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timama [110]
3 years ago
12

En un pozo petrolìfero produce cada dia 2680 barriles de petroleo. Si cada barril contiene 159 litros de petroleo, cuantos litro

s de petroleo produce en un dia y en un mes
Mathematics
1 answer:
Aliun [14]3 years ago
4 0

Answer:

426,120 litros se producen en un día Se producen

12,783,600 litros por mes (para un cálculo de 30 días en un mes)

Step-by-step explanation:

Aquí, debemos calcular la cantidad en litros de petróleo producido en un pozo petrolero.

Se nos dice que se producen 2.680 barriles por día y cada barril contiene 159 litros de petróleo.

La producción diaria en litros es, por lo tanto, de 159 * 2680 = 426,120 litros.

Ahora, para la producción mensual, supongamos que hay 30 días en un mes, la cantidad en litros producidos por mes sería la cantidad diaria multiplicada por la cantidad de días en un mes.

Matemáticamente, eso sería 30 * 426,120 = 12,783,600 litros

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Answer:

Depending on the length and width, it would be the length times the width.    

Step-by-step explanation:

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2 years ago
Find the sample size required to achieve the given margin of error with SD = 51.02 and z - score = 1.96. Round your answer to th
coldgirl [10]

Answer:

The sample size required is 2500.

Step-by-step explanation:

z-score:

In function of the margin of error M, the z-score is given by:

Z = \frac{M}{\frac{\sigma}{\sqrt{n}}} = \frac{M\sqrt{n}}{\sigma}

In this question, we have that:

\sigma = 51.02, M = 2, Z = 1.96

So

Z = \frac{M\sqrt{n}}{\sigma}

1.96 = \frac{2\sqrt{n}}{51.02}

2\sqrt{n} = 51.02*1.96

\sqrt{n} = \frac{51.02*1.96}{2}

(\sqrt{n})^2 = (\frac{51.02*1.96}{2})^2

n = 2500

The sample size required is 2500.

3 0
3 years ago
Find the least common denominator between the two fractions. and 2 12 6
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5 0
3 years ago
H(w)=-5x^2+80 to factored form
algol13

Answer:

- 5(x - 4)(x + 4)

Step-by-step explanation:

Given

- 5x² + 80 ← factor out - 5 from each term

= - 5(x² - 16) ← x² - 16 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 16 = x² - 4² = (x - 4)(x + 4)

Thus

- 5x² + 80 = - 5(x - 4)(x + 4) ← in factored form

3 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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