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Oxana [17]
2 years ago
6

El número de rebanadas de pizza que se comió cada persona en la fiesta de cumpleaños Elian son: 3,4,5,2,1,4,5,3,2,1,0,2,3,4,6. M

edia: Moda: Mediana:
Mathematics
1 answer:
STALIN [3.7K]2 years ago
5 0

Respuesta:

Modo = 2, 3, 4

Mediana = 3

Media = 3

Explicación paso a paso:

Dados los datos:

3,4,5,2,1,4,5,3,2,1,0,2,3,4,6

Datos pedidos: 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6

La media = ΣX / n = 45/15 = 3

Donde n = tamaño de la muestra = 15

Mediana = 1/2 * (n + 1) th término

Mediana = 1/2 (16) término

Mediana = octavo término

Mediana = 3

Modo = 2, 3, 4 (el más alto ocurre con una frecuencia de 3 cada uno)

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omeli [17]

Answer:

The square root of this value is 171

Step-by-step explanation:

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4 0
1 year ago
Jimmy's school is selling tickets to the
natka813 [3]
Children’s tickets are $4, adult tickets are $14




Explanation:

4a+6c=80

If they made $4 more dollars on the second day than the first by selling one extra child ticket then we know a child’s ticket is $4 per ticket.


4a+6*4=80
4a+24=80
4a=56
14=a

So each adult ticket costs $14.


You can check by filling in $14 and $4 for each equation.


4*14+6*4=80

4*14+7*4=84
3 0
2 years ago
Calculus 3 help please.​
Reptile [31]

I assume each path C is oriented positively/counterclockwise.

(a) Parameterize C by

\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}

with -\frac\pi2\le t\le\frac\pi2. Then the line element is

ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt

and the integral reduces to

\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt

The integrand is symmetric about t=0, so

\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt

Substitute u=\sin(t) and du=\cos(t)\,dt. Then we get

\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}

(b) Parameterize C by

\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{3^2+4^2} \, dt = 5\,dt

and

\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt

Integrate by parts with

u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}

\displaystyle \int u\,dv = uv - \int v\,du

\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}

(c) Parameterize C by

\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt

and

\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}

8 0
1 year ago
331 students went on a field trip. Six buses
slamgirl [31]
54 students in each bus. 

Step#1 Subtract 7 from the total of students, the answer is 324
Step#2 Divide 324 by 6, the answer is 54

Therefore the answer is 54 students, hope that helped. <span />
8 0
2 years ago
Whats (c-7)(c+5) , (double brackets)
kherson [118]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

let's evaluate :

  • (c - 7)(c + 5)

  • {c}^{2}  + 5c - 7c - 35

  • {c}^{2}  - 2c - 35
6 0
2 years ago
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