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3 years ago

A snowball has a radius of 7cm. What is the volume of the snowball?

Mathematics

1 answer:

BabaBlast [244]3 years ago

7 0

Answer:

The volume is 1436.8 $in^2$ .

Step-by-step explanation:

-To find the volume of a sphere, you need the formula:

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

-Use the radius for the formula:

$V = \frac{4}{3} \pi 7^3$

-Then, you solve:

$V = \frac{4}{3} \pi 7^3$

$V = \frac{4}{3}$ × $\pi$ × $7^3$

$V = \frac{4}{3} \pi$ × $343$

$V = \frac{1372}{3} \pi = 1436.755$

-Round the answer to the nearest tenth:

$V = 1436.8$

So, therefore the volume is 1436.8 $in^2$ .

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$X \sim N(114.8,13.1)$

Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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:

And we can find this probability using the complement rule:

$P(z>1.924)=1-P(z$

Step-by-step explanation:

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".

Solution to the problem

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

$X \sim N(114.8,13.1)$

Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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: $P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140- 1114.8}{2.6})=P(z>1.924)$ And we can find this probability using the complement rule: