MathGuru

Newest questions and responses by MathGuru
  1. math

    Substitute -1 for x. Simplify. If f(x) = 4x-3, find the following. f(-1)

    asked on February 23, 2007
  2. math

    You have two formulas to use: A = lw -->area = length times width P = 2l + 2w -->perimeter You know the perimeter, which is 200m of fencing. Let length = x Now let's solve the perimeter equation for w, using what we know: 200 = 2x + 2w 200 - 2x = 2w (200 -

    asked on January 17, 2007
  1. Algebra

    Fourth root of 80y cubed

    posted on December 20, 2014
  2. Statistics

    See your most recent post.

    posted on December 18, 2014
  3. STAT- important

    Yes, you are correct. Read about the Central Limit Theorem. It might help to clarify the concepts for you.

    posted on December 18, 2014
  4. Math

    How about D?

    posted on December 18, 2014
  5. Staatistics

    Normal distribution: 99.7% is approximately +/- 3 standard deviations around the mean. 0.3% is outside those limits.

    posted on December 14, 2014
  6. math

    2.82843 from an online calculator. If you do this by hand, find the mean, then the variance. Standard deviation is the square root of the variance. Check your work.

    posted on December 14, 2014
  7. algebra 2

    I'm assuming your problem is this: (-6+i)/(-5+i) If so, then you can multiply both the numerator and denominator by an equivalent of 1: (-6+i)/(-5+i) * (-5 - i)/(-5 - i) = (30 - 5i + 6i - i^2)/(25 - i^2) = (30 + i + 1)/(25 + 1) = (31 + i)/26 Check the

    posted on December 14, 2014
  8. statistics

    Since n = 200, .6 = 120, .3 = 60, and .1 = 20 Therefore, the expected vale of X is: E(X) = sum x*p = 120(.6) + 60(.3) + 20(.1) = 92 To get the variance, subtract the mean (expected value) from each X. Next, square each of those values. Finally, add the

    posted on December 14, 2014
  9. statistics

    Formula: CI95 = mean ± 1.96(sd/√n) You will need to calculate the mean and standard deviation for your data. In the formula, n = sample size. Once you have what you need, substitute into the formula and determine the confidence interval. I hope this

    posted on July 31, 2014
  10. stats

    I would think either one due to the nature of the tests, but check this to be sure.

    posted on July 28, 2014
  11. statistics

    Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is found using a z-table for whatever confidence level is used, when no value is stated in the problem p = .50, q = 1 - p, ^2 means squared, * means to multiply,

    posted on July 28, 2014
  12. Introduction to statistics

    Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is found using a z-table for 99% confidence, p = .50 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = 0.19.

    posted on July 18, 2014
  13. statistics

    Hint: The mean of the sampling distribution is the population mean.

    posted on July 10, 2014
  14. statistics, probability

    Formula (if you don't use the binomial probability table): P(x) = (nCx)(p^x)[q^(n-x)] Values: n = 6 p = 0.4 q = 1 - p = 1 - 0.4 = 0.6 i. Find P(5) ii. Find P(4), P(5), and P(6). Add together for your total probability. iii. I'll let you try this one on

    posted on July 9, 2014
  15. statistics

    Here's one way to do this problem: n = 240 p = .03 q = 1 - p = 1 - .03 = .97 For (a), you will need to find P(0) through P(4). Add together, then subtract the total from 1 for the probability. I'll let you try (b). You can use a binomial probability table

    posted on July 8, 2014
  16. statistics

    Here's one way you might do this: s/[1 + (2.58/√2n)] to s/[1 - (2.58/√2n)] Note: 2.58 represents 99% confidence Find the standard deviation for your data. Use that value for s. Your sample size is 12. I'll let you take it from here.

    posted on July 8, 2014
  17. Statistics

    Substitute t for z in my statements. Sorry for any confusion.

    posted on July 8, 2014
  18. Statistics

    Since your sample size is small, you can use a one-sample t-test formula. With your data: z = (2.8 - 3)/(0.3/√10) Finish the calculation. Next, check a t-table using 9 degrees of freedom (which is n - 1) for a one-tailed test at .05 level of

    posted on July 8, 2014
  19. statistics

    CI95 = 51.4 ± 1.96 (16.66) Calculate to determine the interval.

    posted on July 8, 2014
  20. statistics

    standard error = s/√n Note: The square root of the variance is called the Standard Deviation With your data: 2 = (4.472)/√n Solve for n. I hope this is what you were asking.

    posted on July 8, 2014
  21. stats

    I'll help with the last two: 9. The alternate hypothesis shows a specific direction in a one-tailed test. 10. 2.262 (degrees of freedom would be n - 1)

    posted on July 3, 2014
  22. Statistics (Check Work)

    Looks like you have a good handle on this, Jen. Keep up the good work!

    posted on June 27, 2014
  23. statistics

    You might try a formula like this one: s/(1 + 1.96/√2n) to s/(1 - 1.96/√2n) Substitute and calculate. There may be other variations of similar formulas you can use as well.

    posted on June 26, 2014
  24. Statistics (Check Work)

    I see you figured this out on your own. Good job!

    posted on June 26, 2014
  25. Statistics (Check Work)

    Hi Jen! Just an FYI: I'm a "she" instead of a "he" but thanks for the compliments. I'm always glad to help when and where I can. Let's get started. Both questions are the same type of problem. Question 1: Null: pR = pU Alternate: pR > pU Formula: z = (pR -

    posted on June 26, 2014
  26. Statistics

    PsyDAG answered a similar post. See below under "Related Questions" for the prior posting.

    posted on June 26, 2014
  27. statistics

    95% confidence interval is equivalent to z = 1.96, so if you round to 2, then your calculations for E are almost correct. I think you missed a 0; I get E = 0.04248. The formula used to determine the margin of error is all you should need to answer the

    posted on June 24, 2014
  28. statistics

    Try z-scores. In this case, use the following: z = (x - mean)/(sd/√n) With your data: z = (165 - 150)/(90/√25) I'll let you finish the calculation. Next, check a z-table for the probability. Remember the question says "165 friends or more" when you are

    posted on June 24, 2014
  29. Statistics (Check Answer)

    You are welcome! I'm glad the explanation helped.

    posted on June 17, 2014
  30. Statistics

    Try the binomial probability formula: P(x) = (nCx)(p^x)[q^(n-x)] n = 15 x = 4 p = .20 q = 1 - p = .80 Substitute the values into the formula and calculate your probability.

    posted on June 17, 2014
  31. Statistics

    Formula: n = {[(z-value) * sd]/E}^2 ...where n = sample size, sd = standard deviation, E = maximum error, and ^2 means squared. Using the values you have in your problem: n = {[(1.96) * 15]/3}^2 Calculate for sample size. Round your answer.

    posted on June 16, 2014
  32. Statistics (Check Answer)

    Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust

    posted on June 16, 2014
  33. statistics

    Let's try a binomial proportion one-sample z-test. Ho: p = 0.65 Ha: p does not equal 0.65 Test statistic: z = (0.49 - 0.65)/√[(0.65)(0.35)/100] z = -3.35 The null would be rejected at the .05 level for a two-tailed test (p does not equal 0.65). Use a

    posted on June 13, 2014
  34. statistics

    Formula: n = {[(z-value) * sd]/E}^2 ...where n = sample size, sd = standard deviation, E = maximum error, and ^2 means squared. Using the values you have in your problem: n = {[(1.96) * 15]/3}^2 Calculate for sample size. Round your answer.

    posted on June 13, 2014
  35. Statistics

    The closer it is to 1 or -1, the stronger the correlation. A correlation of 0.3 is a weaker relationship between variables.

    posted on June 10, 2014
  36. Statistics

    Standard deviation = √npq 1) n = 157 p = .16 (for 16%) q = 1 - p = 1 - .16 = .84 2) n = 209 p = .12 (for 12%) q = 1 - p = 1 - .12 = .88 Substitute and calculate.

    posted on June 10, 2014
  37. Statistics

    Try z-scores: z = (x - mean)/sd Your data: 2.33 = (x - 40)/1 Solve for x. Note: 2.33 represents the z-score corresponding to the 1% in your problem.

    posted on June 6, 2014
  38. STATISTICS

    Formula: P(x) = (nCx)(p^x)[q^(n-x)] For (a): x = 5 p = .16 q = 1 - p = 1 - .16 = .84 n = 29 Substitute and calculate for your probability. For (b): x = 0,1,2,3,4 p,q,n stay the same. Add each probability you calculate for the total probability. For (c):

    posted on June 5, 2014
  39. Stats

    Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is 1.96 and is found using a z-table for 95% confidence, p = .5 (when no value is stated), q = 1 - p, ^2 means squared, * means to multiply, and E = .02 (or 2%).

    posted on June 5, 2014
  40. Stats

    Formula: P(x) = (nCx)(p^x)[q^(n-x)] Your data: x = 2 p = .05 q = 1 - p = 1 - .05 = .95 n = 20 Substitute and calculate for your probability.

    posted on June 5, 2014
  41. Stat

    You will need a confidence interval formula for the difference of two population means. Use 1.96 for z (representing 95% confidence). Substitute what you know into the formula and calculate.

    posted on June 2, 2014
  42. statistics

    Using a Poisson distribution: P(x) = (e^-μ) (μ^x) / x! e = 2.71828 µ = 1.4 x = 4 Substitute and calculate.

    posted on May 28, 2014
  43. Statistics

    Correction: Your calculation is correct. z = 5.156 You would still reject the null and accept the alternate hypothesis. Sorry for any confusion.

    posted on May 23, 2014
  44. Statistics

    Use a one-sample z-test for proportions. With your data: z = (.43 - .50)/√(.50)(.50)/1000) = -4.358 Reject the null and accept the alternate hypothesis (p < .50).

    posted on May 23, 2014
  45. Statistics

    Formula: z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size) With your data: z = (26600 - 25000)/(3800/√150) = 4.96 You can reject the null and accept the alternate hypothesis (µ > 25000).

    posted on May 23, 2014
  46. biology

    (B) T cells

    posted on May 22, 2014
  47. Statistics

    Using this formula: P(x) = (nCx)(p^x)[q^(n-x)] Note: q = 1 - p We have: P(x) = (3Cx)(.52^x)[.48^(3-x)]

    posted on May 22, 2014
  48. Statistics

    z = 2.575 for a 99% confidence. Since your sample sizes are large, I would use a two-sample confidence interval formula for large sample sizes. Critical value at .05 using a z-table for a one-tailed test is z = 1.645 If you use a t-table instead to find

    posted on May 22, 2014
  49. statistics

    Hypotheses: Ho: µ = 10 -->null hypothesis Ha: µ < 10 -->alternate hypothesis

    posted on May 22, 2014
  50. please help Human bio

    A. elastin B. collagen C. telomeres Check this!

    posted on May 21, 2014
  51. Statistics

    a) ± 3.1 b) 56.7 - 3.1 = 53.6 56.7 + 3.1 = 59.8 Interval is 53.6 to 59.8

    posted on May 21, 2014
  52. Stats

    Use z-scores. For this problem: z = (x - mean)/(sd/√n) With your data: z = (8.2 - 8)/(2/√100) = 1.00 z = (8.8 - 8)/(2/√100) = 4.00 Answer: .1586 is the probability (check a z-table between z = 1.00 and z = 4.00)

    posted on May 21, 2014
  53. ap stats

    50th percentile

    posted on May 15, 2014
  54. statistics

    Welch's t-test for unequal variances: t = (mean1 - mean2)/√(s^2/n1 + s^2/n2) t = (4.31 - 3.68)/√(0.17^2/10 + 0.22^2/10) t = 0.63/0.08792 = 7.166 (rounded)

    posted on May 15, 2014
  55. Statistics

    If you do a proportional z-test with this data, it would look something like this: z = (.258 - .20)/√[(.20)(.80)/120] Note: 31/120 = .258 Calculating: z = .058/.0365 = 1.589 The null would not be rejected and the conclusion would be that there is no

    posted on May 13, 2014
  56. Statistics

    Answer: That Y increases by 20 units for each unit increase in X.

    posted on May 9, 2014
  57. Statistics (45)

    Formula: n = {[(z-value)(sd)]/E}^2 a) n = [(1.96 * 7)/2]^2 b) n = [(2.575 * 7)/2)^2 Calculate. Round to the next highest whole number.

    posted on May 6, 2014
  58. Statistics (44)

    Formula: n = {[(z-value)(sd)]/E}^2 a) n = [(1.96 * 6.35)/2]^2 b) n = [(1.96 * 6.35)/1)^2 Calculate. Round to the next highest whole number.

    posted on May 6, 2014
  59. Statistics (43)

    Find mean and standard deviation, then use confidence interval formula. CI95 = mean ± 1.96(sd/√n) n = 20 I'll let you take it from here.

    posted on May 6, 2014
  60. Statistics (42)

    I'll get you started. a) CI95 = mean ± 1.96 (sd/√n) mean = 96 sd = 16 n = 60 b) Use the same formula, except n = 120 c) Compare the margin of error in a) and b) to answer this question.

    posted on May 6, 2014
  61. Statistics (39)

    Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust

    posted on May 6, 2014
  62. Statistics (38)

    Mean is 4550.7 (add up the numbers and divide by 10) I'll let you determine standard deviation. Use a calculator, or if you do it by hand, use a formula to calculate the standard deviation.

    posted on May 6, 2014
  63. STAT

    Mean = np = 9 * .17 = 1.53 Use binomial probability formula (or a binomial probability table). Formula: P(x) = (nCx)(p^x)[q^(n-x)] x = 0,1,2 n = 9 p = .17 q = 1 - p = 1 - .17 = .83 For first part: Find P(2) for probability. For second part: Find P(0) and

    posted on May 6, 2014
  64. Stat college

    Mean = np = 9 * .17 = 1.53 Use binomial probability formula (or a binomial probability table). Formula: P(x) = (nCx)(p^x)[q^(n-x)] x = 0,1,2 n = 9 p = .17 q = 1 - p = 1 - .17 = .83 For first part: Find P(2) for probability. For second part: Find P(0) and

    posted on May 6, 2014
  65. statistics

    How about d?

    posted on May 5, 2014
  66. statistics

    How about d?

    posted on May 5, 2014
  67. statistics

    How about d? You have only one sample with a small sample size.

    posted on May 5, 2014
  68. statistics

    Formula: P(x) = (nCx)(p^x)[q^(n-x)] Find P(0), then subtract that value from 1. n = 12 p = .25 q = 1 - p I'll let you take it from here.

    posted on May 5, 2014
  69. statistics

    Formula: n = [(z-value)(p)(q)]/E^2 z-value = 2.33 for 98% confidence p = .5 if no value is stated q = 1 - p E = Maximum error With your data: n = [(2.33)(.5)(.5)]/.03^2 I'll let you take it from here. Round to the next whole number.

    posted on May 5, 2014
  70. Statistics

    CI98 = p ± 2.33 √(pq/n) p = 42/200 = .21 q = 1 - p = 1 - .21 = .79 n = 200 With your data: CI98 = .21 + 2.33 √[(.21)(.79)/200] I'll let you take it from here to finish.

    posted on May 5, 2014
  71. Statistics

    How about a one-sample t-test? Your sample size is fairly small. Fill in what you know into the formula to calculate the test statistic. population mean = 100 sample mean = 115 standard deviation = 30 sample size = 22 I'll let you take it from here.

    posted on May 5, 2014
  72. Statistics

    See later post.

    posted on April 30, 2014
  73. Statistics

    I think you are correct!

    posted on April 30, 2014
  74. statistics

    Using an online calculator: Standard Deviation = 9.21 To find variance, square the standard deviation.

    posted on April 30, 2014
  75. statistics

    True.

    posted on April 29, 2014
  76. statistics

    If you have equal sample sizes and unequal standard deviations, a Welch's t-test might be appropriate. Check degrees of freedom for this type of test before checking the appropriate table to determine critical value.

    posted on April 29, 2014
  77. Statistics

    2.46 A Type I error is rejecting the null when it is true. If the critical value is 2.45, and you have a test statistic of 2.46, you have the least chance among the other choices of rejecting the null and it happens to be true.

    posted on April 29, 2014
  78. statistics

    Try a one-sample t-test since your sample size is small. You will need to calculate the mean and standard deviation. Use a t-table to find the critical value at .05 level of significance for a one-tailed test. Degrees of freedom is (n - 1). Sample size (n)

    posted on April 29, 2014
  79. ap statistics

    CI95 = p ± 1.96[√(pq/n)] p = 20/75 = .27 q = 1 - p = 1 - .27 = .73 n = 75 Substitute into the formula to find the confidence interval.

    posted on April 29, 2014
  80. Statistics

    C. µ1 and µ2 to be unequal.

    posted on April 29, 2014
  81. Statistics

    D. Either (b) or (c).

    posted on April 29, 2014
  82. Statistics

    D. All of the above.

    posted on April 29, 2014
  83. statistics

    Both A and B.

    posted on April 28, 2014
  84. Statistics

    Second choice. Do not reject the null hypothesis because 1.457 lies in the region between -2.326 and 2.326.

    posted on April 28, 2014
  85. ap statistics

    Since the probability of a Type II error is equal to beta, I would say D is not correct.

    posted on April 24, 2014
  86. ap statistics

    B. The probability of rejecting H0 when HA is true.

    posted on April 24, 2014
  87. ap statistics

    The population mean is equal to the mean of the sampling distribution of the sample means. Standard error of the mean is the standard deviation of the sample means.

    posted on April 24, 2014
  88. statistics

    Use a t-table to determine the area. Remember to use the degrees of freedom when looking at the table.

    posted on April 22, 2014
  89. statistics

    Part a) SE of mean = sd/√n = 5/√40 Part b) ME = 1.96 * 5/√40 I'll let you finish the calculations.

    posted on April 22, 2014
  90. statistics

    Part a) ME = 1.96 * sd/√n With your data: ME = 1.96 * 65/√45 Part b) CI95 = mean ± 1.96(sd/n) With your data: CI95 = 273 ± 1.96(65/√45) I'll let you finish the calculations.

    posted on April 22, 2014
  91. statistics

    Try this formula: n = [(z-value * sd)/E]^2 With your data: n = [(1.96 * 40)/10]^2 I'll let you finish the calculation. Remember to round the answer to the nearest whole number.

    posted on April 22, 2014
  92. statistics

    Try this formula: n = [(z-value)^2 * p * (1-p)]/E^2 ...where n = sample size you need, z-value = 1.96 to represent 95% confidence, p and 1-p represent proportions, E = .025 (or 2.5%), and ^ means squared. With your data: n = [(1.96)^2 * .30 * .70]/.025^2

    posted on April 22, 2014
  93. Statistics

    I would think either one due to the nature of the tests, but check this to be sure.

    posted on April 22, 2014
  94. Statistics

    Type I errors result when you reject the null and it's true. Type II errors result when you accept the null and it's false. If you reject the null hypothesis that the subject is guessing and it's true, you have made a Type I error. If you accept the null

    posted on April 16, 2014
  95. Stats

    0.1359 is the area under the normal curve between z = -1 and z = -2 Look at a normal distribution table (z-table) to check.

    posted on April 16, 2014
  96. statistics

    Standard error of the proportion is: √(pq/n) p = .55 q = 1-p = .45 n = sample size I'll let you take it from here.

    posted on April 14, 2014
  97. statistics

    Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust

    posted on April 14, 2014
  98. personal finance

    Common stock

    posted on April 11, 2014