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2 years ago

What is the Median of this?

Mathematics

2 answers:

Arada [10]2 years ago

7 0

The median is 84.5

Hope this helps!

Alexeev081 [22]2 years ago

6 0

First, sort the sumbers in order from the least value to the greatest value.

37, 45, 59, 67, 76, 68, 77, 78, 82, 82, 84, 84, 85, 85, 88, 90, 91, 91, 92, 92, 93, 95, 100, 100

There 24 scores listed. Divide 24 by 2 and you get 12.. Find the 12th number in the ordered list from the least value, which is the number 84. Find the 12th score in the ordered list from the large value, which is 85.
Because there are two central numbers, add the two central numbers and divide it by 2.

84+85=169
169/2=84.5

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The Answer is asking me to find how much female students are on academic teams in

seropon [69]

Answer:

Hi, so unfortunately just knowing the percentages is not enough information to figure out the number of people, let alone females.

Were there problems before this one that provided you with some additional numbers? If so, that is what you need to solve this one.

8 0

3 years ago

Help me: <img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5E%5Cpi%20_0%20sin%28%7Bx%7D%29%20%5C%2C%20dx" id="TexFormula1" title

ikadub [295]

Answer:

$\int^{\pi}_{0}sin(x)dx = 2$

Step-by-step explanation:

To solve for the value of $\int^{\pi}_{0}sin(x)dx$ , we need to first find the antiderivative of $sin(x)$ .

Since the derivative of $cos(x)$ is $-sin(x)$ , therefore the derivative of $-cos(x)$ is $sin(x)$ . With this:

$\int^{\pi}_{0}sin(x)dx = -cos(\pi)-(-cos(0))$

$=-cos(\pi)+cos(0)$

$=-(-1)+1$

$=2$

∴ $\int^{\pi}_{0}sin(x)dx = 2$

Hope this helps :)

5 0

2 years ago

Precal help if you could could you answer both. Me finishing this helps me graduate.

oee [108]

Exercise 1:

The easiest way to compute powers of complex numbers is to write them in the form

$z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))$

In this form, you have

$z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))$

The magnitude of the number is given by

$z=a+bi \implies \rho = \sqrt{a^2+b^2}$

So, we have

$z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2$

As for the angle, we have

$z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}$

So, we have

$z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}$

Finally,

$z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64$

Exercise 2:

You simply have to compute the trigonometric function:

$\cos\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2},\quad \sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$

So, we have

$z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right) = 1-i$

7 0

2 years ago

Please help!!!!

Flauer [41]

The cotangent function is defined as the length of the side adjacent over the length of the side opposite.

Step-by-step explanation:

The sine of an angle is defined as the length of the side opposite over the length of the hypotenuse.

The cosine of an angle is defined as the length of the side adjacent over the length of the hypotenuse.

The tangent of an angle is defined as the length of the side opposite over the length of the side adjacent.

The cotangent function is defined as the length of the side adjacent over the length of the side opposite.

The secant of an angle is defined as the length of the hypotenuse over the length of the side adjacent.

The cosecant of an angle is defined as the length of the hypotenuse over the length of the side opposite.

4 0

3 years ago

Part 4: Use the information provided to write the vertex formula of each parabola.

sergey [27]

Answer: 1. x = (y - 2)² + 8

$\bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1$

3. y = 2(x +9)² + 7

<u>Step-by-step explanation:</u>

Notes: Vertex form is: y =a(x - h)² + k or x =a(y - k)² + h

(h, k) is the vertex
point of vertex is midpoint of focus and directrix: $\dfrac{focus+directrix}{2}$

$\bullet\quad a=\dfrac{1}{4p}$

p is the distance from the vertex to the focus

$focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1$

Now, plug in a = 1 and (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

$focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}$

Now, plug in a = -1/2 and (h, k) = (1, 10) into the equation x =a(y - k)² + h

$\bold{x=-\dfrac{1}{2}(y-10)^2}+1$

***************************************************************************************

$focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)$

Now let's find the a-value:

$p=focus-vertex\\\\p=\dfrac{57}{8}-\dfrac{56}{8}=\dfrac{1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{8})}=\dfrac{1}{\frac{1}{2}}=2$

Now, plug in a = 2 and (h, k) = (-9, 7) into the equation y =a(x - h)² + k

y = 2(x +9)² + 7

4 0

3 years ago