Ask question

Ask question

All categories

3 years ago

7

The theoretical probability of an event occurring is 2/5. Which best describes the experiment probability associated with this e

vent?

2 answers:

pogonyaev3 years ago

5 0

The answer is out of every 5, the desired out come will be approximately 2 times.

V125BC [204]3 years ago

5 0

Answer:

d

Step-by-step explanation:

You might be interested in

what is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube

Dmitrij [34]

Let the least possible value of the smallest of 99 cosecutive integers be x and let the number whose cube is the sum be p, then

$\frac{99}{2} (2x+98)=p^3 \\ \\ 99x+4,851=p^3\\ \\ \Rightarrow x=\frac{p^3-4,851}{99}$

By substitution, we have that $p=33$ and $x=314$ .

Therefore, <span>the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube is 314.</span>

3 0

4 years ago

The question is in the picture

frutty [35]

the answer is third one

5 0

3 years ago

A container in the shape of a square-based prism has a volume of 2744 cm'. What dimensions give the

Vinvika [58]

Answer:

The dimensions that give the minimum surface area are:

Length = 14cm

width = 14cm

height = 14cm

And the minimum surface is:

S = 1,176 cm^2

Step-by-step explanation:

A regular rectangular prism has the measures: length L, width W and height H.

The volume of this prism is:

V = L*W*H

The surface of this prism is:

S = 2*(L*W + H*L + H*W)

If the base of the prism is a square, then we have L = W

Then the equations become:

V = L*L*H = L^2*H

S = 2*(L^2 + 2*H*L)

We know that the volume of the figure is 2744 cm^3

Then:

V = 2744 cm^3 = H*(L^2)

In this equation, we can isolate H.

H = (2744 cm^3)/(L^2)

Now we can replace this on the surface equation:

S = 2*(L^2 + 2*L* (2744 cm^3)/(L^2))

S = 2*L^2 + 4(2744 cm^3)/L

Now we want to minimize the surface area, then we need to find the zeros of the first derivative of S.

S' = 2*(2*L) - 4*(2744 cm^3)/L^2

This is equal to zero when:

0 = 2*(2*L) - 4*(2744 cm^3)/L^2

0 = 4*L*L^2 - 4*(2744 cm^3)

4*(2744 cm^3) = 4*L^3

2744 cm^3 = L^3

∛(2744 cm^3) = L = 14cm

Then the length of the base that minimizes the surface is L = 14.

Then we have:

H = (2744 cm^3)/(L^2) = (2744 cm^3)/(14cm)^2 = 14cm

Then the surface is:

S = 2*(L^2 + 2*L*H) = 2*( (14cm)^2 + 2*(14cm)*(14cm)) = 1,176 cm^2

8 0

3 years ago

a paragraph that consists of information on the coordinate system used in navigation, archaeology, and space.

ANEK [815]

The coordinates system is used to map out and locate objects throughout space. In navigation, satellites use GPS signals to determine where you are relative to a grid placed on earth. This grid typically follows longitudinal and latitudinal lines. Space uses this system in a similar way. Engineers and astronomers use grids to determine the location of planets, stars, and other celestial bodies in the universe. Finally, archaeology uses grids to help map out a site that is being dug. As artifacts are recovered, archaeologists record their location on a coordinate grid because it helps them interpret the living conditions or other human behaviors on the site.

5 0

3 years ago

Find a formula for the least squares solution of ax=b when the columns of a are orthonormal1 .

neonofarm [45]

Let $\mathbf A$ be a rectangular $m\times n$ matrix with column vectors $\mathbf a_1,\ldots,\mathbf a_n$ , i.e.

$\mathbf A=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}$

Then we have

$\mathbf A^\top=\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_n\end{bmatrix}^\top$

and the product of the two is

$\mathbf A^\top\mathbf A=\begin{bmatrix}\mathbf a_1\cdot\mathbf a_1&\mathbf a_1\cdot\mathbf a_2&\cdots&\mathbf a_1\cdot\mathbf a_n\\\mathbf a_2\cdot\mathbf a_1&\mathbf a_2\cdot\mathbf a_2&\cdots&\mathbf a_2\cdot\mathbf a_n\\\vdots&\vdots&\ddots&\vdots\\\mathbf a_n\cdot\mathbf a_1&\mathbf a_n\cdot\mathbf a_2&\cdots&\mathbf a_n\cdot\mathbf a_n\end{bmatrix}$

Because the columns of $\mathbf A$ are orthonormal, we have

$\mathbf a_i\cdot\mathbf a_j=\begin{cases}1&\text{for }i=j\\0&\text{for }i\neq j\end{cases}$

which means $\mathbf A^\top\mathbf A$ reduces to an $n\times n$ matrix with ones along the diagonal and zero everywhere else, i.e.

$\mathbf A^\top\mathbf A=\begin{bmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\end{bmatrix}=\mathbf I_n$

where $\mathbf I$ denotes the identity matrix. This means the solution to $\mathbf{Ax}=\mathbf b$ is given by

$\mathbf A^\top(\mathbf{Ax})=\mathbf A^\top\mathbf b\implies(\underbrace{\mathbf A^\top\mathbf A}_{\mathbf I})\mathbf x=\mathbf A^\top\mathbf b\implies\mathbf x=\mathbf A^\top\mathbf b$

6 0

3 years ago