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Alecsey [184]
3 years ago
13

30men can finish 1/7of a work in 2days.18men will do othe remaining work in ?

Mathematics
1 answer:
Dovator [93]3 years ago
7 0
7/7-1/7=6/7 is the rest of work
1/7 work takes 2days
6*1/7=6/7 work takes 6*2days=12days
30 men takes 12 days to finish the rest of work (6/7)
18 men takes 18*12/30= 7.2days
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En una prueba de duración para determinar la vida útil de un nuevo tipo de lámpara lanzada al mercado, los datos de la muestra s
diamong [38]

Answer:

no habla espenol :((

Step-by-step explanation:

6 0
3 years ago
A long novel is opened to a random page; the probability that the page is numbered something between 21 and 31.
Harrizon [31]

Using the uniform distribution, considering that the novel has 435 pages, the probability that the page is numbered something between 21 and 31 is:

p = \frac{10}{435}

<h3>What is the uniform probability distribution?</h3>

It is a distribution with two bounds, a and b, in which each outcome is equally as likely.

The probability of finding a value between c and d is:

P(c \leq X \leq d) = \frac{d - c}{b - a}

Researching the problem on the internet, it is found that the novel has 435 pages, hence a = 0, b = 435, and the probability that the page is numbered something between 21 and 31 is given by:

P(21 \leq X \leq 31) = \frac{31 - 21}{435 - 0} = \frac{10}{435}

More can be learned about the uniform distribution at brainly.com/question/13889040

#SPJ1

5 0
2 years ago
Lilliana is training for a marathon. She runs the same distance every day for a week. On Monday, Wednesday, and Friday, she runs
Karo-lina-s [1.5K]

Answer: 17 3/4

Step-by-step explanation: because yes

4 0
3 years ago
Read 2 more answers
Suponga que desea utilizar un servicio de correo particular para enviar un paquete que tiene forma de caja rectangular con una s
statuscvo [17]

a) El volumen de la caja en función de su longitud es:

V_{caja}=\frac{x^{3}}{16}-\frac{25x^{2}}{2}+625x

b) El dominio de la ecuación del volumen son todos los números reales.

c) Las dimensiones del paquete con el mayor volumen posible son:

longitud de la caja  x = 30 plg

lado de la sección transversal L = 17.5 plg

a)

Definamos x como la longitud de la caja y L como el lado de la sección transversal, que es cuadrada para este caso.

Sabemos que la suma de su longitud (x) y el perímetro de la sección transversal (P = 4L) es igual a 100 plg.

x+4L=100 (1)

Ahora, el volumen de esta caja rectangular está dada por:

V_{caja}=A_{base}*x

V_{caja}=L^{2}*x (2)

Pero necesitamos expresar el volumen en función de x.

Despejamos L de la ecuación (1) y remplazarlo en (2).

Por lo tanto el volumen en función de x será.

V_{caja}=(\frac{100-x}{4})^{2}*x

V_{caja}=\frac{x^{3}}{16}-\frac{25x^{2}}{2}+625x

b)

Al ser la función un polinomio de orden 3 el dominio de esta función son todos los numeros reales.

c)

Observando la gráfica de esta función, podemos ver que para un valor aproxiamdo de x = 30 plg el valor de V tiene un punto de inflección, es por definición de maximización de una función que podemos usar ese punto para encontrar las dimensiones de la caja.

Por lo tanto si x = 30 plg el valor de L usando la ecuacion (1) sera:

L=\frac{100-x}{4}

L=\frac{100-30}{4}=17.5\: plg

Por lo tanto la máxima dimensión de la caja es:

x = 30 plg

L = 17.5 plg

Puedes aprender más sobre maximizar funciones aquí:

brainly.com/question/16339052

 

     

 

6 0
3 years ago
0-5lbs.-$2. 5-10lbs-$4 10-20lbs-$6
VMariaS [17]
To get to a pound, you must divide how many ounces you have by 16. So 96 divided by 16 = 6lbs. so that pumpkin would be $4     hope I helped :)
4 0
3 years ago
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