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Korvikt [17]
3 years ago
10

What is the volume of the cone with diameter 4 ft and height 4 ft? Round to the nearest cubic foot.

Mathematics
2 answers:
Daniel [21]3 years ago
3 0

Answer:

V = 67.0206 m3

Step-by-step explanation:

ss7ja [257]3 years ago
3 0

Answer: 17ft^3

Step-by-step explanation:

The volume of a cone can be calculated with the following formula:

V_{(cone)}=\frac{1}{3}\pi r^2h

Where "r" is the radius and "h" is the height.

The radius is half the diameter, then "r" is:

r=\frac{4ft}{2}\\\\r=2ft

Since we know that radius and the height, we can substitute them into the formula.

The volume of the cone  to the nearest cubic foot is:

 V_{(cone)}=\frac{1}{3}\pi (2ft)^2(4ft)

 V_{(cone)}=17ft^3

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nexus9112 [7]

Answer:

area of triangle = 135 mm²

Step-by-step explanation:

Area of a triangle = 1/2 × b × h

Area of a triangle = 1/2 × 27 × 10

Area of a triangle = 135 mm²

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Answer:

f(3)=2

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To find f(3), we simply need to find the ordered pair at x=3.

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An electronic device factory is studying the length of life of the electronic components they produced. The manager selects two
Temka [501]

Answer:

The confidence interval will be given by:

200000 \pm 44.72z, in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of 1 - \frac{1 - z}{2}

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In the context of this problems:

It means that the sampling distributions of the sample mean of 500 components will be approximated normal, with mean 200,000 and standard deviation s = \frac{1000}{\sqrt{500}} = 44.72

To build the confidence interval:

The confidence interval for the average length of life of the electronic components they produced will be given by:

200000 \pm 44.72z, in which z is related to the confidence level.

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