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ryzh [129]
3 years ago
8

I’m confused on this one

Mathematics
1 answer:
matrenka [14]3 years ago
4 0

Take a look at triangle OFG, whose hypotenuse is 13 and whose longer side is 12. Its shorter side is 5, via the Pythagorean Theorem. Thus, the width of the trapezoid is 5.

Then the area of the trapezoid is

A = (average of the lengths of the trapezoid)(width of trapezoid)

30+ 40

= -------------- * 5 = 35*5= 175 square units.

2

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Answer:

Mean: 30 students

Standard deviation: 5.2 students

Step-by-step explanation:

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3 years ago
Please help!!!! I’m unsure of the answers
igomit [66]

Answer:

table:

.1, .25, .35, .2, .1

p(x=4) = .1

p(x<2) = .35

p(3≤x≤4)= .55

1.95, 1.12

Step-by-step explanation:

this is kind of hard to read, but i think i've got it

mean:

0*.1+1*.25+2*.35+3*.2+4*.1= 1.95

The second moment:

0²*.1+1²*.25+2²*.35+3²*.2+4²*.1= 5.05

the variance is the second moment minus the first moment squared (first moment is the mean) and then the standard deviation is the square root of the mean

5.05-1.95²= 1.2475 √1.2475= 1.1169 or 1.12

3 0
2 years ago
What is the width of a rectangle with length 5.5 cm and area 220 cm^2
Reika [66]
In the given question it is given that the length of the rectangle is 5.5 cm and the area of the rectangle is 220 cm^2.
Length of the rectangle = 5.5 cm
Area of the rectangle = 220 cm^2
Then we already know that
Area of a rectangle = Length * width
So Width of the rectangle = Area of the rectangle/ Length
                                           = 220/5.5
                                           = 2200/55
                                           = 40
So the width of the rectangle is 40 cm.<span>Hope  that you have got the answer you were looking for.

</span>


3 0
3 years ago
Can you please answer the question?
Roman55 [17]

The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}

\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }

\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}

\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}

\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}

\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}

\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}

\tan \alpha + 1 + \cot \alpha

\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1

\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1

\frac{1}{\cos \alpha \cdot \sin \alpha} + 1

\sec \alpha \cdot \csc \alpha + 1

The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

To learn more on trigonometric expressions: brainly.com/question/10083069

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Answer:

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