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BaLLatris [955]
3 years ago
12

Use the raw data below to create a table that can be used to create a histogram with 5

Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

Five classes

Step-by-step explanation:

1. Sort the data

You get

1.1, 1.3, 1.6, 1.7, 2, 2.6, 2.7, 2.9, 3.2, 3.2, 3.5, 3.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9, 4.9, 5.3, 5.7, 6.4, 6.5, 7.1, 7.5, 7.6, 8.1, 8.2, 9.2, 9.4

2. Calculate the range

Range = Max - Min = 9.4 - 1.1 = 8.3

3. Calculate the class width

Divide the range by the number of classes

8.3/5 = 1.7

Round this up to 2.

4. Decide where to start the histogram

You could use classes: 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 -11

However, even numbers are easier to read.

It would be preferable to use classes: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10

5. Prepare a frequency distribution table

Note: Each class does not include its largest possible value. Thus, a value of 2 goes into Class 2 - 4, (not Class 0 - 2).

\begin{array}{cc}\textbf{Miles run} & \textbf{No. of students} \\0 - 2 & 4 \\2 - 4 & 9 \\4 - 6 & 8 \\6 - 8 & 5 \\8 - 10 & 4 \\\end{array}

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Consider this option (this is not the only way):
1. according to the condition 2345N is divisible by <em>6</em>, it means, that N is even number (0;2;4;6 or 8).
2. <em>6</em>=3*2, then the number 2345N is divisible by 3 also. It means, that the number (2+3+4+5+N)=(14+N) is divisible by 3, where N={0|2|4|6|8}.
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Read 2 more answers
Can someone please help? Thanks!
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Hey, I have already answered these all questions except the last question in your previous question. Make sure to see them.

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