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Ksju [112]
3 years ago
7

PLEASE HELP!!! A.S.A.P!!

Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

142m

Step-by-step explanation:

This problem can be solved by simply using the pythagorean theorem, as you stated at the beginning of the problem, which is: a^{2} +b^{2} =c^{2}

You are given the <em>a</em> side and <em>b</em> side that are needed for this equation, so it's all a matter of plugging in the information you have:

110^{2} +90^{2} =c^{2}

110^{2} =12100

90^{2} =8100

12100+8100=c^{2}

20200=c^{2}

Now, because the <em>c</em> is still squared, you must take the square root of 20200 in order to get the length of just side <em>c</em>:

\sqrt{20200}≈142m

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Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

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c.

Class boundaries

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Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

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