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stellarik [79]
3 years ago
10

Identify the class width, class midpoints, and class boundaries for the given frequency distribution.Daily Low Tempurature Frequ

encey Daily Low Tempurature full data set32-24 1 44-46 735-37 3 47-49 738-40 5 50-52 141-43 11
Mathematics
1 answer:
Natali [406]3 years ago
8 0

Answer:

I . class width is 3

ii. Class midpoints are 33,36,39,42,45,48,51.

iii. Class boundaries are: 31.5-34.5, 34.5-37.5, 37.5-40.5, 40.5-43.5, 43.5-46.5, 46.5-49.5, 49.5-52.5

Step-by-step explanation:

I. The class width is the difference between upper class boundary and lower class boundary.

ii. Class midpoints is the summation of upper and lower class limits, divided by 2.

iii. Class boundaries is the subtraction of 0.5 from the lower classe limits and the addition of 0.5 to upper class limits.

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snow_tiger [21]
For this case, what we must do is fill squares in all the expressions until we find the correct result.
 We have then:
 
 x2 + y2 − 4x + 12y − 20 = 0 x2 + y2  − 4x + 12y = 20
 x2  − 4x + y2 + 12y = 20
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 2x2 + 2y2 − 24x − 16y − 8 = 0
 x2 + y2 − 12x − 8y − 4 = 0
 x2 + y2 − 12x − 8y = 4 
 x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
 x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
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 x2 + y2 + 2x − 12y − 9 = 0
 x2 + y2  + 2x - 12y = 9
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 x2  + 2x + (2/2)^2 + y2 - 12y  + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
 x2  + 2x + (1)^2 + y2 - 12y  + (-6)^2 = 9 + (1)^2 + (-6)^2
 x2  + 2x + 1 + y2 - 12y  + 36 = 9 + 1 + 36
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