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example of equal interval
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Question:I am making a histogram from data that was not taken at equal intervals. For example on the X axis I have 0, 1, 2, 3, 4, 5, 6 but only have data for say, 1, 2, and 5. How wide should the rectangles be? I thought the middle of the rectangle goes where the point is but not sure how wide to make them.
Answers:Personnally, I would show a bar only where you have data at 1, 2 and 5 and make them 1wide and show nothing where you have no data.
Answers:Personnally, I would show a bar only where you have data at 1, 2 and 5 and make them 1wide and show nothing where you have no data.
Question:Would someone mind giving me some guidence with regards to these two?
Use the definition of continuity to determine if f(x) is continuous at x=0
f(x)= absolute value of x+3 where x<0
2x+1 x=0
(x+2)^2 1 x>0
this one is even worse...
Use the definition of continuity to determine over what intervals f(x) is continuous.
f(x) = cos^2 x x< pi/4
(sinx)(cosx) x=pi/4
sin^2 x x>pi/4
I've consulted all my reference books but have no idea where to start with solving them so I am hoping by asking I can at least get a working example of the proper method to approach this. This is a beginner course so we have not covered differentiation yet. Thanks in advance I really appreciate it.
Answers:There are three requirements necessary for continuity to hold for a function f(x) at x = c. 1) lim_{x > c} f(x) exists. 2) f(c) exists. 3) The numbers in 1) and 2) are equal. Condition 1) is equivalent to saying that the left and righthand limits exist and are equal. So, for f(x)= absolute value of x+3 where x<0 2x+1 x=0 (x+2)^2 1 x>0 at c = 0 we have lim_{x > 0} f(x) = lim_{x > 0} x + 3 = lim_{x > 0} x + 3 = 0 + 3 = 3. lim_{x > 0+} f(x) = lim_{x > 0+} (x+2)^2 1 = (0+2)^2  1 = 4  1 = 3. They are equal, thus lim_{x > 0} f(x) exists and is equal to 3. f(0) = 2(0) + 1 = 1. lim_{x > 0} f(x) does not equal f(0). f is not continuous here. Try the same process with your second one. There, observe that if x is not pi/4, the function is continuous, as cos^2 x is continuous everywhere, and hence on the interval x < pi/4. Ditto for the sine squared. Hope this helps.
Answers:There are three requirements necessary for continuity to hold for a function f(x) at x = c. 1) lim_{x > c} f(x) exists. 2) f(c) exists. 3) The numbers in 1) and 2) are equal. Condition 1) is equivalent to saying that the left and righthand limits exist and are equal. So, for f(x)= absolute value of x+3 where x<0 2x+1 x=0 (x+2)^2 1 x>0 at c = 0 we have lim_{x > 0} f(x) = lim_{x > 0} x + 3 = lim_{x > 0} x + 3 = 0 + 3 = 3. lim_{x > 0+} f(x) = lim_{x > 0+} (x+2)^2 1 = (0+2)^2  1 = 4  1 = 3. They are equal, thus lim_{x > 0} f(x) exists and is equal to 3. f(0) = 2(0) + 1 = 1. lim_{x > 0} f(x) does not equal f(0). f is not continuous here. Try the same process with your second one. There, observe that if x is not pi/4, the function is continuous, as cos^2 x is continuous everywhere, and hence on the interval x < pi/4. Ditto for the sine squared. Hope this helps.
Question:A sample of 19 pages was taken without replacement from the 1,591page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured (a display ad is a large block of multicolored illustrations, maps, and text).
Construct a 95% confidence interval. The data are shown below:
0 260 356 350 536 0 268 369 428 536
268 396 469 536 162 338 350 536 536
(in square millimeters)
A. Construct a 95% confidence interval for the true mean.
1. What was the Point Estimate?
2. What was the Margin of error, E =
3. What was the tvalue?
4. Confidence Interval?
(write as 23 << 45)
B. Interpret the confidence interval
C. What sample size, n, would be needed to obtain an error of 10 square millimeters with 99% confidence?
(Use Sample Size for a Mean, Doane, Ch. 8, p. 326 and s in place of )
1. Sample size, n =
2. What is the sample size, n, using the finite formula?
(N = 1,591 pages)
Sample size, n =
Answers:QUESTION A 1. What was the Point Estimate? ANSWER: 352.3 QUESTION: A 2. What was the Margin of error, E = ANSWER: 55.4 Why?? SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION xbar = Point Estimate [352.3] s = Sample standard deviation [166.3] n = Number of samples [9] df = degrees of freedom [9  1 = 8] For confidence level of 95%, twosided interval ("lookup" from Table) "tvalue" [2.306] Margin of error, E = s/SQRT(n): 166.3/SQRT(9) = 55.4 QUESTION A 3. What was the tvalue? ANSWER: 2.306 QUESTION A 4. Confidence Interval? ANSWER: 95% Confidence Interval (224.5 << 480.1) Why?? Resulting Confidence Interval for "true mean" ( ): xbar +/ (t critical value) * s/SQRT(n) = 352.3 +/ 2.306* 55.4 /SQRT(9) = (224.5 << 480.1) QUESTION: B. Interpret the confidence interval ANSWER: In statistics, a confidence interval (CI) or confidence bound is an interval estimate of a population parameter. Instead of estimating the parameter by a single value, an interval likely to include the parameter is given. Thus, confidence intervals are used to indicate the reliability of an estimate. How likely the interval is to contain the parameter is determined by the confidence level or confidence coefficient. Increasing the desired confidence level will widen the confidence interval. For example, a CI can be used to describe how reliable survey results are. In a poll of election votingintentions, the result might be that 40% of respondents intend to vote for a certain party. A 95% confidence interval for the proportion in the whole population having the same intention on the survey date might be 36% to 44%. All other things being equal, a survey result with a small CI is more reliable than a result with a large CI and one of the main things controlling this width in the case of population surveys is the size of the sample questioned. Confidence intervals and interval estimates more generally have applications across the whole range of quantitative studies. Source: http://en.wikipedia.org/wiki/Confidence_interval
Answers:QUESTION A 1. What was the Point Estimate? ANSWER: 352.3 QUESTION: A 2. What was the Margin of error, E = ANSWER: 55.4 Why?? SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION xbar = Point Estimate [352.3] s = Sample standard deviation [166.3] n = Number of samples [9] df = degrees of freedom [9  1 = 8] For confidence level of 95%, twosided interval ("lookup" from Table) "tvalue" [2.306] Margin of error, E = s/SQRT(n): 166.3/SQRT(9) = 55.4 QUESTION A 3. What was the tvalue? ANSWER: 2.306 QUESTION A 4. Confidence Interval? ANSWER: 95% Confidence Interval (224.5 << 480.1) Why?? Resulting Confidence Interval for "true mean" ( ): xbar +/ (t critical value) * s/SQRT(n) = 352.3 +/ 2.306* 55.4 /SQRT(9) = (224.5 << 480.1) QUESTION: B. Interpret the confidence interval ANSWER: In statistics, a confidence interval (CI) or confidence bound is an interval estimate of a population parameter. Instead of estimating the parameter by a single value, an interval likely to include the parameter is given. Thus, confidence intervals are used to indicate the reliability of an estimate. How likely the interval is to contain the parameter is determined by the confidence level or confidence coefficient. Increasing the desired confidence level will widen the confidence interval. For example, a CI can be used to describe how reliable survey results are. In a poll of election votingintentions, the result might be that 40% of respondents intend to vote for a certain party. A 95% confidence interval for the proportion in the whole population having the same intention on the survey date might be 36% to 44%. All other things being equal, a survey result with a small CI is more reliable than a result with a large CI and one of the main things controlling this width in the case of population surveys is the size of the sample questioned. Confidence intervals and interval estimates more generally have applications across the whole range of quantitative studies. Source: http://en.wikipedia.org/wiki/Confidence_interval
Question:How can I understand this meaning. The book said, "Since a sidereal day is shorter than a 24hour day, each sidereal hour passes slightly more quickly. Because the whole day passes four minutes more quickly, each hour passes ten seconds faster. Using your scientific calculator, multiply your GMT birth time by ten seconds to reflect this and write the answer on your worksheet."
My GMT time is 6 hours and 18 minutes.
Answers:What throws you off is that the book says something nonsensical, namely the words "multiply your GMT birth time by ten seconds to reflect this". It should read something like "find the number of sidereal hours ellapsed since midnignt (GMT) on the moment of your birth". Maybe the author of the book has something else in mind but he/she is certainly not saying what it is ;) If you had said (from the context) what the purpose of the calculation was (e.g., determining the position of a star at the time of your birth knowing its position in the sky at midnight GMT) I could have helped you make definite sense out of this. Otherwise, we are left with a mere guess about the intended meaning of a meaningless question... This being said, 6 hours and 18 minutes is 22680 seconds. There are 86164.09 seconds in a sidereal day, so a "sidereal hour" is 1/24 of that, or about 3590.17 s (which is, indeed about 10 s less than a regular hour, as advertised). So, 22680 seconds corresponds to a fraction of a sidereal day equal to 22680/86164.09. If you use the less precise data given in the question, you obtain 22680/86160, which is close enough (let's use that because the question suggests it). 22680 / 86160 is equal to 189/718 of a sideral day. A sideral hour is 1/24 of a sidereal day. Subtract 6 of those (namely 1/4 of a sidereal day) and you have a remainder of 19/1436 which is 114/359 of a sidereal hour, or 6840/359 "sidereal minutes" (a "unit" which nobody ever uses). That's very close to 19 minutes. A quicker approach, without fancy computations, would be to say that 6 sidereal hours are 60 seconds short of 6 hours. So, the 6 sidereal hours are over 1 minute before the 6 hour mark. 18 minutes after that point, you've therefore reached a time corresponding to 6 hours and 19 minutes in "sidereal time".
Answers:What throws you off is that the book says something nonsensical, namely the words "multiply your GMT birth time by ten seconds to reflect this". It should read something like "find the number of sidereal hours ellapsed since midnignt (GMT) on the moment of your birth". Maybe the author of the book has something else in mind but he/she is certainly not saying what it is ;) If you had said (from the context) what the purpose of the calculation was (e.g., determining the position of a star at the time of your birth knowing its position in the sky at midnight GMT) I could have helped you make definite sense out of this. Otherwise, we are left with a mere guess about the intended meaning of a meaningless question... This being said, 6 hours and 18 minutes is 22680 seconds. There are 86164.09 seconds in a sidereal day, so a "sidereal hour" is 1/24 of that, or about 3590.17 s (which is, indeed about 10 s less than a regular hour, as advertised). So, 22680 seconds corresponds to a fraction of a sidereal day equal to 22680/86164.09. If you use the less precise data given in the question, you obtain 22680/86160, which is close enough (let's use that because the question suggests it). 22680 / 86160 is equal to 189/718 of a sideral day. A sideral hour is 1/24 of a sidereal day. Subtract 6 of those (namely 1/4 of a sidereal day) and you have a remainder of 19/1436 which is 114/359 of a sidereal hour, or 6840/359 "sidereal minutes" (a "unit" which nobody ever uses). That's very close to 19 minutes. A quicker approach, without fancy computations, would be to say that 6 sidereal hours are 60 seconds short of 6 hours. So, the 6 sidereal hours are over 1 minute before the 6 hour mark. 18 minutes after that point, you've therefore reached a time corresponding to 6 hours and 19 minutes in "sidereal time".
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