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dimulka [17.4K]
3 years ago
9

A piece of paper is in the shape of a rectangle. The piece of paper is 1 5/8 inches (in.) wide and 8 3/4 in. long. A student cut

s the piece of paper in the following order: The student cuts off inch from the width. The student cuts off inch from the length. The student cuts the remaining piece of paper into 12 equally long pieces of paper. What is the area of each of the 12 equally long pieces of paper? Explain your answer completely and show all your work. Include in your explanation an equation you can use to find the area of each of the 12 equally long pieces of paper.
Mathematics
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

.403645833 square inches

Step-by-step explanation:

1 5/8-1 equals 5/8

8.75-1 equals 7.75

7.75/12 equals .6458

.6458 times 5/8 equals .4 square inches

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Yanka [14]

Answer:

Range=14

\sigma^2 =32.4

\sigma = 5 .7

The standard deviation will remain unchanged.

Step-by-step explanation:

Given

Data: 136, 129, 141, 139, 138, 127

Solving (a): The range

This is calculated as:

Range = Highest - Least

Where:

Highest = 141; Least = 127

So:

Range=141-127

Range=14

Solving (b): The variance

First, we calculate the mean

\bar x = \frac{1}{n} \sum x

\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)

\bar x = \frac{1}{6} *810

\bar x = 135

The variance is calculated as:

\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2

So, we have:

\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]

\sigma^2 =\frac{1}{5}*[162]

\sigma^2 =32.4

Solving (c): The standard deviation

This is calculated as:

\sigma = \sqrt {\sigma^2 }

\sigma = \sqrt {32.4}

\sigma = 5 .7 --- approximately

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Answer:

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Step-by-step explanation:

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2 years ago
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PLEASE HELP! (1/4) - 50 POINTS -
Annette [7]

Answer:

  D)  0.35

Step-by-step explanation:

The table gives the area between z=0 and the given magnitude of z. That is, the area between z = 0 and z = -0.6 is 0.23, as found in the 0.6 column of the table. Similarly, the area between z = 0 and z = 0.3 is 0.12, as found in the 0.3 column of the table.

The area between z = -0.6 and z = +0.3 is the sum of these areas:

  p(-.6<z<.3) = 0.23 +0.12 = 0.35

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