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qwelly [4]
3 years ago
9

Un paracaidista que pesa 70 N es lanzado de un avión. halla su masa​

Mathematics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

<em>La masa del paracaidista es 7.14 Kg</em>

Step-by-step explanation:

<u>Peso y Masa</u>

La masa de un objeto es una medida de la cantidad de material que el objeto posee. Sus unidades más comunes son el Kg, gr, lb, Ton.

El peso de un objeto de masa m es la fuerza con la cual la Tierra atrae al objeto a través de la aceleración de gravedad g. Sus unidades son iguales a las de cualquier fuerza, es decir, Nw, Kgf, Dina. Su fórmula es:

P=m.g

Si tenemos el peso P=70 N, podemos calcular la masa, despejándola de la fórmula:

\displaystyle m=\frac{P}{g}

Sustituyendo los valores:

\displaystyle m=\frac{70}{9.8}=7.14\ Kg

La masa del paracaidista es 7.14 Kg

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
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A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0
3 years ago
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