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natali 33 [55]
3 years ago
7

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 47.3 m

iles per hour.Speed (miles per hour) 42-45 46-49 50-53 54-57 58-61Frequency 21 15 6 4 2The mean of the frequency distribution miles ______ per hour. (Round to the nearest tenth as needed.)Which of the following best describes the relationship between the computed mean and the actual mean?A. The computed mean is not close to the actual mean because the difference between the means is more than 5%.B. The computed mean is not close to the actual mean because the difference between the means is more than 5%.C. The computed mean is not close to the actual mean because the difference between the means is less than 5%.D. The computed mean is close to the actual mean because the difference between the means is less than 5%.
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

47.4167

D. The computed mean is close to the actual mean because the difference between the means is less than 5%

Step-by-step explanation:

Given the distribution :

Speed(m/hr) ___ midpoint(x) ___ F ___fx

42 - 45 _______ 43.5 ________21 __ 913.5

46 - 49 _______47.5 _________15 __712.5

50 - 53 _______ 51.5 _________6 __ 309

54 - 57 _______ 55.5 ________ 4 ___ 222

58 - 61 ________59.5 ________ 2 ___ 119

The midpoint, x= sum of lower and upper boundary divided by 2

For instance, (42 + 45) / 2 = 43.5

The computed mean, Σfx / Σf = 2276 / 48 = 47.4167

Actual mean = 47.3 miles

(47.4167 - 47.3) / 47.3 * 100% = 0.24%

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The equation of line passes through (4,-1) and perpendicular to the graph y=\frac{7}{2}x-\frac{3}{2} is y=\frac{-2x}{7}+\frac{1}{7}

Step-by-step explanation:

Given point is (4,-1) and equation of line is y=\frac{7}{2}x-\frac{3}{2}

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(y-y_1)=m(x-x_1)\\\\(y-(-1))=\frac{-2}{7}(x-4)\\\\y+1=\frac{-2x}{7}-(\frac{2\times -4}{7})\\ \\y+1=\frac{-2x}{7}+\frac{8}{7}\\\\y=\frac{-2x}{7}+\frac{8}{7}-1\\\\y=\frac{-2x}{7}+\frac{8-7}{7}\\\\y=\frac{-2x}{7}+\frac{1}{7}

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