What is the value of X?

Mezclas de alimentos. Un chef repostero creó una solución de azúcar de 50 onzas constituida por 34% de azúcar con una solución d

Kay [80]

Answer:

Las cantidades empleadas para su preparación son: 15 onzas de solución al 20 % y 35 onzas de solución al 40 %.

Step-by-step explanation:

Podemos estimar la proporción de ingredientes mediante el siguiente promedio ponderado:

$0.34= \frac{0.2\cdot m_{A} + 0.4\cdot m_{B}}{m_{A}+m_{B}}$ (1)

Donde:

$m_{A}$ - Masa de la solución al 20 %, en onzas.

$m_{B}$ - Masa de la solución al 40 %, en onzas.

Podemos simplificar la formula como sigue:

$0.2\cdot x + 0.4\cdot (1-x) = 0.34$ (2)

Donde $x$ es la proporción de la solución al 20 % dentro de la solución final, sin unidades.

Ahora resolvemos para $x$ en (2):

$0.4-0.2\cdot x = 0.34$

$0.2\cdot x = 0.06$

$x = \frac{0.06}{0.2}$

$x = 0.3$

Este resultado quiere decir que la solución al 34 % es el resultado de 30 % de la solución al 20 % y 70 % de la solución al 40 %. Si conocemos que la solución final tiene una masa de 50 onzas, entonces las cantidades empleadas para su preparación son: 15 onzas de solución al 20 % y 35 onzas de solución al 40 %.

4 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$