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ch4aika [34]
3 years ago
13

Please help find the mean, median, mode and range.

Mathematics
1 answer:
storchak [24]3 years ago
3 0

<em><u>Mean</u></em>

add up all the numbers, then divide by how many numbers there are.

\frac{425}{14}  = 30.35

<em><u>Median</u></em>

To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost number

\frac{29 + 30}{2}  =  \frac{59}{2}  = 29.5

<em><u>Mode</u></em><em><u> </u></em>

The mode of a data set is the number that occurs most frequently in the set. To easily find the mode, put the numbers in order from least to greatest and count how many times each number occurs.

mode = 38

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em><em>.</em><em>!</em>

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a factory could produce 84 toy robots in one day. They purchased some new equipment. Now they can produce 105 robots in one day.
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Answer:

The answer to the question is

25 % increase in production rate

Step-by-step explanation:

To solve the question, we note that

Initial number of products = 84 toy robots per day

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Percentage increase = 21/84 ×100 = 25 % increase

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The area of the rhombus is 540 cm2; the length of one of its diagonals is 4.5 dm. What is the distance between the point of inte
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1. The area of the rhombus can be found by the formula

A=\dfrac{d_1\cdot d_2}{2}, where d_1,\ d_2 are rhombus's diagonals.

Note that d_1=4.5\ dm=45\ cm, then

540=\dfrac{45\cdot d_2}{2},\\ \\540\cdot 2=45d_2,\\ \\d_2=24\ cm.

2. The diagonals of rhombus are perpendicular and are bisectors of each other. Then the triangle formed with halfs of diagonals is right triangles with legs

\dfrac{d_1}{2}=22.5\ cm,\ \dfrac{d_2}{2}=12\ cm.

The hypotenuse of this triangle is the rhombus's side. By the Pythagorean theorem

\text{rhombus's side}^2=(22.5)^2+12^2=506.25+144=650.25,\\ \\\text{rhombus's side}=25.5\ cm.

3. The distance between the point of intersection of the diagonals and the side of the rhombus is the height of right triangle considered above.

Use twice the Pythagorean theorem to find this height:

\left\{\begin{array}{l}x^2+h^2=12^2\\(25.5-x)^2+h^2=22.5^2,\end{array}\right.

where x is projection of leg 12 cm and h is height.

Subtract the first equation from the second:

(25.5-x)^2+h^2-x^2-h^2=22.5^2-12^2,\\ \\650.25-51x=506.25-144,\\ \\51x=650.25-362.25=288,\\ \\x=\dfrac{96}{17}\ cm.

Then

h^2=144-\left(\dfrac{96}{17}\right)^2=144-\dfrac{9216}{289}=\dfrac{32400}{289},\\ \\h=\dfrac{180}{17}\ cm.

Answer: h=\dfrac{180}{17}\ cm.

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