Which is longer 1/4 foot or 1/4 yard?

En el estante de un negocio hay 2 tipos de tarros de la misma mermelada y marca. El tarro más alto tiene el doble de altura que

Ipatiy [6.2K]

Answer:

tarro corto

Step-by-step explanation:

Aquí tenemos que comprar el frasco que tiene el menor costo por volumen.

$h_1$ = Altura del frasco corto

$h_2$ = La altura del frasco alto es el doble que el del frasco corto. = $2h_1$

$d_1$ = Diámetro del frasco corto

$d_2$ = El diámetro del frasco alto es la mitad del frasco corto = $\dfrac{1}{2}d_1$

El volumen de un cilindro es $\pi \dfrac{d^2}{4}h$

La razón de los volúmenes de los frascos es

$\dfrac{V_1}{V_2}=\dfrac{\pi\dfrac{d_1^2}{4}h_1}{\pi\dfrac{d_2^2}{4}h_2}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{d_2^2h_2}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{(\dfrac{1}{2}d_1)^22h_1}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{d_1^2h_1}{\dfrac{1}{4}d_1^22h_1}\\\Rightarrow \dfrac{V_1}{V_2}=\dfrac{1}{\dfrac{1}{2}}\\\Rightarrow \dfrac{V_1}{V_2}=2\\\Rightarrow V_1=2V_2$

El costo del frasco corto por unidad de volumen es

$\dfrac{8000}{V_1}=\dfrac{8000}{2V_2}=\dfrac{4000}{V_2}$

El costo del frasco alto por unidad de volumen es

$\dfrac{4500}{V_2}=\dfrac{4500}{V_2}$

$\dfrac{4000}{V_2}$

Entonces, el costo del frasco corto por unidad de volumen es menor que el costo por unidad de volumen del frasco alto.

Por lo tanto, deberíamos tomar el frasco corto.

5 0

3 years ago

4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

galben [10]

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$