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3 years ago

Please help with simple probability question. Extra points & brainliest. Thanks in advance.

Mathematics

1 answer:

kodGreya [7K]3 years ago

3 0

The answer is B) 1/3.

There are 6 faces on a rectangular prism, each of which is a rectangle.

The area of the front and back is given by 25(8) = 200 sq. cm.
The area of the bottom and top is given by 25(4) = 100 sq. cm.
The area of the right and left is given by 4(8) = 32.

There are 2 faces out of 6 with an area that is not a multiple of 100; 2/6 = 1/3.

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6x + 14 = 20<br> Show your work <br> Solve the equation

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Answer:

X = 1

Step-by-step explanation:

subtract 14 from both sides

6x = 6

divide both sides by 7

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3 years ago

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Write two subtraction facts related to 7 + 5 =12

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12-5=7

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Step-by-step explanation:

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3 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0

3 years ago

Need help with this question

Andreas93 [3]

The Domain is (-∞, -2) U (-2, 2) U (2, ∞) and Range is (-∞, -2) U [-1, ∞) and x-intercept = (-1.5, 0), (1.5, 0) and y-intercept = (0, -1).

<h3>What is a conic section?</h3>

It is defined as the curve which is the intersection of cone and plane. There are three major conic sections; parabola, hyperbola, and ellipse (circle is a special of type of ellipse).

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

From the graph:

a) The domain: all values of x

The range: all values of y

Except: Values at asymptotes are excluded.

Domain = (-∞, -2) U (-2, 2) U (2, ∞)

Range = (-∞, -2) U [-1, ∞)

x-intercept = (-1.5, 0), (1.5, 0)

y-intercept = (0, -1)

c) Horiznotal asymptotes: y = -2

d)Vetical asymptotes: x = -2, x = 2

e) Oblique asymptotes: There is no oblique asymptotes

Thus, the Domain is (-∞, -2) U (-2, 2) U (2, ∞) and Range is (-∞, -2) U [-1, ∞) and x-intercept = (-1.5, 0), (1.5, 0) and y-intercept = (0, -1).

Learn more about the conic section here:

brainly.com/question/8412465

#SPJ1

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2 years ago

What is the slope of the line shown?? Please help i been stuck on this all day yesterday and now today..

Talja [164]

I hope this helps.if you have any questions regarding what I did you can ask

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4 years ago