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SOVA2 [1]
3 years ago
5

Use the given minimum and maximum data​ entries, and the number of​ classes, to find the class​ width, the lower class​ limits,

and the upper class limits.
minimum​, 19 maximum​, 134,8 classes
Mathematics
1 answer:
Gennadij [26K]3 years ago
6 0

Using proportions and the information given, it is found that:

  • The class width is of 14.375.
  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

  • Minimum value is 19.
  • Maximum value is of 134.
  • There are 8 classes.
  • The classes are all of equal width, thus the width is of:

W = \frac{134 - 19}{8} = 14.375

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

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kykrilka [37]

The solution of the equation is x = -4/3.

<h3>What does it mean to solve an equation?</h3>

An equation represents equality of two or more mathematical expression.

Solutions to an equation are those values of the variables involved in that equation for which the equation is true.

WE have been given an equation as;

|x - 4| = 5x + 12

In an absolute value equation, we solve the original expression as our first equation. Our second one is that we multiply the right side by -1.

Case 1: original equation

|x - 4| = 5x + 12

x - 4 = 5x + 12

x - 5x = 12 + 4

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Case 2: Opposite equation

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x + 5x = -12 + 4

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Check:

|x - 4| = 5x + 12

use x = -4

|-4 - 4| = 5(-4) + 12

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8 = -8              

Thus, it is Not a solution

Now,  |x - 4| = 5x + 12

use x =  -4/3

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16/3 = 16/3

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Learn more about solving equations here:

brainly.com/question/18015090

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Hello, there!


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