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3 years ago

5

What is their vetocity

1 answer:

bixtya [17]3 years ago

5 0

There vetocity is D/t =s is ükùû

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Births are approximately Uniformly distributed between the 52 weeks of the year. They can be said to follow a Uniform distributi

Kaylis [27]

Answer:

a) 27

b) 15

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform distribution is:

$M = \frac{a + b}{2}$

The variance of the uniform distribution is given by:

$V = \frac{(b-a)^{2}}{12}$

The standard deviation is the square root of the variance.

They can be said to follow a Uniform distribution from 1 to 53

This means that $a = 1, b = 53$

a. The mean of this distribution is

$M = \frac{a + b}{2} = \frac{1 + 53}{2} = 27$

b. The standard deviation is:

$s = \sqrt{V} = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(53-1)^{2}}{12}} = 15$

7 0

4 years ago

Jada says that she found three different ways to complete the first question correctly. Do you think this is possible? Explain y

Andrew [12]

Answer:

Yes.

Step-by-step explanation:

Although there is not much info on this question, if you were to assume that there are 3 different answers or 3 different methods of solving that problem, it is definitey possible to assume that this is possible.

7 0

3 years ago

A cone and a cylinder have the same height and their bases are congruent circles

Anettt [7]

Answer:

$30cm^3$

Step-by-step explanation:

the volume of a cylinder is given by:

$v_{cylinder}=\pi r^2 h$

and the volume of a cone is given by:

$v_{cone}=\frac{\pi r^2 h}{3}$

since both have the same height and radius, we can solve each equation for $r^2h$ (because this quantity is the same in both figures) and then match the expressions we find:

from the cylinder's volume formula:

$r^2h=\frac{v_{cylinder}}{\pi}$

and from the cone's volume formula:

$r^2h=\frac{3 v_{cone}}{\pi}$

matching the two previous expressions:

$\frac{v_{cylinder}}{\pi} =\frac{3v_{cone}}{\pi}$

we solve for the volume of a cone $v_{cone}$ :

$v_{cone}=\frac{\pi v_{cylinder}}{3\pi} \\\\v_{cone}=\frac{v_{cylinder}}{3}$

substituting the value of the cylinder's volume $v_{cylinder}=90cm^3$

$v_{cone}=\frac{90cm^3}{3} \\\\v_{cone}=30cm^3$

5 0

4 years ago

Read 2 more answers

Algebra help?<br><br> 343 = (x/6) ^2/3

Vladimir79 [104]

I can only assume that you meant, "Solve for x:"

Apply the exponent 3/2 to both sides of this equation. The result will be

3/2

343 = x/6.

Multiplying both sides by 6 isolates x:

3/2

6*343 = x Since 7^3 = 343, the expression for x

can be rewritten as

3/2

6*(7^3) = x which can be further simplified, as follows:

x = 6^(3/2)*7^(9/2), or:

x = 6^(3/2)*7^(8/2)*√7, or

x = 6^(3/2)*7^4*√7

3 0

4 years ago

Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x w

never [62]

<u>Answer-</u>

$\boxed{\boxed{x=90^{\circ},270^{\circ}}}$

<u>Solution-</u>

The given equation-

$\Rightarrow 2\sin^2 x-\cos^2 x-2=0$

As

$\sin^2 x+\cos^2 x=1\ \ \Rightarrow \cos^2 x=1-\sin^2 x$

Putting this,

$\Rightarrow 2\sin^2 x-(1-\sin^2 x)-2=0$

$\Rightarrow 2\sin^2 x-1+\sin^2 x-2=0$

$\Rightarrow 3\sin^2 x-3=0$

$\Rightarrow 3\sin^2 x=3$

$\Rightarrow \sin^2 x=1$

$\Rightarrow \sin x=\sqrt1$

$\Rightarrow \sin x=\pm 1$

$\Rightarrow \sin x=1,\ \sin x=-1$

$\Rightarrow x=\sin^{-1}(1),\ x=\sin^{-1}(-1)$

$\Rightarrow x=90^{\circ}+n360^{\circ},\ x=270^{\circ}+n360^{\circ}$

Where n=0, 1, 2, ......

As given 0° ≤ x < 360°, so putting n = 0

$\Rightarrow x=90^{\circ},\ x=270^{\circ}$

3 0

4 years ago