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4 years ago

The volume of a sphere is 5,000pi m3. What is the surface area of the sphere to the nearest square meter?

1 answer:

7 0

To get the surface area of the sphere, we need to know the radius. Since the volume is given, we can get the radius.
Volume = $\frac{4}{3}$ π r³
5000πm³ = $\frac{4}{3}$ π r³
5000m³ = $\frac{4}{3}$ r³
5000m³ x 3 = 4r³
$\frac{15000m³}{4}$ = $\frac{4r³}{4}$ <span>
</span>∛3750 = ∛r³

r = 15.536m
r = 15.54m
Since the radius is now known, substitute it directly to the surface area formula
A = 4π x r²
A = 4π x (15.54)²
A= 3,034.67 m² or 3,035 m²

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saul85 [17]

Well if she is making dinner for 10 people and each person want 2 pieces then you need to multiple 10 x 2 = 20. So she is going to need to make 20 pieces.

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4 years ago

The distribution of the number of people in line at a grocery store has a mean of 3 and a variance of 9. A sample of the numbers

Airida [17]

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a) $P(\bar X >4)=P(Z>\frac{4-3}{\frac{3}{\sqrt{50}}}=2.357)$

$P(Z>2.357)=1-P(Z$

b) $P(\bar X \frac{2.5-3}{\frac{3}{\sqrt{50}}}=-1.179)$

$P(Z$

$P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}}$

$P(-1.179$ 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".

Part a

Let X the random variable that represent the number of people of a population, and for this case we know that:

Where $\mu=3$ and $\sigma=\sqrt{9}=3$

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P(\bar X >4)= P(z> \frac{4-3}{\frac{3}{\sqrt{50}}})$

And using a calculator, excel or the normal standard table we have that:

$P(Z>2.357)=1-P(Z$

Part b

$P(\bar X \frac{2.5-3}{\frac{3}{\sqrt{50}}})$

And using a calculator, excel or the normal standard table we have that:

$P(Z$

Part c

For this case we want this probability:

$P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}}$

And using a calculator, excel or the normal standard table we have that:

$P(-1.179$

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3 years ago

In December it snowed 2 days for every 8 days it did not snow what percent of the days in December did it snow

sergey [27]

I think it might be 20%. If in a total of 10 days it only snowed 2, then you could take 2/10, which is .2, then multiply by 100 to make it a percentage... which would be 20%.

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3 years ago