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LUCKY_DIMON [66]
2 years ago
5

Dos agricultores, padre e hijo, tardan 2 horas entre los dos arar un campo. Si solo el padre tarda 6 horas. ¿Cuanto tardara el h

ijo?
Mathematics
1 answer:
OLEGan [10]2 years ago
8 0

Answer:

El hijo solo tardará 3 horas en arar el campo

Step-by-step explanation:

Aquí, se nos dice que ambos pasan dos horas para arar el campo mientras que el padre solo pasa 6 horas.

Queremos saber el tiempo que pasará el hijo si va a trabajar solo

Sea el área del campo x

La tasa de trabajo de ambos es x / 2

La tasa de trabajo del padre será x / 6 Deja que el tiempo que pasa el hijo sea y La tasa de trabajo del hijo es x / y

Sumando la tasa de trabajo del padre a la del hijo da la tasa de trabajo total de ambos Por lo tanto;

x / 6 + x / y = x / 2

Despegar x 1/6 + 1 /y = 1/2

1 / y = 1/2 - 1/6

1 / y= 1/3

entonces y = 3

El hijo solo tardará 3 horas en arar el campo.

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. Over the next two days, Clinton Employment Agency is interviewing clients who wish to find jobs. On the first day, the agency
Vedmedyk [2.9K]

Answer:

4

Step-by-step explanation:

Given that :

Clients are interviewed in groups of 2 on the first day; meaning two persons at a time

Second day, clients are interviewed in groups of 4; meaning 4 persons at a time.

Therefore, if the same number of clients are to be interviewed on each day, the smallest number of clients that could be interviewed each day could be obtained by getting the Least Common Multiple of both numbers: 2 and 4

- - - - 2 - - - 4

2 - - - 1 - - - 2

2 - - - 1 - - - 1

Therefore, the Least common multiple is (2 * 2) = 4

Therefore, the smallest number of clients that could be interviewed each day is 4.

5 0
2 years ago
A tiling company completes two jobs. The first job has $1200 in labor expenses for 40 hours worked, while the second job has $15
daser333 [38]

Answer:

The correct option is;

1560 = 30(52) + b

Step-by-step explanation:

The cost of the first job = $1,200 in labor and expenses

The number of hours worked in the first job = 40

The second job costs $1,560 in labor and expenses

The number of hours worked in the second job = 52

If the cost of labor per hour = L and the expenses = b, we have;

$1,200 = 40×L + b......................(1)

$1,560 = 52×L + b.......................(2)

Subtracting equation (1) from (2), we have;

52×L + b - (40×L + b) =  52×L - 40×L + b - b = $1,560 - $1,200 = $360

12×L = $360

L = $360/12 = $30/hour

From equation (1), we have;

$1,200 = 40×L + X =  40×$30 + b

Therefore, the equation that can be used to calculate the y-intercept of the linear equation is either;

1200 = 40(30) + X or 1560 = 32(52) + b, which gives the correct option as 1560 = 32(52) + b

6 0
2 years ago
According to a nutrition coach, a 100 kg person should eat
alexira [117]

Answer:

0.07%

Step-by-step explanation:

This equation is solving for what percentage of 100 kg is 0.07 kg.

1. Set up the equation

\frac{0.07}{100} = \frac{x}{100}

0.07 kg out of 100 kg is equal to x out of 100 because x represents the percentage and percentages are out of 100.

2. Solve by cross multiplying

100x = 7

3. Solve for x by dividing both sides by 100

x = 0.07

The answer is 0.07%

4 0
2 years ago
Hannah took $50.00 to go shopping and she bought 2 shirts for $17.00 each how change did she have left
Vlad1618 [11]

Answer: Hannah had $16.00 left.

Step-by-step explanation: First, you add $17.00 plus $17.00 or $17.00 times 2 which gives you $34.00. Then, you subtract $50.00 minus $34.00 which gives you $16.00.

4 0
2 years ago
Management is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 perc
Olegator [25]

Answer:

The bonus will be paid on at least 4099 units.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the percentile of this measure.

In this problem, we have that:

The highest 5 percent is the 95th percentile.

Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. This means that \mu = 4000, \sigma = 60.

If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

The least units that the bonus will be paid is X when Z has a pvalue of 0.95.

Z has a pvalue of 0.95 between 1.64 and 1.65. So we use Z = 1.645

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 4000}{60}

X - 4000 = 60*1.645

X = 4098.7

The number of units is discrete, this means that the bonus will be paid on at least 4099 units.

5 0
2 years ago
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