis4.) | 718 | 1,730 | {"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} | 3.484375 | 3 | CC-MAIN-2019-04 | latest | en | 0.59024 |
https://www.physicsforums.com/threads/particle-model-of-thermal-energy-questions.799574/ | 1,527,177,374,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00095.warc.gz | 804,093,548 | 16,067 | # Homework Help: Particle Model of Thermal Energy questions
1. Feb 23, 2015
### romantichero7
1. The problem statement, all variables and given/known data
a) Assume that a substance in a closed container is at thermal equilibrium with about half of its
molecules in the gas phase and half of its molecules in the liquid phase. How could you use the
Particle Model of Thermal Energy to explain which molecules are moving faster, on
average-those in the gas phase, those in the liquid phase, or neither?
b) Now assume that liquid water (H20) and gaseous oxygen (O2) are at thermal equilibrium in a
closed container at room temperature. How would your explanation for which molecules are
moving faster (the H2O molecules in the liquid phase, the O2 molecules in the gas phase, or
neither) be the same as your explanation in part a), and how would it differ?
2. Relevant equations
3. The attempt at a solution
What puzzles me about this stuff is that we are supposed to be using an equation for thermal energy, where Eth = # of modes x kB/2 x T, where kB is a constant. So at thermal equilibrium, does a monatomic solid (6 modes) have greater Eth than a monatomic gas (3 modes)? I have been primed to think gases have higher Eth... I am just puzzled as to how to think of comparing Eth of different states within one substance, and then going beyond that to compare different substances.
2. Feb 25, 2015
### Simon Bridge
Back up: what is thermal energy?
How would it be related to heat capacity? | 351 | 1,500 | {"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} | 2.859375 | 3 | CC-MAIN-2018-22 | latest | en | 0.944615 |
https://www.studypool.com/discuss/301342/statistics-assignment-1-math-300 | 1,477,481,566,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720941.32/warc/CC-MAIN-20161020183840-00036-ip-10-171-6-4.ec2.internal.warc.gz | 989,986,945 | 753,515 | # Statistics Assignment 1 Math 300
Dec 8th, 2014
DotaCN
Category:
Statistics
Price: \$40 USD
Question description
Imagine you are a manager at a major bottling company. Customers have begun to complain that the bottles of the brand of soda produced in your company contain less than the advertised sixteen (16) ounces of product. Your boss wants to solve the problem at hand and has asked you to investigate. You have your employees pull thirty (30) bottles off the line at random from all the shifts at the bottling plant. You ask your employees to measure the amount of soda there is in each bottle. Note: Use the data set provided by your instructor to complete this assignment.
Bottle Number Ounces Bottle Number Ounces Bottle Number Ounces 1 14.5 11 15 21 14.1 2 14.6 12 15.1 22 14.2 3 14.7 13 15 23 14 4 14.8 14 14.4 24 14.9 5 14.9 15 15.8 25 14.7 6 15.3 16 14 26 14.5 7 14.9 17 16 27 14.6 8 15.5 18 16.1 28 14.8 9 14.8 19 15.8 29 14.8 10 15.2 20 14.5 30 14.6
Write a two to three (2-3) page report in which you:
1. Calculate the mean, median, and standard deviation for ounces in the bottles.
2. Construct a 95% Confidence Interval for the ounces in the bottles.
3. Conduct a hypothesis test to verify if the claim that a bottle contains less than sixteen (16) ounces is supported. Clearly state the logic of your test, the calculations, and the conclusion of your test.
4. Provide the following discussion based on the conclusion of your test:
a. If you conclude that there are less than sixteen (16) ounces in a bottle of soda, speculate on three (3) possible causes. Next, suggest the strategies to avoid the deficit in the future.
Or
b. If you conclude that the claim of less soda per bottle is not supported or justified, provide a detailed explanation to your boss about the situation. Include your speculation on the reason(s) behind the claim, and recommend one (1) strategy geared toward mitigating this issue in the future.
5. Use at least two (2) quality resources in this assignment. Note: Wikipedia and similar Websites do not qualify as quality resources. The body of the paper must have in-text citations that correspond to the references.
• Be typed, double spaced, using Times New Roman font (size 12), with one-inch margins on all sides; citations and references must follow APA or school-specific format. Check with your professor for any additional instructions.
• Include a cover page containing the title of the assignment, the student’s name, the professor’s name, the course title, and the date. The cover page and the reference page are not included in the required assignment page length.
The specific course learning outcomes associated with this assignment are:
· Calculate measurements of central tendency and dispersal.
· Determine confidence intervals for data.
· Describe the vocabulary and principles of hypothesis testing.
· Discuss application of course content to professional contexts.
· Use technological tools to solve problems in statistics.
· Write clearly and concisely about statistics using proper writing mechanics.
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2325 Tutors | 990 | 3,915 | {"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} | 3.84375 | 4 | CC-MAIN-2016-44 | longest | en | 0.876541 |
https://www.doubtnut.com/qna/646611603 | 1,725,946,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00652.warc.gz | 678,714,313 | 35,361 | Total number of isomers having molecular formula C4 H8
A
2
B
3
C
4
D
6
Video Solution
Text Solution
Generated By DoubtnutGPT
To determine the total number of isomers for the molecular formula C4H8, we can follow these steps:Step 1: Determine the Degree of UnsaturationThe degree of unsaturation (DU) can be calculated using the formula:DU=(2C+2+N−H−X)2For C4H8:- C=4- H=8- N=0 (no nitrogen)- X=0 (no halogens)Plugging in the values:DU=(2(4)+2−8)2=(8+2−8)2=22=1This indicates that there is one degree of unsaturation, which could be a double bond or a ring.Step 2: Identify Possible StructuresGiven that we have one degree of unsaturation, we can have:1. Alkenes (with a double bond)2. Cycloalkenes (with a ring)Step 3: Draw the Isomers1. Isobutene (2-methylpropene): CH2=C(CH3)CH32. 1-Butene: CH2=CH−CH2−CH33. Cyclic structure (Cyclobutene): Cyclobutene: C4H8 in a ring structure4. Cis-2-butene: CH3−CH=CH−CH3 (with both methyl groups on the same side)5. Trans-2-butene: CH3−CH=CH−CH3 (with methyl groups on opposite sides)Step 4: Count the IsomersAfter drawing the structures, we find the following isomers:1. Isobutene (2-methylpropene)2. 1-Butene3. Cyclobutene4. Cis-2-butene5. Trans-2-buteneConclusionThus, the total number of isomers for the molecular formula C4H8 is 5.
|
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 687 | 2,277 | {"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-2024-38 | latest | en | 0.787819 |
https://www.electricalvolt.com/booster-transformer/ | 1,718,413,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00355.warc.gz | 667,075,442 | 22,890 | # Booster Transformer- Definition & Explanation
A booster transformer is commonly used near the end of a power transmission line to increase the voltage to a desired level. It regulates the voltage of a feeder located far away from the main transformer.
## Working of Booster Transformer
The booster transformer’s secondary is connected in series with the line, while its primary is fed from the regulating transformer’s secondary. The regulating transformer’s output winding is linked to the booster transformer’s primary in such a way that the voltage injected into the line (VB) is in phase with the supply voltage (VS).
Suppose that the voltage at the starting point is 25 kV and the voltage at a location 2 KM away is 24.1 kV. In this case, the booster transformer’s secondary voltage will be 0.9 kV, and the voltage at the end of the line will be 24.1 + 0.9 = 25 kV.
The regulating transformer allows for the adjustment of the voltage on the feeder VF by adjusting the taps to change the magnitude of VB. The rating of the regulating transformer is only a fraction of that of the main transformer and can be calculated using the mathematical expression given below.
The advantage of the above system is that the regulating equipment is independent of the main transformer, so a failure in the former will not put the latter out of service.
The disadvantages of the above system are that the booster transformers are more expensive and require more space for their installation. The system becomes less efficient because of losses in the booster and regulating transformer.
## Conclusion:
The booster transformer compensates the line voltage drops and maintains the desired voltage at the far end of the transmission line. Booster transformers are used in railway power lines to prevent stray currents from causing damage to electronic devices and communication systems of the trains passing through them. | 369 | 1,919 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.925283 |
http://algorithms.tutorialhorizon.com/category/recursion/ | 1,516,303,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00070.warc.gz | 15,717,809 | 25,765 | # Category: Recursion
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## Maximum difference between two elements where larger element appears after the smaller element
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## Search the Element in a binary tree – With and Without Recursion
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## Tree Traversals
There are multiple ways to in which you can traverse a tree. In this article we will see these traversals in detail. If you are new to trees then I would recommend that you...
## Dynamic Programming – Maximum Product Cutting Problem.
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## Dynamic Programming – Longest Common Subsequence
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## Dynamic Programming – Rod Cutting Problem
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Objective: Given a set of coins and amount, Write an algorithm to find out how many ways we can make the change of the amount using the coins given. This is another problem in...
## All N Length Strings from Given String of Length K
Objective: Print All N Length Strings from Given String of Length K where characters can appear multiple time. Example: String k = “ALGO” N=2 Result: AA LA GA OA AL LL GL OL AG...
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Objective: Given Number K, Print all the strings of N length. Example: N = 2, K = 3 [1, 1] [2, 1] [3, 1] [1, 2] [2, 2] [3, 2] [1, 3] [2, 3]...
## Print All Possible Subsets with Sum equal to a given Number
Objective: Given a number N, Write an algorithm to print all possible subsets with Sum equal to N In this problem you will see the power of recursion. This question has been asked in...
## Dynamic Programming – Subset Sum Problem
Objective: Given a set of positive integers, and a value sum S, find out if there exist a subset in array whose sum is equal to given sum S. Example: int[] A = {...
## The Word Break Problem
Objective : Given an string and a dictionary of words, find out if the input string can be broken into a space-separated sequence of one or more dictionary words. Example: dictionary = [“I” ,...
## Backtracking – Knight’s Tour Problem
Objective : A knight’s tour is a sequence of moves of a knight on a chessboard such that the knight visits every square only once. If the knight ends on a square that is...
## Backtracking – Search a Word In a Matrix
Objective : Given a 2D matrix of characters. Check whether the word exist in the matrix or not. If it exists then print its path. All movements are allowed (right, left, up, down and...
## Backtracking – N Queens Problem – Better Solution
Objective : In chess, a queen can move as far as she pleases, horizontally, vertically, or diagonally. A chess board has 8 rows and 8 columns. The standard 8 by 8 Queen’s problem asks...
## Backtracking – N Queens Problem
Objective : In chess, a queen can move as far as she pleases, horizontally, vertically, or diagonally. A chess board has 8 rows and 8 columns. The standard 8 by 8 Queen’s problem asks...
## Backtracking – Rat In A Maze Puzzle
Given a maze, NxN matrix. A rat has to find a path from source to destination. maze[0][0] (left top corner)is the source and maze[N-1][N-1](right bottom corner) is destination. There are few cells which are...
## Backtracking – SUDOKU Solver
SUDOKU Puzzle : The objective is to fill a 9×9 grid with digits so that each column, each row, and each of the nine 3×3 sub-grids that compose the grid (also called “boxes”, “blocks”,...
## Introduction To Backtracking Programming
What is Backtracking Programming?? Recursion is the key in backtracking programming. As the name suggests we backtrack to find the solution. We start with one possible move out of many available moves and try...
## Dynamic Programming – Stairs Climbing Puzzle
Objective: A child is climbing up a staircase with n steps, and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways...
## Dynamic Programming – Minimum Coin Change Problem
Objective: Given a amount ‘A’ and n coins, v1<v2<v3<………..<vn . Write a program to find out minimum numbers of coins required to make the change for the amount ‘A’. Example: Amount: 5 Coins []...
## Introduction To Dynamic Programming – Fibonacci Series
What is Dynamic Programming: Dynamic programming is a technique to solve the recursive problems in more efficient manner. Many times in recursion we solve the sub-problems repeatedly. In dynamic programming we store the solution...
## Find the Deepest Left Node in a Binary Tree.
Objective: – Given a binary tree, Find the deepest left node in it. Approach: This approach is very similar to “Find the Deepest Node in a Binary Tree” with little modification. Take two global...
## Find the Max element in a Given Binary Tree
Objective: – Given a binary tree , Find the max element in it. Example: Approach: Use Recursion. Max will the Max(root, max element in left subtree, max element in rightsubtree) Recursively solve for max...
## Provide the Next Siblings Pointers in a Given Binary Tree
Objective: – Given a binary tree with three pointers left, right and nextSibling). Write the program to provide the nextsibling pointers. Example: Approach:
## Check If One Binary is Mirror Tree of another Binary Tree.
Objective: – Given two binary trees check if they are mirror image of each other. Example: Approach:
## Print All Paths From Root In a Binary Tree Whose Sum is Equal to a Given Number
Objective: – Given a binary tree and X, Print all the paths starting from root so that sum of all the nodes in path equals to a given number. Example:
## Diameter Of a Binary Tree
Objective: – Given a binary tree, find the diameter of it. What is Diameter Of a Tree: Diameter of tree is defined as A longest path or route between any two nodes in a...
## Find the Deepest Node in a Binary Tree.
Objective: – Given a binary tree, Find the deepest node in it. Approach: Take two global variable as “deepestlevel” and “value“. starting with level=0, Do the inorder traversal and whenever you go down one...
## Find the Height of a tree without Recursion
Objective: – Find the Height of a tree without Recursion. In our earlier post “Height of tree” we had used recursion to find it. In this post we will see how to find it...
## Print All The Full Nodes in a Binary Tree
Objective: Given a binary tree, print all nodes will are full nodes. Full Nodes: Nodes Which has both the children, left and right are called Full Nodes Approach: quite simple Solution. Do the any...
## Print All Possible Valid Combinations Of Parenthesis of Given ‘N’
Objective: – Given “n”, generate all valid parenthesis strings of length “2n”. Example: Given n=2 Output: (()) ()() Approach:
## Towers Of Hanoi
The Tower of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The objective of the puzzle...
## Generate All Strings of n bits.
Objective: – Generate All Strings of n bits, consider A[0..n-1] is an array of size n. Example : n = 3 Output: [0, 0, 0] [1, 0, 0] [0, 1, 0] [1, 1, 0]...
## Euclidean algorithm – Greatest Common Divisor(GCD)
The greatest common divisor (GCD) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder. For example, the...
## Print the Bottom View of the Binary Tree.
Objective: – Given a binary tree, print it in Bottom View of it. What is Bottom View: Bottom view means when you look the tree from the bottom the nodes you will see will be...
## Merge Sort in a Linked list
Objective: Given a Linked List, Sort it using merge sort. Example: ->9->3->4->2->5->1 Sorted List: ->1->2->3->4->5->9 Approach: Reference : Merge Sort in array
## AVL Tree – Insertion
What is AVL Tree : AVL tree is widely known as self-balancing binary search tree. It is named after its creator (Georgy Adelson-Velsky and Landis’ tree). In AVL Tree, the heights of child subtrees... | 2,795 | 11,819 | {"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-2018-05 | latest | en | 0.826981 |
http://blog.schoolyourself.org/2013/11/why-airlines-dont-allow-cellphones.html | 1,537,443,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00554.warc.gz | 34,946,511 | 13,135 | ### Why airlines don't allow cell phones
The Federal Aviation Administration (FAA) just announced that most electronic devices can now be used on airplanes, all the way from takeoff to landing. But you still can't make a call on your cell phone during a flight, and all devices must be set to "flight mode." Why can't you make calls on flights? To get a better understanding, let's use some trigonometry.
Airplanes communicate and navigate using a band of radio waves called the airband, which uses frequencies between 108 and 137 MHz (or megahertz). Radio waves are electromagnetic waves, or waves of light, which travel through space like the sine function. Waves with higher frequencies move up and down, or "oscillate," faster, while lower-frequency waves oscillate more slowly. Here are examples of waves that have different frequencies:
How do the frequencies of cell phones compare to those of the airband? Well, it depends on the carrier (Verizon, AT&T, etc.), but for the most part they're between 500 and 2500 MHz. So cell phones send and receive radio waves that are close to the Airband, but are slightly higher in frequency.
If you were to make a call on a plane, then the cell phone's radio waves and the airplane's radio waves would add together. The airplane's waves are probably a lot stronger than the waves coming out of your phone. So what happens if we add a very weak cell phone signal (say, at 700 MHz) to a strong airplane signal at 120 MHz?
The sum of the two waves looks pretty close to the airplane's signal. But what happens if the cell phone signal were a lot stronger?
Suddenly, the sum looks quite different from the airplane's signal, and that's what worries the FAA. While there are mathematical tools that tease apart signals with different frequencies that have been added together, pilots and officials are concerned that cell phone radio waves could still interfere with the communication and navigation of airplanes.
If cell phones instead used frequencies that were really far away from the airband, then the risk of interference would be even smaller (even if the phones emitted really strong signals). Why do you think that is? | 458 | 2,175 | {"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} | 3.15625 | 3 | CC-MAIN-2018-39 | longest | en | 0.952903 |
http://qbasicnews.com/newforum/archive/index.php?thread-9709.html | 1,638,091,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00018.warc.gz | 71,360,306 | 6,484 | # Qbasicnews.com
Full Version: Prime Contest
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Make a program that calculates the value of the 1,000th, the 10,000th and the 100,000th prime. If FB is used the 1,000,000th prime must be calculated too.
The first prime is 2, the second is 3 and so on...
The number of primes up to the integer x can be aproximated by x/(log(x)-1) (QB's log)
The winner is who has the three primes right and faster.
References:
All you wanted to know about primes and never dared to ask http://primes.utm.edu/
The idea is to generate primes and count them up to the required count. There are a lot of optimizations possible in this generation and counting so I hope we will get some interesing sources.
We can open 3 categories according to the speed and memory limitations: Qbasic, Qb4.5 and FreeBASIC
EDITED: Added an additional request for FB, without it even an unoptimized souce takes less than 0.5 second.
EDIT2: For reference, the results are
Code:
```the 1000 th prime is 7919 the 10000 th prime is 104729 the 100000 th prime is 1299709 the 1000000 th prime is 15485863```
of course your program must FIND these results!
err winer? :lol:
Thanks!
Heres my simple FB prime program, using trial divison. It basically combines the only two rules i know, that theres no need to look at even numbers (except 2), and that you only need to check if n is divisible by any prime numbers smaller than sqr(n)
It takes about 0.5 seconds to get the 100,000 prime, and about 12.5 to get the 1,000,000 on my P4 1.8Ghz
I've been reading the link Antoni posted and may try again once i learn some different methods.
Code:
```Dim Shared prime_list(1 To 1000000) As uInteger Dim Shared prime_list_count As uInteger Dim As uInteger prime_val, i, j, n Dim As Integer is_prime prime_list(1) = 2 ' Setup the first two values prime_list(2) = 3 prime_list_count = 2 prime_val = 3 Do prime_val += 2 ' Add 2, we don't need to look at even numbers is_prime = -1 n = prime_val i = 1 j = Int(sqr(n)) + 1 ' Only need to check if n is divisible by any prime smaller than sqr(n) While prime_list(i) < j If (n mod prime_list(i)) = 0 Then is_prime = 0 Exit While End If i += 1 Wend If is_prime Then prime_list_count += 1 prime_list(prime_list_count) = prime_val End If Loop Until prime_list_count = 1000000 Print prime_list(1000) Print prime_list(10000) Print prime_list(100000) Print prime_list(1000000)```
Great, we have one entry!
But sieves are faster...
Heres another one, a fairly standard sieve, using a bit array. I tried it just using a regular array which will take much more memory, and i though would be faster, but this bit based method won in the timings.
I also tried to change IsGood to a macro, but that actually slowed it down, I think thats P4 weirdness.
The only real optimization i did for this was the multiples of two, which because i used a bit array, i just iterated through, and masked out all multiple of two numbers.
This one runs in about 3.5 to 4 seconds, so 1/3 of the time of my first attempt, and i'm sure there are better methods yet...
Code:
```Const MaxPrime = 1000000 ' The max value for P Const MaxVal = MaxPrime * 16 ' The max value for N (a bit of a cheat using * 16...) Const Max32 = (MaxVal \ 32) + 1 ' The number of 32-bit vars needed to store bit array Dim Shared BitsArray(0 To Max32) As uInteger #macro MarkGood(n) Scope Dim As uInteger p = (n) \ 32, o = (n) mod 32 BitsArray(p) = BITRESET(BitsArray(p), o) End Scope #endmacro #macro MarkBad(n) Scope Dim As uInteger p = (n) \ 32, o = (n) mod 32 BitsArray(p) = BITSET(BitsArray(p), o) End Scope #endmacro Function IsGood(ByVal n As uInteger) As Integer Dim As uInteger p = n \ 32, o = n mod 32 Return NOT BIT(BitsArray(p), o) End Function Dim As uInteger i, n1, n2, count, mask = &H55555555 For i = 0 To Max32 - 1 ' Mark off all multiples of two quickly using a mask of 10... BitsArray(i) = mask Next i MarkGood(2) ' Restore 2 as a prime MarkBad(1) ' Make 1 not a prime count = 1 ' start count offset at 1 to account for 2 being a prime For n1 = 3 To MaxVal If IsGood(n1) Then count += 1 If count = 1000 Then Print n1 If count = 10000 Then Print n1 If count = 100000 Then Print n1 If count = 1000000 Then Print n1 Exit For ' We've found the 1 millionth prime, we can quit End If For n2 = (n1 + n1) To MaxVal Step n1 ' work from n+n to max marking off multiples MarkBad(n2) Next n2 End If Next n1```
This is an interesting source by Rich Geldreich anyone can find at ABC packets. It uses an original idea and it's probably the only way to do it in QBasic 1.1, because of the 160K memory limits.
It ran in 14 secondes in QB1.1 and in 2 seconds compiled in QB4.5
Code:
```'Prime tally using a moving window version of the Erathostenes' Sieve 'Antoni Gual 10/2006 for the comtest at QBN. Qbasic1.1 version '---------------------------------------------------------------- 'A true bit sieve would be faster, but the memory sizes in QB1.1 'require bold ideas. This one was created for QB by Rich Geldreich in '1992 from an idea in Donald Knuth's TAOCP. ' 'In a normal sieve each prime found is used in turn to mark all its 'composites thus the complete sieve must be hold in memory tor the final 'tally. In Rich's version all primes found so far are used at the same 'time to mark composites in the same moving slice of ths sieve, the 'numbers left unmarked are primes,and they can be counted as the slice 'progresses. 'In fact there is no data representing the sieve slice...only a priority 'queue that keeps the primes and it's factors used in the present sieve 'slice. This queue has to be dimensioned to hold all primes up to the 'square root of the maximum prime, the present size of 4096 would allow 'for primes up to 2^31. 'Additional optimizations; ' Multiples of 2 and 3 are skipped ' A prime p starts to sieve at p*p, because p*a for a<p will be found ' by a. ' The heap is an udt but is kept in separate arrays for speed. DEFINT A-Z DECLARE SUB PutPrime (a&) DECLARE FUNCTION GetPrime& () CONST heapsize = 4096 'Priority queue DIM heapq(1 TO heapsize) AS LONG DIM HeapQ1(1 TO heapsize) AS LONG DIM HeapQ2(1 TO heapsize) AS LONG DIM SHARED n AS LONG DIM t AS LONG DIM Q AS LONG, Q1 AS LONG, Q2 AS LONG DIM TQ AS LONG, TQ1 AS LONG DIM u AS LONG, primepos AS LONG, cnt AS LONG primepos = 1000 n = 5 d = 2 r = 1 t = 25 heapq(1) = 25 HeapQ1(1) = 10 HeapQ2(1) = 30 cnt = 2 DO DO Q = heapq(1) Q1 = HeapQ1(1) Q2 = HeapQ2(1) TQ = Q + Q1 TQ1 = Q2 - Q1 '***Insert Heap(1) into priority queue i = 1 DO j = i * 2 IF j <= r THEN IF j < r THEN IF heapq(j) > heapq(j + 1) THEN j = j + 1 END IF END IF IF TQ > heapq(j) THEN heapq(i) = heapq(j) HeapQ1(i) = HeapQ1(j) HeapQ2(i) = HeapQ2(j) i = j ELSE EXIT DO END IF ELSE EXIT DO END IF LOOP heapq(i) = TQ HeapQ1(i) = TQ1 HeapQ2(i) = Q2 '*** LOOP UNTIL n <= Q DO WHILE n < Q cnt = cnt + 1 IF cnt < heapsize THEN heapq(cnt - 2) = n IF cnt = primepos THEN PRINT USING "The ####### th prime is ######### "; primepos; n IF primepos = 100000 THEN PRINT "Ended": SYSTEM primepos = primepos * 10 END IF n = n + d d = 6 - d LOOP IF n = t THEN u = heapq(r + 1) t = u * u '***Find location for new entry j = r + 1 DO i = j \ 2 IF i = 0 THEN EXIT DO END IF IF heapq(i) <= t THEN EXIT DO END IF heapq(j) = heapq(i) HeapQ1(j) = HeapQ1(i) HeapQ2(j) = HeapQ2(i) j = i LOOP '*** heapq(j) = t IF (u MOD 3) = 2 THEN HeapQ1(j) = 2 * u ELSE HeapQ1(j) = 4 * u END IF HeapQ2(j) = 6 * u r = r + 1 END IF n = n + d d = 6 - d LOOP```
I've just been looking at some of the previous posts about primes, seeing what methods other people used. I found a lot by you Antoni!, and some other interesting things I may try and add to my program.
I came across this too which made me laugh
http://members.surfeu.fi/kklaine/primebear.html
Thats a nice one Antoni, works fast, i'm still trying to understand how it works.
I improved my second one, its a bit faster now, but I still need to learn more to make it go even faster. Some of the code wasn't necessary, and i even forgot to ignore multiples of 2.
Code:
```Const MaxPrime = 1000000 ' The max value for P Const MaxVal = MaxPrime * 16 ' The max value for N (a bit of a cheat using * 16...) Const Max32 = (MaxVal \ 32) + 1 ' The number of 32-bit vars needed to store bit array Dim Shared BitsArray(0 To Max32) As uInteger Dim As uInteger n1, n2, count, p, o, n1x2, n1x3 Dim As uInteger steps(0 To 47) = { _ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, _ 6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6, _ 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10 _ } ' This lookup table is used to calculate the step, to avoid multiples of 2, 3, 5, and 7 ' any more that that and the table becomes very large (the one inc. 11 is 480 entrys) Dim As Integer curr_step Dim As Integer prime_to_find = 1000 count = 4 ' start count offset to account for 2, 3, 5 and 7 being prime n1 = 11 ' start at 11 due to count starting at 4 While n1 <= MaxVal p = n1 shr 5 ' \ 32 'p is integer postition in bitarray o = n1 and 31 ' mod 32 'o is bit offset If NOT BIT(BitsArray(p), o) Then ' If the bit isn't set then it hasn't been struck out count += 1 If count = prime_to_find Then Print Using "###,###,###th prime - ###,###,###"; prime_to_find; n1 If prime_to_find = 1000000 Then Exit While prime_to_find *= 10 End If If n1 <= (sqr(MaxVal) + 1) Then ' Only strike out multiples of primes <= sqr(MaxVal) n1x2 = n1 + n1 ' (the +1 is just to account for any rounding, may not n1x3 = n1x2 + n1 ' be needed?) ' we don't need to step by n1, as that will wastefully look at even numbers (odd+odd=even) ' same goes for start pos, n1 is odd, so 2*n1 not needed, start at 3*n1 For n2 = n1x3 To MaxVal Step n1x2 ' work from 3n to max marking off multiples p = n2 shr 5 ' \ 32 'p is integer postition in bitarray o = n2 and 31 ' mod 32 'o is bit offset BitsArray(p) = BITSET(BitsArray(p), o) ' Set the bit to show its bad Next n2 End If End If ' we can step by set amounts, to avoid multiples of 2, 3 and 5 n1 += steps(curr_step) curr_step += 1 If curr_step = 48 Then curr_step = 0 Wend```
EDIT: added a check for n1 <= sqr(maxval)
Quote:I came across this too which made me laugh
http://members.surfeu.fi/kklaine/primebear.html
Ok,Arktinen Krokotiili Projekti is the winner! Let's close the contest, nothing more more can be done.. :rotfl: :rotfl: :rotfl:
EDITED:
Their javascript prime finding algorithm is a little slow..
Code:
```function is_x_prime_number(x) { var limit=0; var div=3; var x_limit = Math.sqrt(x); while (x%div!=0 && div<x_limit)div+=2; is_prime = (x%div==0 && x!=div)*1 return is_prime; }```
A simple trial division...
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http://kiloubox.com/percent-error/percent-error-significant-figures.html | 1,511,401,523,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00724.warc.gz | 165,407,589 | 5,406 | Home > Percent Error > Percent Error Significant Figures
# Percent Error Significant Figures
## Contents
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Jumeirah College Science 68.533 weergaven 4:33 Scientific Notation: Multiplication and Division - Duur: 5:30. Whether error is positive or negative is important. The most significant digit is the leftmost digit (not counting any leading zeros which function only as placeholders and are never significant digits.) If you are rounding off to n significant This is helpful when looking for systematic errors in a measurement technique. news
## How To Find Percent Error Chemistry
Peter Blake 1.475 weergaven 7:59 How to Chemistry: Percent error - Duur: 4:39. e.g. 7.10 g + 3.10 g = 10.20 g 3 sig. When you have estimated the error, you will know how many significant figures to use in reporting your result.
Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. Yes - I like it chilly No - I need it hot vote now UniMatch course search Find your perfect uni place go Useful resources Make your revision easierDon't miss out Page last modified: 12.08.2010 This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. ⇐index | units | measurement error sig figs reliability | simple statistics See here for information Can Percent Error Be Negative Suppose, for example, that you wanted to achieve a precision of 1 part in a 100 (1%) in a measurement of the period of a pendulum.
An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. Percentage Error Formula Meer weergeven Laden... Percent Error (Percent Deviation, Relative Error) Accuracy When scientists need to compare the results of two different measurements, the absolute difference between the values is of very little use. https://answers.yahoo.com/question/?qid=20071104183800AAhtRB7 Ans. % error = (2.50 liters - 2.38 liters) x 100% 2.50 liters = (.12 liters) x 100% 2.50 liters = .048 x 100% = 4.8%
Taal: Nederlands Contentlocatie: Nederland Beperkte modus: Uit Geschiedenis Help Laden... Negative Percent Error It is used in chemistry and other sciences to report the difference between a measured or experimental value and a true or exact value. To better reflect this fact, one might list the population (in an atlas, for example) as 157,900 or even 158,000. You just need to create an account in order to submit the post Username this can't be left blank that username has been taken, please choose another Forgotten your password?
## Percentage Error Formula
Bozeman Science 175.902 weergaven 7:05 Uncertainty and Error Introduction - Duur: 14:52.
Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR Groups associated with this forum: View associated groups Study resources Everything for Physics P3 in Video's How To Find Percent Error Chemistry Over Pers Auteursrecht Videomakers Adverteren Ontwikkelaars +YouTube Voorwaarden Privacy Beleid & veiligheid Feedback verzenden Probeer iets nieuws! Percent Error Calculator Q&A with Paralympian Jack Rutter Who would you like to thank?
Implied uncertainty If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 http://kiloubox.com/percent-error/percent-or-error.html For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. Tyler DeWitt 242.978 weergaven 13:26 Accuracy and Precision (Part 2) - Duur: 9:46. Notice that this has nothing to do with the "number of decimal places". Percent Error Definition
Bezig... values quoted to 5 sig. For sake of this example, suppose the uncertainty is 0.1 s. have a peek at these guys Stated differently, how do I meet a criterion for a particular number of significant figures?
This is a personal judgment. What Is A Good Percent Error The value “3.3 g” suggests an implied uncertainty of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the
## Normally you don't round intermediate answers.
The fact that no digit is certain here is an artifact of decimal notation. Add your answer Source Submit Cancel Report Abuse I think this question violates the Community Guidelines Chat or rant, adult content, spam, insulting other members,show more I think this question violates Rules for rounding off numeric values The standard rules for rounding off are well known. Percent Error Chemistry Definition Im bored and can't sleep Started by: starfab Forum: Chat Replies: 36 Last post: 1 Hour Ago Most times you've ever masturbated in a day Started by: Coonflakes Forum: Chat Replies:
Please enter a title Please enter a message Post Thanks for posting! like actually see it.? 6 answers What is po po? 5 answers More questions The number of protons and neutrons in an atom is also the ______ number.? 25 answers Is Which of these would you be justified in dismissing immediately? check my blog Dit beleid geldt voor alle services van Google.
The goal is generally to reduce the percentage errors. Powered by Blogger. This will give you a decimal number. Convert the decimal number into a percentage by multiplying it by 100. Add a percent or % symbol to report your percent error value.Percent Error Example Started by: Becca1993x Forum: Relationships Replies: 10 Last post: 1 Hour Ago Is anyone else glad black lives matter isn't in the UK?
Question: Why would I use a graph of residuals, and how do I create such a graph in Logger Pro?Answer: The process is described in L04. To express the magnitude of the error (or deviation) between two measurements scientists invariably use percent error . How many significant digits should there be in the reported measurement? In one extreme case the first volume could be 24.9 cm3 and the second volume 24.9 cm3 which would give a total volume of 48.8 cm3.
We can also say that the relative uncertainty is 100/157900, which we can also express as 1 part in 1579, or 1/1579 = 0.000633, or about 0.06 percent. Trending Now Philip Rivers Billy Bush Gabrielle Union Diane Kruger Shania Twain 2016 Crossovers Truman Capote Auto Insurance Quotes Samsung Galaxy Dating Sites Answers Best Answer: % error = (experimental error The accuracy of a measurement cannot be reported if an accepted value is unavailable. Started by: Rashina Forum: A-levels Replies: 0 Last post: 11 minutes ago What subjects do I need to study in Advanced Levels if I want to become an Engineer?
fig.) which would imply that you know the value to one part in a thousand. You calculate the density of the block of aluminum to be 2.68 g/cm3. In this case, all you have to go on is the number of digits contained in the data. Trending What is 0.000000075 in scientific notation? 14 answers Can hydrogen be observed under a microscope?
We just need to check something in your message and will publish it as soon as we can. The same kind of thing could happen if the original measurement was 9.98 ± 0.05 g. The following example will clarify these ideas. About.com Autos Careers Dating & Relationships Education en Español Entertainment Food Health Home Money News & Issues Parenting Religion & Spirituality Sports Style Tech Travel 1 How To Calculate Percent Error
The art of estimating these deviations should probably be called uncertainty analysis, but for historical reasons is referred to as error analysis. Sign in Psst… Don't have an account yet? | 1,864 | 8,176 | {"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-2017-47 | latest | en | 0.835424 |
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1. ### Set of continuous bounded functions.
Yes, the Heine-Borel theorem and compactness is clear to me. I do have a question though about the limit of f_n "not being" a limit in any metric space. This part is not clear to me. Thanks again though!
2. ### Set of continuous bounded functions.
This is exactly what I had on my mind. Thanks for clarifying the same. It is very helpful. I understood the part of boundedness and the fact that the set of functions {x^n} is also not compact as there is a sequence {fn} which has limit point {f} that is not in C(X). The only confusion was about...
3. ### Set of continuous bounded functions.
I know what that closed set have all its limit points. You don't need to explain that part to me. By the way, I think you should read rudinreader's comments. They are right on dot!
4. ### Set of continuous bounded functions.
It forms a closed sphere with radius =1. Also, the limit points of the sequence of functions is either 0 or 1 and both are contained in the range of fn(x). Yes, I stand corrected. I realized that the sequence of functions are all continuous. It is just the f(x) to which the sequence tends to...
5. ### Set of continuous bounded functions.
Homework Statement This is not a homework question. I am solving this from the lecture notes that one of my friend's has got from last year. If C(X) denotes a set of continuous bounded functions with domain X, then if X= [0,1] and fn(x) = x^n. Does the sequence of functions {fn} closed ...
6. ### Pointwise vs. Uniform Convergence.
I re-read your statement on pointwise convergence of the previous post and referred to the text book. I understood what you were saying about pointwise convergence. Thanks for further clarifying the same and for your example. Appreciate it.
7. ### Pointwise vs. Uniform Convergence.
Well, I understood the part of finding the sup using derivatives. And yes you are right that the domain is set of Real number (R). But, now I am a bit more confused about the difference between pointwise and uniform convergence. I am under the impression (and correct me if am wrong) that the...
8. ### Pointwise vs. Uniform Convergence.
Homework Statement I need to understand as to why the following series fn(x) = x/(1+n*x^2) is point wise convergent (as mentioned in the book of Ross) and not [obviously] uniformly convergent. Homework Equations The relevant equation used is that lim (n-> infinity) sup|(fn(x) -...
9. ### Every sequence of bounded functions that is uniformly converent is uniformly bounded
That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded. Thanks, again.
10. ### Every sequence of bounded functions that is uniformly converent is uniformly bounded
That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded. Thanks, again.
11. ### Every sequence of bounded functions that is uniformly converent is uniformly bounded
Homework Statement Prove that every sequence of bounded functions that is uniformly convergent is uniformly bounded. Homework Equations Let {fn} be the sequence of functions and it converges to f. Then for all n >= N, and all x, we have |fn -f| <= e (for all e >0). ---------- (1)...
12. ### Geometric interpretation of Generalized MVT
Homework Statement I am trying to see the geometric interpretation of the generalized MVT. It is not a homework problem, but would like to know how to interpret the equation Homework Equations [f(b)- f(a)]* g'(x) = [g(b)- g(a)]* f'(x) The Attempt at a Solution On...
13. ### Simple question on continuity
Well, how is this solution then. I am badly confused. So please go through this one. If X is bounded non empty subset in R (usual) and f:X->R is uniformly continuous function. Prove that f is bounded. Since X is bounded in R, it has a supremum and infimum. Also, we can have a...
14. ### Simple question on continuity
Homework Statement If X is bounded non empty subset in R (usual) and f:X->R is uniformly continuous function. Prove that f is bounded. Homework Equations The Attempt at a Solution Since X is bounded in R, it is a subset of cell. And all cells in R are compact.All bounded sub...
15. ### A simple problem in Real Analysis
Thanks for the detailed response. Really appreciate it.
16. ### Simple Question on continuity
Homework Statement 1) If f is a continuous mapping from a matric space X to metric space Y. A E is a subset of X. The prove that f(closure(E)) subset of closure of f(E). 2) Give an example where f(closure (E)) is a proper subset of closure of f(E). Homework Equations The...
17. ### Simple question in topology (finite vs. infinite)
Homework Statement I always get confused between countably many vs. uncountable. I suppose if one can index the points of a set , then it is countable. 1)So, anything that is finitie is countable. Anything that is infinite is also countable? Then what is uncountable, something that...
18. ### Very badly stuck (Prove that X is compact)
Thank you very much!
19. ### Very badly stuck (Prove that X is compact)
I think I do see what you are saying. If I understand this correctly, "p" is in one of the Gn's as you said and Gn is open. Hence, all points of Gn are internal points. In other words, one can find a neighborhood Nr(p) such that it is a subset of a Gn. But E contains points of Fn that are not in...
20. ### Very badly stuck (Prove that X is compact)
Homework Statement If X is a metric space such that every infinite subset has a limit point, then prove that X is compact. Homework Equations Hint from Rudin: X is separable and has a countable base. So, it has countable subcover {Gn} , n=1,2,3..... Now, assume that no finite sub...
21. ### Please comment. (Topology question on separability)
Homework Statement If X be a metric space in which every infinite subset has a limit point, then X is separable. This is a question from Rudin but I am having some difficulty just understanding how to use the hint. Homework Equations The hint as in the book is . Fix delta >0, and...
22. ### A simple problem in Real Analysis
No problem. Have a good day!
23. ### A simple problem in Real Analysis
Thank you very much to start with. I think I now understand the problem better. Just before you posted your approach, I had mine as follows. I think a part of my approach is similar to your. But yours is much more cleaner. I will appreciate if you can read the following and understand my...
24. ### A simple problem in Real Analysis
Homework Statement I am having somewhat a difficult time just understanding a simple concept. I am trying to prove that every open subset G of a separable metric space X is the union of a sub collection {Vi} such that for all x belongs to G, x belongs to some Vi (subset of G). I am...
25. ### A metric space having a countable dense subset has a countable base.
I suppose I wanted to say that the collection of Vi is countable and not finitely many. Anyways...
26. ### A metric space having a countable dense subset has a countable base.
1. Homework Statement Every separable metric space has countable base, where base is collection of sets {Vi} such that for any x that belongs to an open set G (as subset of X), there is a Vi such that x belongs to Vi. 2. Homework Equations Hint from the book of Rudin: Center the point...
27. ### Every separable Metric space has countable base.
Homework Statement Every separable metric space has countable base, where base is collection of sets {Vi} such that for any x that belongs to an open set G (as subset of X), there is a Vi such that x belongs to Vi. Homework Equations Hint from the book of Rudin: Center the point in a...
28. ### Prove that the closure is the following set.
Does not make any sense to me Edit: Yup, it makes sense now. For some reason, the class on sequence is chapter 3 in Rudin but the homework problem given to me is after Chapter 2. so, it was not not easy understanding the subsequence part. However, I solved it in a different way.
29. ### Prove that the closure is the following set.
Homework Statement Suppose (X,d) is a metric space. For a point in X and a non empty set S (as a subset of X), define d(p,S) = inf({(d(p,x):x belongs to S}). Prove that the closure of S is equal to the set {p belongs to S : d(p,S) =0} Homework Equations Closure of S = S U S' , where S'...
30. ### Open n-cell is open?
I understand that to prove a point x as internal point, I do not need to prove all points are internal points. I can just take x and find a neigborhood of x that is contained in the cell. And if all points of cell are internal points, then the set/cell is open. If that was not apparent from my... | 2,203 | 9,022 | {"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} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.967682 |
www.dclick.cl | 1,537,666,466,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.65/warc/CC-MAIN-20180923000827-20180923021227-00257.warc.gz | 308,377,894 | 14,888 | # Force on a hinge
force on a hinge 0 kg beam is attached to a wall with a hinge and its far end is supported by a cable. Scott1 and Gregory L. The ladder makes a 60 ° angle with the horizontal. com: Helping you understand friction hinge technology - from Reell Precision Manufacturing Corporation The Hinge Joint groups together two Rigidbodies A component that allows a GameObject to be affected by simulated gravity and other forces. Structural Stability and Determinacy (axial force, shear force, Member Reaction Joint Internal Hinge Reaction Dosi 5 reviews of G-Force Hinge "This was my first time building a fence and gate. Hinges and pins are about transferring shear forces. Ryan. If you push with a force of 200 newtons and r is 0. Description. I'm having trouble with the direction of force acting on hinges. How to calculate the torque required for detent or friction hinges on a door with horizontal axis: A. Its weight is a downward force. m change without notice. Determine the cable force in CD needed to just start lifting vertical components of force at the hinge (pin) at A. Open the door gradually; do not “jerk” it open. Find quality door hinges online or in store. The door will be 20 feet wide and 10 feet tall, there will be one hinge at each end of the door and the hinges will be 11. Innovative Connections Hinge and Pin Connections. Case Solution Chapter: 1. So the study of hinge moments is important to be able to predict the forces (moments) required by the human pilot or the hydraulic or electric actuator system. For example, start with all Waterson hinges at a scale value of 0. Designed to eliminate the need for gas shocks or hatch springs in some applications; Friction is built into the barrel of the hinge using patented technology from GEM Self closing door hinges, gate hardware suppliers & manufacturers. i will choose best answer. Reddit gives you the best of the internet in one place. A spring is a component of a hinge, that applies force to secure a hinge closed or keep a hinge opened. An engineer's explanation of the different forces the building, cylinders and hinges will receive from a a one-piece hydraulic door design. Determine the minimum force that must be applied at the hinge to hold the gate shut. Thank you to G-Force Hinge and owner Eric, i was able to successfully build both by myself. Interesting Finds Updated Daily. is a designer and manufacturer of friction hinges used in many commercial and industrial applications. Force, closing speed and final snap are all individually adjustable according to your needs. Probably the best self-closing hinge on the market! Manufactured with stainless steel and other rustproof components for outdoor use. For example, I could apply a force of 10 N on the door very close to the hinge, or very close to the edge. Ryanb) The moment-rotation behavior of force-based frame elements is expressed as This article shows you how to install pin hinges to make a box with a lid that opens traditionally or a box with a lid that swivels open. Large Selection of Liquid Force gear in stock. There are a variety of types of forces. This will be explored further in the next section. provides businesses and schools with security solutions by being a leading supplier of commercial doors, frames, hinges, locks & more, Hinge frictional force and reaction due to door mass centre which creates moment during closing was considered and Torque of this parameter was evaluated and energy due to friction and Inertia was estimated. 5. Select best furniture hinges for cabinet doors from wide range of our products. 32 Self lubricating concealed bearings per hinge for Adds higher degree of security over standard hinges; Low operating force Cabinet Door Hinges to Reface your Kitchen. A 15 m uniform ladder weighing 500 N rests against a frictionless wall. A community of people looking for real connections. Here is the problem: I am not sure how to deal with the 2 kN point force at $C$ ($C$ and $E$ Amazon. H. This document describes procedures and methods for measuring door closing force (DCF) and striker bolt load (SBL). This force, Y h, is equal to zero when the load moves between A an h, and hence, no shear or moment will be induced in portion hC. Playmaker II FourcePoint increases flexion Playmaker II with FourcePoint hinge technology provides It increases the posterior tibial force placed on the PCL Start studying Chapter 7 physics. Interior doors should require no more than 5 lbs. S. The tension force of the wire, the weight force of the beam and the force of the hinge which can be thought of as a sum of two forces, a force acting in the vertical direction on the beam and a force acting to the right on the beam. Express the result as a Cartesian Vectors. 5 reviews of G-Force Hinge "This was my first time building a fence and gate. 1. Shear Force and Bending Moment in Beams For example, support “A” is pinned, like a hinge, so the symbol is a triangle. The technical name for the force required to deflect a control surface is the hinge moment. Damaged Steel Doors: Repair or Replace? that kind of force often leads to a compromised Clamp or hold the hinge reinforcement in the original place and plug Force Method for Analysis of Indeterminate Structures Number of unknown Reactions or Internal forces > Number of equilibrium equations Note: Most structures in the real world are statically indeterminate. Scotta) and Keri L. The first is the resultant pressure force that is a distance ℓ above the hinge and the opposing force, FH that is 4 m above the hinge. Therefore a hinge consists of horizontal and vertical support reaction (see Problem 2). A simple rectangular door will have its CG at its center. Buy Liquid Force Hinge Classic Women's CGA Vest 2019 at the best price at Miami Nautique. VESTS. Size #5 = 11 pounds of closing force, minimum; Size #6 = 14 pounds of closing force, minimum; Many of our most popular door closers have an adjustable Spring/Power setting. Includes G-Force Hinge Reviews, maps & directions to G-Force Hinge in Gulf Breeze and more from Yahoo US Local Shop through a wide selection of Gate Hinges at Amazon. WWHardware's friction catches and latches help easily secure cabinet doors and entry ways. LIQUID FORCE 2019 HINGE CGA VEST. - References for Plastic Theory of Bending with worked examples Best Page for: TWO COMMON MISTAKES RELATING TO WRIST HINGE IN YOUR GOLF SWING VIDEO >> So, wrist set is the idea of hinging the wrist around about 90 degrees during the backswing. Thus, H+ Pcos = 0 =)H+ (270:13)(0:77715) = 0 =)H= 209:93 The negative sign indicates the force His pointing left. This is an example problem showing how to calculate support reactions for a beam with a hinge. A pinned support may have vertical and Friction Hinges. PAWL HANDLE Prosecution and defense attorneys are at odds over whether a use-of-force expert should be allowed to testify at the trial of a man charged in a fatal stabbing outside a Guttenberg restaurant last year. Previously in this lesson, a variety of force types were placed into two broad category headings on the basis of whether the force resulted from the contact or What is the part number for the screw(s) that affix the media door hinge assembly? I need to get the part number for the hinges on the media door. Designed to blend in or make a statement, these hinges add sleekness and originality to any cabinet. Product - Amerock 0. I want to. The Armor frame is a denser thicker aluminum frame recommended for contact sport athletes. They are available in a wide range of options, meaning even your specialty cabinet hinges will match the rest of the kitchen. Eric Elowsky; Videos; Our G-Force or Infinity Hinges allow How to build a single or double wood gate in minutes by using the GForce wood gate hinge, Unlike traditional strap-style hinges, the flanges of the GForce Hinge grab the top and bottom of the gate's 2-by rail, using the gate's weight to compress the wood grain and limit sagging. a spring-loaded hinge made to provide assistance in the closing or the opening of the hinge leaves. Shear and moment at section S 1 and S 2 To determine the shear at s 1, remove the shear resistance of the beam at the section by inserting a support that does not resist shear, but maintains axial force and bending moment resistance (see the inserted support in Figure 4). "Moment-Rotation Behavior of Force-Based Plastic Hinge Elements", All Videos Tagged hinge The Hinge force in Kangaroo can produce some very realistic sheet bending behaviour! Tags: developable, bending Codes and Compliance: Demystifying the Door Opening forces, and many other Door and gate spring hinges shall be adjusted so that from the open Home; Success of Joint Sahel Counter-terrorism Force Hinges on Cooperation, Funding to Fill Capacity Gaps, Key Peacekeeping Official Tells Security Council The gate height, and the distance between the hinges, is 4'6'. 1 Answer to Find the force exerted on the strut by the hinge at A for the arrangement in Figure if (a) The strut is weightless, and (b) The strut weighs 20N. Shown in Figure 1. A force is a push or pull acting upon an object as a result of its interaction with another object. The hinges can also be used to compensate for slight grade changes making it possible to install the gate on a hill, yet keep the bottom […] What is the horizontal component of the force exerted by the bar on the hinge at B? Express your answer with the appropriate units. Passionate about something niche? Hallo all, I am having trouble visualising the reaction forces on a hinge. There is also a reaction force along the axis 2. LaForce Inc. This is a typical problem in statics and mechanics. We can modify a large variety of hinges with customized width, heights, hole patterns, finishing options and materials. In the case of three-hinged arch, we have three hinges: two at the support and one at the crown thus making it statically determinate structure. This does not apply to the initial force needed to overcome the weight of a motionless door. WMNS HINGE CLASSIC CGA. Neglect the weight of the gate and the friction at the hinge. Designed to eliminate the need for gas shocks or hatch springs in some applications; Friction is built into the barrel of the hinge using patented technology from GEM “Hinge moment” is the technical name for the force required to deflect a control surface. Shop door hinges in the door hardware section of Lowes. Figure 1(a) shows a door and the forces on the door are shown in Figures 1(b). G-Force Hinge is the best wood gate anti-sag hinge available to prevent your wood gate from sagging over time. In this case, choosing the hinge as the point to take torques around eliminated both components of the support force at the hinge. Moment-Rotation Behavior of Force-Based Plastic Hinge Elements Michael H. I understand there would be a Force doesn't depend on the distance you need to push. Hallo all, I am having trouble visualising the reaction forces on a hinge. 9-54). SL11 Concealed Geared Continuous Hinge 01 SELECT Pruts Limite Printe in U. 40 m from the bottom each support half the door's weight. and K. European hinges, How to Adjust Self Closing Kitchen Cabinet Hinges hinge. If the spring-loaded hinge does not apply enough tension, the door will not close properly. Fully Concealed Hinge, will force the door to 'overlay' (cover) the frame of the cabinet by 3/8" ON THE HINGE SIDE. Read or share reviews of the Liquid Force Hinge CGA Classic Wakeboard Vest - Women's 2016 or shop similar Vests In the case of three-hinged arch, we have three hinges: two at the support and one at the crown thus making it statically determinate structure. Free Shipping & Lowest Price Guarantee! The Liquid Force Hinge Classic CGA Wakeboard Vest 2018 is in stock now. Friction Force X* X* $6,550. A 28. 00 Firehouse Innovations X X X$6,500. Addition to this latch energy was estimated as latch forces and distance travelled by pawl along the striker was known. CLIP top BLUMOTION hinges are also available in onyx black. From OpenSeesWiki. 4 Example 1. From my experience after seeing many broken hinge pins, double shear is an optimistic approach, although it is the usual way it is presenred in the engineering books and where the reaction forces are taken at the middle of the padeye. A force tends to move the body in the direction of its action. com: liquid force hinge. A 27. And they come in different styles. You're thinking of work. Motion analysis invites me to plot reaction forces in joints. Unfortunately, 4 Adjust a Spring-Loaded Door Hinge; Analysis of Rotational Column with Plastic Hinge Michael Long, actuator applied no axial force on the column. Torques produce rotations in the same way that forces produce translations. I understand there would be a Support 2: Standard Hinge Click to view movie (229k) The second support is a standard hinge. Hanging cabinet doors in perfect alignment can be a very difficult task. The closer relies on hydraulic power to swing the door closed with enough force so that it will latch. ENIGMA COMP. Force doesn't depend on the distance you need to push. Chapter 2 Exercises . 75 inches from the top o 3 1. AF=- K2 (3) At WWHardware, we offer exclusive bulk-order discounts on concealed cabinet hinges from brands like Blum and Salice. The plastic hinge, detailed in Figure 2, Describes Bending above the yield Stress for elastic materials (Mild Steel). The HINGE CGA is an excellent all-around vest for all of your watersports activities! Wakeboarding at the cable, wakesurfing on the lake, or running through the waterpark– the HINGE CGA will move with you and respond as needed. 11 Your force far from the hinges produces a large torque. The hinge prevents sagging and splitting wood by embracing the 2x4 over the top and bottom. LEECO Tech. However, application of these definitions, developed for a horizontal beam, to a frame structure will require some adjustments. Bill is right about the forces going south if someone wants to ride the gate. L. 5 meters, what’s the torque you see in the figure? In diagram A, you push on the hinge, so your distance from the pivot point is zero, which means the lever arm is zero. 38 in. Fenves2 Abstract: A new plastic hinge integration method overcomes the problems with nonobjective response caused by strain-softening Use the sum of the forces in the vertical direction to calculate the other reaction force. Example: If a door closer is listed with a Spring Size of 3-6 then it is adjustable through the Power range of #3 (5 pounds of force) to #6 (14 pounds of force). Measure the maximum force required to pull the door open. of force to open. Adjustable Elastic Force Elbow Hinge/Elbow Brace is for conservative treatment or post-op external immobilization for forearm and elbow injuries;prevention for elbow contracture and its joint movement training Donjoy Knee Brace Hinge Adjustment DonJoy Black Force Point Hinge Adjustable Defiance ACL, PCL CI Left Knee Brace. Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door. 2 In horizontal load, the strength per unit of length is constant. force on the door produce a large force on the shell? E. One method will be discussed for measuring door-closing force (DCF) and two methods for measuring striker bolt load (SBL). Amazon Try Prime All 5 reviews of G-Force Hinge "This was my first time building a fence and gate. The standard 4-way latch provides versatility for retrofit applications and makes ordering easier. 40m from the bottom each support half the door's weight (Fig. In case of external hinge applied load /force is distributed in both horizontal and vertical direction and also if we apply moment there then equal and opposite torque/reaction is provided by hinge. Write the equation that states that the net force on the beam is zero. Buy storage chest hinges or contact us to find the right part for you! A hinge 0. 00 * The complete function or simulator is Patented or Patent Pending or uses a device that is, to perform such function. Multiply the forces generated by the door and attachments (F door) by the distance from the hinge to the center of gravity of the door (d 1) and divide by the distance between the hinges (dhinge): Rradial 2 = F door • d 1 / (dhinge) 3. DonJoy Performance Webtech knee braces delivers center of gravity at G. This moment causes rotation of the hinge, resisted by the axial and bending forces within the fuse-bars. Contact us today for a custom quote Page 2 of 26 Snap-Fit Joints for Plastics - A Design Guide. The product of the force and the perpendicular distance to a pivot (or hinge) is called the torque or the moment. Features steel construction. In the case of a spherical/ball joint, I can guess that the force acts at the point which The force of the door against a sealing gasket, if one is present, is an out-of-plane horizontal stress on the hinges. Plastic Hinge Integration Methods for Force-Based Beam–Column Elements Michael H. Consider a three hinged arch subjected to a concentrated force P as shown in Fig 32. Force Method for Analysis of Indeterminate Structures Number of unknown Reactions or Internal forces > Number of equilibrium equations Note: Most structures in the real world are statically indeterminate. 40m from the top and another hinge 0. 188 COMBINED PLASTIC BENDING AND COMPRESSION (b) To get the hinge’s horizontal support H, we can use fact that the forces must balance (otherwise the system would move). Mass: It is a measure of the inertia of a body, which is its resistance to a change of velocity. Frictionhinge. Largest Manufacturing Technology Community on the Web That's force * distance, so Beam With Hinges Element. As you get farther from the hinge, you can apply a smaller force to make the door swing. The distance from the surface of the liquid to the bottom of the liquid is yc + 4 m from the diagram. Contact us today for a custom quote 4. The Müller Breslau Principle states that the ordinate value of apply a shear force, The moment resistance is eliminated by inserting a hinge in the An intenal hinge is a hinge linkage located between ends of two separate beams at which there may be discontinuous slope from one side of the hinge linkage to the other, and the allowed displacement on both sides of the hinge linkage separating the two beams is continuous. Buy the Liquid Force Hinge Classic The Liquid Force Hinge Classic is a USCG approved vest with neoprene making it a comfortable fit. If the beam is inclined at an angle of θ = 23. SP-0931 Information subect to P: 00-3-11 Fa: 00-3-10 ww. Get a constantly updating feed of breaking news, fun stories, pics, memes, and videos just for you. If you push near the hinge, Semi-Rigid Main Rotors It gets its name from the fact that it does not have a lead-lag hinge, when each blade moves up until centrifugal force balances lift. Price Products Hinges Continuous Geared Hinges. 00 The Challenger® X X X* X* X* X X X* X* X $7,360. 900B (3/8" Overlay) Fully Concealed. A regular platform$ABCD$of weight$200N$is smoothly hinged, along its edge$AB$, to a verical wall, The platform is kept horizontal by two parallel chains inclined at$45^o$to the horizontal Figure 4: (a) Motion of the hinge is due to moment about pin in center. The horizontal forces are: H, the horizontal support from the hinge and P x, the horizontal support from the brace. The first force is Get high-quality gas springs and dampers from Grainger to help A gas spring uses compressed gas to exert the force needed for such Hinge (5) Hinge Moment-Rotation Behavior of Force-Based Plastic Hinge Elements Michael H. HINGE CLASSIC CGA. Friction hinges are alternately known as “constant torque hinges,” which is descriptive of the mechanism of a friction hinge’s operation. com. Force Method for Trusses – One Redundant Force • A truss is a structural system that satisfies the following requirements: (a) The members are straight, slender, and prismatic (b) The joints are frictionless pins (internal hinges) (c) The loads are applied only at the joints • Check for indeterminacy: # of unknowns > # of equations. Barrel hinge a sectional barrel secured by a pivot. is 0 Magnitude and direction of the force exerted by the wall on a rod smoothly hinged at its end on the wall. Free Shipping, no sales tax and our best price promise on the 2018 Liquid Force Hinge Classic CGA Life Jacket at WakeMAKERS, the trusted name for boat accessories. Easy to install with no welding required, no other hinges to buy or mounting brackets needed. Force of Friction Lab Report Force of Hand –now the 2 hinge parts are sliding across a skull fracture . The script creates two mesh triangles with t… stick force, the elevator hinge moment, and the tab hinge moment, when the system is in equilibrium, is given by the formula AH. Shop a wide selection of Liquid Force Men's Hinge Classic Life Vest at DICKS Sporting Goods and order online for the finest quality products from the top brands you trust. Buy storage chest hinges or contact us to find the right part for you! this pulling force is balanced/equal to the resisting frictional force. 2 is a beam with two internal hinges. Now solve for the reaction force R A=40kN!12kN=28kN. Free shipping and free returns on Prime eligible items. Two types of hinges are there such as external hing and internal hing . 00 The Enforcer X X X X$7,500. The DonJoy Armor Knee Brace with FourcePoint Hinge is the strongest and most supportive off-the-shelf knee brace on the market for contact sports. The wall, in turn exerts the same on the hinge. - 432302 2. (a) Find the horizontal and vertical forces that the ground exerts on the base of the ladder when an 800 N firefighter is 4 meters from the bottom. I understand there would be a Frictionhinge. Again, the hinge force counterbalances this by having a component in the opposite direction. 1: Beam with internal hinges Fig. Friction Hinges. skull radiograph (PA and the presence of a skull fracture is an indication that severe force has been ‘Hinge’ fractures occur when the The gravity type self-closing hinge set has no springs to wear out. Ryanb) The moment-rotation behavior of force-based frame elements is expressed as Hager’s 3300 Series Grade 3 tubular leverset is field reversible, with a thru-bolt design for ease of installation. com: Helping you understand friction hinge technology - from Reell Precision Manufacturing Corporation Hinge - A hinge resists horizontal and vertical translation but allows rotation. 5° with respect to horizontal. Depending on the seal configuration, the loads can be substantial enough to bend the door away from the gasket, defeating its purpose. Inset Hinge - Set of 2. hinge. Basics: 2 the coordinate axes that are perpendicular to the axis of the hinge. HINGE SELECTION CHART CALCULATE DOOR WEIGHT TO ACTUAL DOOR MASS the lateral force on the top hinge, it is recommended that the hinges are spaced as shown. Are you curious about the simple physics of seesaws? we mean the act of exerting a force on an object over a distance. Whether you're searching for swimming pool gate hardware for a repair or new build, HardwareSource is sure to have what you need. Diagram the locations and directions of the various forces. The effect of the shear force is usually less important, and is also more difficult to incorporate. G. 75 inches from the top o Vectors in equilibrium - The Door The forces in the hinges of a door is a useful demonstration of three vectors in equilibrium. 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Plasticity may be associated with force-displacement behaviors (axial and shear) or moment-rotation (torsion and bending). The length of the each hinge is also specified by the user: While the integration of distributed-plasticity force-base elements distributes the gauss points along the entire element length, the beamWithHinges element localizes the integration points in the hinge regions. Product Image. LOWEST PRICE - Donjoy Armor Knee Brace w/ FourcePoint Hinge technology provides enhanced ACL/PCL support for contact sports. Full-Text Paper (PDF): Plastic Hinges and Inertia Forces in RC Beams under Impact Loads surface. The motor will apply a force up to a maximum force to achieve the target velocity in degrees per second. 1 Hinges are strongest in horizontal stress when the forces are applied perpendicular to the hinge pin. Continuous 32 Self lubricating concealed bearings per hinge for heavy duty Low operating force helps opening to Homework # 3 . Statics Applications knob and a larger force nearer the hinge as shown. Shop our selection of Gate Hinges in the Hardware Department at The Home Depot. Hanaya, Inc. but the self-closing hinges on ours would sometimes snap the door closed so Continuous Geared Hinges. 2 hinges you are @ 2165# of force on the hinge. com! Axial Force, Shear Force and Bending Moment Diagrams for Plane Frames Previous definitions developed for shear forces and bending moments are valid for both beam and frame structures. I have a door that will hang on a vertical hinge. By raising the bottom hinge set you can increase the force at which the gate will close. 188 COMBINED PLASTIC BENDING AND COMPRESSION For steel members, brittle hinges are based on minimum strengths, which are the lower bounds of material properties, and are recommended for force-controlled actions. The FullFource and the Armor with Fourcepoint Hinges have the same hinges. Force : Force is the action of one body on another. A hinge has two unknown reaction forces and two unknown reaction moments along the coordinate axes that are perpendicular to the axis of the hinge. Self Closing Hinges. sin300(9 sin200) + 200 Page 2 of 26 Snap-Fit Joints for Plastics - A Design Guide. In this case, let's make counter-clockwise negative and clockwise positive. Common features Snap joints are a very simple, deflection force is given largely by torsion of its A hinge is an internal reaction that doesn't allow moments and can only transfer shear force (summing moments about a hinge traditional Structural Analysis, bellows must accommodate or the force that must be used to Axial Lateral Angular slotted hinge can be installed across the individual bellows elements 9 STATICS AND TORQUE force is applied is another factor in determining whether or not equilibrium is achieved. Browse our collection today! LOWEST PRICE - Donjoy Armor Knee Brace w/ FourcePoint Hinge technology provides enhanced ACL/PCL support for contact sports. A. Problem: Draw the bending moment and shear force diagrams for the beam in Fig. Combines DonJoy's proven 4-Points-of-Leverage System™ with our patented FourcePoint™ hinge technology, the most powerful and clinically-proven solutions available to protect the ACL. Call the force from the wire F1 and the force from the hinge F2. Get expert support and fast shipping. Box hinges and supports from HardwareSource are available in brass, steel, and zinc alloy formats. thanks so much to anyone who can help. Notice that in spite of all this, the $mg\sin{\theta}$ component still manages to provide angular acceleration to (in other words, rotate) the rod. Hinges are weakest in vertical stress when the forces are applied parallel to the pin. Hinge and pin connections are some of the simplest steel to steel connections that we find. Torque is the measure of the force necessary to cause an object to rotate. Hinges may be assigned (uncoupled) to any of the six DOF. In small, relatively slow airplanes, hinge moments are not very large; pilots move the controls with ease. Ryanb) The moment-rotation behavior of force-based frame elements is expressed as Time: It is a measure of the succession of events and is a basic quantity in dynamics. In the case of a spherical/ball joint, I can guess that the force acts at the point which Heavier and wider doors, therefore, generate greater horizontal forces on the hinges. 2 Equilibrium of Rigid Bodies component is connected to a fixed hinge by a pin The underlying support transmits a reaction force roller Ry pin hinge R M. Find G-Force Hinge in Gulf Breeze with Address, Phone number from Yahoo US Local. Forces have magnitude and direction; pick upwards as positive, so !F y=0=R A"40kN+12kN. Should you have any additional questions please call us at 800-553-6019. Learn vocabulary, terms, and more with Two forces produce equal torques on a door about the door hinge. 2. Box 5005 Navarre, FL 32566 United States Ph: 1-850-916-0570 Hallo all, I am having trouble visualising the reaction forces on a hinge. Membership is free. GForce Hinge P. I have thought about the hinges on a door. (b) Using small angle theory, rotation of the top plate can be modeled The stiffness of the hinge is related to the moment acting about the pin in the center. A spring-loaded door hinge contains a cylinder that houses a spring to provide tension to the hinge to push, or hold, a door closed. I have been attempting to solve a question where there is a point force acting on the hinge of a beam. The door does not move. "Moment-Rotation Behavior of Force-Based Plastic Hinge Elements", Multiple hinges may also coincide at the same location. The force exerted on the door by the hinge… A. FREE Shipping. 2 Beam with internal hinges. You can check your answer by solving the sum of the moments about point B. For reinforcement in concrete members, minimum strengths are currently being used for both deformation-controlled and force-controlled hinges. Further, the feedback of this required force or moment to the pilot is an additional cue to help her/him to fly the aircraft. This is the magnitude of the horizontal component of the force exerted by the hinge on the beam. Common features Snap joints are a very simple, deflection force is given largely by torsion of its Monroe Hinge and Stamping specializes in the manufacturing of all varieties of hinges including mil-spec hinges, piano hinges, custom hinges and many more. Showing 40 of 131 results that match your query. FREE ground shipping included. •This type is identical in construction to the general-purpose basic switch except that it has two pai rs of Low-force hinge lever W44 : Long hinge Problem 2: A uniform beam of mass M and length L is attached at one end to a wall via a hinge. and the force exerted by the hinge, . 5. High Holding Force One-way Hinge With torque from 50 to 100 lb-in the PH35 is the largest Reell catalog hinge, and Reell's 'One-way' torque technology eliminates the torque required to move the hinge in one direction, without compromising the holding force in the other. ANSWER: P a r t B The moment-rotation behavior of force-based frame elements is expressed as a function of plastic hinge length and moment-curvature parameters for two types of plastic hinge integration under the representative loading condition of antisymmetric bending. Three screws sit on the hinge in the interior of the cabinet, allowing you to subtly shift the orientation of the cabinet door. seet-inges. Jump to: navigation, Scott, M. The beam has 3 vertical forces. offers standard lines of friction hinges, torque hinges & positioning hinges. I'm struggling to get this basic script to work so that I can understand the basic principles of the hinge force. The hinges are defined by assigning to each a previously-defined section. A hinge 0. A hinged panel in front of the kitchen sink is a great place to store the drain plug etc. 4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. Waterson stainless steel 3 in 1 hinge (door closer, floor spring, door stop). Passionate about something niche? I have a door that will hang on a vertical hinge. Search Product Result. 40 m from the top and another hinge 0. Check out our Spring Torque Calculator at CustomHinges. A regular platform $ABCD$ of weight $200N$ is smoothly hinged, along its edge $AB$, to a verical wall, The platform is kept horizontal by two parallel chains inclined at $45^o$ to the horizontal Conforce® constant force spring, constant tension springs, and custom constant force springs with high force output in a small flat spring How to Design the Optimum Hinge by Christie L. E. Monroe Hinge and Stamping specializes in the manufacturing of all varieties of hinges including mil-spec hinges, piano hinges, custom hinges and many more. 25" from the hinge, and at a 170 degree angle from the door position you will need much more force than a cylinder mounted at a 90 degree angle from the face of the door and 15 inches from the hinge. (including two force components at the hinge), and six equations are avail- 310 Analysis and Design of Beams for Bending tic section modulus of the beam, Motion analysis invites me to plot reaction forces in joints. Buy swimming pool gate hinges online here! The free-body diagram to the left shows that there are two forces on the gate. The angle between the beam and the cable is 90o. All; Mens; Womens; Kids; GHOST COMP. O. Self closing door hinges can save 50% of opening force by door. 3. © 2018 Liquid Force Wakeboards. For example: Why is Cx's and Cy's direction so in parts 2 and 3? And isn't H a hinge as well why is there only a force acting in th Shop a wide selection of Liquid Force Women's Hinge Classic Life Vest at DICKS Sporting Goods and order online for the finest quality products from the top brands you trust. StrikeMaster II Door Frame and Hinge Reinforcement keeps your door from splitting during kick-in attempts. The HINGE CLASSIC CGA can really do it all and then some! A true performer for quite some time, the HINGE CLASSIC CGA is there when you need it to provide protection and the safety. The hinge exerts a leftward normal reaction on the rod. Internal Roller - This is the same has a roller that is used has an external support since it allows rotation and horizontal translation. Jones, Market Development Manager WHITE PAPER SPIROL International Corporation, Danielson, CT, U. Answer to Determine the moment of the force F about the door hinge at A. If the cylinder is attached . force on a hinge | 7,733 | 35,115 | {"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} | 2.796875 | 3 | CC-MAIN-2018-39 | latest | en | 0.894834 |
https://www.johndcook.com/blog/2020/06/03/relatively-prime-determinants/ | 1,721,780,864,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00342.warc.gz | 734,411,123 | 10,940 | # Relatively prime determinants
Suppose you fill two n×n matrices with random integers. What is the probability that the determinants of the two matrices are relatively prime? By “random integers” we mean that the integers are chosen from a finite interval, and we take the limit as the size of the interval grows to encompass all integers.
Let Δ(n) be the probability that two random integer matrices of size n have relatively prime determinants. The function Δ(n) is a strictly decreasing function of n.
The value of Δ(1) is known exactly. It is the probability that two random integers are relatively prime, which is well known to be 6/π². I’ve probably blogged about this before.
The limit of Δ(n) as n goes to infinity is known as the Hafner-Sarnak-McCurley constant [1], which has been computed to be 0.3532363719…
Since Δ(n) is a decreasing function, the limit is also a lower bound for all n.
## Python simulation
Here is some Python code to experiment with the math discussed above. We’ll first do a simulation to show that we get close to 6/π² for the proportion of relatively prime pairs of integers. Then we look at random 2×2 determinants.
from sympy import gcd
from numpy.random import randint
from numpy import pi
def coprime(a, b):
return gcd(a, b) == 1
def random_int(N):
return randint(-N, N)
def random_det(N):
a, b, c, d = randint(-N, N, 4)
return a*d - b*c
count = 0
N = 10000000 # draw integers from [-N, N)
num_reps = 1000000
for _ in range(num_reps):
count += coprime(random_int(N), random_int(N))
print("Simulation: ", count/num_reps)
print("Theory: ", 6*pi**-2)
This code printed
Simulation: 0.607757
Theory: 0.6079271018540267
when I ran it, so our simulation agreed with theory to three figures, the most you could expect from 106 repetitions.
The analogous code for 2×2 matrices introduces a function random_det.
def random_det(N):
a, b, c, d = randint(-N, N, 4, dtype=int64)
return a*d - b*c
I specified the dtype because the default is to use (32 bit) int as the type, which lead to Python complaining “RuntimeWarning: overflow encountered in long_scalars”.
I replaced random_int with random_det and reran the code above. This produced 0.452042. The exact value isn’t known in closed form, but we can see that it is between the bounds Δ(1) = 0.6079 and Δ(∞) = 0.3532.
## Theory
In [1] the authors show that
This expression is only known to have a closed form when n = 1.
## Related posts
[1] Hafner, J. L.; Sarnak, P. & McCurley, K. (1993), “Relatively Prime Values of Polynomials”, in Knopp, M. & Seingorn, M. (eds.), A Tribute to Emil Grosswald: Number Theory and Related Analysis, Providence, RI: Amer. Math. Soc., ISBN 0-8218-5155-1. | 737 | 2,695 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.858673 |
http://mail.scipy.org/pipermail/scipy-user/2007-April/011799.html | 1,441,124,535,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645195983.63/warc/CC-MAIN-20150827031315-00129-ip-10-171-96-226.ec2.internal.warc.gz | 143,174,201 | 2,047 | # [SciPy-user] filling array without loop...
fred fredmfp@gmail....
Mon Apr 23 08:06:40 CDT 2007
Anne Archibald a écrit :
> On 22/04/07, fred <fredmfp@gmail.com> wrote:
>
>
>> Forget that is a convolution (because in fact, it is really not),
>> simply a scalar product (inner product says scipy) in each cells of an
>> array, as I wrote in my example.
>>
>> Does it changes something ?
>>
>
> Uh, maybe I'm confused -
Ok, so let me explain a little more...
I have two 2D matrices, say A et B, with same dims = 571x876.
B is to be computed, A is known.
Each cell of the matrix B is computed from a "scalar product" with a few
cells of the matrix A and
a "weights" vector W:
B[i,j] = \sum_0^8 w_n a_n
where
- W = [w_n] is a 1D vector (computed from solve(a,b)) with dim = 9
- a_n = A[i,j]
a_n are selected from a submatrix of A (much smaller than A), say A':
a_2 -- a_5 -- a_8
| | |
A' = a_1 -- a_4 -- a_7
| | |
a_0 -- a_3 -- a_6
My idea is to write B[i,j] as a scalar product between W and A'
flattened to a 1D vector:
B[i,j] = dot (W,A')
The problem, from my own opinion, is once again, the two (three in
general case) loops to compute each cell
of the matrix B.
So my question is : is it possible to fill the B matrix without loops ?
I hope I more clear. | 408 | 1,299 | {"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} | 3.140625 | 3 | CC-MAIN-2015-35 | latest | en | 0.919558 |
https://phys.org/news/2013-08-mathematician-event-cloaking-device-metamaterials.html | 1,561,034,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00130.warc.gz | 558,887,416 | 20,918 | # Mathematician designs event cloaking device without using metamaterials
(Phys.org) —Miguel A. Lerma a mathematician at Northwestern University has uploaded a paper to the preprint server arXiv, in which he describes the design of an event cloaking device that doesn't require the use of metamaterials. In his design, events are cloaked using mirrors.
An event cloaking device, also known as a time cloak or an that hides time is a mechanism that causes what appears to an observer, to be a lapse in time during which events have occurred but have not been observed. The idea is based on first slowing down light, then speeding it up again. Doing so causes a lag or gap during which events cannot be observed. Researchers have built such devices using that are able to speed up light or slow it down. But those devices have proven to be complicated and expensive. In contrast, in this new method introduced by Lerma, can be manipulated with mirrors by causing it to travel longer or shorter distances before striking an object.
The basic idea involves using multiple mirrors, half of which can be manipulated on-demand to either allow light to pass through or to reflect. The mirrors are arranged in such a way as to first cause light to "slow" by sending it to another mirror instead of directly to an object, which then reflects it to the object. Following that, the light is caused to take a direct route to the source, effectively causing it to speed up again. Doing so creates a "lag" in time, which to an observer would appear as a lapse. If the object were a regular wall clock for example, the observer might see the clock jump from 12:05 to 12:07, if the lag were two minutes. The duration of the lag is dependent on the distance of the mirrors from the , thus lags of minutes, hours or days could be caused by placing mirrors on distant planets or .
Lerma hasn't actually created an event cloaking device, but now that he's discovered a way for doing it cheaply, it's likely others will very soon set to work developing such devices as well as for ways to use it. At this time, it's not really clear if such devices could be used to deceive people, such as jewelry thieves as the device would only cloak objects in a direct line of sight.
More information: A Mirror Based Event Cloaking Device, arXiv:1308.2606 [physics.optics] arxiv.org/abs/1308.2606
Abstract
We propose a way of implementing an event cloaking device without the use of metamaterials. Rather than slowing down and speeding up light, we manipulate an obscurity gap by diverting the light through paths of appropriate length with an arrangement of switchable transreflective mirrors.
via Arxiv Blog
Journal information: arXiv
Feedback to editors
Aug 15, 2013
I wouldn't call this an event cloaking device, I'd call it an illusion. A magic trick. Smoke and mirrors without the smoke. Not really anything I didn't invent when I was 10 years old. no seriously. I'm serious.
Aug 15, 2013
No you didn't - because when you were 10 you didn't know that there were mirrors that could switch that fast.
Aug 15, 2013
My God this is stupid easy.
I thought of shit like this ages ago....I thought this stuff was obvious...
Aug 15, 2013
No you didn't - because when you were 10 you didn't know that there were mirrors that could switch that fast.
antialias, I don't think you understand the theoretical device. It doesn't even require fast mirrors. Only long distances. You could be operating the mirrors by hand, and it would still work you moron. Learn to read.
Aug 15, 2013
You probably thought of a mirror arrangement around an object as a kid, but that is not what this is (Hint: look at the image).
The thing this solves is that just redirecting the signal would mean that at some point you'd get either a gap or double the photons fom the light source - which would be noticeable. The double (more precise: alternate) redirect to compensate for the runtime of light is the trick here.
It doesn't even require fast mirrors.
You need to switch the first set of mirrors PRECICESLY when you switch the second set. No. You can't do this by hand.
No. You didn't think of this as a 10 year old. And even as an adult you didn't understand the article.
you moron
Right back at ya.
Aug 15, 2013
I think he might be pulling your leg.
Aug 15, 2013
You need to switch the first set of mirrors PRECICESLY when you switch the second set. No. You can't do this by hand..
If light takes X minutes to traverse the "short route", it may take X+2 minutes to traverse the "long route" (in the example given in the article). Switching the mirrors fast is not the issue, unless you claim you meant switching mirrors "fast enough" simply to fool the eye, in some sort of actual magic trick or something. But that would be a practical matter involving how skilled the magician is, and has nothing to do with HOW this experiment WORKS. Any 10 year old should be able to understand this thought experiment. I think that "research paper" may actually be a hoax, btw.
Aug 15, 2013
If light takes X minutes to traverse the "short route"
There is no short route in this setup. The length of the route is always the same (when you take the lomng path in the first set you take the short path in the second set of mirrors and vice versa) When you switch from the long way in the first part you have to switch to the short way in the second path. Otherwise you'd get gaps in your observation (e.g. times when no photons arrive at all).
Just found the link to the paper. It describes exactly what I've been saying all along (down to noting the precise timing needed for the mirror switching.)
http://arxiv.org/...2606.pdf
Any 10 year old should be able to understand this thought experiment
Obviously not.
Q.E.D.
Aug 15, 2013
There is no short route in this setup.
Search for the word "transparent" in the paper. Light is taking a shorter path when it passes directly thru, rather than being reflected. duh?
Since we are talking about minutes of light travel, not milliseconds, you could have the mirrors controlled by motors that took a full second to rotate, and every aspect of this experiment would still work. You would only need fast mirrors to conduct this experiment at very short distances.
Aug 15, 2013
I'm not sure the mirrors between the light source and the object is useful. Unless the light source itself is time-dependent and thus contains information, there is no need to delay it, since it always provide the same light.
Aug 16, 2013
I'm not sure the mirrors between the light source and the object is useful.
You need it, because when you switch to the long path after the object then you need the light travelling already in that long path before the object to fill in, what would otherwise be a visible gap of the timelength it takes the light to travel the long path after the object.
If you were to switch the mirrors only to semi transparent you'd register a drop in brightness.
You could, conceivably, alleviate this by playing with the brightness of the light source - but since the temperature/brightness curve of a light source isn't linear that would be noticeable as well.
But the real point is to mask something that is happening at the object for a time. If you don't have those mirrors before the object you can't do that.
Aug 16, 2013
I'm not sure the mirrors between the light source and the object is useful. Unless the light source itself is time-dependent and thus contains information, there is no need to delay it, since it always provide the same light.
If you think about it, this is a classical "ray tracing" scenario where you follow a beam of light as it bounces off an object and onto a camera. Interestingly with the mirrors space-time itself is used to create a sort of 'video buffer' (FIFO frame queue), of fixed time-length. There are two "caches" of image data: 1) ABCD is a "cache", and 2) EFGH, is a "cache". Transparent AD or EH essentially controls whether each cache is active or inactive at any given time.
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# Oxford MAT 2013/2014 Watch
1. I've been having a mental battle with Q1 G on the same paper (2011), thinking you have to somehow translate/transform the graph, then find the area underneath using some means. However finally looking at the solutions, I have no idea what they are talking about and how they have got to certain inequalities. Anyone can explain?
2. (Original post by TheFuture001)
Yeah thanks, I got it! I had ended with a quadratic earlier on, but I discarded it as meaningless since I thought the P's and A's were just x's and y's in the end. I didn't expect to do that.
@dutchmaths, I'm not sure I really like your method aha. For example, to my eyes the first option after subbing in looks like it could probably be true, just by thinking of how large the number could get.
Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
if we say the rectangle has sides x and y, then the P^3 is
(2x+2y)^3
= 8x^3 + 24x^2y + 24xy^2 + 8y^3
i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
3. (Original post by MelvyW)
Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
if we say the rectangle has sides x and y, then the P^3 is
(2x+2y)^3
= 8x^3 + 24x^2y + 24xy^2 + 8y^3
i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
No, that inequality is not always true.
If we set then
and
So we can get provided - that is provided
4. Does anyone know where the answering space for the questions (that are not multiple choice) will be?
I usually take up a lot of space for my working out.
And, what does A(theta) and B(pi/2 * theta) mean on question 4?
http://www.maths.ox.ac.uk/system/fil...e/0/test07.pdf
Is A and B a constant, function or what?
5. (Original post by Noble.)
No, that inequality is not always true.
If we set then
and
So we can get provided - that is provided
Ahh okay i see. Thanks!
I very much doubt i could something like that out in exam though -.-
6. (Original post by Sayonara)
Does anyone know where the answering space for the questions (that are not multiple choice) will be?
I usually take up a lot of space for my working out.
And, what does A(theta) and B(pi/2 * theta) mean on question 4?
http://www.maths.ox.ac.uk/system/fil...e/0/test07.pdf
Is A and B a constant, function or what?
From what i understand A(theta) just means the area A. The area is dependent on the value of theta, and hence A(theta)
7. (Original post by MelvyW)
Ahh okay i see. Thanks!
I very much doubt i could something like that out in exam though -.-
Well, it is somewhat intuitive. When you see a potential answer that says "If you cube the perimeter, it'll always be greater than the area" the first thing that should spring to mind is "Well, what if the perimeter is less than 1? Then when we cube it, it's going to be even smaller". This is partially why I, subconsciously, chose to use in the notation because is typically used to denote a very small value (it doesn't necessarily have to be, but in analysis the arguments are only really interesting for small ).
This is why you can instantly discount (b) as a potential answer, as once you apply some thought to it you realise how utterly absurd it is. It's suggesting as a possible answer, yet you can make an area of less than 1 with a perimeter of more than one (i.e. a 3x0.25 box has area <1 yet perimeter 6.5) and clearly a number less than one squared is also less than one, yet the number on the RHS is bounded below by 1.
8. (Original post by MelvyW)
Hey guys, Im quite new to this, and i have managed this question, but i think there is potential mistake in the question.
if we say the rectangle has sides x and y, then the P^3 is
(2x+2y)^3
= 8x^3 + 24x^2y + 24xy^2 + 8y^3
i would think that this is greater than the area xy for all values of x and y, which makes the first inequality P^3 > A seem like the answer
Ive probably made a mistake somewhere, cause the people who design these papers arent exactly stupid
Hey, I think you misinterpreted my comment as claiming that (a) was the answer, which is perhaps what made you think there was a mistake or something. I meant to say that I prefer pursuing the question as the guy I initially quoted prompted me to do so, since there's no ambiguity as you get the final answer dead on.
I'd suggest you try this method too, I'll hint you. With the two equations you have, eliminate either x or y and see what happens. And don't be scared off seeing x's/y's and P's and A's in the same equation like I did
Q1 G in the following paper http://www.mathshelper.co.uk/Oxford%...est%202011.pdf I am having trouble with. The solutions claim -1 <= x^2 - 1 <= 0 . How they came to this conclusion, and what follows, I can not understand. I would appreciate at least a hint in the correct understanding, thanks!
9. (Original post by Noble.)
Well, it is somewhat intuitive. When you see a potential answer that says "If you cube the perimeter, it'll always be greater than the area" the first thing that should spring to mind is "Well, what if the perimeter is less than 1? Then when we cube it, it's going to be even smaller". This is partially why I, subconsciously, chose to use in the notation because is typically used to denote a very small value (it doesn't necessarily have to be, but in analysis the arguments are only really interesting for small ).
This is why you can instantly discount (b) as a potential answer, as once you apply some thought to it you realise how utterly absurd it is. It's suggesting as a possible answer, yet you can make an area of less than 1 with a perimeter of more than one (i.e. a 3x0.25 box has area <1 yet perimeter 6.5) and clearly a number less than one squared is also less than one, yet the number on the RHS is bounded below by 1.
Youve sat this exam so could you explain the format of the answer book please.
I realise you are expected to do all workings for the multiple choice in the space underneath each part. So how are you expected to answer the 15 mark questions.
Is there a question paper bookl and and answer book. are you given say x pages for all of the longer questions. or is it y pages for each of the 15 mark questions. and do they give a certain amount of space for each part of a question dependednt on how much space they think it should take up, or a set amount. Or is there 1 big space for the whole of the q
cheers
10. (Original post by IceKidd)
Youve sat this exam so could you explain the format of the answer book please.
I realise you are expected to do all workings for the multiple choice in the space underneath each part. So how are you expected to answer the 15 mark questions.
Is there a question paper bookl and and answer book. are you given say x pages for all of the longer questions. or is it y pages for each of the 15 mark questions. and do they give a certain amount of space for each part of a question dependednt on how much space they think it should take up, or a set amount. Or is there 1 big space for the whole of the q
cheers
You're just given a massive answer booklet with your answers to the multiple choice on the front. I was worried about not having enough room to write as I usually use a shed load of paper, but they provide more than enough.
11. (Original post by TheFuture001)
The solutions claim -1 <= x^2 - 1 <= 0 . How they came to this conclusion, and what follows, I can not understand. I would appreciate at least a hint in the correct understanding, thanks!
What are the limits of the integral -- and so the bounds for x?
12. Hi, does anyone have any tips on how to approach questions, because sometimes I just look at questions and don't even know where to start! Any help would be much appreciated! Thanks XD!
13. (Original post by MelvyW)
From what i understand A(theta) just means the area A. The area is dependent on the value of theta, and hence A(theta)
Thanks for the reply but I find your explanation very unclear.
You said that A and B would be dependent on theta, but I don't understand how it is used in this context:
A(theta) = B(0.5pi - theta)
I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
14. (Original post by Sayonara)
I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
There is nothing more unclear about writing A(theta) than there is f(x) or y(t).
15. (Original post by Sayonara)
Thanks for the reply but I find your explanation very unclear.
You said that A and B would be dependent on theta, but I don't understand how it is used in this context:
A(theta) = B(0.5pi - theta)
I find that this notation is very unclear and most likely there was quite a few people who sat this MAT paper losing marks due to this.
Think of it as just a function, as much as it may seem like you can't find an equation to give the area for those sections, you actually can. I fell into the same trap.
16. (Original post by RichE)
What are the limits of the integral -- and so the bounds for x?
Yes, but when you square the bounds and take off 1 you don't end with the same limits.
I think I understand now, the graph shows for x=-1, f(x)=0, so if you square f(x), 0, and take off 1 the limit is -1, yes? Using the same logic for x=0 and x=1, the upper limit becomes 0. Is this the correct way to think about this?
17. (Original post by TheFuture001)
I think I understand now, the graph shows for x=-1, f(x)=0, so if you square f(x), 0, and take off 1 the limit is -1, yes? Using the same logic for x=0 and x=1, the upper limit becomes 0. Is this the correct way to think about this?
At no point should you be squaring f(x). All you need to understand is what the function f(x^2-1) equals.
x is a number between -1 and 1.
So x^2 is a number between 0 and 1
and x^2 - 1 is a number between -1 and 0.
We now need to work out what f(x^2-1) is. The function f is made up of different "rules" depending on the input. From the graph we can see that f is the rule "add 1" for inputs between -1 and 0.
So f(x^2-1) = (x^2-1) + 1 = x^2.
So it is x^2 you need to integrate between -1 and 1.
18. (Original post by fizzy pops)
Hi, does anyone have any tips on how to approach questions, because sometimes I just look at questions and don't even know where to start! Any help would be much appreciated! Thanks XD!
I've always found it helpful to write down everything you know - all the given information and what you can infer from it. Also, with MAT questions, reading the full question (especially trying to use previous results or given hints) is a good idea.
Haha might not have been that helpful, just my two cents. Could you give an example of one of those questions?
19. (Original post by RichE)
.
(Original post by Noble.)
.
Hi, at most cambridge colleges they have a small 1 hour paper on the day of your interview, and your solutions are discussed during your interview and new problems are approached.
Since Oxford has the MAT already is this the same. Is there also a small test for any of the colleges so that some of your interview involves discussing your answers to some of these problems. Or is there no test/problem sheet for any college?
20. (Original post by RichE)
At no point should you be squaring f(x). All you need to understand is what the function f(x^2-1) equals.
x is a number between -1 and 1.
So x^2 is a number between 0 and 1
and x^2 - 1 is a number between -1 and 0.
We now need to work out what f(x^2-1) is. The function f is made up of different "rules" depending on the input. From the graph we can see that f is the rule "add 1" for inputs between -1 and 0.
So f(x^2-1) = (x^2-1) + 1 = x^2.
So it is x^2 you need to integrate between -1 and 1.
Cheers, I see the required thought process now!
Reply
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 3,448 | 13,699 | {"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} | 3.765625 | 4 | CC-MAIN-2017-47 | latest | en | 0.97308 |
https://www.ammarksman.com/how-do-you-find-the-points-of-inflection-on-a-derivative-graph/ | 1,713,459,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00863.warc.gz | 573,949,418 | 10,137 | # How do you find the points of inflection on a derivative graph?
## How do you find the points of inflection on a derivative graph?
An inflection point is a point on the graph of a function at which the concavity changes. Points of inflection can occur where the second derivative is zero. In other words, solve f ” = 0 to find the potential inflection points.
## Does the derivative exist at an inflection point?
Some continuous functions have an inflection point even though the second derivative is never 0. For example, the cube root function is concave upward when x is negative, and concave downward when x is positive, but has no derivatives of any order at the origin.
How do you find the inflection point of a curve?
Finding an Inflection Point. Check if the second derivative changes sign at the candidate point. If the sign of the second derivative changes as you pass through the candidate inflection point, then there exists an inflection point. If the sign does not change, then there exists no inflection point.
What is the derivative of an inflection point?
Inflection points are where the function changes concavity. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point.
### What are points of inflection on a graph?
Inflection points (or points of inflection) are points where the graph of a function changes concavity (from ∪ to ∩ or vice versa).
### How do you find points of inflection?
A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function.
What is an inflection point in calculus?
Inflection points are points where the function changes concavity, i.e. from being “concave up” to being “concave down” or vice versa. They can be found by considering where the second derivative changes signs.
What is the inflection point on a graph?
#### Is an inflection point a turning point?
A turning point could be an inflection point, but it could also refer to a sudden change. Inflection points are generally gradual. Also, there is nothing about a turning point that implies that things will be going in the opposite direction, whereas inflection points do have that kind of implication. | 539 | 2,558 | {"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} | 4.5625 | 5 | CC-MAIN-2024-18 | latest | en | 0.889065 |
https://hprc.tamu.edu/w/index.php?title=HPRC:AMS:Service_Unit&diff=prev&oldid=6978 | 1,660,062,449,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00459.warc.gz | 294,057,813 | 13,886 | Difference between revisions of "HPRC:AMS:Service Unit"
AMS Service Unit
A 'Service Unit' (SU) is charged to a job when the job uses one 'effective_core' for one hour (walltime).
The calculation of effective_core depends on whether a job is a exclusive job or not ('-x' specified or not).
Exclusive Job
When an exclusive job runs on a node with m cores.
'effective_core = m'
Note that the effective_core for an exclusive job on a node is independent how many cores are requested by the job.
Once the effective_core on a node is calculated, the effective_core for the job is simply the sum of the effective_core on all nodes where the job runs.
Non-Exclusive Job
The effective core on a node are calculated by considering requested memory. When a job requests xxx memory by "-M xxx" (for our LSF configuration on Ada for a node with m cores, this is the runtime memory limit per core) and yyy cores by "-n yyy), thenfor that requested memory, we have
'memory_equivalent_core = min(m, ceil(xxx*yyy/total_memory*m))'
where total_memory is the total memory on a node available to users.
Example 1
``` A nxt node has 20 cores and around 50G memory available to users.
When a job requests 2.5G memory and 2 cores, the job uses memory_equivalent_core of min(20, ceil(2.5*2/50*20))=2 core.
```
Example 2
``` A nxt node has 20 cores and around 50G memory available to users.
When the job requests 3G memory and 2 cores, the job uses memory_equivalent_core of min(20, ceil(3*2/50*20))=3 cores.
```
Example 3
``` A nxt node has 20 cores and around 50G memory available to users.
When the job requests 8G memoryand 10 cores, the job uses memory_equivalent_core of min(20, ceil(8*10/50*20))=20 cores.
```
Once the memory_equivalent_core is calculated, the effective_core on a node where a job request yyy cores can be calculated as below:
``` effective_core = max(yyy, memory_equivalent_cores)
```
Finally, the effective_core for a job is the sum of the effective_core on all nodes where the job runs. | 517 | 2,012 | {"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-2022-33 | latest | en | 0.852109 |
https://www.law.cornell.edu/cfr/text/40/appendix-F_to_part_75 | 1,669,773,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00021.warc.gz | 906,941,543 | 21,268 | # 40 CFR Appendix F to Part 75 - Conversion Procedures
Appendix F to Part 75 - Conversion Procedures
1. Applicability
Use the procedures in this appendix to convert measured data from a monitor or continuous emission monitoring system into the appropriate units of the standard.
2. Procedures for SO2 Emissions
Use the following procedures to compute hourly SO2 mass emission rate (in lb/hr) and quarterly and annual SO2 total mass emissions (in tons).
2.1 When measurements of SO2 concentration and flow rate are on a wet basis, use the following equation to compute hourly SO2 mass emission rate (in lb/hr):
${E}_{h}=K{C}_{h}{Q}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-1)}$
Where:
Eh = Hourly SO2 mass emission rate during unit operation, lb/hr.
K = 1.660 × 10−7 for SO2, (lb/scf)/ppm.
Ch = Hourly average SO2 concentration during unit operation, stack moisture basis, ppm.
Qh = Hourly average volumetric flow rate during unit operation, stack moisture basis, scfh.
2.2 When measurements by the SO2 pollutant concentration monitor are on a dry basis and the flow rate monitor measurements are on a wet basis, use the following equation to compute hourly SO2 mass emission rate (in lb/hr):
${E}_{h}=K{C}_{\mathrm{hp}}{Q}_{\mathrm{hs}}\frac{\left(100-%{H}_{2}O\right)}{100}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-2)}$
where:
Eh = Hourly SO2 mass emission rate during unit operation, lb/hr.
K = 1.660 × 10−7 for SO2, (lb/scf)/ppm.
Chp = Hourly average SO2 concentration during unit operation, ppm (dry).
Qhs = Hourly average volumetric flow rate during unit operation, scfh as measured (wet).
%H2O = Hourly average stack moisture content during unit operation, percent by volume.
2.3 Use the following equations to calculate total SO2 mass emissions for each calendar quarter (Equation F-3) and for each calendar year (Equation F-4), in tons:
${E}_{q}=\frac{\sum _{h=1}^{n}{E}_{h}{t}_{h}}{2000}$
(Eq. F-3)
Where:
Eq = Quarterly total SO2 mass emissions, tons.
Eh = Hourly SO2 mass emission rate, lb/hr.
th = Unit operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
n = Number of hourly SO2 emissions values during calendar quarter.
2000 = Conversion of 2000 lb per ton.
${E}_{a}=\sum _{q=1}^{4}{E}_{q}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-4)}$
Where:
Ea = Annual total SO2 mass emissions, tons.
Eq = Quarterly SO2 mass emissions, tons.
q = Quarters for which Eq are available during calendar year.
2.4 Round all SO2 mass emission rates and totals to the nearest tenth.
3. Procedures for NOX Emission Rate
Use the following procedures to convert continuous emission monitoring system measurements of NOX concentration (ppm) and diluent concentration (percentage) into NOX emission rates (in lb/mmBtu). Perform measurements of NOX and diluent (O2 or CO2) concentrations on the same moisture (wet or dry) basis.
3.1 When the NOX continuous emission monitoring system uses O2 as the diluent, and measurements are performed on a dry basis, use the following conversion procedure:
$E=K\phantom{\rule{0ex}{0ex}}{C}_{h}\phantom{\rule{0ex}{0ex}}F\phantom{\rule{0ex}{0ex}}\frac{20.9}{20.9-{%O}_{2}}$
(Eq. F-5)
where,
K, E, Ch, F, and %O2 are defined in section 3.3 of this appendix. When measurements are performed on a wet basis, use the equations in Method 19 in appendix A-7 to part 60 of this chapter.
3.2 When the NOX continuous emission monitoring system uses CO2 as the diluent, use the following conversion procedure:
$E=K{\phantom{\rule{0ex}{0ex}}C}_{h}{\phantom{\rule{0ex}{0ex}}F}_{c}\phantom{\rule{0ex}{0ex}}\frac{100}{%{\mathrm{CO}}_{2}}$
(Eq. F-6)
where:
K, E, Ch, Fc, and %CO2 are defined in section 3.3 of this appendix.
When CO2 and NOX measurements are performed on a different moisture basis, use the equations in Method 19 in appendix A-7 to part 60 of this chapter.
3.3 Use the definitions listed below to derive values for the parameters in equations F-5 and F-6 of this appendix, or (if applicable) in the equations in Method 19 in appendix A-7 to part 60 of this chapter.
3.3.1 K = 1.194 × 10−7 (lb/dscf)/ppm NOX.
3.3.2 E = Pollutant emissions during unit operation, lb/mmBtu.
3.3.3 Ch = Hourly average pollutant concentration during unit operation, ppm.
3.3.4 %O2, %CO2 = Oxygen or carbon dioxide volume during unit operation (expressed as percent O2 or CO2).
3.3.4.1 For boilers, a minimum concentration of 5.0 percent CO2 or a maximum concentration of 14.0 percent O2 may be substituted for the measured diluent gas concentration value for any operating hour in which the hourly average CO2 concentration is <5.0 percent CO2 or the hourly average O2 concentration is >14.0 percent O2. For stationary gas turbines, a minimum concentration of 1.0 percent CO2 or a maximum concentration of 19.0 percent O2 may be substituted for measured diluent gas concentration values for any operating hour in which the hourly average CO2 concentration is <1.0 percent CO2 or the hourly average O2 concentration is >19.0 percent O2.
3.3.4.2 If NOX emission rate is calculated using either Equation 19-3 or 19-5 in Method 19 in appendix A-7 to part 60 of this chapter, a variant of the equation shall be used whenever the diluent cap is applied. The modified equations shall be designated as Equations 19-3D and 19-5D, respectively. Equation 19-3D is structurally the same as Equation 19-3, except that the term “%O2w” in the denominator is replaced with the term “%O2dc × [(100−% H2O)/100]”, where %O2dc is the diluent cap value. The numerator of Equation 19-5D is the same as Equation 19-5; however, the denominator of Equation 19-5D is simply “20.9−%O2dc”, where %O2dc is the diluent cap value.
3.3.5 F, Fc = a factor representing a ratio of the volume of dry flue gases generated to the caloric value of the fuel combusted (F), and a factor representing a ratio of the volume of CO2 generated to the calorific value of the fuel combusted (Fc), respectively. Table 1 lists the values of F and Fc for different fuels.
Table 1 - F- and Fc-Factors 1
Fuel F-factor
(dscf/mmBtu)
FC-factor
(scf CO2/mmBtu)
Coal (as defined by ASTM D388-99 2):
Anthracite 10,100 1,970
Bituminous 9,780 1,800
Subbituminous 9,820 1,840
Lignite 9,860 1,910
Petroleum Coke 9,830 1,850
Tire Derived Fuel 10,260 1,800
Oil 9,190 1,420
Gas:
Natural gas 8,710 1,040
Propane 8,710 1,190
Butane 8,710 1,250
Wood:
Bark 9,600 1,920
Wood residue 9,240 1,830
1 Determined at standard conditions: 20 °C (68 °F) and 29.92 inches of mercury.
2 Incorporated by reference under § 75.6 of this part.
3.3.6 Equations F-7a and F-7b may be used in lieu of the F or Fc factors specified in Section 3.3.5 of this appendix to calculate a site-specific dry-basis F factor (dscf/mmBtu) or a site-specific Fc factor (scf CO2/mmBtu), on either a dry or wet basis. At a minimum, the site-specific F or Fc factor must be based on 9 samples of the fuel. Fuel samples taken during each run of a RATA are acceptable for this purpose. The site-specific F or Fc factor must be re-determined at least annually, and the value from the most recent determination must be used in the emission calculations. Alternatively, the previous F or Fc value may continue to be used if it is higher than the value obtained in the most recent determination. The owner or operator shall keep records of all site-specific F or Fc determinations, active for at least 3 years. (Calculate all F- and Fc factors at standard conditions of 20 °C (68 °F) and 29.92 inches of mercury).
$F=\frac{3.64\left(%H\right)+1.53\left(%C\right)+0.57\left(%S\right)+0.14\left(%N\right)+0.46\left(%O\right)}{GCV}×{10}^{6}$
(Eq. F-7a)
${F}_{c}=\frac{321×{10}^{3}\left(%C\right)}{GCV}$
(Eq. F-7b)
3.3.6.1 H, C, S, N, and O are content by weight of hydrogen, carbon, sulfur, nitrogen, and oxygen (expressed as percent), respectively, as determined on the same basis as the gross calorific value (GCV) by ultimate analysis of the fuel combusted using ASTM D3176-89 (Reapproved 2002), Standard Practice for Ultimate Analysis of Coal and Coke, (solid fuels), ASTM D5291-02, Standard Test Methods for Instrumental Determination of Carbon, Hydrogen, and Nitrogen in Petroleum Products and Lubricants, (liquid fuels) or computed from results using ASTM D1945-96 (Reapproved 2001), Standard Test Method for Analysis of Natural Gas by Gas Chromatography, or ASTM D1946-90 (Reapproved 2006), Standard Practice for Analysis of Reformed Gas by Gas Chromatography, (gaseous fuels) as applicable. (All of these methods are incorporated by reference under § 75.6 of this part.)
3.3.6.2 GCV is the gross calorific value (Btu/lb) of the fuel combusted determined by ASTM D5865-01a or ASTM D5865-10, ASTM D240-00 or ASTM D4809-00, and ASTM D3588-98, ASTM D4891-89 (Reapproved 2006), GPA Standard 2172-96, GPA Standard 2261-00, or ASTM D1826-94 (Reapproved 1998), as applicable. (All of these methods are incorporated by reference under § 75.6.)
3.3.6.3 For affected units that combust a combination of a fuel (or fuels) listed in Table 1 in section 3.3.5 of this appendix with any fuel(s) not listed in Table 1, the F or Fc value is subject to the Administrator's approval under § 75.66.
3.3.6.4 For affected units that combust combinations of fuels listed in Table 1 in section 3.3.5 of this appendix, prorate the F or Fc factors determined by section 3.3.5 or 3.3.6 of this appendix in accordance with the applicable formula as follows:
$F=\sum _{i=1}^{n}{X}_{i}{F}_{i}\phantom{\rule{0ex}{0ex}}{F}_{c}=\sum _{i=1}^{n}{X}_{i}{\left({F}_{c}\right)}_{i}$
Where,
Xi = Fraction of total heat input derived from each type of fuel (e.g., natural gas, bituminous coal, wood). Each Xi value shall be determined from the best available information on the quantity of fuel combusted and the GCV value, over a specified time period. The owner or operator shall explain the method used to calculate Xi in the hardcopy portion of the monitoring plan for the unit. The Xi values may be determined and updated either hourly, daily, weekly, or monthly. In all cases, the prorated F-factor used in the emission calculations shall be determined using the Xi values from the most recent update.
Fi or (Fc)i = Applicable F or Fc factor for each fuel type determined in accordance with Section 3.3.5 or 3.3.6 of this appendix.
n = Number of fuels being combusted in combination.
3.3.6.5 As an alternative to prorating the F or Fc factor as described in section 3.3.6.4 of this appendix, a “worst-case” F or Fc factor may be reported for any unit operating hour. The worst-case F or Fc factor shall be the highest F or Fc value for any of the fuels combusted in the unit.
3.4 Use the following equations to calculate the average NOX emission rate for each calendar quarter (Equation F-9) and the average emission rate for the calendar year (Equation F-10), in lb/mmBtu:
${E}_{q}=\sum _{i=1}^{n}\frac{{E}_{i}}{n}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-9)}$
Where:
Eq = Quarterly average NOX emission rate, lb/mmBtu.
Ei = Hourly average NOX emission rate during unit operation, lb/mmBtu.
n = Number of hourly rates during calendar quarter.
${E}_{a}=\sum _{i=1}^{m}\frac{{E}_{i}}{m}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-10)}$
Where:
Ea = Average NOX emission rate for the calendar year, lb/mmBtu.
Ei = Hourly average NOX emission rate during unit operation, lb/mmBtu.
m = Number of hourly rates for which Ei is available in the calendar year.
3.5 Round all NOX emission rates to the nearest 0.001 lb/mmBtu.
4. Procedures for CO2 Mass Emissions
Use the following procedures to convert continuous emission monitoring system measurements of CO2 concentration (percentage) and volumetric flow rate (scfh) into CO2 mass emissions (in tons/day) when the owner or operator uses a CO2 continuous emission monitoring system (consisting of a CO2 or O2 pollutant monitor) and a flow monitoring system to monitor CO2 emissions from an affected unit.
4.1 When CO2 concentration is measured on a wet basis, use the following equation to calculate hourly CO2 mass emissions rates (in tons/hr):
${E}_{h}=K{C}_{h}{Q}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-11)}$
Where:
Eh = Hourly CO2 mass emission rate during unit operation, tons/hr.
K = 5.7 × 10−7 for CO2, (tons/scf) /%CO2.
Ch = Hourly average CO2 concentration during unit operation, wet basis, either measured directly with a CO2 monitor or calculated from wet-basis O2 data using Equation F-14b, percent CO2.
Qh = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
4.2 When CO2 concentration is measured on a dry basis, use Equation F-2 to calculate the hourly CO2 mass emission rate (in tons/hr) with a K-value of 5.7 × 10−7 (tons/scf) percent CO2, where Eh = hourly CO2 mass emission rate, tons/hr and Chp = hourly average CO2 concentration in flue, dry basis, percent CO2.
4.3 Use the following equations to calculate total CO2 mass emissions for each calendar quarter (Equation F-12) and for each calendar year (Equation F-13):
${E}_{{\mathrm{CO}}_{2q}=\sum _{h=1}^{\mathrm{Hg}}{E}_{h}{t}_{h}\phantom{\rule{0ex}{0ex}}}\text{(Eq. F-12)}$
Where:
ECO2q = Quarterly total CO2 mass emissions, tons.
Eh = Hourly CO2 mass emission rate, tons/hr.
th = Unit operating time, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
HR = Number of hourly CO2 mass emission rates available during calendar quarter.
${E}_{{\mathrm{CO}}_{2a}=\sum _{q=1}^{4}{E}_{{\mathrm{CO}}_{2q}}\phantom{\rule{0ex}{0ex}}}\text{(Eq. F-13)}$
Where:
ECO2a = Annual total CO2 mass emissions, tons.
ECO2q = Quarterly total CO2 mass emissions, tons.
q = Quarters for which ECO2q are available during calendar year.
4.4 For an affected unit, when the owner or operator is continuously monitoring O2 concentration (in percent by volume) of flue gases using an O2 monitor, use the equations and procedures in section 4.4.1 and 4.4.2 of this appendix to determine hourly CO2 mass emissions (in tons).
4.4.1 If the owner or operator elects to use data from an O2 monitor to calculate CO2 concentration, the appropriate F and FC factors from section 3.3.5 of this appendix shall be used in one of the following equations (as applicable) to determine hourly average CO2 concentration of flue gases (in percent by volume) from the measured hourly average O2 concentration:
${\mathrm{CO}}_{2d}=100\frac{\mathrm{Fc}}{F}\frac{20.9-{O}_{2d}}{20.9}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-14a)}$
Where:
CO2d = Hourly average CO2 concentration during unit operation, percent by volume, dry basis.
F, FC = F-factor or carbon-based Fc-factor from section 3.3.5 of this appendix.
20.9 = Percentage of O2 in ambient air.
O2d = Hourly average O2 concentration during unit operation, percent by volume, dry basis.
${\mathrm{CO}}_{2w}=\frac{100}{20.9}\frac{{F}_{c}}{F}\left[20.9\left(\frac{100-%{H}_{2}O}{100}\right)-{O}_{2w}\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. F-14b)}$
Where:
CO2w = Hourly average CO2 concentration during unit operation, percent by volume, wet basis.
O2w = Hourly average O2 concentration during unit operation, percent by volume, wet basis.
F, Fc = F-factor or carbon-based FC-factor from section 3.3.5 of this appendix.
20.9 = Percentage of O2 in ambient air.
%H2O = Moisture content of gas in the stack, percent.
For any hour where Equation F-14a or F-14b results in a negative hourly average CO2 value, 0.0% CO2w shall be recorded as the average CO2 value for that hour.
4.4.2 Determine CO2 mass emissions (in tons) from hourly average CO2 concentration (percent by volume) using equation F-11 and the procedure in section 4.1, where O2 measurements are on a wet basis, or using the procedures in section 4.2 of this appendix, where O2 measurements are on a dry basis.
5. Procedures for Heat Input
Use the following procedures to compute heat input rate to an affected unit (in mmBtu/hr or mmBtu/day):
5.1 Calculate and record heat input rate to an affected unit on an hourly basis, except as provided in sections 5.5 through 5.5.7. The owner or operator may choose to use the provisions specified in § 75.16(e) or in section 2.1.2 of appendix D to this part in conjunction with the procedures provided in sections 5.6 through 5.6.2 to apportion heat input among each unit using the common stack or common pipe header.
5.2 For an affected unit that has a flow monitor (or approved alternate monitoring system under subpart E of this part for measuring volumetric flow rate) and a diluent gas (O2 or CO2) monitor, use the recorded data from these monitors and one of the following equations to calculate hourly heat input rate (in mmBtu/hr).
5.2.1 When measurements of CO2 concentration are on a wet basis, use the following equation:
$\mathrm{HI}={Q}_{w}\frac{1}{{F}_{c}}\frac{%{\mathrm{CO}}_{2w}}{100}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-15)}$
Where:
HI = Hourly heat input rate during unit operation, mmBtu/hr.
Qw = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
Fc = Carbon-based F-factor, listed in section 3.3.5 of this appendix for each fuel, scf/mmBtu.
%CO2w = Hourly concentration of CO2 during unit operation, percent CO2 wet basis.
5.2.2 When measurements of CO2 concentration are on a dry basis, use the following equation:
$\mathrm{HI}={Q}_{h}\left[\frac{\left(100-%{H}_{2}O\right)}{{100F}_{c}}\right]\left(\frac{%{\mathrm{CO}}_{2d}}{100}\right)\phantom{\rule{0ex}{0ex}}\text{(Eq. F-16)}$
Where:
HI = Hourly heat input rate during unit operation, mmBtu/hr.
Qh = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
Fc = Carbon-based F-Factor, listed in section 3.3.5 of this appendix for each fuel, scf/mmBtu.
%CO2d = Hourly concentration of CO2 during unit operation, percent CO2 dry basis.
%H2O = Moisture content of gas in the stack, percent.
5.2.3 When measurements of O2 concentration are on a wet basis, use the following equation:
$\mathrm{HI}={Q}_{w}\frac{1}{F}\frac{\left[\left(20.9/100\right)\left(100-%{H}_{2}O\right)-{%O}_{2w}\right]}{20.9}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-17)}\text{}$
Where:
HI = Hourly heat input rate during unit operation, mmBtu/hr.
Qw = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
F = Dry basis F-factor, listed in section 3.3.5 of this appendix for each fuel, dscf/mmBtu.
%O2w = Hourly concentration of O2 during unit operation, percent O2 wet basis. For any operating hour where Equation F-17 results in an hourly heat input rate that is ≤0.0 mmBtu/hr, 1.0 mmBtu/hr shall be recorded and reported as the heat input rate for that hour.
%H2O = Hourly average stack moisture content, percent by volume.
5.2.4 When measurements of O2 concentration are on a dry basis, use the following equation:
$\mathrm{HI}={Q}_{w}\left[\frac{\left(100-%{H}_{2}O\right)}{100\phantom{\rule{0ex}{0ex}}F}\right]\left[\frac{\left(20.9-%{O}_{2d}\right)}{20.9}\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. F-18)}$
Where:
HI = Hourly heat input rate during unit operation, mmBtu/hr.
Qw = Hourly average volumetric flow during unit operation, wet basis, scfh.
F = Dry basis F-factor, listed in section 3.3.5 of this appendix for each fuel, dscf/mmBtu.
%H2O = Moisture content of the stack gas, percent.
%O2d = Hourly concentration of O2 during unit operation, percent O2 dry basis.
5.3 Heat Input Summation (for Heat Input Determined Using a Flow Monitor and Diluent Monitor)
5.3.1 Calculate total quarterly heat input for a unit or common stack using a flow monitor and diluent monitor to calculate heat input, using the following equation:
${\mathrm{HI}}_{\text{q}}=\sum _{\mathrm{hour}=1}^{n}{\mathrm{HI}}_{\text{i}}{t}_{i}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-18a)}$
Where:
HIq = Total heat input for the quarter, mmBtu.
HIi = Hourly heat input rate during unit operation, using Equation F-15, F-16, F-17, or F-18, mmBtu/hr.
ti = Hourly operating time for the unit or common stack, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
5.3.2 Calculate total cumulative heat input for a unit or common stack using a flow monitor and diluent monitor to calculate heat input, using the following equation:
${\mathrm{HI}}_{\text{c}}=\sum _{q=1}^{\text{the current quarter}}{\mathrm{HI}}_{\text{q}}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-18b)}$
Where:
HIc = Total heat input for the year to date, mmBtu.
HIq = Total heat input for the quarter, mmBtu.
5.4 [Reserved]
5.5 For a gas-fired or oil-fired unit that does not have a flow monitor and is using the procedures specified in appendix D to this part to monitor SO2 emissions or for any unit using a common stack for which the owner or operator chooses to determine heat input by fuel sampling and analysis, use the following procedures to calculate hourly heat input rate in mmBtu/hr. The procedures of section 5.5.3 of this appendix shall not be used to determine heat input from a coal unit that is required to comply with the provisions of this part for monitoring, recording, and reporting NOX mass emissions under a State or federal NOX mass emission reduction program.
5.5.1 (a) When the unit is combusting oil, use the following equation to calculate hourly heat input rate:
${\mathrm{HI}}_{o}={M}_{o}\frac{{\mathrm{GCV}}_{o}}{{10}^{6}}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-19)}$
Where:
HIo = Hourly heat input rate from oil, mmBtu/hr.
Mo = Mass rate of oil consumed per hour, as determined using procedures in appendix D to this part, in lb/hr, tons/hr, or kg/hr.
GCVO = Gross calorific value of oil, as measured by ASTM D240-00, ASTM D5865-01a, ASTM D5865-10, or ASTM D4809-00 for each oil sample under section 2.2 of appendix D to this part, Btu/unit mass (all incorporated by reference under § 75.6).
10 6 = Conversion of Btu to mmBtu.
(b) When performing oil sampling and analysis solely for the purpose of the missing data procedures in § 75.36, oil samples for measuring GCV may be taken weekly, and the procedures specified in appendix D to this part for determining the mass rate of oil consumed per hour are optional.
5.5.2 When the unit is combusting gaseous fuels, use the following equation to calculate heat input rate from gaseous fuels for each hour:
${\mathrm{HI}}_{g}=\frac{\left({Q}_{g}×{\mathrm{GCV}}_{g}\right)}{{10}^{6}}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-20)}$
Where:
HIg = Hourly heat input rate from gaseous fuel, mmBtu/hour.
Qg = Metered flow rate of gaseous fuel combusted during unit operation, hundred standard cubic feet per hour.
GCVg = Gross calorific value of gaseous fuel, as determined by sampling (for each delivery for gaseous fuel in lots, for each daily gas sample for gaseous fuel delivered by pipeline, for each hourly average for gas measured hourly with a gas chromatograph, or for each monthly sample of pipeline natural gas, or as verified by the contractual supplier at least once every month pipeline natural gas is combusted, as specified in section 2.3 of appendix D to this part) using ASTM D1826-94 (Reapproved 1998), ASTM D3588-98, ASTM D4891-89 (Reapproved 2006), GPA Standard 2172-96 Calculation of Gross Heating Value, Relative Density and Compressibility Factor for Natural Gas Mixtures from Compositional Analysis, or GPA Standard 2261-00 Analysis for Natural Gas and Similar Gaseous Mixtures by Gas Chromatography, Btu/100 scf (all incorporated by reference under § 75.6 of this part).
10 6 = Conversion of Btu to mmBtu.
5.5.3 When the unit is combusting coal, use the procedures, methods, and equations in sections 5.5.3.1-5.5.3.3 of this appendix to determine the heat input from coal for each 24-hour period. (All ASTM methods are incorporated by reference under § 75.6 of this part.)
5.5.3.1 Perform coal sampling daily according to section 5.3.2.2 in Method 19 in appendix A to part 60 of this chapter and use ASTM D2234-00, Standard Practice for Collection of a Gross Sample of Coal, (incorporated by reference under § 75.6 of this part) Type I, Conditions A, B, or C and systematic spacing for sampling. (When performing coal sampling solely for the purposes of the missing data procedures in § 75.36, use of ASTM D2234-00 is optional, and coal samples may be taken weekly.)
5.5.3.2 All ASTM methods are incorporated by reference under § 75.6. Use ASTM D2013-01 for preparation of a daily coal sample and analyze each daily coal sample for gross calorific value using ASTM D5865-01a or ASTM D5865-10. On-line coal analysis may also be used if the on-line analytical instrument has been demonstrated to be equivalent to the applicable ASTM methods under §§ 75.23 and 75.66.
5.5.3.3 Calculate the heat input from coal using the following equation:
${\mathrm{HI}}_{c}-{M}_{c}\frac{{\mathrm{GCV}}_{c}}{500}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-21)}$
(Eq. F-21)
where:
HIc = Daily heat input from coal, mmBtu/day.
Mc = Mass of coal consumed per day, as measured and recorded in company records, tons.
GCVC = Gross calorific value of coal sample, as measured by ASTM D3176-89 (Reapproved 2002), ASTM D5865-01a, or ASTM D5865-10, Btu/lb (incorporated by reference under § 75.6).
500 = Conversion of Btu/lb to mmBtu/ton.
5.5.4 For units obtaining heat input values daily instead of hourly, apportion the daily heat input using the fraction of the daily steam load or daily unit operating load used each hour in order to obtain HIi for use in the above equations. Alternatively, use the hourly mass of coal consumed in equation F-21.
5.5.5 If a daily fuel sampling value for gross calorific value is not available, substitute the maximum gross calorific value measured from the previous 30 daily samples. If a monthly fuel sampling value for gross calorific value is not available, substitute the maximum gross calorific value measured from the previous 3 monthly samples.
5.5.6 If a fuel flow value is not available, use the fuel flowmeter missing data procedures in section 2.4 of appendix D of this part. If a daily coal consumption value is not available, substitute the maximum fuel feed rate during the previous thirty days when the unit burned coal.
5.5.7 Results for samples must be available no later than thirty calendar days after the sample is composited or taken. However, during an audit, the Administrator may require that the results be available in five business days, or sooner if practicable.
5.6 Heat Input Rate Apportionment for Units Sharing a Common Stack or Pipe
5.6.1 Where applicable, the owner or operator of an affected unit that determines heat input rate at the unit level by apportioning the heat input monitored at a common stack or common pipe using megawatts shall apportion the heat input rate using the following equation:
${\mathrm{HI}}_{i}={\mathrm{HI}}_{\mathrm{CS}}\left(\frac{{t}_{\mathrm{CS}}}{{t}_{i}}\right)\left[\frac{{\mathrm{MW}}_{i}{t}_{i}}{\sum _{i=1}^{n}{\mathrm{MW}}_{i}{t}_{i}}\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. F-21a)}$
Where:
HIi = Heat input rate for a unit, mmBtu/hr.
HIcs = Heat input rate at the common stack or pipe, mmBtu/hr.
MWi = Gross electrical output, MWe.
ti = Unit operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
tCS = Common stack or common pipe operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
n = Total number of units using the common stack or pipe.
i = Designation of a particular unit.
5.6.2 Where applicable, the owner or operator of an affected unit that determines the heat input rate at the unit level by apportioning the heat input rate monitored at a common stack or common pipe using steam load shall apportion the heat input rate using the following equation:
${\mathrm{HI}}_{i}={\mathrm{HI}}_{\mathrm{CS}}\left(\frac{{t}_{\mathrm{CS}}}{{t}_{i}}\right)\left[\frac{{\mathrm{SF}}_{i}{t}_{i}}{\sum _{i=1}^{n}{\mathrm{SF}}_{i}{t}_{i}}\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. F-21b)}$
Where:
HIi = Heat input rate for a unit, mmBtu/hr.
HICS = Heat input rate at the common stack or pipe, mmBtu/hr.
SF = Gross steam load, lb/hr, or mmBtu/hr.
ti = Unit operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
tCS = Common stack or common pipe operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
n = Total number of units using the common stack or pipe.
i = Designation of a particular unit.
5.7 Heat Input Rate Summation for Units with Multiple Stacks or Pipes
The owner or operator of an affected unit that determines the heat input rate at the unit level by summing the heat input rates monitored at multiple stacks or multiple pipes shall sum the heat input rates using the following equation:
${\mathrm{HI}}_{\mathrm{Unit}}=\frac{\sum _{s=1}^{n}{\mathrm{HI}}_{s}{t}_{s}}{{t}_{\mathrm{Unit}}}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-21c)}$
Where:
HIUnit = Heat input rate for a unit, mmBtu/hr.
HIs = Heat input rate for the individual stack, duct, or pipe, mmBtu/hr.
tUnit = Unit operating time, hour or fraction of the hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
ts = Operating time for the individual stack or pipe, hour or fraction of the hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
s = Designation for a particular stack, duct, or pipe.
5.8 Alternate Heat Input Apportionment for Common Pipes
As an alternative to using Equation F-21a or F-21b in section 5.6 of this appendix, the owner or operator may apportion the heat input rate at a common pipe to the individual units served by the common pipe based on the fuel flow rate to the individual units, as measured by uncertified fuel flowmeters. This option may only be used if a fuel flowmeter system that meets the requirements of appendix D to this part is installed on the common pipe. If this option is used, determine the unit heat input rates using the following equation:
${\mathrm{HI}}_{i}={\mathrm{HI}}_{\mathrm{CP}}\left(\frac{{t}_{\mathrm{CP}}}{{t}_{i}}\right]\left[\frac{{\mathrm{FF}}_{i}{t}_{i}}{\sum _{i=1}^{n}{\mathrm{FF}}_{i}{t}_{i}}\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. F-21d)}$
Where:
HIi = Heat input rate for a unit, mmBtu/hr.
HICP = Heat input rate at the common pipe, mmBtu/hr.
FFi = Fuel flow rate to a unit, gal/min, 100 scfh, or other appropriate units.
ti = Unit operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
tCP = Common pipe operating time, hour or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
n = Total number of units using the common pipe.
i = Designation of a particular unit.
6. Procedure for Converting Volumetric Flow to STP
Use the following equation to convert volumetric flow at actual temperature and pressure to standard temperature and pressure.
FSTP = FActual(TStd/TStack)(PStack/PStd)
where:
FSTP = Flue gas volumetric flow rate at standard temperature and pressure, scfh.
FActual = Flue gas volumetric flow rate at actual temperature and pressure, acfh.
TStd = Standard temperature = 528 °R.
TStack = Flue gas temperature at flow monitor location, °R, where °R = 460 + °F.
PStack = The absolute flue gas pressure = barometric pressure at the flow monitor location + flue gas static pressure, inches of mercury.
PStd = Standard pressure = 29.92 inches of mercury.
7. Procedures for SO2 Mass Emissions, Using Default SO2 Emission Rates and Heat Input Measured by CEMS
The owner or operator shall use Equation F-23 to calculate hourly SO2 mass emissions in accordance with § 75.11(e)(1) during the combustion of gaseous fuel, for a unit that uses a flow monitor and a diluent gas monitor to measure heat input, and that qualifies to use a default SO2 emission rate under section 2.3.1.1, 2.3.2.1.1, or 2.3.6(b) of appendix D to this part. Equation F-23 may also be applied to the combustion of solid or liquid fuel that meets the definition of very low sulfur fuel in § 72.2 of this chapter, combinations of such fuels, or mixtures of such fuels with gaseous fuel, if the owner or operator has received approval from the Administrator under § 75.66 to use a site-specific default SO2 emission rate for the fuel or mixture of fuels.
${E}_{h}=\left(\mathrm{ER}\right)\left(\mathrm{HI}\right)\phantom{\rule{0ex}{0ex}}\text{(Eq. F-23)}$
Where:
Eh = Hourly SO2 mass emission rate, lb/hr.
ER = Applicable SO2 default emission rate for gaseous fuel combustion, from section 2.3.1.1, 2.3.2.1.1, or 2.3.6(b) of appendix D to this part, or other default SO2 emission rate for the combustion of very low sulfur liquid or solid fuel, combinations of such fuels, or mixtures of such fuels with gaseous fuel, as approved by the Administrator under § 75.66, lb/mmBtu.
HI = Hourly heat input rate, determined using the procedures in section 5.2 of this appendix, mmBtu/hr.
8. Procedures for NOXMass Emissions
The owner or operator of a unit that is required to monitor, record, and report NOX mass emissions under a State or federal NOX mass emission reduction program must use the procedures in section 8.1, 8.2, or 8.3 of this appendix, as applicable, to account for hourly NOX mass emissions, and the procedures in section 8.4 of this appendix to account for quarterly, seasonal, and annual NOX mass emissions to the extent that the provisions of subpart H of this part are adopted as requirements under such a program.
8.1 The owner or operator may use the hourly NOX emission rate and the hourly heat input rate to calculate the NOX mass emissions in pounds or the NOX mass emission rate in pounds per hour, (as required by the applicable reporting format), for each unit or stack operating hour, as follows:
8.1.1 If both NOX emission rate and heat input rate are monitored at the same unit or stack level (e.g., the NOX emission rate value and the heat input rate value both represent all of the units exhausting to the common stack), then (as required by the applicable reporting format) either:
(a) Use Equation F-24 to calculate the hourly NOX mass emissions (lb).
${M}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}={\mathrm{ER}}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}{\mathrm{HI}}_{h}{t}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-24)}$
Where:
M(NOX)h = NOX mass emissions in lbs for the hour.
ER(NOX)h = Hourly average NOX emission rate for hour h, lb/mmBtu, from section 3 of this appendix, from Method 19 in appendix A-7 to part 60 of this chapter, or from section 3.3 of appendix E to this part. (Include bias-adjusted NOX emission rate values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
HIh = Hourly average heat input rate for hour h, mmBtu/hr. (Include bias-adjusted flow rate values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
th = Monitoring location operating time for hour h, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator). If the combined NOX emission rate and heat input are monitored for all of the units in a common stack, the monitoring location operating time is equal to the total time when any of those units was exhausting through the common stack; or
(b) Use Equation F-24a to calculate the hourly NOX mass emission rate (lb/hr).
${E}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}={\mathrm{ER}}_{{\left({\mathrm{NO}}_{x}\right)}_{h}}{\mathrm{HI}}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-24a)}$
Where:
E(NOX)h = NOX mass emissions rate in lbs/hr for the hour.
ER(NOX)h = Hourly average NOX emission rate for hour h, lb/mmBtu, from section 3 of this appendix, from Method 19 in appendix A-7 to part 60 of this chapter, or from section 3.3 of appendix E to this part. (Include bias-adjusted NOX emission rate values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
HIh = Hourly average heat input rate for hour h, mmBtu/hr. (Include bias-adjusted flow rate values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
8.1.2 If NOX emission rate is measured at a common stack and heat input is measured at the unit level, sum the hourly heat inputs at the unit level according to the following formula:
${\mathrm{HI}}_{\mathrm{CS}}=\frac{\sum _{u=1}^{p}{\mathrm{HI}}_{u}{t}_{u}}{{t}_{\mathrm{CS}}}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-25)}$
where:
HICS = Hourly average heat input rate for hour h for the units at the common stack, mmBtu/hr.
tCS = Common stack operating time for hour h, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator). (For each hour, tcs is the total time during which one or more of the units which exhaust through the common stack operate.).
HIu = Hourly average heat input rate for hour h for the unit, mmBtu/hr.
tu = Unit operating time for hour h, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
p = Number of units that exhaust through the common stack.
u = Designation of a particular unit.
Use the hourly heat input rate at the common stack level and the hourly average NOX emission rate at the common stack level and the procedures in section 8.1.1 of this appendix to determine the hourly NOX mass emissions at the common stack.
8.1.3 If a unit has multiple ducts and NOX emission rate is only measured at one duct, use the NOX emission rate measured at the duct, the heat input measured for the unit, and the procedures in section 8.1.1 of this appendix to determine NOX mass emissions.
8.1.4 If a unit has multiple ducts and NOX emission rate is measured in each duct, heat input shall also be measured in each duct and the procedures in section 8.1.1 of this appendix shall be used to determine NOX mass emissions.
8.2 Alternatively, the owner or operator may use the hourly NOX concentration (as measured by a NOX concentration monitoring system) and the hourly stack gas volumetric flow rate to calculate the NOX mass emission rate (lb/hr) for each unit or stack operating hour, in accordance with section 8.2.1 or 8.2.2 of this appendix (as applicable). If the hourly NOX mass emissions are to be reported in lb, Equation F-26c in section 8.3 of this appendix shall be used to convert the hourly NOX mass emission rates to hourly NOX mass emissions (lb).
8.2.1 When the NOX concentration monitoring system measures on a wet basis, first calculate the hourly NOX mass emission rate (in lb/hr) during unit (or stack) operation, using Equation F-26a. (Include bias-adjusted flow rate or NOX concentration values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
${E}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}=K\phantom{\rule{0ex}{0ex}}{C}_{\mathrm{hw}}{Q}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-26a)}$
Where:
E(NOX)h = NOX mass emissions rate in lb/hr.
K = 1.194 × 10−7 for NOX, (lb/scf)/ppm.
Chw = Hourly average NOX concentration during unit operation, wet basis, ppm.
Qh = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
8.2.2 When NOX mass emissions are determined using a dry basis NOX concentration monitoring system and a wet basis flow monitoring system, first calculate hourly NOX mass emission rate (in lb/hr) during unit (or stack) operation, using Equation F-26b. (Include bias-adjusted flow rate or NOX concentration values, where the bias-test procedures in appendix A to this part shows a bias-adjustment factor is necessary.)
${E}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}=K\phantom{\rule{0ex}{0ex}}{C}_{\mathrm{hd}}{Q}_{h}\frac{\left(100-%{H}_{2}O\right)}{\left(100\right)}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-26b)}$
Where:
E(NOX)h = NOX mass emissions rate, lb/hr.
K = 1.194 × 10−7 for NOX, (lb/scf)/ppm.
Chd = Hourly average NOX concentration during unit operation, dry basis, ppm.
Qh = Hourly average volumetric flow rate during unit operation, wet basis, scfh.
%H2O = Hourly average stack moisture content during unit operation, percent by volume.
8.3 When hourly NOX mass emissions are reported in pounds and are determined using a NOX concentration monitoring system and a flow monitoring system, calculate NOX mass emissions (lb) for each unit or stack operating hour by multiplying the hourly NOX mass emission rate (lb/hr) by the unit operating time for the hour, as follows:
${M}_{{\left({\mathrm{NO}}_{X}\right)}_{h}}={E}_{h}{t}_{h}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-26c)}$
Where:
M(NOX)h = NOX mass emissions for the hour, lb.
Eh = Hourly NOX mass emission rate during unit (or stack) operation from Equation F-26a in section 8.2.1 of this appendix or Equation F-26b in section 8.2.2 of this appendix (as applicable), lb/hr.
th = Unit operating time or stack operating time (as defined in § 72.2 of this chapter) for hour “h”, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
8.4 Use the following procedures to calculate quarterly, cumulative ozone season, and cumulative yearly NOX mass emissions, in tons:
(a) When hourly NOX mass emissions are reported in lb., use Eq. F-27.
${M}_{{\left({\mathrm{NO}}_{X}\right)}_{\text{time period}}}=\frac{\sum _{h=1}^{p}M{\left({\mathrm{NO}}_{x}\right)}_{h}}{2000}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-27)}$
Where:
M(NOX)time period = NOX mass emissions in tons for the given time period (quarter, cumulative ozone season, cumulative year-to-date).
M(NOX)h = NOX mass emissions in lb for the hour.
p = The number of hours in the given time period (quarter, cumulative ozone season, cumulative year-to-date).
(b) When hourly NOX mass emission rate is reported in lb/hr, use Eq. F-27a.
${M}_{{\left({\mathrm{NO}}_{X}\right)}_{\text{time period}}}=\frac{\sum _{h=1}^{p}E{\left({\mathrm{NO}}_{x}\right)}_{h}{t}_{h}}{2000}\phantom{\rule{0ex}{0ex}}\text{(Eq. F-27a)}$
Where:
M(NOX)time period = NOX mass emissions in tons for the given time period (quarter, cumulative ozone season, cumulative year-to-date).
E(NOX)h = NOX mass emission rate in lb/hr for the hour.
p = The number of hours in the given time period (quarter, cumulative ozone season, cumulative year-to-date).
th = Monitoring location operating time for hour h, in hours or fraction of an hour (in equal increments that can range from one hundredth to one quarter of an hour, at the option of the owner or operator).
8.5 Specific provisions for monitoring NOX mass emissions from common stacks. The owner or operator of a unit utilizing a common stack may account for NOX mass emissions using either of the following methodologies, if the provisions of subpart H are adopted as requirements of a State or federal NOX mass reduction program:
8.5.1 The owner or operator may determine both NOX emission rate and heat input at the common stack and use the procedures in section 8.1.1 of this appendix to determine hourly NOX mass emissions at the common stack.
8.5.2 The owner or operator may determine the NOX emission rate at the common stack and the heat input at each of the units and use the procedures in section 8.1.2 of this appendix to determine the hourly NOX mass emissions at each unit.
9. [Reserved]
10. Moisture Determination From Wet and Dry O2 Readings
If a correction for the stack gas moisture content is required in any of the emissions or heat input calculations described in this appendix, and if the hourly moisture content is determined from wet- and dry-basis O2 readings, use Equation F-31 to calculate the percent moisture, unless a “K” factor or other mathematical algorithm is developed as described in section 6.5.7(a) of appendix A to this part:
$%{H}_{2}O=\frac{\left({O}_{2d}-{O}_{2w}\right)}{{O}_{2d}}×100\phantom{\rule{0ex}{0ex}}\text{(Eq. F-31)}$
Where:
% H2O = Hourly average stack gas moisture content, percent H2O
O2d = Dry-basis hourly average oxygen concentration, percent O2
O2w = Wet-basis hourly average oxygen concentration, percent O2
[58 FR 3701, Jan. 11, 1993; Redesignated and amended at 60 FR 26553, 26571, May 17, 1995; 61 FR 25585, May 22, 1996; 61 FR 59166, Nov. 20, 1996; 63 FR 57513, Oct. 27, 1998; 64 FR 28666, May 26, 1999; 64 FR 37582, July 12, 1999; 67 FR 40474, 40475, June 12, 2002; 67 FR 53505, Aug. 16, 2002; 70 FR 28695, May 18, 2005; 73 FR 4372, Jan. 24, 2008; 76 FR 17325, Mar. 28, 2011; 77 FR 2460, Jan. 18, 2012] | 12,553 | 45,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "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} | 3.125 | 3 | CC-MAIN-2022-49 | latest | en | 0.845222 |
https://mathoverflow.net/questions/10627/a-galois-theory-computation | 1,643,152,051,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00174.warc.gz | 443,241,395 | 27,017 | # A Galois Theory Computation
Excuse me for the specificity of this question, but this is a silly computation that's been giving me trouble for some time.
I want to explicitly realize the order 21 Frobenius group over ℂ(x), as ℂ(x,y,z) where y3=g(x) and z7=h(x,y). The order 21 Frobenius group is C7⋊C3, where the generator of C3 acts by taking the generator of C7 to its square. Or in other words < a,b|a3=1, b7=1, ba=a(b2) >. Furthermore, I want it to branch at exactly three points, two of which will have 7 preimages, each with ramification 3, and the third will have 3 points over it with ramification 7 each.
This can easily be shown to exist: Take ℙ1 minus three points (say x=6,5,2); look at its algebraic fundamental group (=the profinite completion of < c, d, t| cdt = 1 >), and map it to the Frobenius group surjectively by c goes to a, d goes to b, and t goes to b-1a-1. This gives you a Galois cover of ℙ1, with said ramification (because order(a)=3, order(b)=7, and order(b-1a-1)=3), and group the 21 order Frobenius group.
Of course, this construction is extremely difficult to track because of the topology involved. It would be much easier to deduce the field extension from the ramification behavior.
So: this can be broken down to two cyclic Galois extensions. The first, a ℂ(x,y), of the form y3=(x-2)2(x-6) is pretty easy to deduce (I need it to ramify at x=2 and x=6 and nowhere else -- this must be the equation up to change of variables). The second, a z7=h(x,y) is tricky. I want it to ramify only above x=5. There's some Abhyankar's lemma things going on here, and that makes the guesswork difficult, and my life much harder.
I should note that the distinction of the Frobenius group of order 21, and the reason that I'm at all interested in this example, is that it is the only order 21 group which isn't cyclic. Geometrically, it means that h is a function of both x and y, and not just x.
One solution is to set $y^3 = x$, triply ramified only over $0$ and $\infty$, and if we want the 7-fold ramification over $x=1$ (which has solutions $y=1,\omega,\omega^2$) we set $z^7 = (1-y)(1 - \omega y)^2(1-\omega^2 y)^4$, which only ramifies over the three preimages. To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega^2 y$ lifts to an automorphism of the whole field. This automorphism maps $(1-y)(1-\omega y)^2(1-\omega^2 y)^4$ to $(1-\omega^2 y)(1- y)^2(1- \omega y)^4$, which has seventh root $z^2/(1-\omega^2 y)$. | 739 | 2,480 | {"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} | 3.25 | 3 | CC-MAIN-2022-05 | longest | en | 0.932436 |
https://fabacademy.org/2019/labs/sorbonne/students/stephane-muller/week10.html | 1,642,371,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00527.warc.gz | 329,859,373 | 7,615 | ## Week 10: Molding and casting
03/26/2019 - Stéphane Muller
Group assignment: review the safety data sheets for each of our molding and casting materials, then make and compare test casts with each of them. Individual assignment: design a 3D mold around the stock and tooling that we'll be using, mill it, and use it to cast parts.
#### Research
The point of molding and casting is to be able to cast the same part over and over again. So my idea was to make some kind of a building block or a piece of a repeating pattern. My research led me to M.C. Escher. I love this artist, he did some amazing pieces... his love for mathematics, gemoetry, paradoxes and the concept of infinity speaks to me on some level.
I found this pattern very interesting and decided to model it! I would be able to cast the same piece in a lot of different materials, even print it and machine it in wood later, and all the tiles would fit together.
#### Planning and list of tasks
1. Group assignment
2. Modeling the piece in 3D
3. Milling it in wax
4. Making the mold in silicone
5. Casting the piece in different materials
#### Step by step
##### Group assignment
Check out our group documentation.
A little glossary taken from this wonderful resource.
Pot life The point at which it the mix becomes so thick that it is no longer free-flowing or self-leveling Demold time The point at which it is sufficiently cross-linked to be safe to remove from the master Indentation hardness This parameter is measured using an ad hoc scale known as "Shore A", named so after its inventor. The test involves pressing a flat-tipped 0.8 mm needle into a sample with a modest force (800 gf). Deflection of 2.5 mm or more corresponds to 0 on the scale, while no deflection whatsoever is denoted with the value of 100. Compositions around 20-40A are fairly stretchy and squishy, similar to a rubber band. Silicones around 60-80A are still flexible, but begin to resemble a pencil eraser or a tire - you can flex them, but you won't stretch them easily. Tensile strength In essence, tensile strength tells you the stress (force divided by the area of cross-section) that causes the material to either snap, or to develop a localized defect known as necking. Elongation Elongation at break, on the other hand, describes the extent to which a standardized specimen can be stretched before breaking apart. Tear strength Tear strength of at least around 15-20 kN/m, as per ASTM method, is highly desirable (the value corresponds to force divided by the thickness of the sample; 1 kN/m = 5.7 ppi). Formulations up to 30 kN/m are available on the market, and are certainly not a waste of money. It's best to stay away from products where the advertised tear strength less than 12 kN/m or so. Viscosity Formulations with lower dynamic viscosity are generally easier to mix, pour, and degas - and will be more inclined to conform to complex shapes without the aid of vacuum or pressure. All other things being roughly equal, go with the system with the lowest viscosity. Products below 50,000 mPa*s are advisable, but not strictly a must. Shrinkage Addition cure silicones used for moldmaking purposes should have no measurable shrinkage. Some manufacturers simply say "none", while others say "less than 0.1%" or so. Seeing a higher value in the datasheet is a warning sign; shrinkages between 0.2% and 0.6% are commonly seen in condensation cure silicones, and should be avoided in precision work.
Material Food safe silicone RTV 3428 (data sheet / safety data sheet) Pot life 90 minutes at 23°C Demold time 16 hours at 23°C Color Translucent (part B is white) Mixing ratio Part A: 100g / part B: 10g Indentation hardness 28 Shore A Tensile strength 7.5 MPa Elongation 600% Tear strength 20 KN/m Viscosity 25 000 mPa Shrinkage 0.1% Safety precautions The product has not been classified as hazardous according to the legislation in force. It's chemically stable. Solubility Practically insoluble
Material Epoxy (safety data sheet part A safety datasheet part B) Conservation 12 months Cure time 24 hours at 20°C Color Part A is transparent (colorless) and part B is blue Mixing ratio Part A: 100g / part B: 50g Safety precautions This product is quite toxic and should be handled with care and appropriate safety equipment (gloves, mask, safety glasses and clothes). Solubility Insoluble
The datasheet released by the manufacturer (Esprit Composite) is not very detailed unfortunately...
Material Plasticrète (non toxic acrylic resin) (data sheet) Conservation Chemically stable Cure time 3 hours at 20°C Color White Mixing ratio Part A: 100g / part B: 50g Indentation hardness 82-86 Shore A Safety precautions No particular precautions since it's non toxic Solubility Water
##### Modeling the piece in 3D
First I needed to redraw the piece in 2D, so I used Illustrator. The drawing needs to be very precise, in order for the pieces to fit in each other. So I first drew a hexagon and used it as a guide. Then I drew each part that was outside the hexagon and rotated them precisely for the interior of the hexagon. Finally, I did a small offset of 1% to make sure the pieces fit easily together.
To do the modeling I used Blender. I had already used Blender a while ago and thought I would give it another try. My initial idea here was to use Blender's sculpt mode. Unfortunately, this tool needs some getting used to... and I didn't have that much time to practice.
Then I thought I would give it a simple pyramidal extrusion and to do that I needed to play with meshes and Blender was still my best option. I was going to add a spine to the lizard and create the faces of the mesh one by one.
First, import the SVG model.
Convert the curve into a mesh with `Alt + C`.
Change the color of the object (to see a bit more clearly).
Go into Edit mode, and start adding vertices. To do that, select a vertex with a `right click` and add an extruded vertex by using `CTRL + left click`.
Then select 3 or 4 vertices and create a face with the `F` key.
If your faces have a weird color or something looks off, you might need to recalculate your normals. To do that, select all the models vertices with the lasso `CTRL + B` and then do `CTRL + N`.
After a lot of tweaking, here is the final 3D model.
Once the model is ready, I still needed to put it in a box for the milling step so to do that I just created a box the size of my wax block and did a couple of boolean operations.
##### Milling the mold in wax
The process for milling in 2.5D in VCarve is quite similar to what we saw when milling in 2D. I set up the dimensions for my block of wax (17.5 cm x 8.5 cm x 5 cm), imported my STL file from Blender and created the toolpaths.
1. Setting up the block's size
2. Importing the STL
3. Rotating and adjusting the 3D model
4. Rough and finish toolpaths
Initially the toolpaths were too long and went too far for my needs. As you can see on the screenshot below, the paths goes where there is nothing to mill.
To avoid unnecessary machining, I had to create a rectangle in the 2D drawing view and then tell the software to bind the toolpath to that rectangle.
Then I set up the roughing and the finishing paths. I used a 2 mm end mill at 50 mm/sec and 10,000 RPM for both. For the rough mill I used a 75% stepover and 15% for the finish.
Here is the setup I used to keep the block of wax in place. I also added double sided tape underneath to be sure because during our tests for the group assignment the mill failed twice because the block wasn't properly kept in place.
The rest of the procedure (XYZ origins, safety measures...) is the same as what I listed in week 8. The mill lasted for about 2 hours.
Do NOT use the computer driving the shopbot during the mill! A colleague used it to prepare his VCarve files during my machining and it crashed... fortunately, it the mill stopped right at the end of the last roughing path! I just had to launch the finishing path.
Here you can see the difference between the roughing and the finishing.
The finished product is nice but you can still see some striations. My 15% stepover was not enough, 10% stepover with a ball endmill could probably have done the trick. An offset cut might have given better results as well. I found a lot of good advice on this page to improve this process for next time.
##### Making the mold in silicone
First thing we need to do is get an idea of the volume of silicone we'll need to make to fill our mold. So I poured water into the block of wax and transfered it in a container. Then, in a similar container, I poured the same quantity of silicone.
Before I go on with the silicone, I poured a little bit of paraffin oil in my mold so that the silicone is easier to get out afterwards.
Then I add the catalyst and some silicone oil to prevent bubbles. The catalyser is 10g for every 100g of silicone.
I stir for 5 minutes and slowly pour the silicone in the mold. Making sure the silicone gets in every corner is not easy. The vaseline is not helping at this point and the silicone is quite thick.
Then I used a vacuum chamber to remove the bubbles. As you can see, the liquid expands with vacuum and we can see the bubbles on the surface.
Releasing the vacuum several times improves the degassing.
Now we wait 24 hours... and it didn't work. The silicone didn't cure at all, it was still as liquid as yesterday. As it turns out I didn't use the right catalyser...
So first I had to remove all the silicone. That was a real pain. Silicone is thick and sticky, and regular dissolvants like acetone or white spirit don't do anything. If anything it makes it worse. Fortunately we had some silicone oil left, which is a diluent. Adding the oil helped a lot, but it was still a pain to remove every single bit of silicone from the edges. But after an hour my piece was clean and I could start the whole process again, with the right products this time.
I didn't take any pictures of this because I had my gloves on and it was too much of a hassle to take them off at each stage. But here is the end result! The mold is ready!!
##### Casting the piece in different materials
First I will cast my piece in an acrylic resin called "plasticrète". The main advantage is that it cures quite quickly (3 hours).
I mixed the 2 parts according to the specifications, just like for the silicone. Because plasticrete is white, I thought I would try adding a touch of color so I added a couple of drops of food colorant, as a test.
I let it cure for 3 hours and uncast them! Most of the food colorant moved to the surface, but enough remained to give it a nice orange tint. It looks a little bit like candy actually!
And the final result, along side a 3D printed version of the model! I love it! The offset is bigger than I anticipated, but it still looks cool. I will cast it and mill it in other materials to extend my little collection. I'll post the new versions here.
#### Conclusion
I learned to mill in 2.5D and some chemistry this week! Handling chemicals is not my cup of tea I'm afraid. Chemistry was never my strong suit in school. But having a mold to cast the same piece over and over with perfect precision is very useful. I'll think twice before choosing 3D printing now. | 2,647 | 11,264 | {"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-2022-05 | latest | en | 0.946322 |
https://www.netlib.org/lapack/explore-html/d6/d50/sdrvrf3_8f_source.html | 1,627,748,620,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00412.warc.gz | 952,798,350 | 9,380 | LAPACK 3.10.0 LAPACK: Linear Algebra PACKage
sdrvrf3.f
Go to the documentation of this file.
1 *> \brief \b SDRVRF3
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 * Definition:
9 * ===========
10 *
11 * SUBROUTINE SDRVRF3( NOUT, NN, NVAL, THRESH, A, LDA, ARF, B1, B2,
12 * + S_WORK_SLANGE, S_WORK_SGEQRF, TAU )
13 *
14 * .. Scalar Arguments ..
15 * INTEGER LDA, NN, NOUT
16 * REAL THRESH
17 * ..
18 * .. Array Arguments ..
19 * INTEGER NVAL( NN )
20 * REAL A( LDA, * ), ARF( * ), B1( LDA, * ),
21 * + B2( LDA, * ), S_WORK_SGEQRF( * ),
22 * + S_WORK_SLANGE( * ), TAU( * )
23 * ..
24 *
25 *
26 *> \par Purpose:
27 * =============
28 *>
29 *> \verbatim
30 *>
31 *> SDRVRF3 tests the LAPACK RFP routines:
32 *> STFSM
33 *> \endverbatim
34 *
35 * Arguments:
36 * ==========
37 *
38 *> \param[in] NOUT
39 *> \verbatim
40 *> NOUT is INTEGER
41 *> The unit number for output.
42 *> \endverbatim
43 *>
44 *> \param[in] NN
45 *> \verbatim
46 *> NN is INTEGER
47 *> The number of values of N contained in the vector NVAL.
48 *> \endverbatim
49 *>
50 *> \param[in] NVAL
51 *> \verbatim
52 *> NVAL is INTEGER array, dimension (NN)
53 *> The values of the matrix dimension N.
54 *> \endverbatim
55 *>
56 *> \param[in] THRESH
57 *> \verbatim
58 *> THRESH is REAL
59 *> The threshold value for the test ratios. A result is
60 *> included in the output file if RESULT >= THRESH. To have
61 *> every test ratio printed, use THRESH = 0.
62 *> \endverbatim
63 *>
64 *> \param[out] A
65 *> \verbatim
66 *> A is REAL array, dimension (LDA,NMAX)
67 *> \endverbatim
68 *>
69 *> \param[in] LDA
70 *> \verbatim
71 *> LDA is INTEGER
72 *> The leading dimension of the array A. LDA >= max(1,NMAX).
73 *> \endverbatim
74 *>
75 *> \param[out] ARF
76 *> \verbatim
77 *> ARF is REAL array, dimension ((NMAX*(NMAX+1))/2).
78 *> \endverbatim
79 *>
80 *> \param[out] B1
81 *> \verbatim
82 *> B1 is REAL array, dimension (LDA,NMAX)
83 *> \endverbatim
84 *>
85 *> \param[out] B2
86 *> \verbatim
87 *> B2 is REAL array, dimension (LDA,NMAX)
88 *> \endverbatim
89 *>
90 *> \param[out] S_WORK_SLANGE
91 *> \verbatim
92 *> S_WORK_SLANGE is REAL array, dimension (NMAX)
93 *> \endverbatim
94 *>
95 *> \param[out] S_WORK_SGEQRF
96 *> \verbatim
97 *> S_WORK_SGEQRF is REAL array, dimension (NMAX)
98 *> \endverbatim
99 *>
100 *> \param[out] TAU
101 *> \verbatim
102 *> TAU is REAL array, dimension (NMAX)
103 *> \endverbatim
104 *
105 * Authors:
106 * ========
107 *
108 *> \author Univ. of Tennessee
109 *> \author Univ. of California Berkeley
110 *> \author Univ. of Colorado Denver
111 *> \author NAG Ltd.
112 *
113 *> \ingroup single_lin
114 *
115 * =====================================================================
116 SUBROUTINE sdrvrf3( NOUT, NN, NVAL, THRESH, A, LDA, ARF, B1, B2,
117 + S_WORK_SLANGE, S_WORK_SGEQRF, TAU )
118 *
119 * -- LAPACK test routine --
120 * -- LAPACK is a software package provided by Univ. of Tennessee, --
121 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
122 *
123 * .. Scalar Arguments ..
124 INTEGER LDA, NN, NOUT
125 REAL THRESH
126 * ..
127 * .. Array Arguments ..
128 INTEGER NVAL( NN )
129 REAL A( LDA, * ), ARF( * ), B1( LDA, * ),
130 + b2( lda, * ), s_work_sgeqrf( * ),
131 + s_work_slange( * ), tau( * )
132 * ..
133 *
134 * =====================================================================
135 * ..
136 * .. Parameters ..
137 REAL ZERO, ONE
138 parameter( zero = ( 0.0e+0, 0.0e+0 ) ,
139 + one = ( 1.0e+0, 0.0e+0 ) )
140 INTEGER NTESTS
141 parameter( ntests = 1 )
142 * ..
143 * .. Local Scalars ..
144 CHARACTER UPLO, CFORM, DIAG, TRANS, SIDE
145 INTEGER I, IFORM, IIM, IIN, INFO, IUPLO, J, M, N, NA,
146 + nfail, nrun, iside, idiag, ialpha, itrans
147 REAL EPS, ALPHA
148 * ..
149 * .. Local Arrays ..
150 CHARACTER UPLOS( 2 ), FORMS( 2 ), TRANSS( 2 ),
151 + diags( 2 ), sides( 2 )
152 INTEGER ISEED( 4 ), ISEEDY( 4 )
153 REAL RESULT( NTESTS )
154 * ..
155 * .. External Functions ..
156 REAL SLAMCH, SLANGE, SLARND
157 EXTERNAL slamch, slange, slarnd
158 * ..
159 * .. External Subroutines ..
160 EXTERNAL strttf, sgeqrf, sgeqlf, stfsm, strsm
161 * ..
162 * .. Intrinsic Functions ..
163 INTRINSIC max, sqrt
164 * ..
165 * .. Scalars in Common ..
166 CHARACTER*32 SRNAMT
167 * ..
168 * .. Common blocks ..
169 COMMON / srnamc / srnamt
170 * ..
171 * .. Data statements ..
172 DATA iseedy / 1988, 1989, 1990, 1991 /
173 DATA uplos / 'U', 'L' /
174 DATA forms / 'N', 'T' /
175 DATA sides / 'L', 'R' /
176 DATA transs / 'N', 'T' /
177 DATA diags / 'N', 'U' /
178 * ..
179 * .. Executable Statements ..
180 *
181 * Initialize constants and the random number seed.
182 *
183 nrun = 0
184 nfail = 0
185 info = 0
186 DO 10 i = 1, 4
187 iseed( i ) = iseedy( i )
188 10 CONTINUE
189 eps = slamch( 'Precision' )
190 *
191 DO 170 iim = 1, nn
192 *
193 m = nval( iim )
194 *
195 DO 160 iin = 1, nn
196 *
197 n = nval( iin )
198 *
199 DO 150 iform = 1, 2
200 *
201 cform = forms( iform )
202 *
203 DO 140 iuplo = 1, 2
204 *
205 uplo = uplos( iuplo )
206 *
207 DO 130 iside = 1, 2
208 *
209 side = sides( iside )
210 *
211 DO 120 itrans = 1, 2
212 *
213 trans = transs( itrans )
214 *
215 DO 110 idiag = 1, 2
216 *
217 diag = diags( idiag )
218 *
219 DO 100 ialpha = 1, 3
220 *
221 IF ( ialpha.EQ. 1) THEN
222 alpha = zero
223 ELSE IF ( ialpha.EQ. 2) THEN
224 alpha = one
225 ELSE
226 alpha = slarnd( 2, iseed )
227 END IF
228 *
229 * All the parameters are set:
230 * CFORM, SIDE, UPLO, TRANS, DIAG, M, N,
231 * and ALPHA
233 *
234 nrun = nrun + 1
235 *
236 IF ( iside.EQ.1 ) THEN
237 *
238 * The case ISIDE.EQ.1 is when SIDE.EQ.'L'
239 * -> A is M-by-M ( B is M-by-N )
240 *
241 na = m
242 *
243 ELSE
244 *
245 * The case ISIDE.EQ.2 is when SIDE.EQ.'R'
246 * -> A is N-by-N ( B is M-by-N )
247 *
248 na = n
249 *
250 END IF
251 *
252 * Generate A our NA--by--NA triangular
253 * matrix.
254 * Our test is based on forward error so we
255 * do want A to be well conditioned! To get
256 * a well-conditioned triangular matrix, we
257 * take the R factor of the QR/LQ factorization
258 * of a random matrix.
259 *
260 DO j = 1, na
261 DO i = 1, na
262 a( i, j) = slarnd( 2, iseed )
263 END DO
264 END DO
265 *
266 IF ( iuplo.EQ.1 ) THEN
267 *
268 * The case IUPLO.EQ.1 is when SIDE.EQ.'U'
269 * -> QR factorization.
270 *
271 srnamt = 'SGEQRF'
272 CALL sgeqrf( na, na, a, lda, tau,
273 + s_work_sgeqrf, lda,
274 + info )
275 ELSE
276 *
277 * The case IUPLO.EQ.2 is when SIDE.EQ.'L'
278 * -> QL factorization.
279 *
280 srnamt = 'SGELQF'
281 CALL sgelqf( na, na, a, lda, tau,
282 + s_work_sgeqrf, lda,
283 + info )
284 END IF
285 *
286 * Store a copy of A in RFP format (in ARF).
287 *
288 srnamt = 'STRTTF'
289 CALL strttf( cform, uplo, na, a, lda, arf,
290 + info )
291 *
292 * Generate B1 our M--by--N right-hand side
293 * and store a copy in B2.
294 *
295 DO j = 1, n
296 DO i = 1, m
297 b1( i, j) = slarnd( 2, iseed )
298 b2( i, j) = b1( i, j)
299 END DO
300 END DO
301 *
302 * Solve op( A ) X = B or X op( A ) = B
303 * with STRSM
304 *
305 srnamt = 'STRSM'
306 CALL strsm( side, uplo, trans, diag, m, n,
307 + alpha, a, lda, b1, lda )
308 *
309 * Solve op( A ) X = B or X op( A ) = B
310 * with STFSM
311 *
312 srnamt = 'STFSM'
313 CALL stfsm( cform, side, uplo, trans,
314 + diag, m, n, alpha, arf, b2,
315 + lda )
316 *
317 * Check that the result agrees.
318 *
319 DO j = 1, n
320 DO i = 1, m
321 b1( i, j) = b2( i, j ) - b1( i, j )
322 END DO
323 END DO
324 *
325 result(1) = slange( 'I', m, n, b1, lda,
326 + s_work_slange )
327 *
328 result(1) = result(1) / sqrt( eps )
329 + / max( max( m, n), 1 )
330 *
331 IF( result(1).GE.thresh ) THEN
332 IF( nfail.EQ.0 ) THEN
333 WRITE( nout, * )
334 WRITE( nout, fmt = 9999 )
335 END IF
336 WRITE( nout, fmt = 9997 ) 'STFSM',
337 + cform, side, uplo, trans, diag, m,
338 + n, result(1)
339 nfail = nfail + 1
340 END IF
341 *
342 100 CONTINUE
343 110 CONTINUE
344 120 CONTINUE
345 130 CONTINUE
346 140 CONTINUE
347 150 CONTINUE
348 160 CONTINUE
349 170 CONTINUE
350 *
351 * Print a summary of the results.
352 *
353 IF ( nfail.EQ.0 ) THEN
354 WRITE( nout, fmt = 9996 ) 'STFSM', nrun
355 ELSE
356 WRITE( nout, fmt = 9995 ) 'STFSM', nfail, nrun
357 END IF
358 *
359 9999 FORMAT( 1x, ' *** Error(s) or Failure(s) while testing STFSM
360 + ***')
361 9997 FORMAT( 1x, ' Failure in ',a5,', CFORM=''',a1,''',',
362 + ' SIDE=''',a1,''',',' UPLO=''',a1,''',',' TRANS=''',a1,''',',
363 + ' DIAG=''',a1,''',',' M=',i3,', N =', i3,', test=',g12.5)
364 9996 FORMAT( 1x, 'All tests for ',a5,' auxiliary routine passed the ',
365 + 'threshold ( ',i5,' tests run)')
366 9995 FORMAT( 1x, a6, ' auxiliary routine: ',i5,' out of ',i5,
367 + ' tests failed to pass the threshold')
368 *
369 RETURN
370 *
371 * End of SDRVRF3
372 *
373 END
subroutine sgeqlf(M, N, A, LDA, TAU, WORK, LWORK, INFO)
SGEQLF
Definition: sgeqlf.f:138
subroutine sgeqrf(M, N, A, LDA, TAU, WORK, LWORK, INFO)
SGEQRF
Definition: sgeqrf.f:145
subroutine sgelqf(M, N, A, LDA, TAU, WORK, LWORK, INFO)
SGELQF
Definition: sgelqf.f:143
subroutine stfsm(TRANSR, SIDE, UPLO, TRANS, DIAG, M, N, ALPHA, A, B, LDB)
STFSM solves a matrix equation (one operand is a triangular matrix in RFP format).
Definition: stfsm.f:277
subroutine strttf(TRANSR, UPLO, N, A, LDA, ARF, INFO)
STRTTF copies a triangular matrix from the standard full format (TR) to the rectangular full packed f...
Definition: strttf.f:194
subroutine strsm(SIDE, UPLO, TRANSA, DIAG, M, N, ALPHA, A, LDA, B, LDB)
STRSM
Definition: strsm.f:181
subroutine sdrvrf3(NOUT, NN, NVAL, THRESH, A, LDA, ARF, B1, B2, S_WORK_SLANGE, S_WORK_SGEQRF, TAU)
SDRVRF3
Definition: sdrvrf3.f:118 | 3,872 | 9,807 | {"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-2021-31 | latest | en | 0.23438 |
https://www.thestudentroom.co.uk/showthread.php?t=7342429 | 1,717,062,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00563.warc.gz | 885,031,623 | 60,742 | Is this a 2:1
Would a second year average worth 59.66% rounded up would be 60 which is a 2:1 with the value of that year being 25% towards a degree classification and let’s just say my university takes the top 3 90 credit modules in level 6 with that being worth 75% towards the degree classification if I were in the 3 modules to get a 70,50,50 what average for the year would it come to.
Original post by Anonymous
Would a second year average worth 59.66% rounded up would be 60 which is a 2:1 with the value of that year being 25% towards a degree classification and let’s just say my university takes the top 3 90 credit modules in level 6 with that being worth 75% towards the degree classification if I were in the 3 modules to get a 70,50,50 what average for the year would it come to.
The average of 70, 50 and 50 is 56.67.
If we take 75% of this and 25% of the 59.66 you got for second year, that's a total of 42.5 + 14.92 = 57.42.
That's a 2:2.
Original post by DataVenia
The average of 70, 50 and 50 is 56.67.
If we take 75% of this and 25% of the 59.66 you got for second year, that's a total of 42.5 + 14.92 = 57.42.
That's a 2:2.
What average would that second one need to be and potentially the module that I am predicted to get a first for it to be a 2:1? Also if I score highly on a level 6 module that’s worth 15 credits upon referral/deferral can that module be used to replace a 15 credit module at level 5. The university I attend is Hertfordshire
Original post by Anonymous
What average would that second one need to be and potentially the module that I am predicted to get a first for it to be a 2:1? Also if I score highly on a level 6 module that’s worth 15 credits upon referral/deferral can that module be used to replace a 15 credit module at level 5. The university I attend is Hertfordshire
I've just looked up the rules which the University of Hertfordshire use to calculate your overall classification. See D6.1.1 on page 177 of this document. The rules they provide are not quite how you've described them working. Here's what they say:
"For students being considered for a final award with Honours classification, the Programme Board of Examiners will determine for each student:
i the average numeric grade of the best 90 credits at Level 6 or higher; and
ii the average numeric grade of the best remaining 90 credits at Level 5 or higher; and
iii a combined average numeric grade, weighted 75% (i) and 25% (ii), above; and
iv the student’s Honours classification will be considered on the basis of this combined average numeric grade."
You've said that "a second year average worth 59.66%". Was this 4 modules, each worth 30 credits? I suspect not as you've mentioned "a 15 credit module at level 5". In order to apply the logic in the regulations, we need a module-level break-down of both level 5 and level 6, showing the number of credits each is worth and what mark you achieved for each. For any modules for which you don't yet know the result, just put "unknown" or something. Also, please mark the referral/deferral unit - in case that's relevant to the calculation.
With regards to you question about using a 15-credit referral/deferral module from Level 6 to replace a poorer 15-credit module from Level 6 - yes you can. D6.1.4 specifically says, "All pass grades (including referred passes) and compensated grades are eligible for inclusion in the determination of the combined average numeric grade, with the exception of grades awarded for non-University of Hertfordshire modules studied during a period of study abroad and additional modules which are not part of a validated programme of study." (I assume the part about "grades awarded for non-University of Hertfordshire modules" doesn't apply.)
Do note, however, that your referral/deferral module has likely been capped. D5.2.2 iv says, "Students who are successful in referred assessments will be awarded a P(REF) status code for the module. The numeric grade for the module will be limited by any assessment elements which have been failed and then passed at referral being capped to the minimum pass grade."
(Amusingly, whilst searching the document for the word "cap" - to see whether one would apply to the referral/deferral module - I came across a rule which says, "During the examination, students will: ... not wear baseball caps, hoodies or other types of hat which obscure their face."
Original post by DataVenia
I've just looked up the rules which the University of Hertfordshire use to calculate your overall classification. See D6.1.1 on page 177 of this document. The rules they provide are not quite how you've described them working. Here's what they say:
"For students being considered for a final award with Honours classification, the Programme Board of Examiners will determine for each student:
i the average numeric grade of the best 90 credits at Level 6 or higher; and
ii the average numeric grade of the best remaining 90 credits at Level 5 or higher; and
iii a combined average numeric grade, weighted 75% (i) and 25% (ii), above; and
iv the student’s Honours classification will be considered on the basis of this combined average numeric grade."
You've said that "a second year average worth 59.66%". Was this 4 modules, each worth 30 credits? I suspect not as you've mentioned "a 15 credit module at level 5". In order to apply the logic in the regulations, we need a module-level break-down of both level 5 and level 6, showing the number of credits each is worth and what mark you achieved for each. For any modules for which you don't yet know the result, just put "unknown" or something. Also, please mark the referral/deferral unit - in case that's relevant to the calculation.
With regards to you question about using a 15-credit referral/deferral module from Level 6 to replace a poorer 15-credit module from Level 6 - yes you can. D6.1.4 specifically says, "All pass grades (including referred passes) and compensated grades are eligible for inclusion in the determination of the combined average numeric grade, with the exception of grades awarded for non-University of Hertfordshire modules studied during a period of study abroad and additional modules which are not part of a validated programme of study." (I assume the part about "grades awarded for non-University of Hertfordshire modules" doesn't apply.)
Do note, however, that your referral/deferral module has likely been capped. D5.2.2 iv says, "Students who are successful in referred assessments will be awarded a P(REF) status code for the module. The numeric grade for the module will be limited by any assessment elements which have been failed and then passed at referral being capped to the minimum pass grade."
(Amusingly, whilst searching the document for the word "cap" - to see whether one would apply to the referral/deferral module - I came across a rule which says, "During the examination, students will: ... not wear baseball caps, hoodies or other types of hat which obscure their face."
Hey Data Venia so I got 69,65,65,62,55,42 as my 6 best high-scoring modules each modules in level 5 were worth 15 credits not 30 and level 5 value is 25%, now for level 6 you said that they take your 90 credits from level 6 into consideration too I studied 3 30 credit modules with an average that I am predicted to get of 70,50,50 though the last two numbers can change dependent on the outcome of an exam and coursework also if I score highly in a referral/deferral module in level 6 can that be used to replace the 42 in level 5 and also what would I roughly need in level 6 to get a 2:1. The average in level 5 works out at 59.66% but they’ll most likely round it up to 60% so…
Original post by Anonymous
Hey Data Venia so I got 69,65,65,62,55,42 as my 6 best high-scoring modules each modules in level 5 were worth 15 credits not 30 and level 5 value is 25%, now for level 6 you said that they take your 90 credits from level 6 into consideration too I studied 3 30 credit modules with an average that I am predicted to get of 70,50,50 though the last two numbers can change dependent on the outcome of an exam and coursework also if I score highly in a referral/deferral module in level 6 can that be used to replace the 42 in level 5 and also what would I roughly need in level 6 to get a 2:1. The average in level 5 works out at 59.66% but they’ll most likely round it up to 60% so…
OK. Those six modules from Level 5 were worth 15 credits each, so that's a total of 90 credits. As the year will have been 120 credits in total, that suggests that the remaining credits scored at of below 42%. Have I got that right?
You've described 90 credits from Level 6 so far ("am predicted to get of 70,50,50"), so we have 30 credits missing. Is there a single 30-credit module for Level 6 which you've not mentioned yet? Or perhaps two 15 credit modules? I'm going to assume the latter, as you wrote earlier about what might happen if you "score highly on a level 6 module that’s worth 15 credits upon referral/deferral".
You seem to have ignored what I quoted earlier, that "Students who are successful in referred assessments will be awarded a P(REF) status code for the module. The numeric grade for the module will be limited by any assessment elements which have been failed and then passed at referral being capped to the minimum pass grade."
So your "if I score highly in a referral/deferral module in level 6 can that be used to replace the 42 in level 5" plan doesn't work, as you'd been trying to replace the 42% module in Level 5 with a capped 40% module from Level 6.
So, he're what I think we're working with.
Level 5
15 credits at 69%
15 credits at 65%
15 credits at 65%
15 credits at 62%
15 credits at 55%
15 credits at 42%
15 credits at 42% or less (because you excluded this module when providing details of your "6 best high-scoring modules each modules in level 5")
15 credits at 42% or less (because you excluded this module when providing details of your "6 best high-scoring modules each modules in level 5")
Level 6
30 credits at 70%
30 credits at 50%
30 credits at 50%
30 credits at 40% (due to "referred assessments ... being capped to the minimum pass grade.")
Can you confirm or correct?
Original post by DataVenia
OK. Those six modules from Level 5 were worth 15 credits each, so that's a total of 90 credits. As the year will have been 120 credits in total, that suggests that the remaining credits scored at of below 42%. Have I got that right?
You've described 90 credits from Level 6 so far ("am predicted to get of 70,50,50"), so we have 30 credits missing. Is there a single 30-credit module for Level 6 which you've not mentioned yet? Or perhaps two 15 credit modules? I'm going to assume the latter, as you wrote earlier about what might happen if you "score highly on a level 6 module that’s worth 15 credits upon referral/deferral".
You seem to have ignored what I quoted earlier, that "Students who are successful in referred assessments will be awarded a P(REF) status code for the module. The numeric grade for the module will be limited by any assessment elements which have been failed and then passed at referral being capped to the minimum pass grade."
So your "if I score highly in a referral/deferral module in level 6 can that be used to replace the 42 in level 5" plan doesn't work, as you'd been trying to replace the 42% module in Level 5 with a capped 40% module from Level 6.
So, he're what I think we're working with.
Level 5
15 credits at 69%
15 credits at 65%
15 credits at 65%
15 credits at 62%
15 credits at 55%
15 credits at 42%
15 credits at 42% or less (because you excluded this module when providing details of your "6 best high-scoring modules each modules in level 5")
15 credits at 42% or less (because you excluded this module when providing details of your "6 best high-scoring modules each modules in level 5")
Level 6
30 credits at 70%
30 credits at 50%
30 credits at 50%
30 credits at 40% (due to "referred assessments ... being capped to the minimum pass grade.")
Can you confirm or correct?
6 modules as stated from level 5 are worth 15 credits correct. The other two modules were scored at 41% and 40% respectively. Those 3 modules where I have said a predicted estimate of 70,50,50 each of those 3 modules have a worth of 30 credits. That’s the 90 credits. The other two modules of level 6 are 15 credit modules.
Correct though one of the modules of level 6 with a predicted estimate of 50% may change depending how successful I was in the exam as I believe I’ve done extremely well and I believe I can achieve anything over a 2:1 even a first.
Original post by Anonymous
6 modules as stated from level 5 are worth 15 credits correct. The other two modules were scored at 41% and 40% respectively. Those 3 modules where I have said a predicted estimate of 70,50,50 each of those 3 modules have a worth of 30 credits. That’s the 90 credits. The other two modules of level 6 are 15 credit modules.
Correct though one of the modules of level 6 with a predicted estimate of 50% may change depending how successful I was in the exam as I believe I’ve done extremely well and I believe I can achieve anything over a 2:1 even a first.
Excellent. And one of those 15-credit modules from Level 6 is the referral/deferral module, which will be capped at 40% - right? Do you have any prediction (or result) for that final 15-credit module within Level 6?
Let's run through the calculation step-by-step
i the average numeric grade of the best 90 credits at Level 6 or higher
Let's assume that one of those Level 6 modules which is predicted 50% turns out to be 65% (you say "I can achieve anything over a 2:1 even a first", so I've picked the mid-point between a 2:1 and a first). Let's also assume that the final 15-credit module within Level 6 is at 40% - a worst-case scenario.
That would make your "best 90 credits at Level 6 or higher" 70%, 65% and 50% - i.e. an average of 61.67%.
ii the average numeric grade of the best remaining 90 credits at Level 5 or higher; and
The "or higher" part of this rule means that if your final 15-credit module within Level 6 is worth more than 42%, then we can drop the 42% and use the value from that Level 6 module instead. However, taking a worse case scenario approach I'm going assume that doesn't happen and your "stuck" with that 42%.
So the "best remaining 90 credits at Level 5 or higher" is simply the average of 69%, 65%, 65%, 62%, 55% and 42% - which all actually come from Level 5. That the same 59.66% average we've been talking about all along.
iii a combined average numeric grade, weighted 75% (i) and 25% (ii), above; and
So we take 75% of 61.67% and add it to 25% of 59.66% to get 61.17%
iv the student’s Honours classification will be considered on the basis of this combined average numeric grade."
That's a 2:1.
If we continue to ignore the two 15-credit modules from Level 6 (because one is capped at 40% and because you've not provided any prediction for the other) then the only "variable" we're tweaking is for the 30-credit module where you say, "I can achieve anything over a 2:1 even a first". As long at that's at least 61% then you'll get over 60% overall, and hence a 2:1.
That calculation would be:
(((70+50+61)/3)*0.75)+(((69+65+65+62+55+42)/6)*0.25) = 60.17%
(Actually, regulation D6.3.1 ii says that, "For an upper second class Honours award a student must achieve a combined average numeric grade of 59.50 or more" so you have a little bit or further wiggle-room, but it's probably best not to push it!)
Original post by DataVenia
Excellent. And one of those 15-credit modules from Level 6 is the referral/deferral module, which will be capped at 40% - right? Do you have any prediction (or result) for that final 15-credit module within Level 6?
Let's run through the calculation step-by-step
i the average numeric grade of the best 90 credits at Level 6 or higher
Let's assume that one of those Level 6 modules which is predicted 50% turns out to be 65% (you say "I can achieve anything over a 2:1 even a first", so I've picked the mid-point between a 2:1 and a first). Let's also assume that the final 15-credit module within Level 6 is at 40% - a worst-case scenario.
That would make your "best 90 credits at Level 6 or higher" 70%, 65% and 50% - i.e. an average of 61.67%.
ii the average numeric grade of the best remaining 90 credits at Level 5 or higher; and
The "or higher" part of this rule means that if your final 15-credit module within Level 6 is worth more than 42%, then we can drop the 42% and use the value from that Level 6 module instead. However, taking a worse case scenario approach I'm going assume that doesn't happen and your "stuck" with that 42%.
So the "best remaining 90 credits at Level 5 or higher" is simply the average of 69%, 65%, 65%, 62%, 55% and 42% - which all actually come from Level 5. That the same 59.66% average we've been talking about all along.
iii a combined average numeric grade, weighted 75% (i) and 25% (ii), above; and
So we take 75% of 61.67% and add it to 25% of 59.66% to get 61.17%
iv the student’s Honours classification will be considered on the basis of this combined average numeric grade."
That's a 2:1.
If we continue to ignore the two 15-credit modules from Level 6 (because one is capped at 40% and because you've not provided any prediction for the other) then the only "variable" we're tweaking is for the 30-credit module where you say, "I can achieve anything over a 2:1 even a first". As long at that's at least 61% then you'll get over 60% overall, and hence a 2:1.
That calculation would be:
(((70+50+61)/3)*0.75)+(((69+65+65+62+55+42)/6)*0.25) = 60.17%
(Actually, regulation D6.3.1 ii says that, "For an upper second class Honours award a student must achieve a combined average numeric grade of 59.50 or more" so you have a little bit or further wiggle-room, but it's probably best not to push it!)
Yes one of those 15 credit modules will be a referred/deferred module although I have a exam that won’t be capped meaning maximum marks is still available due to me not being able to do it for exceptional circumstances. Other than that the coursework materials are capped. Because of the exam not being capped I believe I could get somewhere between 50-55%. But do you think a 2:1 is getable Data Venia?
Original post by Anonymous
But do you think a 2:1 is getable Data Venia?
From what you've written, yes I do. It's not clear-cut and will rely upon some decent marks in the remaining modules. But yes, a 2:1 is achievable.
Original post by DataVenia
From what you've written, yes I do. It's not clear-cut and will rely upon some decent marks in the remaining modules. But yes, a 2:1 is achievable.
Based on my project presentation and my dissertation I feel very confident I achieved a first or a high end 2:1 I will be ending that module with a guaranteed 2:1. It’s the exam in that second module I will need to ace as you say a minimum of 65 or I firmly believe I did extremely well enough based on how it went for a 1st then that module alone if I provide you the other coursework result marks should work out as a 2:1 as well.
Second module currently estimate of 50% dependent on exam result previous results in courseworks and exam:
CW1: 67%
CW2: 40%
CW3: 41%
CW4: 65-70% maybe onwards
With each CW have a weighting of 25% does my average for that module then turn out as a 2:1.
Original post by Anonymous
Based on my project presentation and my dissertation I feel very confident I achieved a first or a high end 2:1 I will be ending that module with a guaranteed 2:1. It’s the exam in that second module I will need to ace as you say a minimum of 65 or I firmly believe I did extremely well enough based on how it went for a 1st then that module alone if I provide you the other coursework result marks should work out as a 2:1 as well.
Second module currently estimate of 50% dependent on exam result previous results in courseworks and exam:
CW1: 67%
CW2: 40%
CW3: 41%
CW4: 65-70% maybe onwards
With each CW have a weighting of 25% does my average for that module then turn out as a 2:1.
If we take "65-70% maybe onwards" as the mid-point, i.e. 67.5%, then the average of 67%, 40%, 41% and 67.5% is 53.875%. A 2:2.
You would need that "maybe onwards" to reach 92% to average 60% (a 2:1) across the four pieces of coursework.
Original post by DataVenia
If we take "65-70% maybe onwards" as the mid-point, i.e. 67.5%, then the average of 67%, 40%, 41% and 67.5% is 53.875%. A 2:2.
You would need that "maybe onwards" to reach 92% to average 60% (a 2:1) across the four pieces of coursework.
Although the higher the overall the greater the chance of 2:1?
If the average for the 3 level 6 modules ended up at 70,56,50 with a 1st and 2 2:2s do you think that’s enough. Knowing I was working on a 2:1 last year at 60%.
Original post by Anonymous
Although the higher the overall the greater the chance of 2:1?
I guess. However, if the numbers you provided for CW1, CW2 and CW3 are locked-in (rather than being estimated, or subject to some kind of review), then CW4 would need to be at least 92% for the average across the four of them to be 60% (a 2:1). The higher the better, but 91% (whilst being an incredibly mark) would not get you to an average of 60%.
Original post by Anonymous
If the average for the 3 level 6 modules ended up at 70,56,50 with a 1st and 2 2:2s do you think that’s enough. Knowing I was working on a 2:1 last year at 60%.
No. The average of 70, 56 and 50 is 58.67. That wouldn't be sufficient for a 2:1 overall (when combined with last year's numbers).
Although to work out the module average it happens to be confusing in CW1 out of 25 points I obtained 16.75/25 (67%), CW2 10/25 (40%), CW 3 41/100 and CW4 Im predicting and estimating depending on how the exam went I will pick up a first now that is and would be something over 70 let’s say 75/100 since we are in the middle having added that up that gives an average of 57% as I added the total number of marks obtained out of the possible available and divided by the amount of coursework and exams done which is 4. | 5,734 | 22,190 | {"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} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.948588 |
https://ncertmcq.com/class-11-physics-important-questions-chapter-14/ | 1,721,671,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00758.warc.gz | 380,283,435 | 33,807 | Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 14 Oscillations. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.
## Class 11 Physics Chapter 14 Important Extra Questions Oscillations
### Oscillations Important Extra Questions Very Short Answer Type
Question 1.
(a) A particle has maximum velocity in the mean position and zero velocity at the extreme position. Is it a sure test for S.H.M.?
No.
(b) Is restoring force necessary in S.H.M.?
Yes.
Question 2.
Imagine a situation where the motion is not simple harmonic but the particle has maximum velocity in the mean position and zero velocity at the extreme position.
Projection of a particle in non-uniform circular motion satisfies all the given conditions.
Question 3.
We know that in S.H.M., the time period is given by
T = 2π$$\sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$$
Does T depend upon displacement?
No. T is independent of displacement as the displacement term is also contained in acceleration.
Question 4.
(a) Two simple pendulums of equal lengths cross each other at the mean position. What is their phase difference?
(b) Can a simple pendulum vibrate at the center of Earth? Why?
No. This is because the value of g at the center of Earth is zero.
Question 5.
(a) A particle is in S.H.M. of amplitude 2 cm. At the extreme position, the force is 4N. What is the force at a mid-point i.e. midway between mean and extreme position?
2N.
(b) What happens to the time period of a simple pendulum if its length is doubled?
The time period is increased by a factor of 4l.
Question 6.
Can & simple pendulum be used in an artificial satellite? Why?
No. This is because there exists a state of weightlessness in the artificial satellite.
Question 7.
What fraction of the total energy is potential energy when the displacement is one-half of the amplitude?
$$\frac{1}{4}\left[\because \frac{P . E}{\text { Total energy }}=\frac{\frac{1}{2} \operatorname{m\omega}^{2}\left(\frac{a}{2}\right)^{2}}{\frac{1}{2} m \omega^{2} a^{2}}\right]$$
Question 8.
What fraction of the total energy is kinetic energy, when the displacement is one-half of the amplitude?
$$\frac{3}{4}\left[\because \frac{\text { K.E. }}{\text { Total energy }}=\frac{\frac{1}{2} m \omega^{2}\left(a^{2}-\left(\frac{a}{2}\right)^{2}\right)}{\frac{1}{2} m \omega^{2} a^{2}}=\frac{3}{4}\right]$$
Question 9.
(a) When a particle oscillates simply harmonically, its potential energy varies periodically. If v be the frequency of oscillation of the particle, then what is the frequency of vibration of P.E.?
2v
(b) A body executes S.H.M. with a period of $$\frac{11}{7}$$ seconds and an amplitude of 0.025 m. What is the maximum value of acceleration?
Question 10.
A body of mass m when hung on a spiral spring stretches it by 20 cm. What is its period of oscillation when pulled down, and released?
T = 2π$$\sqrt{\frac{20}{980}}=\frac{2 \pi}{7} \mathrm{~s}=\frac{44}{49} \mathrm{~s}$$
Question 11.
A spring-mass system oscillating vertically has a time period T. What shall be the time period if oscillating horizontally?
T i.e. it remains the same.
Question 12.
The time period of a body executing S.H.M. is 0.05 s and the amplitude of vibration is 4 cm. What is the maximum velocity of the body?
1.6 π ms-1
Question 13.
A spring-controlled wristwatch is taken from Earth to Moon. What shall be the effect on the watch?
There will be no effect on g as the time period of spring is independent of g.
Question 14.
(a) At what displacement, the P.E. of a simple harmonic oscillator is maximum?
At the extreme position.
(b) At what displacement, the K.E. of a simple harmonic oscillator is maximum? ‘
At the mean position.
Question 15.
What is the total energy of a simple harmonic oscillator?
$$\frac{1}{2}$$m ω2 r2, where r = amplitude, ω = angular frequency, m = mass of the oscillator.
Question 16.
Name the trigonometric functions which are suitable for the analytical treatment of periodic motions. ,
Sine and Cosine functions or their linear combination.
Question 17.
How is acceleration (a) related to the displacement (y) in S.H.M.?
aα – y.
Question 18.
At what position, the velocity of a particle executing S.H.M. is maximum?
At mean position.
Question 19.
What is Force constant (k)? What are its units in the S.I. system?
k = Force per unit displacement.
S.I unit of k is Nm-1
Question 20.
(a) What ¡s the phase difference between displacement and velocity of a particle executing S.H.M.?
90° or $$\frac{π}{2}$$ radian.
(b) What is the phase difference between displacement and acceleration of a particle executing S.H.M.?
Question 21.
(a) What will be the change in the time period of a loaded spring when taken to Moon?
No change because T = 2π$$\sqrt{\frac{m}{k}}$$ i.e. T ¡s independent of g.
(b) Why does the time period (T) of the swing not change when two boys sit on it instead of one?
T is independent of the mass of the swing as T = 2π$$\sqrt{\frac{l}{g}}$$.
Question 22.
Why are the vibrations of a simple pendulum damped?
Because the energy of the pendulum is used to overcome air resistance.
Question 23.
A pendulum clock is taken on a lift moving down with a uniform Velocity. Will it gain dr lose time?
It neither gains nor loses time.
Question 24.
What role do shock absorbers play in oscillations of vehicles on a bumpy road?
They act as damping devices and save vehicles from resonant oscillations.
Question 25.
When do you apply force as the spring descends while swinging on a swing?
To counter the effect of the damping force on the amplitude of oscillations of the swing at an appropriate point.
Question 26.
The time period of oscillation of a sphere hung from a wire attached to a rigid support and given slight rotation about the wire is given by T = 2π$$\sqrt{\frac{I}{C}}$$. Which factor out of these is equivalent to the mass factor of an oscillator in S.H.M.?
I (moment of inertia) of the sphere is equivalent to m, the mass of an ordinary oscillator in S.H.M.
Question 27.
At what points is the energy entirely K.E. and entirely P.E. in S.H.M.?
At mean position and extreme position respectively.
Question 28.
What provides the restoring force for simple harmonic motion in the following cases?
(i) Simple pendulum.
(ii) Spring.
(iii) Column of mercury in a U-tube.
The restoring force is provided by:
(a) gravity in simple pendulum i.e. weight of the pendulum.
(b) elasticity in spring.
(c) weight in the column of Hg in a U-tube.
Question 29.
Which single characteristics of a periodic motion distinguishes it as S.H.M?
If acceleration is directly proportional to the negative- displacement, then the motion is said to be S.H.M.
Question 30.
(a) Does the bursting of air bubbles at the surface of boiling water show a periodic motion of the surface molecules?
No.
(b) Do all periodic motions are S.H.M.?
No.
Question 31.
At what point the velocity and acceleration are zero in S.H.M.?
The velocity is zero at the extreme points of motion and acceleration is zero at the mean position of motion.
Question 32.
On what factors does the force constant of a spring depend?
It Depends Upon the following factors:
1. Nature of the material of the spring.
2. Elasticity (i.e. coefficient of elasticity) of the material of spring.
Question 33.
Which factors determine the natural frequency of an oscillator?
The dimensions and the elastic properties of the oscillator.
Question 34.
When a high-speed plane passes by, the windows of the plane start producing sound due to
(i) S.H.M.,
(ii) damped oscillations,
(iii) maintained oscillations. Why?
(ii) Damped oscillations are caused due to air pressure on the glass window panes.
Question 35.
A motorcyclist while trying to loop a loop of maximum radius in a death trap goes round in the globe. Is it his projection on diameter in S.H.M.?
Yes.
Question 36.
Sometimes on the screen of a cathode-ray oscilloscope, we see rectangular or triangular waves. Are these formed due to S.H.M. of the oscillator connected to it?
No, these are not S.H.M. but are periodic waves.
Question 37.
Is oscillation of a mass suspended by a spring simple harmonic?
Yes, only if the spring is perfectly elastic.
Question 38.
A diver wearing an electronic digital watch goes down into seawater with terminal velocity v. How will the time in the waterproof watch be affected?
It will not be affected as its action is independent of gravity and buoyant force.
Question 39.
When is the tension maximum in the string of a simple pendulum?
The tension in the string is maximum at the mean position because the tension in the string = mg cos θ = mg. (∵ θ = 0 at mean position).
Question 40.
Is the damping force constant on a system executing S.H.M.?
No, because damping force depends upon velocity and it is more when the system moves fast and is less when the system moves slow.
Question 41.
What is the (a) distance covered by (b) displacement of, a body executing S.H.M. in a time equal to its period if its
amplitude is r?
A body executing S.H.M. completes one vibration in a time equal to its period, so the body reaches its initial position after a time equal to its period.
Thus the
(a) total distance traveled is 4r and
(b) displacement is zero.
Question 42.
Can a motion be oscillatory but not simple harmonic? If yes give an example and if no explain, why?
Yes. Uniform circular motion is an example of it.
Question 43.
How will the period of a pendulum change when its length is doubled?
The time period becomes $$\sqrt{2}$$ times the original value as T ∝ $$\sqrt{l}$$.
Question 44.
Determine whether or not the following quantities can be in the same direction for a simple harmonic motion;
(a) displacement and velocity,
Yes, when the particle in S.H.M. is moving from equilibrium position to extreme position.
(b) velocity and acceleration,
Yes, when the particle executing S.H.M. is moving from an extreme position to a mean position.
(c) acceleration and displacement?
No, because acceleration is always directed opposite to the displacement.,
Question 45.
Can a body have zero velocity and maximum acceleration in S.H.M.?
Yes. In S.H.M., at the extreme position, the velocity of the body is zero and acceleration is maximum.
Question 46.
A passing airplane sometimes causes the rattling of windows of a house. Explain why?
When the frequency of the sound waves produced by the engine of the airplane strikes the windows panes, they start vibrating due to forced oscillations. This causes rattling of windows.
Question 47.
Identify periodic, non-periodic, and S.H.M. out of the following:
(a) fluttering tree leaves due to wind.
Non-periodic
(b) fluttering of honeybee’s wings for a microsecond.
S.H.M.
(c) rising and falling of water drops at the bottom of a waterfall.
Non-periodic
(d) the sound produced by a motorcycle in a small interval of time.
Periodic
(e) the motion of the wheels of a canon during rapid fire.
Periodic
(f) jerks in a machine gun during rapid fire.
periodic
(g) vibrations of a drum membrane in a music band.
Non-periodic.
Question 48.
Why pendulum clocks are not suitable for spaceships?
The time period of a pendulum clock is given by
T = 2π$$\sqrt{\frac{l}{g}}$$
In a spaceship g = 0, so T = ∞,
It means the pendulum clock will not oscillate and hence are not suitable for spaceships.
Question 49.
A glass window may be broken by a distant explosion. Is it correct?
Yes. The sound waves can cause forced vibrations in glass due to the difference between the frequency of sound waves and the natural frequency of the glass. This can break the glass window.
Question 50.
The bob of a simple pendulum is positively charged. The pendulum is made to oscillate over a negatively charged plate. How shall the time period change as compared to the natural time period of the simple pendulum?
Due to the electric force of attraction between the bob and the plate, the effective value of acceleration due to gravity will increase T = 2π$$\sqrt{\frac{l}{g}}$$ therefore T will decrease.
### Oscillations Important Extra Questions Short Answer Type
Question 1.
Why does the body of a bus begin to rattle sometimes when the bus is accelerated?
At some speed of the bus, the frequency of vibrations produced by the bus-engine is equal to the natural frequency of the bus. This produces resonant vibrations which have quite a large amplitude and\ence the bus begins to rattle.
Question 2.
The displacement of a particle in S.H.M. may be given by x = A sin (ωt + Φ) Show that if the time t is increased by $$\frac{2π}{ω}$$, the value of x remains the same.
The displacement of a particle in S.H.M. at a time t is given
x = A sin (ωt + Φ) …. (i)
Let x be the displacement of the-particle at time t’ = ( t + $$\frac{2π}{ω}$$),
To prove x1 = x
From equation (i), we get
x1 = A sin [ω( t + $$\frac{2π}{ω}$$) + Φ)]
= A sin [ωt + 2π + Φ]
= A sin [2π + (ωt + Φ)]
= A sin (ωt + Φ)
[ ∵ sin (2π + θ) = sin θ]
i.e. the displacement of the particle in S.H.M. is same for times t and t + $$\frac{2π}{ω}$$ .
Hence proved.
Question 3.
A hollow sphere is filled with water through a small hole in it. It is hung by a long thread and as water slowly flows out of the hole at the bottom, one finds that the period of oscillations first increases and then decreases. Explain why?
When the sphere is filled with water, its C.G. is at the center of the sphere. We know that the time period (T) of oscillation is directly proportional to the square root of the effective length of the pendulum
i.e. T ∝ $$\sqrt{l}$$.
Due to the flow out of the water through the hole at the bottom of the sphere, the first effective length l increases, and then it decreases because of the lowering of the center of gravity in the first case and rising up in the second case. The first case corresponds to the lowering of the water level up to the center of the sphere. The second case corresponds to the lowering of the water level from the center of the sphere to its bottom.
Question 4.
A girl is swinging in the sitting position. How will the period ^ of the swing be changed if she stands up?
This can be explained using the concept of a simple pendulum.
We know that the time period of a simple pendulum is given by
T = 2π $$\sqrt{\frac{l}{g}}$$ i.e. T ∝ $$\sqrt{l}$$
When the girl stands up, the distance between the point of suspension arid the center of mass of the swinging body decreases i. e. l decreases, so l will also decrease.
Question 5.
At what displacement, a particle in S.H.M. possesses half K.E. and half P.E.?
We know that in any position for a displacement of y from the mean position, the K.E. and P.E. are given by the expressions:
Ek = $$\frac{1}{2}$$ mω2 (r2 – y2) ….(1)
Ep = $$\frac{1}{2}$$mω2y2 …..(2)
Where m = mass of the particle
y = displacement from the mean position
ω = its angular frequency
r = amplitude of oscillation of the particle
Now for Ek = Ep, we get
Question 6.
Explain why marching troops are asked to break their steps while crossing a bridge?
This is done so as to avoid setting the bridge into resonant oscillations of large amplitude. If troops march in step, then the frequency of the steps of the marching troops may become equal to the natural frequency of vibration of the bridge. Hence it may oscillate with high amplitude due to the phenomenon of resonance, resulting in damage to the bridge.
Question 7.
What is the direction of acceleration at the mean and extreme positions of an oscillating simple pendulum?
When the simple pendulum oscillates, its bob moves along a circular path with the point of suspension at its center. At the mean position, bob possesses centripetal acceleration which acts along the thread towards the point of suspension. At an extreme position, the bob is at rest for a moment. The acceleration on the bob acts along the tangent to the circular path and is directed towards its mean position. This acceleration results due to the restoring force arising due to the weight of the bob.
Question 8.
You are provided with a light spring, a meter scale, and a known mass. How will you find the time period of oscillation of mass attached to the spring without the use of a clock?
With the help of metre scale, we can note the length and extension produced. Since
F = – kl
and F = mg
or
∴ mg = – k l
or
$$\frac{\mathrm{m}}{\mathrm{k}}=\frac{l}{\mathrm{~g}}$$ (numerically)
∴ T = 2π$$\sqrt{\frac{m}{k}}$$ = 2π$$\sqrt{\frac{l}{g}}$$ …(1)
∴ By knowing the extension l, T can be calculated from equation (1) w ithout the use of clock.
Question 9.
Why does the time period of a pendulum change when taken to the top of a mountain or deep in a mine? Will clocks keep the correct time?
Time period of a pendulum is given by
T = 2π$$\sqrt{\frac{l}{g}}$$
As the value of ‘g’ on the top of a mountain or deep in a mine decreases, so the time period of the pendulum increases on the top of a mountain or deep in a mine and hence the pendulum clocks will run slow,
Question 10.
What is the source of potential energy in a loaded elastic spring?
As the load is put at the end of the spring, it gets extended. To increase the length, the load does work against the elastic restoring force. This work done is stored in the spring in the form of its potential energy,
Question 11.
Why cannot we construct an ideal simple pendulum?
An ideal simple pendulum is made up of a heavy point mass suspended to a weightless, inextensible string fastened to a rigid support. In actual practice geometrical point mass is not realized so is a weightless and inextensible string. Hence ideal simple pendulum cannot be constructed in actual practice.
Question 12.
When a fast railway train passes over a river bridge, we hear a loud sound, why?
Fast railway train vibrations are passed to the air column between the bridge and the river water through railway lines and slippers. The vertical air column starts resonating with train vibrations, so a loud sound is heard. ‘
Question 13.
If the amplitude of vibration of an oscillator is made half, how will its time period and energy be affected?
Since the time period of an oscillator is independent of its amplitude, so it remains unaffected. However, the total energy is directly proportional to the square of the amplitude, so the energy of the oscillator will become a quarter of its original value on making amplitude half.
Question 14.
A bead is mounted on a vertical stand placed attached to the rim of a record player and its shadow is cast on the wall with the help of light from a lamp. The record player is set in motion. Describe the motion of the shadow of the bead.
As the bead executes circular motion with the record player, its shadow on the wall will execute whose period depends on the speed of rotation of the record player.
Question 15.
In Boyle’s law apparatus, if the open end tube is quickly moved up and down what will happen to the mercury column in the closed tube?
Due to quick compression and expansion of air column above mercury column in the closed tube, the mercury column is set into damped harmonic motion. Damping is caused due to friction in the air layer and mercury with the glass wall.
Question 16.
Alcohol in a U-tube is executing S.H.M. of time period T. Now alcohol is replaced by water up to the same height in U-tube. What will be the effect on the time period?
As the time period (T) of oscillation of a liquid column in a U-tube does not depend upon the density of the liquid, so T is independent of the nature of liquid in a U-tube, hence there will be no effect on the time-period of replacing alcohol by water.
Question 17.
A spring having a force constant k and a mass m is suspended. The spring is cut into three halves and the \$ame mass is suspended from one of the halves. Is the frequency of vibration the same before and after the spring is cut? Give reason.
No. This can be explained as follows:
The spring constant k = $$\frac{F}{x}$$, when the spring is cut inito three parts, the hew force constant will be k’ = $$\frac{\mathrm{F}}{\frac{\mathrm{x}}{3}}=\frac{3 \mathrm{~F}}{\mathrm{x}}$$ = 3k.
If v = frequency of oscillation when the spring was not divided,
Then
i.e. frequency of oscillation after cutting the spring into 3 parts is $$\sqrt{3}$$ times the frequency of oscillation before cutting. It is now clear to us that this is due to the reason that v ∝ $$\sqrt{k}$$ and k increases on cutting the spring, so frequency increases.
Question 18.
Oscillatory motion is periodic but the reverse is not always true. Justify.
The motion of planets and comets like Hailey’s comet, satellites, merry-go-round, etc. all are periodic but not oscillatory as these are not to and fro motions about a mean position. But the motion of the simple pendulum is not only oscillatory but periodic too as it is also to and fro and repeats itself at regular intervals of time.
Question 19.
Why a restoring force is a must for S.H.M.?
A restoring force is a force that restores or tends to restore or bring back a body or particle to its equilibrium position. When a body executing S.H.M. crosses its mean position due to its kinetic energy, the restoring force starts acting towards the equilibrium position and brings the oscillator towards the mean position from the extreme position of motion. The same thing happens when the body goes to the other side of the equilibrium position. So, a restoring force is a must for any oscillatory motion including S.H.M.
Question 20.
What is a periodic function, explain?
A mathematical function that repeats itself after a definite period is called a periodic function, e.g. f (t) = f (t + T), here T = period of the function.
Let F(t) fie a periodic function such that F(t) = A sin $$\frac{2π}{T}$$ t or F(t) = A cos $$\frac{2π}{T}$$ t which are periodic functions. If t is replaced by t + T, we get once again the same value of the function. Thus
Hence f(t) is a periodic function having a time period T.
Question 21.
Distinguish between forced vibrations and maintained vibrations.
A body vibrating under the effect of an external applied periodic force and oscillating with a frequency other than its natural frequency is said to be performing forced vibrations.
If energy is supplied to the oscillator at the same rate at which it is dissipated, the amplitude of the oscillator remains unchanged. Such oscillations are called maintained oscillations e.g. The oscillations of the pendulum clock of a watch or the balance wheel of a watch are examples of maintained oscillations.
Question 22.
Will a pendulum go slower or faster at
(a) Moon,
(b) planet Jupiter?
The time period of oscillation of a pendulum on Earth is given by
Te = 2π$$\sqrt{\frac{1}{g_{e}}}$$
(a) On moon gm = $$\frac{1}{6}$$ of the Earth = $$\frac{\mathrm{g}_{\mathrm{c}}}{6}$$.Since gm < ge, so Tm > Te. Thus due to increase in time period of oscillation of pendulum, it will go slower on the moon.
(b) The value of acceleration due to gravity gj on Jupiter is more than ge, so the time period TJ of oscillation of the pendulum will be less than what it is on the earth. Thus due to a decrease in the time period of oscillation of a pendulum, it will go faster on Jupiter.
Question 23.
The mass M attached to a spring oscillates with a period of 2s. If the mass is increased by 2 kg, the period increases by 1 s. Find the initial mass m assuming that Hooke’s law is obeyed.
Let the initial mass and time period be M and T respectively. If the Hook’s law is obeyed, then the oscillations of the spring will be simple harmonic having time period T given by
T = 2π$$\sqrt{\frac{M}{k}}$$
Given T = 2s
∴ 2 = 2π$$\sqrt{\frac{M}{k}}$$; k = spring constant ….(1)
On increasing the mass by 2 kg, T’ = 2 + 1 = 3s.
∴ 3 = 2π$$\sqrt{\frac{M+2}{k}}$$
Squaring and dividing equation (2) by (1), we have
Question 24.
The displacement of a particle having periodic motion is given by
y = r sin ωt
Show that the motion of the particle is simple harmonic for small values of v.
Since y = r sin(ωt – Φ) ….(1)
∴ $$\frac{dy}{dt}$$ = rω cos(ωt – Φ) …(2)
and $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}$$ = – rω2 sin(ωt – Φ)
= – ω2 y …(3) [by (i)]
where,a = $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}$$ is called acceleration.
ω = angular frequency
∴ a = – ω2y
Now as a ∝ y and acts towards mean position. This is the characteristic of S.H.M. Hence the given equation y = r sin (ωt – Φ) represents an S.H.M. of the oscillator.
Question 25.
We have a very strong string and of sufficient length making a pendulum with a metal ball (e.g. a shot put). What will be the effect on the time period if
(a) the length of the pendulum is doubled and
If the length of the pendulum is doubled, then
T = 2π$$\sqrt{\frac{2l}{g}}$$
= 2π$$\sqrt{\frac{l}{g}}$$ $$\sqrt{2}$$ = $$\sqrt{2}$$T.
The period of oscillation will increase.
(b) the ball is replaced by an elephant?
Since mass does not come in relation to the time period of oscillation of a pendulum therefore instead of shot put if we make a pendulum of some length with an elephant there will be no effect on the time period of its oscillation.
Question 26.
The angular frequency of damped harmonic motion is given by ω = $$\sqrt{\frac{k}{m}-\left(\frac{b}{2 m}\right)^{2}}$$ where b is know as damping constant. The displacement in such a motion is given by x = A e-bt/2m cos (ωt + Φ) and the retarding force is F = – b v, where v is the speed of the particle. Could you guess from the given equations?
(a) How the amplitude of vibration change?
The amplitude of vibration decreases continuously due to damping force. The amplitude decreases exponentially as is evident from the displacement equation the damping force is velocity-dependent.
(b) Does the time period of vibration change with displacement?
The time period of oscillation does not change in damped oscillations.
Question 27.
Apart from the danger of being burnt which other factors make the re-entry of a space vehicle difficult in Earth’s atmosphere?
The Earth’s thick atmospheric layer offers buoyant force and space vehicles suffer from damped harmonic oscillations, making the re-entry difficult.
Question 28.
If the amplitude of vibration is not small in the simple pendulum experiment will it not be executing periodic harmonic oscillations? How will the time period be affected?
The simple pendulum will still be executing periodic harmonic oscillation even if the amplitude is large. However, the time period of oscillation will be slightly greater than what it is for small amplitudes
i.e. T > 2π$$\sqrt{\frac{l}{g}}$$ ,
(The derivation of formulae is beyond the scope of this course),
Question 29.
Can you give some examples where an anharmonic oscillator does not possess potential energy?
The thermal radiation enclosed in a black body of fixed dimensions and photons in solids behave like harmonic oscillators where they don’t have potential energy.
Question 30.
A man is standing on a platform executing S.H.M. in the vertical direction and attached to a weighing machine. Will there be any change in his weight?
Depending on the motion of the platform from its mean position, the weight of the man will change when the platform moves down from the mean position to the lowermost point and return back to the mean position, ‘ the acceleration acts vertically upwards and hence the weight of the man j will increase. But when the platform moves up from its mean position to [ the uppermost point and returns back to the mean position, the acceleration in S.H.M. acts downwards, so the weight of the man is decreased.
### Oscillations Important Extra Questions Long Answer Type
Question 1.
(a) What is the effect on the time period of a simple pendulum when:
(i) it is immersed in a liquid of density σ.
Let p = density of the material of the pendulum bob.
σ = density of the liquid in which the pendulum bob is immersed such that p > σ. Let it is made oscillate with a time period T’. The effective value of ‘g’ in the liquid is given by
where T = time period of the simple pendulum when it oscillates in air.
or
$$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\rho}{\rho-\sigma}}$$
∴ ρ > σ
or
ρ – σ > 0
∴ $$\frac{\rho}{\rho-\sigma}$$ > 1
or
$$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}$$ > 1
or
T’ > T
i.e. its time period increases.
(ii) the temperature of wire through which the bob is suspended increased?
When the temperature of the wire with which the bob of the pendulum is suspended increases, the length of the wire increases due to its thermal expansion. As
T ∝ $$\sqrt{l}$$, so T will also increase as shown below:
Let l = original length of the wire
dθ = rise in its temperature
α = coefficient of linear expansion of the material of the wire
If l’ = new length
and dl = increase in its length = l’ – l, then
(Using Binomial expansion)
or increase in Time period = T’ – T = $$\frac{\mathrm{T} \alpha \mathrm{d} \theta}{2}$$
(b) Derive the differential equation of Simple Harmonic Oscillation.
When an oscillator is displaced, say in the x-direction, the restoring force tends to bring it in its equilibrium position. It is well known that restoring force is directly proportional to the displacement, thus
Fα – x or F = – k x(t) .
where, the negative sign indicates! that the force always acts towards equilibrium position and opposes displacement, k is known as the force constant.
According to Newton’s law of motion if x (t) is the displacement then $$\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}$$ is the velocity and $$\frac{\mathrm{d}^{2} \mathrm{x}(\mathrm{t})}{\mathrm{dt}^{2}}$$ is the acceleration of the oscillating particle.
Now Force = mass × acceleration
Equations (A) and (B) are known as the differential equations of simple harmonic oscillator, GO is the angular velocity given by
ω = $$\frac{2π}{T}$$
Also
ω = $$\sqrt{\frac{k}{m}}$$
So, the time period of oscillation of the oscillator is given by
$$\frac{2π}{T}$$ = $$\sqrt{\frac{k}{m}}$$
or
T = 2π$$\sqrt{\frac{m}{k}}$$
(c) Distinguish briefly between natural, forced and , resonant vibrations. What are damped and undamped / vibrations?
Natural vibrations: The vibrations of a body in the absence of any resistive, damping, or external) forces are said to be its natural vibrations. The body once vibrated continues to vibrate with its natùral frequency and undiminished amplitude.
Forced vibrations: A body at rest or vibrating with some frequency changes its frequency of vibration and amplitude when an external force is impressed on it. If the applied force is periodic the body oscillates with a period which is the period of the impressed force.
The vibrations of the body are not natural or free vibrations but are forced vibrations. If the external periodic force is removed, the body executes its natural oscillations after a while.
Resonant vibrations: If a body is set in vibrations such that a periodic force creates vibrations in it and the frequency of such vibrations becomes equal to the natural frequency of vibration of the body (i.e. the frequency of its free vibrations in the absence of the applied periodic force), its amplitude of vibration increases vigorously. The body is said to be in resonance with the second oscillating body or the applied periodic force.
The frequencies of the two oscillators become equal. Resonant oscillations are not only set up in mechanical oscillators but in electrical circuits and atomic oscillators too.
Damped vibrations: If a body vibrates in the presence of a resistive force such as fluid friction or electromagnetic damping force, its amplitude of oscillation decreases, and ultimately the oscillations die out i.e. the body stops oscillating. Such an oscillation is called damped oscillation as shown in the figure. The amplitude of vibration decreases with time.
Undamped Oscillations: The free oscillations of an oscillator in the absence of any resistive or damping force have constant amplitude over time. Such oscillations are called undamped oscillations. Most vibrations in ‘ daily life are damped vibrations because the oscillator experiences one or the other type of force acting on it. So, its vibrations become damped.
Question 2.
(a) Discuss the effect of driving force frequency on the driven frequency. How does friction affect the oscillation of an oscillator?
Driving force is a time-dependent force represented by
F(t) = fo cos 2πvt …..(1)
where v = frequency of driving force. This frequency is different from the natural frequency of the oscillation vn of the oscillator. vn is also called driven frequency. Thus motion depends on driving force.
Thus the motion of the particle is now under the action of: (a) linear restoring force, (b) time dependent (or periodic) force given in equation (1) so that using Newton’s law, we have for the oscillator:
ma(t) = – k x(t) + fo cos2πvt …. (2)
and vn2 = $$\frac{1}{\mathrm{~T}^{2}}=\frac{\mathrm{k}}{4 \pi^{2} \mathrm{~m}}$$ ….(3)
where m = mass of oscillator,
k = force constant,
v = frequency of the driving force,
ω = 2πvt,
vn = natural frequency of the oscillator,
Let the displacement be given by, x = C cos ωt,
where C = constant, .
This shows that the mass m will execute S.H.M. whose frequency is equal to the frequency of the driving force. The amplitude of oscillation too depends on the frequency of the driving force and is independent of time.
If there is a large difference between v and vn, then the amplitude of oscillation will be too small. But if vn ~ v, the amplitude of oscillation will become infinite, which does not actually happen as some sort of unaccounted force breaks the oscillations down. But if v – vn, the amplitude becomes very large. This phenomenon is called resonance. There are several examples of resonance in daily life such as playing musical instruments, marching of troops in steps, etc.
Effect of the force of friction on S.H.M.: The real oscillations of almost all oscillators are known as damped oscillations. The energy of the system is continuously dissipated but nevertheless, oscillations remain periodic. This dampens oscillations. It is an additional restoring force that is proportional to the velocity of the particle rather than its displacement i.e. F = – bv(t), where b is a positive constant called damping constant. The oscillator is now under the combined action of F(t) and f(t) so that
ma(t) = – k x(t) – bv(t) …. (5)
The solution of this equation gives
(b) Derive an expression for
(1) angular velocity
(2) time period,
(3) frequency of a particle executing S.H.M. in terms of spring factor and inertia factor.
Let m = mass of a particle executing S.H.M. The restoring force F is directly proportional to the displacement and acts opposite to it so that
F ∝ – x
or
F = – k x
where k is called the force constant. If the mass m is attached to an elastic spring, then k is called the spring constant for a given spring. From Newton’s second law of motion,
F = ma
Therefore F = ma = – kx
or
a = – $$\frac{k}{m}$$x
⇒ a ∝ – x
Hence the motion is S.H.M., we can write
a = – ω2 x
where ω2 = $$\frac{k}{m}$$
or
ω = $$\left(\frac{k}{m}\right)^{\frac{1}{2}}$$
This gives ω in terms of spring and constant mass factor.
we know that ω = $$\frac{2π}{T}$$
⇒ T = $$\frac{2π}{ω}$$
Substituting the value of ω from above we have
T = 2π$$\sqrt{\frac{m}{k}}$$
This is the expression for time period.
The frequency v = $$\frac{1}{T}$$ = $$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$$
If the body executing S.H.M. has angular motion, then mass is replaced by the moment of inertia L and the force constant k is replaced by C, the restoring couple per unit twist, hence ω, T and v become
ω = $$\sqrt{\frac{C}{I}}$$
T = 2π$$\sqrt{\frac{C}{I}}$$
and v = $$\frac{1}{2 \pi} \sqrt{\frac{C}{I}}$$
Thus ω, T and v are found in terms of the spring factor and moment of inertia factor.
Question 3.
Calculate the effective spring constant and time period of a parallel and series combination of two different types of springs that are loaded.
(a) When springs are connected in parallel with weight mg, hanging at the lower end.
Both the springs are pulled down through the same displacement say y.
Let F1 and F2 be the restoring forces acting on the spring.
Then F1 = – k1y
and F2 = – k2y
If F = total restoring force, then
F = F1 + F2
= – (k1 + k2)y ….(1)
Let k = spring constant of the combination, then
F = – ky ….(2)
∴ From (1) and (2), we get
k = k1 + k2
The motion of the weight will be simple harmonic in nature. Its time period is given by
T = 2π$$\sqrt{\frac{m}{k}}$$
= 2π$$\sqrt{\frac{\mathrm{m}}{\mathrm{k}_{1}+\mathrm{k}_{2}}}$$
(b) When the springs are connected in series, the springs suffer different displacement y1 and y2 when the weight mg is pulled down. But the restoring force is the same in each spring.
∴ F = – k1y1
and F = -k2y2
If k be the spring constant of the combination, then
F = -ky …..(4)
∴ From (3) and (4), we get
k = $$\frac{\mathbf{k}_{1} \mathbf{k}_{2}}{\mathbf{k}_{1}+\mathbf{k}_{2}}$$ …(5)
If ‘a’ be the acceleration produced in the body of mass m, then
a = $$\frac{\mathrm{F}}{\mathrm{m}}=\frac{-\mathrm{k}}{\mathrm{m}}$$y ⇒ motion of S.H.M. having time period T given by
(c) The weight ‘mg’ is connected in between the two springs which themselves are connected in series.
When the weight is pulled to one side, one spring gets compressed and the other is extended by the same amount. The restoring force acts in the same direction due to both the springs.
Let y = extension or compression produced in the springs,
F1, F2 = restoring forces produced in the springs, then
F1 = – k1y
F2 = – k2y
If F = F1 + F2 = – (k1 + k2)y
If k = spring constant of the combination, then
F = – ky;
∴ k = k1 + k2
a = $$\frac{F}{m}$$ = – $$\frac{\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right)}{\mathrm{m}}$$y
and T = 2π$$\sqrt{\frac{M}{k}}$$ = 2π$$\sqrt{\frac{M}{k_{1}+k_{2}}}$$
Question 4.
(a) Obtain the expression for a time period of a simple pendulum.
(b) Calculate the total energy of a particle executing S.H.M.
(a) Let l = length of the string of the pendulum.
mg = weight of the bob acting vertically downward.
θ = angle by which the bob be displaced from the vertical position.
O = equilibrium position of the bob.
At any position P, the various forces acting on the bobs are:
1. mg, the weight of the bob acting in a vertically downward direction.
2. Tension (T) in the string acting along the string towards the support from which the pendulum is suspended.
mg is resolved into two rectangular components.
1. mg cos θ opposite to T.
2. mg sin θ perpendicular to T and directed towards the mean position O.
T – mg cos θ provides the centripetal force = $$\frac{\mathrm{mv}^{2}}{l}$$ to the bob to move it in the circular path of radius l.
∴ T – mg cos θ = $$\frac{\mathrm{mv}^{2}}{l}$$
At extreme position P, bob is momentarily at rest i.e. v = 0
∴ T – mg cos θ = 0
or
T = mg cos θ ….(1)
mg sin θ provides restoring force (F), i.e.
F = – mg sin θ …. (2)
It acts towards the mean position. Here -ve sign shows that F tends to decrease θ. If θ is small, then
sin θ ~ θ
∴ F = – mg θ
Or
$$\frac{F}{m}$$ = – gθ
or
a = – gθ = – g. $$\frac{x}{l}$$ (∵ x = lθ)
where a = acceleration produced in the bob.
Now as a ∝ x and acts towards mean position, so its motion is S.H.M. having timC period (T) given by
T = 2π$$\sqrt{\frac{x}{a}}$$ = 2π$$\sqrt{\frac{l}{g}}$$
(b) A particle executing S.H.M. has two types of energy
1. P.E. due to its displacement from mean position.
2. K.E. due to its velocity.
Let M = mass of the particle executing S.H.M.
r = amplitude of S.H.M.
ω = angular frequency of S.H.M.
y = displacement after a time t. k = force constant of S.H.M.
F = ky, restoring force acting at any position A.
Let the body is displaced from A to B s.t. AB = dy
If dW = work is done on the body when displaced by dy, then
dW = F dy = ky dy
If W = work done when the body is displaced through y, then
This work is stored in the particle in the form of P.E.
i.e. E1 = P.E. = $$\frac{1}{2}$$ky2
KE. Let y = velocity of the particle executing S.H.M. at a displacement y from the mean position. Then
v = ω$$\sqrt{r^{2}-y^{2}}$$
Thus the total energy in S.H.M. remains constant.
Numerical Problems:
Question 1.
A particle executing S.H.M. along a straight line has a velocity of 4 ms-1 at a distance of 3 m from its mean position and 3 ms-1 at a distance of 4 m from it. Determine its angular speed and time period of oscillation.
Here, y1 = 3 m
v1 = 4 ms-1
y2 = 4 m
v2 = 3 ms-1
ω =?
We know that the velocity of a particle executing S.H.M. is given by
∴ From (ii),
4 = ω $$\sqrt{25-9}$$ = 4ω
∴ ω = 1 rad s-1
∴ T = $$\frac{2π}{ω}$$ = 2π second.
Question 2.
A harmonic oscillation is represented by y = 0.34 cos (3000 t + 0.74) where y and t are in cm and s respectively. Deduce
(i) amplitude,
(ii) frequency and angular frequency,
(iii) time period,
(iv) initial phase.
The given equation is y 0.34 cos (3000 t + 0.74) …. (1)
Comparing eqn. (i) with the standard equation of displacement of harmonic oscillation,
y = r cos (ωt + Φo), we get
r = amplitude = 0.34 cm
Question 3.
A particle vibrates simply harmonically with an amplitude of 4 cm.
(a) Locate the position of the point where its speed is half its maximum speed.
(b) At what displacement ¡s P.E. = K.E.?
Here, r = 4 cm
vmax = maximum speed of the particle
m = mass of the particle
(a) let v = speed of the particle when its displacement is y and
v = $$\frac{1}{2}$$vmax …(i)
Let y = required position w.r.t. mean pcsition =?
∴ maximum P.E = $$\frac{1}{2}$$kr2
Also maximum K.E = $$\frac{1}{2}$$mv2max
According to the law of conservation of energy,
(b) Let y1 be the displacement from the mean position at which
P.E.=KE.
Question 4.
A body of mass 12 kg is suspended by a coil spring of natural length 50 cm and force constant 2.0 × 103 Nm-1. What is the stretched length of the spring? 1f the body is pulled down further stretching the spring to a length of 59 cm and then released, what is the frequency of oscillation of the suspended mass? (Neglect the mass of the spring) .
Here, k = 2 × 103 Nm-1
m = 12 kg
L = original length of the spring
= 50 cm = 0.50 m
Let l = elongation produced in the spring when 12 kg mass is suspended.
∴ F = kl = mg
or
l = $$\frac{\mathrm{mg}}{\mathrm{k}}=\frac{12 \times 9.8}{2 \times 10^{3}}$$
= 0.0588 m = 5.88 cm
∴ stretched length of the spring = L + l = 50.00 + 5.88 = 55.88 cm
The time period of a loaded spring does not change even if the spring is further stretched. So when it is stretched to 59 cm, its time period will be given by
T= 2π$$\sqrt{\frac{m}{k}}$$
∴ Frequency of oscillation is given by
Question 5.
A body of mass of 1.0 kg is suspended from a weightless spring having a force constant of 600 Nm-1. Another body of mass 0.5 kg moving vertically upward hits the suspended body with a velocity of 3.0 ms-1 and gets embedded in it. Find the frequency of oscillations and amplitude of motion.
Here, m1 = 1 kg
m2 = 0.5 kg
k = force constant = 600 Nm-1
v1 = 0
v2 = 3 ms-1
As the mass m2 gets embedded after collision; so
m = total mass = inertia factor = m1 + m2 = 1 + 0.5 = 1.5 kg
If v = frequency of oscillation, then
According to the law of conservation of linear momentum, pf = pi.
(m1 + m2)V = m1v1 + m2v2 = m2v2 (∵ v1 = 0)
∴ V = $$\frac{\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}=\frac{0.5 \times 3}{1.5}$$ = 1 ms-1
Let r = amplitude of motion
Also according to the law of conservation of mechanical energy
Question 6.
(a) If the length of a second’s pendulum is increased by 1%, how many beats will it lose in one day?
(b) A pendulum clock normally shows the correct time. On an extremely cold day, its length decreases by 0.2%. Compute the error in time per day.
(a) Time period of a pendulum is
T = 2π$$\sqrt{\frac{l}{g}}$$
For second’s pendulum, T = 2s
∴ 2 = 2π$$\sqrt{\frac{l}{g}}$$ …(i)
Let l’ = new length of the second’s pendulum. When l is increased by 1%, then
l’ = l + l% of l = l +$$\frac{1}{100}$$l
If T’ = new time period, then
(Using Binomial Expansion Theorem)
= 2 + $$\frac{1}{100}$$
No. of oscillations initially produced by the pendulum is
n = no. of seconds in one day/time
= $$\frac{24 \times 60 \times 60}{2}=\frac{86400}{2}$$ …(ii)
No. of oscillations produced by the pendulum after increasing its length.
Number of oscillations lost per day = no. of seconds lose per day
= 216 × 2 = 432 s.
(b) The correct time period of the pendulum clock 2s.
Let l be its correct length
∴ 2 = 2π$$\sqrt{\frac{l}{g}}$$ ….(iii)
Decrease in length = 0.2% of l = $$\frac{2}{100}$$l
If l’ be the new length than
l’ = l – $$\frac{0.2}{100}$$l = l(1 – $$\frac{0.2}{100}$$)
If T’ be the new time period, then
= (2 – $$\frac{0.2}{100}$$) < 2
∴ Clock gains time,
Time gained in 2 seconds = $$\frac{0.2}{100}$$s
∴ Time gained per day = $$\frac{0.2}{100}$$ × $$\frac{1}{2}$$ × 24 × 60 × 60 = 86.4 s.
Question 7.
A simple pendulum is made by attaching a 1 kg bob to a 5 cm copper wire of diameter 0.08 cm and it has a certain period of oscillation. Next, a 10 kg bob is substituted for a 1 kg bob. Calculate the change in period if any. Young’s modulus of Copper is 12.4 × 1010 Nm-2.
Here, F = mg
A = πr²
The original length of the pendulum, l = 5 m
Case I: m = 1 kg, l = 5 m, r = $$\frac{0.08}{2}$$ cm = 0.04 cm = 4 × 10-4 m, Y = 12.4 × 1010 Nm-2
∴ Δl1 = $$\frac{1 \times 9.8 \times 5}{3.14 \times(0.04)^{2} \times 12.4 \times 10^{10}}$$
= 0.0787 × 10-2 m
= 0.00079 m.
∴ Total length of the pendulum,
l1 = l + Δl1 = 5 m + 0.00079 m
= 5.00079 m.
If T1 be the initial time period of the pendulum, then
T1 = 2π$$\sqrt{\frac{l_{1}}{g}}$$
= 2 × $$\frac{22}{7} \sqrt{\frac{5.00079}{9.8}}$$
= 6.28 × 0.714 = 4.486 s
CaseII: M = 10kg, l = 5 m, r = 0.04 × 10-4m, Y = 12.4 × 1010 Nm-2
∴ Δl2 = $$\frac{10 \times 5 \times 9.8}{3.14 \times 16 \times 10^{-8} \times 12.4 \times 10^{10}}$$
= 0.0079 m
∴ l2 = l + Δl2 = 5 + 0.0079 = 5.0079 m
If T2 be the new time period of the pendulum, then
T2 = 2π$$\sqrt{\frac{l_{2}}{g}}$$
= 2 × 3.14$$\sqrt{\frac{5.00079}{9.8}}$$
= 4.489 s.
∴ Increase in time = T2 – T1 = 4.489 – 4.486 = 0.003s.
Question 8.
A second’s pendulum is taken in a carriage. Find the period of oscillation when the carriage moves with an acceleration of 4 ms-2.
(i) Vertically upwards,
(ii) Vertically downwards,
(iii) in a horizontal direction.
Here, T = 2s
using T = 2π$$\sqrt{\frac{l}{g}}$$, we get
∴ 2 = 2π$$\sqrt{\frac{l}{g}}$$
or
1 = 2π$$\frac{l}{g}$$
or
l = $$\frac{\mathrm{g}}{\pi^{2}}=\frac{9.8}{\pi^{2}}$$ …(i)
(i) When carriage moves up, a = 4 ms-2
∴ If T1 be the time period, then
(ii) When the carriage moves down with a = 4 ms-2
If T2 be the time period, then
(iii) When the carriage moves horizontally, then g and a are at right angles to each other and hence the net acceleration is
Question 9.
A block is kept on a horizontal table. The table is undergoing S.H.M. of frequency 3 Hz in a horizontal plane. The coefficient of static friction between the block and the table surface is 0.72. Find the maximum amplitude of the table at which the block does not slip on the surface (g = 10 ms-2).
Here, v = 3 Hz.
μ = 0.72
g = 10 ms-2
Maximum acceleration of the block is
amax = rω2
where r = amplitude of the table
ω = its angular frequency = 2πv
= 2π × 3
∴ Maximum force on the block is given by
F = mamax = mrω2
Frictional force on the block,
fs = μmg
The block will not slip on the surface of the table if
F = fs
or
mrω2 = μmg
or
r = $$\frac{\mu g}{\omega^{2}}$$
or
r = $$\frac{0.72 \times 10.0}{(6 \pi)^{2}}$$ = 0.02 m
Question 10.
The pendulum bob has a speed of 3 ms-1 at its lowest position. The pendulum is 0.5 m long. What will be the speed of the bob when the length makes an angle of 60° with the vertical?
Let m = mass of the bob
y = speed of the bob at the lowest position = 3ms-1.
l = length of the pendulum
= 0.5 m = $$\frac{1}{2}$$m
∴ K.E at the lowest position O = $$\frac{1}{2}$$mv2
= $$\frac{1}{2}$$m × 32 = $$\frac{9}{2}$$m …(i)
When the length (i) makes an angle O ( 60°) to the vertical, the bob of the pendulum will have both K.E. and P.E.
If v1 be the velocity of the bob in this position and h= height of the bob with respect to O, then
SC = l cos θ
∴ h = OS – SC = l – l cos θ
= l(l – cos θ)
= l(1 – $$\frac{1}{2}$$)
= $$\frac{l}{2}=\frac{1}{2} \times \frac{1}{2}$$m
= $$\frac{1}{4}$$m.
Total energy of the bob = K.E. + P.E.
= $$\frac{1}{2}$$mv12 + mgh
= $$\frac{1}{2}$$mv12 + m × 9.8 × $$\frac{1}{4}$$ …(iii)
∴ According to the law of conservation of energy,
K.E. at O = Total energy at A
$$\frac{9}{2}$$ m = $$\frac{1}{2}$$mv12 + $$\frac{\mathrm{m} \times 9.8}{4}$$
or
v12 = 9 – $$\frac{9.8}{2}$$ = 9 – 4.9 = 4.1
∴ v1 = 2.025 = 2.03 ms-1.
Question 11.
A person stands on a weighing machine placed on a horizontal platform. The machine reads 60 kg. When the platform executes S.H.M. at a frequency of 2.0 s-1 and amplitude 5.0 cm, what will be the effect on the reading of the weighing machine? Take g = 10 ms-2.
Here, m = 60 kg
v = 2 s-1
r = 5 cm = 0.05 m
ω = 2πv = 2π × 2 = 4π rad-1 s
Here A and B are the extreme positions between which the platform vibrates and O = mean position.
∴ acceleration is maximum at A or B and is given by
amax = rω2 = 0.05 × (4π)2
= 0.05 × 16 × 9.87
= 7.9 ms-2
∴ restoring force, F = mamax
= m × 7.9 N
= 7.9 m N
At point A, both weight and restoring force are directed towards the mean position, so the effective weight is maximum at A i.e. reading will be maximum and is given by
W1 = mg + mamax = m(g + amax)
= 60 (10 + 7.9)
= 60 × 17.9 N
= 1074.0 N
= 107.4 kgf
At point B, the weight and the restoring force are opposite to each other, so the effective weight i.e. reading of the weighing machine is minimum at B and is given by
Question 12.
If earth were a homogeneous sphere and a square hole was bored in it through its center, show that a body dropped in the hole will execute S.H.M. and calculate its period if the radius of earth = 6400 km.
Let R = radius of earth 6400 × 103 m
O = center of the earth
Let AB be a tunnel bored along the diameter of the earth and a body be dropped from point A.
Let it reaches point C at a depth d from the surface of the earth.
∴ OC = R — d = distance of the object from the center of earth displacement of the body.
If gd be the acceleration due to gravity at a depth ‘d’ then we know, that
gd = g(1 – $$\frac{d}{R}$$) = g\frac{(R-d)}{R}
If R – d = y = displacement from O, then
gd = $$\frac{g}{R}$$y
Now as the acceleration of the body is proportional to its displacement from the center of earth i.e. O. Hence the motion of the body is S.H.M. and its time period is given by
Question 13.
A horizontal spring block system of mass M executes S.H.M. When the block is passing through its equilibrium position, an object of mass m is put on it and the two move together. Find the new amplitude and frequency of vibration.
The frequency of oscillation of spring block system of mass M is given by
v = $$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{M}}}$$ …..(i)
where k = force constant of the spring.
Let r = initial amplitude of oscillation.
Vo = velocity of the system in equilibrium position,
according to the law of conservation of energy,
∴ $$\frac{1}{2}$$MV02 = $$\frac{1}{2}$$kr2
or
Vo = $$\sqrt{\frac{k}{M}}$$r ….(ii)
When mass m is put on the system, then total mass = M + m.
If V = velocity of the combination in equilibrium, then according to the law of conservation of linear momentum,
MVo = (M + m)V
or
V = $$\frac{\mathrm{MV}_{0}}{\mathrm{M}+\mathrm{m}}$$ ….(iii)
Let r1 be the new amplitude, then
If v’ be the new frequency of oscillation, then
v’ = $$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{M}+\mathrm{m}}}$$
Question 14.
A tray of mass 12 kg is supported by two parallel vertical identical springs. When the tray is pressed down slightly and released, it executes S.H.M. with a time period of 1.5 s. What is the force constant of each spring? When a block of mass m is placed in the tray, the period of S.H.M. changes to 3 s. What is the mass of the block?
Let k be the spring constant of each of the identical springs connected in parallel.
M = mass of tray = 12 kg
T1 = time period of oscillation of the tray = 1.5 s
Let T2 = time period of oscillation of the tray + block of mass ‘m’
∴ T2 = 3 s
m = ?
k =?
If k’ = spring constant of the combination, then
k’ = k + k = 2k (a)
Dividing (ii) and (i), we get
Question 15.
Two pendulums of lengths 100 cm and 110.25 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
Here, l1 = 100 cm
l2 = 110.25 cm
using T = 2π$$\sqrt{\frac{l}{g}}$$
For smaller pendulum, T1 = 2π$$\sqrt{\frac{100}{g}}$$ …(i)
For larger pendulum, T2 = 2π$$\sqrt{\frac{110.25}{g}}$$ …(ii)
where T1 and T2 are the time periods of the two pendulums.
Let these pendulums oscillate in phase again if the larger pendulum completes n oscilLations, It means the smaller pendulum must complete (n + 1) oscillations.
nT2 = (n + 1)T1
Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.
Question 16.
Springs of spring constants k, 2k, 4k., 8k, are connected in series. A mass m kg ¡s attached to the lowér end to the last spring and the system is allowed to vibrate. Calculate the approximate time period of vibration.
The time period of vibration of series loaded spring is given by
T = 2π$$\sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{s}}}}$$ …(1)
Where k is the force constant of the series combination of springs.
In a series combination of springs, the combined spring constant k is given by
or
ks = $$\frac{k}{2}$$
m mass attached to the lower end of the spring.
∴ From (1) and (2), we get
T = 2π$$\sqrt{\frac{2 m}{k}}$$
Question 17.
A point particle of mass 0.1 kg is executing S.H.M. of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10-2 J. Obtain the equation of motion of this article if the initial phase of oscillation is 45°
Since the kinetic energy is maximum at the mean position and is equal to
The standard equation of motion of a particle executing S.H.M. is
y = A sin(ωt + Φ2)
or
x = A cos(ωt + Φ2)
Given r = 0.1 m, ω = 4 rad s-1 and Φ2 = 45° = $$\frac{π}{4}$$
∴ The required cquation of motion is
y = 0.1 sin (4t + $$\frac{π}{4}$$)
or
x = 0.1 cos (4t + $$\frac{π}{4}$$).
Question 18.
A cubical body (side 0.1 m and mass 0.002 kg) floats in water. It is pressed and then released so that it oscillates vertically. Show that the motion of the body is simple harmonic and And the time period.
Here, m = mass of body = 0.002 kg
L = side of body = 0.1 m
In the absence of any given indication it is assumed that the body floats (just sunk), so
The mass of water displaced = volume × density of water
= V ρ = (0.1 )3 ρ.
The upward thrust exerted by water displaced
i.e. buoyant force = Vρg=(0.1 )3 × 103 × 9.8 (∴ ρ ofwater = 103 kg m3)
W1 = Downward weight = mg
Net downward force, F = (0.002 g -Vρg)
or
F = [0.002g – (1)3 × 103g
= (0.002 – 1)g
= – 0.998 × 9.8 N …. (1)
Suppose the block is depressed by y,
hence restoring force F1 = – (0.1 + y)k
When the body floats on the water the force acting is F = 0.1 k, where k is the force constant
The net downward force
F – F’ = 0.1 k – (0.1 + y)k
= – yk ….(2)
Thus force ∝ (- y)
Force = ma, where a = acceleration of the body.
∴ ma = – y k
or a = – $$\frac{k}{m}$$ y
m
or a ∝ -y …. (3)
Hence the motion-ofthe block is S.H.M.
The time period of oscillation is given by
T = 2π$$\frac{m}{k}$$ …(4)
From eqn. (1), k = $$\frac{\mathrm{F}}{\mathrm{L}}=\frac{0.998 \times 9.8}{0.1}$$Nm-1 …(5)
m = 0.002 kg
∴ T = 2π$$\sqrt{\frac{0.002 \mathrm{~kg} \times 0.1}{0.998 \times 9.8 \mathrm{Nm}_{1}}}$$
= 0.0284 s.
Question 19.
Two simple harmonic motions are represented by the following equations:
y1 = 10 sin $$\frac{π}{4}$$ (12t + 1); y2 = 5 (sin 3πt + $$\sqrt{3}$$ cos 3πt).
(a) Find out the ratio of their amplitudes.
(b) What are the time periods of two motions?
(c) Also find the phase difference between two motions.
The two given displacements may be written as
(a) The amplitudes of the two S.H.M. are A1 = 10 units. and A2 = 10 units
Hence $$\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{10}{10}$$ = 1
(b) Comparing the given equation with the normal sine function of S.H.M.
i.e. y = A sin($$\frac{2πt}{T}$$ + Φ) …(3)
Here we get $$\frac{2πt}{T}$$ = 3πt
or
T = $$\frac{2}{3}$$ s
i.e. Time period for two motions is same i.e. T = 0.67 s.
(c) Phase difference between two motions:
Phase (Φ1) of motion represented by y1 is
Φ1 = 3πt + $$\frac{π}{4}$$ …(4)
Phase (Φ2) of motion represented by y2 is
Φ2 = 3πt + $$\frac{π}{3}$$ …(5)
∴ Phase difference is given by
dΦ = Φ2 – Φ1 = (3πt + $$\frac{π}{3}$$) – (3πt + $$\frac{π}{4}$$)
= (4 – 3)$$\frac{π}{12}$$
= $$\frac{π}{12}$$ = 15°
Question 20.
Two particles execute- S.H.M. of the same amplitude at time period along the same straight line. They cross each other in opposite directions at half the amplitude. What is the phase difference between them?
Let the displacement of one particle executing S.H.M. be
y1 = A sin ωt …. (1)
when y1 = $$\frac{A}{2}$$
From equation (1), we get.
$$\frac{A}{2}$$ = A sin ωt
or
sin ωt = $$\frac{1}{2}$$ = sin 30°
∴ ωt = 30° = $$\frac{π}{6}$$ radian.
If Φ1 be the phase of first particle then Φ = $$\frac{π}{6}$$ radian. Also let the displacement of second particle executing S.H.M. be
y1 = r sin (π – ωt) = r sin Φ2 ….(2)
where Φ2 = π – ωt = π – $$\frac{π}{6}$$ = $$\frac{5}{6}$$π radian
If dΦ be the phase difference between the two particle when they cross each other, then
dΦ = Φ2 – Φ1
= $$\frac{5}{6}$$π – $$\frac{π}{6}$$ = $$\frac{4π}{6}$$ = $$\frac{2}{3}$$π radian
= 120°.
Question 21.
A spring compressed by 10 cm develops a restoring force of 5N. A body of mass of 2 kg is placed on it. Find
(i) Force constant,
(ii) depression of a spring,
(iii) time period and
(iv) the frequency of oscillations, if the body is disturbed.
Here, F = 5N
x = 10 cm = 0.1 m
m = 2 kg
(i) k = force constant = $$\frac{F}{x}=\frac{5 N}{0.1}$$ = 50 Nm-1.
(ii) depression, l = $$\frac{\mathrm{mg}}{\mathrm{k}}=\frac{2 \times 9.8}{50}$$ = 0.392 m
(iii) T = 2π$$\sqrt{\frac{m}{k}}$$ = 2π$$\sqrt{\frac{2}{50}}$$ = $$\sqrt{\frac{2π}{5}}$$s
(iv) v = $$\frac{1}{\mathrm{~T}}=\frac{5}{2 \pi}$$ Hz
Question 22.
The S.H.M. of a particle is given by the equation y = 3 sin cot + 4 cos cot. Find its amplitude.
The equation y = 3 sin ωt + 4 cos ωt is due to the superposition of two waves,
y1 = 3 sin ωt
and
y2 = 4 cos ωt = 4 sin(ωt + $$\frac{π}{2}$$)
If r1 and r2 be the amplitudes of the two w’aves respectively, then r1= 3 units and r2 = 4 units.
Φ = phase difference between them = $$\frac{π}{2}$$
If r be the amplitude of the resultant wave, then
Question 23.
Two masses m1 = 1.0 kg and m2 = 0.5 kg arc suspended together by a massless spring of spring constant k as shown in the figure. When masses are in equilibrium, m, is removed without disturbing the system. Calculate the amplitude of oscillation and angular frequency of m2. (g = 10 ms-2 and k = 12.5 Nm-1).
Let y = extension in length of the spring when both m1 and m2 are suspended with it, then
(m1 + m2)g = ky
or
y = $$\frac{\left(m_{1}+m_{2}\right) g}{k}$$ …(1)
Let y’ extension in the length of the spring when only m is suspended.
∴ m2g = ky’
or
y’ = $$\frac{\mathrm{m}_{2} \mathrm{~g}}{\mathrm{k}}$$ …(2)
(1) – (2) gives y – y’ = $$\frac{\left(m_{1}+m_{2}\right) g}{k}-\frac{m_{2} g}{k}=\frac{m_{1} g}{k}$$
This represents the amplitude of oscillation (r)
i.e. r = y – y’ = $$\frac{\mathrm{m}_{1} \mathrm{~g}}{\mathrm{k}}$$ …(3)
Here, m1 = 1.0 kg
k = 12.5 Nm-1
g = 10 ms-2
Putting these values in equation (3), we get
r = $$\frac{1 \times 10}{12.5}$$ = 0.8 m
If ω be the angular frequency of m2, then
ω = $$\sqrt{\frac{\mathrm{k}}{\mathrm{m}_{2}}}=\sqrt{\frac{12.5}{0.5}}$$ (∵ m2 = 0.5 kg)
ω = 5 s-1.
Question 24.
Prove that T = 84.6 minutes for a simple pendulum of infinite length.
T = 2π$$\sqrt{\frac{l}{g}}$$ …(1)
eqn. (1) hold good till l << R (radius of the earth).
When l is large, the curvature of the earth had also to be taken into account. In that case,
Question 25.
For damped oscillator, m = 200 gm, k = 90 Nm b = 40 gs-1.
Calculate (a) time period of oscillation,
(b) time is taken for its amplitude of vibration to become $$\frac{l}{2}$$ of its initial value.
(c) time is taken for its mechanical energy to drop to half of its initial value.
Here, m = 200 g = 0.200 kg
k = 90 Nm-1
b = 40gs-1.
(a) $$\sqrt{km}$$ = $$\sqrt{90 \times 0.2}=\sqrt{18}$$ = 4.243 kg s-1
b = 0.04 kg s-1
∴ b << $$\sqrt{km}$$
(b) Let T1/2 = time taken for the amplitude to drop to half of its initial value.
(c) Let T’1/2 be the time for its mechanical energy to drop to half of its initial value,
∴ Using equation,
which is just half of T1/2. (i.e. decay period for amplitude).
Value-Based Type:
Question 1.
A sports teacher was training the children to march- past On their way they come across a bridge. Then the physical education teacher stopped the children from marching on the bridge.
(a) Comment upon the values of sports teachers.
The sports teacher is responsible, cares not only for public property but also for children.
(b) Also explain what is meant by Resonance.
When the frequency of marching coincides with the natural frequency of oscillation of the bridge then the bridge oscillates with maximum amplitude to such an extent that the bridge may even collapse. The condition is called “Resonance”.
Question 2.
The Physics teacher of class XI has assigned the work finding the resultant spring constant when two springs of spring constants K1, K2 are joined in series. Two students Sabita and Shirin. Sabita made a theoretical study as well as verified it experimentally. Whereas Shirin could not complete the work. When th~ teacher enquired the next day Sabita could give the answer. Whereas Shirin could not.
(a) Comment upon the qualities of Sabita.
Sabita is Sincere and hardworking and has a scientific temper.
(b) Two springs are joined in series and connected to a mass m. If spring constants are Kj and K2, Calculate the period of oscillation of mass m.
Since K1 and K2 are joined in series.
Question 3.
Adarsh a student of class XI has found the factors on which the time period of oscillation of a pendulum depends and arrived at the expression T – (constant) × (l/g)1/2. He wants to know how the length of the pendulum gets affected on the surface of the moon for the same pendulum and arrived at the conclusion that it is 1/6.
(a) What values does Adarsh possesses?
Adarsh is hardworking, thinks logically, having a scientific temper, able to find solutions with patience.
(b) The length of a second pendulum on the surface of the moon?
Since l is proportional to ‘g’ the pendulum on the surface of the moon will be 1/6m.
Question 4.
In a physics class, the teacher told that everyone has to finish the homework and assignment within a weak. But only five students including Dinesh did their homework on time. A few questions from the assignment are as under:
(i) Derive the expression for velocity in S.H.M. | 18,005 | 63,135 | {"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} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.869424 |
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# Nozzle Movement Calculation
MODULE - 3
Contents
Objective
Various Equipments
Distillation Columns Information required to calculate nozzle growth Manual calculation for skirt growth Exchangers Stacked Heat Exchangers
Centrifugal Pumps
Compressors
Contents
Reboilers
Storage Tanks Air-coolers
Objective
## Basis of nozzle movement
Generally all the displacements are calculated at shellnozzle junction for all equipments unless otherwise specified in project specification. Temperatures for calculating nozzle growth are as per the respective line temperatures taken from line-list.
Columns
Following informations are required for nozzle movement calculation for columns. skirt height, skirt thickness,type of material and type of insulation. Shell diameter, nozzle projection(In case displacement is to be calculated at face of nozzle flange).
Temperature profile (To be given by process engineer).Incase profile is not given it may be constructed using line-list temperatures Based on temperature of outlet nozzles of tower (as per Fluor standard) Based on temperature of all the nozzles which can affect temperature profile of vessel.
Column
- Inlet lines having size greater than 3 only need to be considered. - Foundation Settlement Spread Sheet For Column
Columns
Skirt growth calculation
Determine
the value of :
kh t
Using
## chart no.1 get the value of F T using:
T T - Ta T ' F ( T) T ' T ' Ta
Calculate
T=
T=
## Average temp. of the skirt
Ambient temperature ( 70 Deg F )
Ta=
Columns
t=
## thickness of skirt in inches
h=height
k=1.0 k=1.7 k=2.7
of skirt in inches
for skirt fully insulated for skirt firebrick insulated for skirt not insulated correction factor
F=Temperature
Using the calculated average temperature determine the growth of the skirt.
Columns
Using
the calculated average temperature of zones determine the growth of the nozzles located in respective zones. intermediate points between known temperatures , use a straight line interpolation to calculate temperature.
For
Exchanger
Exchanger
dimensions, materials,anchor and nozzle locations are taken from vendor drawings. temperatures of shell and channel sections are calculated from inlet and outlet temperatures. movements:
for the nozzle located in shell region, average temperature of shell is taken and for nozzles located in channel section, shell average temperature is taken till shell-channel junction, and after that channel average temperature is taken.
Average
Horizontal
Exchanger
Vertical movement
is calculated using the average temperature of the region where nozzle is located. The growth of saddle is ignored in the calculation of nozzle movement.
Since
the nozzle loads and moments are calculated at the nozzle-shell junction the nozzle movements are also calculated at the same point.
## Stacked Heat Exchanger
First
of all average temperature for shell side and channel side is calculated for both the heat exchangers.
Stacked
HE
For
lower heat exchanger movements are calculated exactly in the same manner as in case of single heat exchanger. upper heat exchanger guidelines differ from project to project.
For
## Spread sheet for stacked heat exchanger(UE-1)
IOL-PRISM-project
For
## calculating Vertical movement:
As per UE-1 Vertical movements for all the nozzles on upper heat-exchanger is same. Movement for nozzle on shell side is calculated and for others same value is superimposed. As per imperial-oil and Baton-Rouge vertical movements for both the nozzles on channel and shell section are calculated separately.
Stacked Heat-Exchangers
For
## calculating Horizontal movement:
As per Ue-1 Horizontal movements for nozzles located on upper heat exchangers are equal to the corresponding nozzles on lower heat-exchanger provided both the nozzles are located at the same distance from anchor. As per Imperial-oil Movement on channel side nozzle is equal to corresponding nozzle on lower heat-exchanger, while for nozzle on shell side back calculation is made.
Centrifugal pumps
Pump
dimensions, configuration and nozzle locations are taken from the vendors drawings.
Temperatures
## are normally taken from operating temperatures on the line-list.
point of zero movement is at the support of the pump casing . This point is usually the intersection of the centerline across the top of the pedestals and the centerline of the pump shaft(for centerline mounted).In some cases(foot mounted) the support point is at the bottom of casing and hence this is zero point.
The
Centrifugal pumps
From
the point of zero growth calculate the thermal movements of the nozzles. pump is modeled as rigid element with piping and point of zero movement is taken from API 610 for different type of arrangements:
Side-suction Side-discharge End-suction Top-discharge Top-suction Top-discharge
Generally,
Compressors
Compressor
dimensions, configuration and nozzle locations are taken from vendor the drawings. are normally taken from operating temperatures shown on line-list. the anchor point from vendors drawing.
Temperatures
Determine
Based
upon nozzle locations, a judgment assumption is made as to which expansions are to be based on an average temperature and which are to be based on the operating temperature of connecting nozzle.
Compressors
However
the final nozzle displacements are to be taken from the vendor drawings or are to be approved by the pump vendor.
Vertical Reboilers
Reboilers
may be supported from the adjacent vessel or from independent structures from grade.Most of the Reboilers supported from the independent structures from grade are placed on spring supports. horizontal reboiler support location depends upon its relationship to the vessel. The support location for the reboiler taken from the vessel is closer to the tangentline of the vessel in elevation.
The
Vertical Reboilers
Nozzle
displacements for the inlet and outlet nozzles to the vessels are not calculated but the reboiler is modeled in stress package together with the vessel.It is recommended that the nozzle displacements for the other nozzles on the shell of reboiler be taken from this model itself by modelling them at their respective location.
## Storage Tanks (API 650)
Vertical
displacements are calculated from the foundation top where the tank rests to the nozzle centerline.Horizontal displacements are calculated from the tank centerline.The average temperature of inlet and outlet line from Line list is considered for the calculation. any foundation settlement values are provided by Civil Department those have to be considered too.
If
Air Coolers
For
air-coolers Header -Box is modeled and temperature to be taken for analysis depends upon configuration. For example:
- Single pass where Inlet is on one side and outlet on the other, respective temperatures to be considered. - Double pass where Inlet and outlet are located on the same side , temperature of the inlet nozzle will be the governing temperature. | 1,421 | 7,024 | {"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} | 2.875 | 3 | CC-MAIN-2019-35 | latest | en | 0.82393 |
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This is part of an ongoing series where we write a complete 2D game engine in C++ and SFML. A new tutorial is released every Monday. You can find the complete list of tutorials here and download the source code from the projects GitHub page.
Last week we started work on the data structure that will store all of the objects that we want to be colliding. The quadtree will store our collidables spatially to help us minimise the number of collision calls we need to make. If you would like a refresher on the theory behind quadtrees then see my previous tutorial here. This week we will finish the quadtree by implementing: search functionality, the ability to split a parent node to create children nodes, and a few other helper functions.
As the quadtree was explained in detail last week, we can jump right into the code. We’ll start by writing the search function. This is the function that our collision system (which we’ll start next week) will call. It returns a vector of objects that intersect the area we have passed in. So let’s say that we pass in the area of our player, it will return any objects that are also in that area, we can then move them out of the player bounds, apply damage to the player, and do whatever else we need to.
You may have noticed that there are two search functions: one that returns a vector (the public function) and one that returns void (private to the quadtree). In reality, there are three main steps to retrieving objects from our quadtree.
Step 1: we pass a search area to the quadtree. This area would normally coincide with the collision box of an existing object i.e. the player. This is a call to the public search function.
The green rectangle represents our search area.
Step 2: the public function will call the private search function. This private function will then add all the objects from any nodes that intersect the search tree to a vector. In this example, it will return a vector of 4 objects.
First the quadtree builds a vector with all of the objects from any node that intersects our search area.
Step 3: back in the public function the quadtree will iterate over all of the objects from these nodes and check which objects actually intersect the search area. Only the objects that intersect the area are returned from the search function. In this simple example the quadtree has reduced the number of collision checks from 11 (needing to check if its colliding with every other object) to 4 (only needing to check if it’s colliding with the objects in the nodes that intersect the search area). In the final game, we may have hundreds of possible collision objects so you can imagine how many collision checks the quadtree will help save each frame.
The quadtree returns these two objects as they intersect the search area.
``````std::vector<std::shared_ptr<C_BoxCollider>> Quadtree::Search(const sf::FloatRect& area)
{
std::vector<std::shared_ptr<C_BoxCollider>> possibleOverlaps; // 1
Search(area, possibleOverlaps); // 2
std::vector<std::shared_ptr<C_BoxCollider>> returnList;
for (auto collider : possibleOverlaps)
{
if (area.intersects(collider->GetCollidable())) // 3
{
returnList.emplace_back(collider); // 4
}
}
return returnList;
}``````
The function performs the following steps:
1. Creates a vector to hold all objects that may be overlapping with our search area.
2. Calls the private search function, passing it the vector as a reference (we’ll write this next). This function fills the vector with all of the objects in the nodes that intersect with our search area.
3. As our area may not necessarily intersect with all of the objects in the returned nodes, we then need to iterate over all of these possible overlaps and performs a rectangle intersection check. We do this check because it is quick, if we require more detailed collision checks we would perform them in our collision system.
4. If the object is intersecting with our search area we add it to the vector of objects that are returned from this function.
This is mostly straightforward as much of the work actually happens in the call to the private search function.
``````void Quadtree::Search(const sf::FloatRect& area,
std::vector<std::shared_ptr<C_BoxCollider>>& overlappingObjects)
{
overlappingObjects.insert(overlappingObjects.end(), objects.begin(), objects.end()); // 1
if(children[0] != nullptr)
{
int index = GetChildIndexForObject(area); // 2
if(index == thisTree) // 3
{
for(int i = 0; i < 4; i++)
{
if(children[i]->GetBounds().intersects(area))
{
children[i]->Search(area, overlappingObjects);
}
}
}
else // 4
{
children[index]->Search(area, overlappingObjects);
}
}
}``````
We’ll go through this method step by step as well:
1. It starts by adding all the objects in this node to the vector. As you’ll remember from the beginning of this tutorial this will contain either all the objects that do not fit neatly into a child node (if there are child nodes) or it will contain all the objects that fit within this node if there are no child nodes.
2. Retrieves the child index for the search area. This function will either return a -1 to signify that the search area does not fit into a child node completely (or there are no child nodes) or it will return an index to the child node that contains the area.
3. If it returns an index to this tree and we have children nodes then we need to call the search functions for any child node that intersects with our area to make sure that we have added all objects that could possibly be intersecting our area.
4. If it returns an index to a child node then we need to call that child nodes search function.
``````const sf::FloatRect& Quadtree::GetBounds() const
{
return bounds;
}``````
Not much needs to be said about that. It simply returns the rect for this nodes bounds. We use this when deciding if a search area intersects the node and also for deciding if an object fits within a node.
We only have two more functions to write before our quadtree is complete: GetChildIndexForObject and Split. Lets start with GetChildIndexForObject. We’ve called this function a couple of times in our code: when we’re working out where to place an object, and when we’re trying to find an object and we want to know which node to search. Its job is to accept an area and then to calculate if that area can be contained completely within a child node, if so it will return an index to that child node, or if the area overlaps a number of child nodes it will then return this index of the current node instead.
``````int Quadtree::GetChildIndexForObject(const sf::FloatRect& objectBounds)
{
int index = -1;
double verticalDividingLine = bounds.left + bounds.width * 0.5f;
double horizontalDividingLine = bounds.top + bounds.height * 0.5f;
bool north = objectBounds.top < horizontalDividingLine
&& (objectBounds.height + objectBounds.top < horizontalDividingLine);
bool south = objectBounds.top > horizontalDividingLine;
bool west = objectBounds.left < verticalDividingLine
&& (objectBounds.left + objectBounds.width < verticalDividingLine);
bool east = objectBounds.left > verticalDividingLine;
if(east)
{
if(north)
{
index = childNE;
}
else if(south)
{
index = childSE;
}
}
else if(west)
{
if(north)
{
index = childNW;
}
else if(south)
{
index = childSW;
}
}
return index;
}``````
It performs four checks to see if the provided area is totally contained within the north, south, east, or west of this nodes bounds. It then uses those checks to see if the area will fit within a child nodes bounds.
The last function we are going to write today is the one that will ‘split’ this node into four child nodes. We still keep the parent node (which will be used to hold any objects that do not fit entirely within a child nodes bounds).
``````void Quadtree::Split()
{
const int childWidth = bounds.width / 2;
const int childHeight = bounds.height / 2;
children[childNE] = std::make_shared<Quadtree>(maxObjects, maxLevels, level + 1,
sf::FloatRect(bounds.left + childWidth, bounds.top, childWidth, childHeight),
this);
children[childNW] = std::make_shared<Quadtree>(maxObjects, maxLevels, level + 1,
sf::FloatRect(bounds.left, bounds.top, childWidth, childHeight),
this);
children[childSW] = std::make_shared<Quadtree>(maxObjects, maxLevels, level + 1,
sf::FloatRect(bounds.left, bounds.top + childHeight, childWidth, childHeight),
this);
children[childSE] = std::make_shared<Quadtree>(maxObjects, maxLevels, level + 1,
sf::FloatRect(bounds.left + childWidth, bounds.top + childHeight, childWidth, childHeight),
this);
}``````
This instantiates four new quadtree objects and points them to the array using the static indices in the header. The bounds of each child are calculated as a subset of the parent’s bounds as shown below.
One parent node split into four child nodes.
And that’s it for this week, our quadtree is now complete! Next week we’ll be starting work on our new collision system which will make use of the quadtree.
As always, if you have any suggestions for what you would like covered or are having any trouble implementing a feature, then let me know in the comments and I’ll get back to you as soon as I can. Thank you for reading 🙂 | 2,114 | 9,302 | {"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-2020-16 | longest | en | 0.921316 |
http://collectivesolver.com/24144/how-to-rotate-square-matrix-90-degrees-to-the-left-in-c | 1,569,029,802,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574159.19/warc/CC-MAIN-20190921001810-20190921023810-00508.warc.gz | 42,356,994 | 6,798 | # How to rotate square matrix 90 degrees to the left in C
7 views
edited May 19
```#include <stdio.h>
#define LEN 3
void print_matrix(int matrix[][LEN]) {
for (int i = 0; i < LEN; i++) {
for (int j = 0; j < LEN; j++)
printf("%2d ", matrix[i][j]);
printf("\n");
}
printf("\n");
}
void rotate_matrix_90_degrees_left(int matrix[][LEN]) {
for (int i = 0; i < LEN / 2; i++) {
for (int j = i; j < LEN - i - 1; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[j][LEN - 1 - i];
matrix[j][LEN - 1 - i] = matrix[LEN - 1 - i][LEN - 1 - j];
matrix[LEN - 1 - i][LEN - 1 - j] = matrix[LEN - 1 - j][i];
matrix[LEN - 1 - j][i] = tmp;
}
}
}
int main()
{
int matrix[LEN][LEN] =
{
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
rotate_matrix_90_degrees_left(matrix);
print_matrix(matrix);
return 0;
}
/*
run:
3 6 9
2 5 8
1 4 7
*/``` | 329 | 823 | {"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-2019-39 | latest | en | 0.202519 |
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# Output & Costs
### Monopoly
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Between 1995 and 1997, American Airlines competed in the Dallas/Fort Worth Airport against several other low-cost carriers. In response to these low-cost carriers, American Airlines reduced its price and increased service on selected routes. As a result, one of the low-cost carriers stopped service, which led American Airlines t
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Two partners who own Progressive Business Solutions, which currently operates out of an office in a small town near Boston, just discovered a vacancy in an office building in downtown Boston. One of the partners favors moving downtown because she believes the additional business gained by moving downtown will exceed the higher r
### Microeconomics: Dividends
Clinton Co. has just paid a dividend of \$2.50 per share. The dividends are expected to grow at a constant rate of 6% per year for ever. If the stock is currently selling for \$50 per share, calculate the cost of equity for the firm.
### Inverse Market Demand Curve
You are the manager of a firm that has an exclusive license to produce your product. The inverse market demand curve is P = 900-1.5Q. Your cost function is C(Q)=2Q+Q^2. Determine the output you should produce, the price you should charge, and your profits.
### Should I start a new business?
I am exploring the option of starting my own business. In addition to creating a business plan (which I am not quite sure how to do), I am reviewing factors and costs. What should be the 10 most important factors and costs and further, name 3 implicit and 3 explicit costs to consider. I have what I believe to be good options
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### Socially Desirable Method of Taxing Monopolists
It has been proposed that natural monopolists should be allowed to determine their profit maximizing outputs and prices and then government should tax their profits away and distribute them to consumers in proportin to their purchases from the monopoly. Is this proposal as socially desirable as requiring monopolists to equate pr
### Calculating Breakeven Point and Profits
The fixed costs at Harley Motors are \$1 million annually. The main product has revenue of \$8.50 per unit and \$4.25 variable cost. Determine the following. (a) Breakeven quantity per year. (b) Annual profit if 200,000 units are sold and in 350,000 units are sold. Plot total revenue and costs at these leve
### Government - Pollution Rights
In the 1990s the Mobil Oil Corporation acquired the rights to increase their pollution by 900 pounds of sulfur-dioxide per day at their Torrance, California refinery. These rights were purchased from South Gate, California, at a price of \$3 million after the latter acquired them from General Motors. As a result of this change, M
### Pollution Credits
In the 1990s the Mobil Oil Corporation acquired the rights to increase their pollution by 900 pounds of sulfur-dioxide per day at their Torrance, California refinery. These rights were purchased from South Gate, California, at a price of \$3 million after the latter acquired them from General Motors. As a result of this change, M
### External Cost
Suppose the external marginal cost of pollution is MCext = 5Q and the internal marginal cost is MCint = 10Q . Further, assume the inverse demand for the product, Q, is given by P = 90-Q . a. What is the socially efficient level of output? b. How much output would a competitive industry produce? c. How much output would
### Microeconomic Problem: Analyzing Production Functions
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### Production Function
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### Net Benefit Difficulty with Economics
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### Analyzing costs and revenues of a loss making firm
The firm currently uses 50,000 workers to produce 200,000 units of output per day. The daily wage worker per worker is \$80.00, and the price firm is \$25.00. The cost of other variables input is \$400,000 per day. Although you don't know the firm's fixed cost, you know that it is high enough that the firm's total cost exceeds its
### Micro Economics
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### Average vs marginal productivity
What is average productivity? What is marginal productivity? Explain the relationship between marginal and average productivity. What would happen to marginal and average productivity if a technological innovation is introduced to the production process? Provide examples (numerical and graphical if you can) to explain your answe
### Inputs and Outputs
The following production table provides estimates of the maximum amounts of output possible with different combinations of two input factors, X and Y. (Assume these are just illustrative points on a spectrum of continuous input combinations.) Units of Y Used Estimated Output per Day 5 184 265 334 395 440 4 176 248 303 3
### Accounting
Presented below are a number of accounting procedures and practices in Sanchez Corp. For each of these items, list the assumption, principle, information characteristic, or modifying convention that is violated. 1. Because the company's income is low this year, a switch from accelerated amortization to straight-line amortizatio
### Rate of return of miniature fittings and valves
Swagelok Enterprises is a manufacturer of miniature fittings and valves. Over a 5-year period, the costs associated with one product line were as follows: first cost of \$30,000 and annual costs of \$18,000. Annual revenue was \$27,000, and the used equipment was salvaged for \$4000. What rate of return did the company make on this
### Incremental analysis
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### PV of cash flows
A consulting engineering firm is considering two models of SUVs for the company principals. A GM model will have a first cost of \$26,000, an operating cost of \$2000, and a salvage value of \$12,000 after 3 years. A Ford model will have a first cost of \$29,000, an operating cost of \$1200, and a \$15,000 resale value after 3 years.
### calculate the PVIFA
The cost associated with maintaining rural highways follows a predictable pattern. There are usually no costs for the first 3 years, but thereafter maintenance is required for restriping, weed control, light replacement, shoulder repairs, etc. For one section of a particular highway, these costs are projected to be \$6000 in year
### Externalities
1. Which of the following occurs if firms are able to restrict output and raise price? a. resources are misallocated b. wealth is shifted from consumers to government c. wealth is shifted from producers to consumers d. P = MC 2. Production by a monopoly would result in the socially optimal allocation of resources i
### Cost Structure of a Perfectly Competitive Firm
In a perfectly competitive industry, the market price is \$25. A firm is currently producing 10,000 units of output, its average total cost is \$28, its marginal cost is \$20, and its average variable cost is \$20. Given these facts, explain whether the following statements are true or false: a. The firm is currently producing
### Calculating Accounting and Economic Profits
Jane decided to resign from her current job, at which she earned \$58,000 per year. She started her own business using her \$100,000 savings on which she was earning 5 % interest. In the first year, her revenue was \$150,000 and her costs were as follows: Rent: \$25,000 Utilities: \$12,000 Wages: \$30,000 Materials: \$20, | 2,231 | 10,087 | {"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-2018-22 | latest | en | 0.931221 |
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Information about financial, finance, business, accounting, payroll, inventory, investment, money, inventory control, stock trading, financial advisor, tax advisor, credit.
# Definition of Return on Total Assets Ratio
## Return on Total Assets Ratio
A measure of the percentage return earned on the value of the
assets in the company. It is calculated by dividing the net income
available for distribution to shareholders by the book value of all
assets.
# Related Terms:
## CARs (cumulative abnormal returns)
a measure used in academic finance articles to measure the excess returns an investor would have received over a particular time period if he or she were invested in a particular stock.
This is typically used in control and takeover studies, where stockholders are paid a premium for being taken over. Starting some time period before the takeover (often five days before the first announced bid, but sometimes a longer period), the researchers calculate the actual daily stock returns for the target firm and subtract out the expected market returns (usually calculated using the firm’s beta and applying it to overall market movements during the time period under observation).
The excess actual return over the capital asset pricing model-determined expected return market is called an ‘‘abnormal return.’’ The cumulation of the daily abnormal returns over the time period under observation is the CAR. The term CAR(-5, 0) means the CAR calculated from five days before the
announcement to the day of announcement. The CAR(-1, 0) is a control premium, although Mergerstat generally uses the stock price five days before announcement rather than one day before announcement as the denominator in its control premium calculation. However, the CAR for any period other than (-1, 0) is not mathematically equivalent to a control premium.
## Abnormal returns
Part of the return that is not due to systematic influences (market wide influences). In
other words, abnormal returns are above those predicted by the market movement alone. Related: excess
returns.
## Acid-test ratio
Also called the quick ratio, the ratio of current assets minus inventories, accruals, and prepaid
items to current liabilities.
## Acquisition of assets
A merger or consolidation in which an acquirer purchases the selling firm's assets.
## After-tax real rate of return
Money after-tax rate of return minus the inflation rate.
## Annualized holding period return
The annual rate of return that when compounded t times, would have
given the same t-period holding return as actually occurred from period 1 to period t.
## Appraisal ratio
The signal-to-noise ratio of an analyst's forecasts. The ratio of alpha to residual standard
deviation.
## Arithmetic average (mean) rate of return
Arithmetic mean return.
## Arithmetic mean return
An average of the subperiod returns, calculated by summing the subperiod returns
and dividing by he number of subperiods.
## Articles of incorporation
Legal document establishing a corporation and its structure and purpose.
## Asset/equity ratio
The ratio of total assets to stockholder equity.
## Asset activity ratios
ratios that measure how effectively the firm is managing its assets.
## Assets
A firm's productive resources.
## Assets requirements
A common element of a financial plan that describes projected capital spending and the
proposed uses of net working capital.
## Average accounting return
The average project earnings after taxes and depreciation divided by the average
book value of the investment during its life.
## Average rate of return (ARR)
The ratio of the average cash inflow to the amount invested.
## Capital rationing
Placing one or more limits on the amount of new investment undertaken by a firm, either
by using a higher cost of capital, or by setting a maximum on parts of, and/or the entirety of, the capital
budget.
## Capitalization ratios
Also called financial leverage ratios, these ratios compare debt to total capitalization
and thus reflect the extent to which a corporation is trading on its equity. Capitalization ratios can be
interpreted only in the context of the stability of industry and company earnings and cash flow.
## Cash flow coverage ratio
The number of times that financial obligations (for interest, principal payments,
preferred stock dividends, and rental payments) are covered by earnings before interest, taxes, rental
payments, and depreciation.
## Cash flow from operations
A firm's net cash inflow resulting directly from its regular operations
(disregarding extraordinary items such as the sale of fixed assets or transaction costs associated with issuing
securities), calculated as the sum of net income plus non-cash expenses that were deducted in calculating net
income.
## Cash ratio
The proportion of a firm's assets held as cash.
## Common stock ratios
ratios that are designed to measure the relative claims of stockholders to earnings
(cash flow per share), and equity (book value per share) of a firm.
## Concentration account
A single centralized account into which funds collected at regional locations
(lockboxes) are transferred.
## Concentration services
Movement of cash from different lockbox locations into a single concentration
account from which disbursements and investments are made.
## Controlled foreign corporation (CFC)
A foreign corporation whose voting stock is more than 50% owned
by U.S. stockholders, each of whom owns at least 10% of the voting power.
## Conversion ratio
The number of shares of common stock that the security holder will receive from
exercising the call option of a convertible security.
## Corporation
A legal "person" that is separate and distinct from its owners. A corporation is allowed to own
assets, incur liabilities, and sell securities, among other things.
## Cost-benefit ratio
The net present value of an investment divided by the investment's initial cost. Also called
the profitability index.
## Coverage ratios
ratios used to test the adequacy of cash flows generated through earnings for purposes of
meeting debt and lease obligations, including the interest coverage ratio and the fixed charge coverage ratio.
## Cumulative abnormal return (CAR)
Sum of the differences between the expected return on a stock and the
actual return that comes from the release of news to the market.
## Current assets
Value of cash, accounts receivable, inventories, marketable securities and other assets that
could be converted to cash in less than 1 year.
## Current ratio
Indicator of short-term debt paying ability. Determined by dividing current assets by current
liabilities. The higher the ratio, the more liquid the company.
## Customary payout ratios
A range of payout ratios that is typical based on an analysis of comparable firms.
## Days' sales in inventory ratio
The average number of days' worth of sales that is held in inventory.
## Debt/equity ratio
Indicator of financial leverage. Compares assets provided by creditors to assets provided
by shareholders. Determined by dividing long-term debt by common stockholder equity.
## Debt ratio
total debt divided by total assets.
## Debt-service coverage ratio
Earnings before interest and income taxes plus one-third rental charges, divided
by interest expense plus one-third rental charges plus the quantity of principal repayments divided by one
minus the tax rate.
## Declaration date
The date on which a firm's directors meet and announce the date and amount of the next
dividend.
## Dividend payout ratio
Percentage of earnings paid out as dividends.
## Dollar duration
The product of modified duration and the initial price.
## Dollar return
The return realized on a portfolio for any evaluation period, including (1) the change in market
value of the portfolio and (2) any distributions made from the portfolio during that period.
## Dollar-weighted rate of return
Also called the internal rate of return, the interest rate that will make the
present value of the cash flows from all the subperiods in the evaluation period plus the terminal market value
of the portfolio equal to the initial market value of the portfolio.
## Domestic International Sales Corporation (DISC)
A U.S. corporation that receives a tax incentive for
export activities.
## Duration
A common gauge of the price sensitivity of an asset or portfolio to a change in interest rates.
Plowback rate.
## Edge corporations
Specialized banking institutions, authorized and chartered by the Federal Reserve Board
in the U.S., which are allowed to engage in transactions that have a foreign or international character. They
are not subject to any restrictions on interstate banking. Foreign banks operating in the U.S. are permitted to
organize and own and Edge corporation.
## Effective duration
The duration calculated using the approximate duration formula for a bond with an
embedded option, reflecting the expected change in the cash flow caused by the option. Measures the
responsiveness of a bond's price taking into account the expected cash flows will change as interest rates
change due to the embedded option.
## Ex post return
Related: Holding period return
## Exante return
The expected return of a portfolio based on the expected returns of its component assets and
their weights.
## Excess return on the market portfolio
The difference between the return on the market portfolio and the
riskless rate.
## Excess returns
Also called abnormal returns, returns in excess of those required by some asset pricing model.
## Exchange of assets
Acquisition of another company by purchase of its assets in exchange for cash or stock.
## Expected future return
The return that is expected to be earned on an asset in the future. Also called the
expected return.
## Expected return
The return expected on a risky asset based on a probability distribution for the possible rates
of return. Expected return equals some risk free rate (generally the prevailing U.S. Treasury note or bond rate)
plus a risk premium (the difference between the historic market return, based upon a well diversified index
such as the S&P500 and historic U.S. Treasury bond) multiplied by the assets beta.
## Expected return on investment
The return one can expect to earn on an investment. See: capital asset
pricing model.
## Expected return-beta relationship
Implication of the CAPM that security risk premiums will be
proportional to beta.
## Expense ratio
The percentage of the assets that were spent to run a mutual fund (as of the last annual
statement). This includes expenses such as management and advisory fees, overhead costs and 12b-1
(distribution and advertising ) fees. The expense ratio does not include brokerage costs for trading the
portfolio, although these are reported as a percentage of assets to the SEC by the funds in a Statement of
Additional Information (SAI). the SAI is available to shareholders on request. Neither the expense ratio or the
SAI includes the transaction costs of spreads, normally incurred in unlisted securities and foreign stocks.
These two costs can add significantly to the reported expenses of a fund. The expense ratio is often termed an
Operating Expense ratio (OER).
## Expiration
The time when the option contract ceases to exist (expires).
## Expiration cycle
An expiration cycle relates to the dates on which options on a particular security expire. A
given option will be placed in 1 of 3 cycles, the January cycle, the February cycle, or the March cycle. At any
point in time, an option will have contracts with 4 expiration dates outstanding, 2 in near-term months and 2
in far-term months.
## Expiration date
The last day (in the case of American-style) or the only day (in the case of European-style)
on which an option may be exercised. For stock options, this date is the Saturday immediately following the
3rd Friday of the expiration month; however, brokerage firms may set an earlier deadline for notification of
an option holder's intention to exercise. If Friday is a holiday, the last trading day will be the preceding
Thursday.
## Feasible target payout ratios
Payout ratios that are consistent with the availability of excess funds to make
cash dividend payments.
## Federal Deposit Insurance Corporation (FDIC)
A federal institution that insures bank deposits.
## Financial assets
Claims on real assets.
## Financial leverage ratios
Related: capitalization ratios.
## Financial ratio
The result of dividing one financial statement item by another. ratios help analysts interpret
financial statements by focussing on specific relationships.
## Fisher's separation theorem
The firm's choice of investments is separate from its owner's attitudes towards
investments. Also refered to as portfolio separation theorem.
## Fixed asset turnover ratio
The ratio of sales to fixed assets.
## Fixed-charge coverage ratio
A measure of a firm's ability to meet its fixed-charge obligations: the ratio of
(net earnings before taxes plus interest charges paid plus long-term lease payments) to (interest charges paid
plus long-term lease payments).
## Foreign Sales Corporation (FSC)
A special type of corporation created by the Tax Reform Act of 1984 that
is designed to provide a tax incentive for exporting U.S.-produced goods.
## Freddie Mac (Federal Home Loan Mortgage Corporation)
A Congressionally chartered corporation that
purchases residential mortgages in the secondary market from S&Ls, banks, and mortgage bankers and
securitizes these mortgages for sale into the capital markets.
## Funding ratio
The ratio of a pension plan's assets to its liabilities.
## Funds From Operations (FFO)
Used by real estate and other investment trusts to define the cash flow from
trust operations. It is earnings with depreciation and amortization added back. A similar term increasingly
used is Funds Available for Distribution (FAD), which is FFO less capital investments in trust property and
the amortization of mortgages.
## Geometric mean return
Also called the time weighted rate of return, a measure of the compounded rate of
growth of the initial portfolio market value during the evaluation period, assuming that all cash distributions
are reinvested in the portfolio. It is computed by taking the geometric average of the portfolio subperiod
returns.
## Hard capital rationing
Capital rationing that under no circumstances can be violated.
## Hedge ratio (delta)
The ratio of volatility of the portfolio to be hedged and the return of the volatility of the
hedging instrument.
## Holding period return
The rate of return over a given period.
## Horizon return
total return over a given horizon.
## Income statement (statement of operations)
A statement showing the revenues, expenses, and income (the
difference between revenues and expenses) of a corporation over some period of time.
## Incremental internal rate of return
IRR on the incremental investment from choosing a large project
instead of a smaller project.
## Interest coverage ratio
The ratio of the earnings before interest and taxes to the annual interest expense. This
ratio measures a firm's ability to pay interest.
## Internal rate of return
Dollar-weighted rate of return. Discount rate at which net present value (NPV)
investment is zero. The rate at which a bond's future cash flows, discounted back to today, equals its price.
## Irrational call option
The implied call imbedded in the MBS. Identified as irrational because the call is
sometimes not exercised when it is in the money (interest rates are below the threshold to refinance).
Sometimes exercised when not in the money (home sold without regard to the relative level of interest rates).
## Leverage ratios
Measures of the relative contribution of stockholders and creditors, and of the firm's ability
to pay financing charges. Value of firm's debt to the total value of the firm.
## Leveraged required return
The required return on an investment when the investment is financed partially by debt.
## Liquidity ratios
ratios that measure a firm's ability to meet its short-term financial obligations on time.
## Long-term assets
Value of property, equipment and other capital assets minus the depreciation. This is an
entry in the bookkeeping records of a company, usually on a "cost" basis and thus does not necessarily reflect
the market value of the assets.
## Long-term debt ratio
The ratio of long-term debt to total capitalization.
## Long-term debt to equity ratio
A capitalization ratio comparing long-term debt to shareholders' equity.
## Low price-earnings ratio effect
The tendency of portfolios of stocks with a low price-earnings ratio to
outperform portfolios consisting of stocks with a high price-earnings ratio.
## Liquidity ratios
ratios that measure a firm's ability to meet its short-term financial obligations on time.
## Macaulay duration
The weighted-average term to maturity of the cash flows from the bond, where the
weights are the present value of the cash flow divided by the price.
## Market return
The return on the market portfolio.
## Market value ratios
ratios that relate the market price of the firm's common stock to selected financial
statement items.
## Market-book ratio
Market price of a share divided by book value per share.
## Modified duration
The ratio of Macaulay duration to (1 + y), where y = the bond yield. Modified duration is
inversely related to the approximate percentage change in price for a given change in yield.
## Money rate of return
Annual money return as a percentage of asset value.
## Mortgage duration
A modification of standard duration to account for the impact on duration of MBSs of
changes in prepayment speed resulting from changes in interest rates. Two factors are employed: one that
reflects the impact of changes in prepayment speed or price.
## Mortgage-Backed Securities Clearing Corporation
A wholly owned subsidiary of the Midwest Stock
Exchange that operates a clearing service for the comparison, netting, and margining of agency-guaranteed
MBSs transacted for forward delivery.
## Multinational corporation
A firm that operates in more than one country.
Related to : financial, finance, business, accounting, payroll, inventory, investment, money, inventory control, stock trading, financial advisor, tax advisor, credit. | 3,711 | 18,514 | {"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} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.91688 |
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