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svlad2 [7]
2 years ago
6

The number of students in each class is shown on the line plot.

Mathematics
1 answer:
Tatiana [17]2 years ago
3 0

Answer:

Your answer is 24.

Step-by-step explanation:

The data set is 22,23,23,23,24,24,25,26,26,26,28

The median of this set is 24

The mean is 24.54

The mode is 23,26

The range is 6

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The bar chart shows the number of students with each letter grade in a certain professor's math classes. Since students need at
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Answer:

Since students need at least a C for the grade to transferdes that are less than C into one category instead of grouping D's and F's separately. How many ...

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Evaluate the function for x = –2a, if a = 3.<br> f(x)= -4x+7
wariber [46]
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7 0
3 years ago
Which equation of the line that is parallel to the line 5x + 2y =12 and passes through point -2,4
Gnesinka [82]

Answer:

The equation of the line would be y = -5/2x - 1

Step-by-step explanation:

In order to find the equation of the line, we first need to find the slope of the original line. We can do that by solving for y.

5x + 2y = 12

2y = -5x + 12

y = -5/2x + 6

Now that we have a slope of -5/2, we know the new slope will be the same since parallel lines have the same slope. So we can use it along with the point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - 4 = -5/2(x + 2)

y - 4 = -5/2x - 5

y = -5/2x - 1

3 0
3 years ago
4y³+2y²-2y²-2-3y-3y simplify
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5 0
3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
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