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3 years ago

¿Cuál es el volumen en pulgadas cúbicas de una esfera cuya circunferencia mide 64 pulgadas?

Mathematics

1 answer:

pochemuha3 years ago

6 0

Answer:

4445.18 pulgadas cúbicas

Step-by-step explanation:

Paso 1

La fórmula para la circunferencia de una esfera = 2πr

Circunferencia de una esfera = 64 pulgadas

Por eso,

64 = 2πr

Por lo tanto, encontramos r

Dividir ambos lados por 2π

64 / 2π = 2πr / 2π

r = 10.185916358 pulgadas

Radio r = 10.2 pulgadas

Paso 2

Volumen de una esfera

= 4/3 × π × r³

r = 10.2 pulgadas

Por lo tanto, 4/3 × π × (10.2) ³

= 4445.18 pulgadas cúbicas

Por lo tanto, el volumen en pulgadas cúbicas de una esfera = 4445.18 pulgadas cúbicas

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1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several ti

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1) $0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

b) $ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom, given by:

$df=n-1=5-1=4$

The Confidence level is 0.95 or 95%, and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom: $t_{\alpha/2}=2.776$

Now we have everything in order to replace into formula (1):

$0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588$

$0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219$

Part 2

The original margin of error is given by:

$ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653$

And we want 2/3 of the margin of error so then would be: $2/3 ME = 0.00001111$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (2)

Replacing we got:

$n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

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