# MathGuru

Popular questions and responses by MathGuru
1. ## math

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

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 -

1. ## Algebra

Fourth root of 80y cubed

posted on December 20, 2014
2. ## Statistics

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

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

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

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

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

posted on May 5, 2014
66. ## statistics

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