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yarga [219]
3 years ago
9

2,631 dividido entre 24

Mathematics
1 answer:
Ivan3 years ago
5 0

Answer:

2631 divided by 24 = 109.625

Step-by-step explanation:

The result of 2631/24 is a terminating decimal with 3 digits to the right of the decimal point.

2631 divided by 24 in decimal = 109.625

2631 divided by 24 in fraction = 2631/24

2631 divided by 24 in percentage = 10962.5%

Note that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or decimals, including 2631 and 24, of course.

Repetends, if any, are denoted in ().

The conversion is done automatically once the nominator, e.g. 2631, and the denominator, e.g. 24, have been inserted.

SPANISH:

El resultado de 2631/24 es un decimal final con 3 dígitos a la derecha del punto decimal.

2631 dividido por 24 en decimal = 109,625

2631 dividido por 24 en fracción = 2631/24

2631 dividido por 24 en porcentaje = 10962,5%

Tenga en cuenta que puede utilizar nuestra calculadora de última generación anterior para obtener el cociente de dos enteros o decimales cualesquiera, incluidos 2631 y 24, por supuesto.

Los repetends, si los hay, se indican en ().

La conversión se realiza automáticamente una vez que el nominador, por ejemplo, 2631, y el denominador,

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Step-by-step explanation:

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The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop
Anettt [7]

Answer:

a. \frac{1}{15}

b. \frac{2}{5}

c. \frac{14}{15}

d. \frac{8}{15}

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

a. Favorable cases for Both the laptops to be selected = _2C_2 = 1

Total number of cases = 15

Required probability is \frac{1}{15}.

b. Favorable cases for both the desktop machines selected = _4C_2=6

Total number of cases = 15

Required probability is \frac{6}{15} = \frac{2}{5}.

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

2. Both desktop:

Favorable cases = _4C_2=6

Total number of favorable cases = 8 + 6 = 14

Required probability is \frac{14}{15}.

d. 1 desktop and 1 laptop:

Favorable cases = _2C_1\times _4C_1 = 8

Total number of cases = 15

Required probability is \frac{8}{15}.

8 0
4 years ago
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