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3 years ago

Finish this for points fractions to decimals

Mathematics

1 answer:

timofeeve [1]3 years ago

7 0

1: 0.7142857143

2: 0.04761904762

3: 0.2142857143

4: 0.8666666667

5: 0.2727272727

6: 0.3888888889

7: 0.3333333333

8: 0.4444444444

9: 0.3333333333

10: 0.3571428571

11: 0.36

12: 0.8

13: 0.2307692308

14: 0.875

15: 0.8

16: 0.04761904762

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Which line is parallel to the line y = 5/2x - 9

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What is the simplified value of the expression below?

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An orange has 30 calories more than a peach

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3 years ago

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

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3 years ago

The plan for the Thorne’s backyard play area, shaped like a trapezoid, is shown. The area of the play area is 286.56 square feet

lorasvet [3.4K]

The length of material needed for the border is the perimeter of the backyard play area

<h3>How to calculate the length of material needed </h3>

The area of the play area is given as:

$Area = 286.56$

The area of a trapezoid is calculated using:

$Area = 0.5 * (L1 + L2) * H$

Where L1 and L2, are the parallel sides of the trapezoid and H represents the height.

The given parameter is not enough to solve the length of material needed.

So, we make use of the following assumed values.

Assume that the parallel sides are: 25 feet and 31 feet long, respectively.

While the other sides are 10.2 feet and 8.2 feet

The length of material needed would be the sum of the above lengths.

So, we have:

$Length = 25 + 31 + 10.2 + 8.2$

$Length = 74.4$

Using the assumed values, the length of material needed for the border is 74.4 feet