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3 years ago

A can holds 4 golf balls. The diameter of each golf ball is 5 cm. What is the volume of the empty space inside the can?

Mathematics

1 answer:

forsale [732]3 years ago

7 0

Answer:

Volume of the empty space in the can is 327.05 $cm^3$

Step-by-step explanation:

Given:

Number of golf ball the can holds = 4

Diameter of the golf ball = 5 cm

To Find:

Volume of the empty space in the can = ?

Solution:

Step 1: Finding the volume of one golf ball

Ball is shape of the sphere

So lets use the volume of the sphere formula

Volume of the golf ball = $\frac{4}{3}\pi r^3$

Radius = $\frac{5}{2}$ = 2.5 cm

Substituting the values

Volume of the golf ball

= $\frac{4}{3} \pi \times(2.5)^3$

= $\frac{4}{3} \pi \times (15.625)$

= $\frac{4}{3} \times 49.06$

= $\frac{196.24}{3}$

= 65.41

Step 2: Finding the volume of empty space of the can

volume of empty space of the can = volume of 5 golf ball

= 5 X 65.41

= 327.05

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The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th

Nikitich [7]

Answer:

a) $\mu_{\bar X} =13.35$

e.none of the above

b) $\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02$

a.0.0200

c) $z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5$

$P(\bar X< 13.32)= P(z$

a.0.0668

d) $z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5$

$z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5$

$P(13.30$

c.0.6853

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

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

$X \sim N(13.35,0.12)$