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2 years ago

A baseball has a 45 cm diameter. what is the volume of the contents of the ball

1 answer:

3 0

The volume of a sphere is found from the equation V=4/3πr
r is the radius.
If the radius is 45 cm and pi is about 3.14, you can rewrite and equation as V=4/3(3.14)(45)
Then just solve by multiplying all the terms!
4/3(3.14) = 4.19
4.19(45) = 188.55
So the volume of the ball is 188.55 cm cubed.

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Given AC ≅ FE and CB ≅ ED which statement is correct?

Kisachek [45]

Answer:

D is the correct answer.

Step-by-step explanation:

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2 years ago

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Byron has 1 7/10 kilograms of black pepper. He uses 7/8 of the black pepper and splits between 7 pepper shakers. How much pepper

IrinaVladis [17]

Amount of black pepper = 1 7/10 kg = 17/10 kg.
Amount used is
(7/8)*(17/10) = 119/80 kg

This amount is split between 7 shakers. Each shaker has
(119/80)/7 = 17/80 kg = 0.2125 kg

Answer: 17/80 kg (or 0.2125 kg)

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2 years ago

Judy’s brother Sam has a collection of 96, but what are the 10 way Sam can divide his comic books equal groups

Ierofanga [76]

He can divide by 2,3,4,6,8,12,16,24,32, and 48 but

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2 years ago

The projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is

NeX [460]

S(t) = -4.9t^2 + 19.6t + 24.5

when it reaches the ground s(t) = 0

so we have -4.9t^2 + 19.6t + 24.5 = 0

solve this for t:-

-4.9(t^2 - 4t - 5) = 0

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2 years ago

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Tesla wanted to determine the average miles per kWh that their vehicles get across all models and variations. They took a sample

LiRa [457]

Answer:

$\mu_{\bar x} = \mu = E(X) =30KWh$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3$

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".

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".

Solution to the problem

For this case we select a sample of n =100

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

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

So then the sample mean would be:

$\mu_{\bar x} = \mu = E(X) =30KWh$

And the standard deviation would be:

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3$

6 0

2 years ago