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3 years ago

The surface area of a sphere is 113.04 mm2. What is the volume of the sphere? Use 3.14 for π and round to the nearest hundredth.

2 answers:

5 0

By definition, the surface area of the sphere is given by:
$A = (4) * ( \pi ) * (r ^ 2)
$
Where,
r: sphere radio
Clearing r we have:
$r = \sqrt{A / (4 * \pi )} 

$
Substituting values:
$r = \sqrt{113.04 / (4 * 3.14 )}$
$r = 3 mm
$
Then, the volume of the sphere is given by:
$V = (4/3) * (\pi) * (r ^ 3)
$
Substituting values we have:
$V = (4/3) * (3.14) * (3 ^ 3)

 V = 113.04 mm ^ 3$
Answer:
the volume of the sphere is:
$V = 113.04 mm ^ 3$

ElenaW [278]3 years ago

5 0

Answer:

volume of the sphere will be 113.04 mm³

Step-by-step explanation:

Surface area of a sphere is represented by

A = 4πr²

r² = $\frac{A}{4\pi }$

$r=\sqrt{\frac{A}{4\pi } } =\sqrt{\frac{113.04}{4*(3.14)} }=\sqrt{9}=3mm$

and volume of the sphere is

V = $\frac{4}{3}\pi r^{3}$

= $\frac{4}{3}\pi (3)^{3}$

= 4π × 9

= 36π

= 36 (3.14)

= 113.04 mm³

Therefore, volume of the sphere will be 113.04 mm³

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Suppose the heights of a population of people are normally distributed with a mean of 65.5 inches and a standard deviation of 2.

kupik [55]

Answer:

a) $P(64.2$

b) 69.764

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(65.5,2.6)$

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(64.2$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(64.2$

And we can find this probability on this way:

$P(-0.50$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.50$

3) Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

$P(X$