Write 127.90 in expanded form

Gus has a fish tank that holds 471047104710 inches^3 3 cubed of water. He is using a cylinder shaped bucket with a radius of 555

tester [12383]

Answer:

Gus must fill the cylindrical bucket three times to get it filled for the tank

Step-by-step explanation:

Extracting essential details from the query:-

*** A fish tank has a volume of 4710 inches^3

*** The tank owner utilizes a cylindrical bucket for filling the tank

*** The bucket has a radius of 5 inches and a height of 20 inches

*** We need to compute how many times Gus has to fill the bucket to fully fill the fish tank.

Since the container used for filling the tank is cylindrical, we will apply the formula for finding a cylinder's volume to derive the bucket's volume used for filling the tank.

Volume of a cylinder = π × r^2 × h. The radius of the cylindrical bucket is stated as 5 inches while the height is mentioned as 20 inches.

The bucket's volume is therefore calculated as 3.14 × 5^2 × 20 = 1570 inches^3

Next, it is given that the fish tank’s capacity is 4710 inches^3. To find out how many times the bucket can fill the tank, we must divide the tank's volume by the bucket's volume:

4710/1570 = 3 full buckets.

Thus, Gus has to fill the bucket three times for filling the fish tank completely without spillage.

7 0

2 months ago

A software engineer creates a LAN game where an 8 digit code made up of 1,2,3,4,5,6,7,8 has to be decided on, universal code . T

PIT_PIT [12445]

Respuesta:

0.0143 = 1.43% de probabilidad de que los primeros 4 dígitos del código sean números pares.

Explicación paso a paso:

Arreglos de elementos n:

El número de arreglos posibles de n elementos se determina por:

$A_{n} = n!$

Probabilidad:

La probabilidad se calcula dividiendo el número de resultados deseados por el número total de resultados.

En esta pregunta:

Resultados deseados:

Los primeros cuatro dígitos son pares (2, 4, 6 y 8 son pares, así que se consideran 4 elementos).

Los últimos cuatro dígitos son impares (1, 3, 5 y 7 son impares, por lo que los últimos cuatro elementos son también arreglos de 4 elementos). Entonces

$D = 4!*4! = 24*24 = 576$