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2 years ago

8

problem 5 answer based of the following distribution. what would be the appropriate measures of center of spread circle one answ

er

1 answer:

pentagon [3]2 years ago

8 0

Answer:

b

Step-by-step explanation:

You might be interested in

en un colegio mixto hay x estudiantes distribuidos en c cursos. Si hay m niños por curso, ¿Cuantas niñas hay en el colegio?

Tomtit [17]

La ecuación que nos dice cuantas niñas hay en el colegio es:

A = x - c*m

¿Como encontrar una expresión para la cantidad de niñas?

Sabemos que hay un total de x estudiantes, definimos las variables:

A = cantidad total de niñas
B = cantidad total de niños.

Entonces tenemos A + B = x

Tambien sabemos que los estudiantes estan divididos en c cursos, de tal forma que hay m niños por curso.

Entonces c por m es igual a la cantidad total de niños:

c*m = B

Reemplazando esto en la ecuación de arriba podemos obtener:

A + c*m = x

A = x - c*m

Esta es la ecuación que nos da el numero total de niñas en el colegio.

Sí quieres aprender más sobre ecuaciones, puedes leer:

brainly.com/question/18168483

4 0

1 year ago

What is the area of a circle if the circumference is 29

Kruka [31]

Answer:

67 square units

Step-by-step explanation:

Circumference/pi = diameter

29/3.14 = 9.24

Radius = 9.24/2 = 4.62

A = pi * r^2

A = 3.14 * 4.62^2 = 67 square units

8 0

2 years ago

Which of the following is equivalent to √-90?<br> A. -3i√10<br> B. 3i√10<br> C. 9i√10<br> D. 9√-10

erastova [34]

The answer to this question is B. I hope this helps

7 0

2 years ago

Read 2 more answers

You can buy DVDs at a local store for $15.49 each. You can buy them at an online store for $13.99 each plus $6 for shippimg. How

NeTakaya

15.49x = 13.99x + 6
15.49x - 13.99x = 6
1.50x = 6
x = 6 / 1.50
x = 4 <=== u can buy 4 DVD's for the same amount

15.49 * 4 = 61.96
13.99(4) + 6 = 61.96

6 0

2 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago