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3 years ago

Which three-dimensional figure has the greatest number of faces?

Mathematics

2 answers:

Nataly [62]3 years ago

7 0

Answer:

Three-dimensional figure that has the greatest number of faces is octagonal pyramid.

Step-by-step explanation:

we have to choose the three-dimensional figure has the greatest number of faces.

Hexagonal prism has 8 faces.

Octagonal pyramid has 9 faces(8 triangular and one bottom)

Rectangular prism has 6 rectangular faces.

A pentagonal pyramid has five faces and one base.

Hence, three-dimensional figure that has the greatest number of faces is octagonal pyramid.

Option 2 is correct.

irina1246 [14]3 years ago

6 0

ITS B , octagonal pyramid

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The ____ axis is vertical.

4vir4ik [10]

Answer:

Step-by-step explanation:

The y-axis is the the secondary or vertical axis of a system of coordinates, points along which have a value of zero for all other coordinates.

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4 years ago

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Need some help please!!

ella [17]

Answer:

The other point on the diameter which is also on the circle is; (-3, 2)

Step-by-step explanation:

The center of the circle, also a point where the diameter crosses, is (3, 6)

The diameter touches the circle at point (9, 10)

The displacement is (9 - 3, 10 - 6) = (6, 4)

This means that we move, +6 units along the x-axis and +4 units along the y-axis , from the center of the circle in order to reach point (9, 10)

If we want to reach the other point on the diameter which also on the circle, we move -6 units along the x-axis and -4 units along the y-axis to get;

(3, 6) - (6, 4) = (-3, 2)

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3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago

Answer these series of events

Alchen [17]

Answer: A= 34

B= 25

C= -30

D= -25

E= -1.3

F= 0.6 repeating or rounded to the nearest tenth place 0.7

Hope this helps!

Step-by-step explanation:

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3 years ago

Read 2 more answers

I am having trouble with number 2. Please show how the answer was gotten.

Andrews [41]

We know how we got 90, since a right angle is 90 degrees,

180 degrees are in a triangle.

180-90=90

Find the pair that equals 90 so you get 180 degrees!

If you have more questions don’t have hesitation to ask!

Or add all number to get 180 degrees!

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3 years ago