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Harrizon [31]
4 years ago
9

Marti is filling a 10– inch diameter ball with sand to make a medicine ball that can be used for exercising. To determine if the

medicine ball will be too heavy after it is completely full of sand, she did some research and found that there is approximately 100 pounds of sand per cubic foot. How heavy will the medicine ball be after it is filled with sand, rounded to the nearest pound? A.30 pounds B.58 pounds C.24 pounds D.83 pounds
Mathematics
1 answer:
Mice21 [21]4 years ago
6 0

Answer:

Option A. 30\ pounds

Step-by-step explanation:

step 1

Find the volume of the sphere ( medicine ball)

The volume is equal to

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

we have

r=10/2=5\ in ----> the radius is half the diameter

Convert inches to feet

Remember that

1 ft=12 in

r=5\ in=5/12\ ft

assume

\pi=3.14

substitute

V=\frac{4}{3}(3.14)(5/12)^{3}

V=0.3029\ ft^{3}

step 2

Find the weight of the ball

Multiply the volume in cubic foot by 100

0.3029*100=30.29\ pounds

Round to the nearest pound

30.29=30\ pounds

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answer
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If the diameter of a microscope's field is 0.12mm, how wide is an object that fills one-quarter of the field? (Explain how you'v
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Step-by-step explanation:

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Mobile users in India have gone up by 20% percent in a year. There are 540 million mobile users today.
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The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A
Aloiza [94]

Answer:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

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 annual income of a population, and for this case we know the following info:

\mu=37 and \sigma=28  and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

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( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

4 0
3 years ago
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