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Fynjy0 [20]
3 years ago
9

A container in form of a frustum of a cone is 16 cm in diameter at the open end and 24 cm diameter at the bottom. If the vertica

l depth of the container is 8 cm calculate the capacity of the container.
Mathematics
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

The capacity of the container is 2546.78 cm³.

Step-by-step explanation:

The volume of the frustum of a cone is:

\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]

The information provided is:

r = 16/2 = 8 cm

R = 24/2 = 12 cm

h = 8 cm

Compute the capacity of the container as follows:

\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]

            =\frac{\pi\cdot8}{3}\cdot[(12)^{2}+(12\cdot 8)+(8)^{2}]\\\\=\frac{8\pi}{3}\times [144+96+64]\\\\=\frac{8\pi}{3}\times304\\\\=2546.784445\\\\\approx 2546.78

Thus, the capacity of the container is 2546.78 cm³.

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mestny [16]

Answer:

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Step-by-step explanation:

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3 years ago
Suppose that public opinion in a large city is 55% in favor of increasing taxes to support renewable energy and 45% against such
Harman [31]

Answer:

The approximate probability that more than 360 of these people will be against increasing taxes is P(Z> <u>0.6-0.45)</u>

                                              √0.45*0.55/600

The right answer is B.

Step-by-step explanation:

According to the given data we have the following:

sample size, h=600

probability against increase tax p=0.45

The probability that in a sample of 600 people, more that 360 people will be against increasing taxes.

We find that P(P>360/600)=P(P>0.6)

The sample proposition of p is approximately normally distributed mith mean p=0.45

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If x≅N(u,σ∧∧-2), then z=(x-u)/σ≅N(0,1)

Now, P(P>0.6)=P(<u>P-P</u>   >     <u>0.6-0.45)</u>

                             σ          √0.45*0.55/600

=P(Z> <u>0.6-0.45)</u>

       √0.45*0.55/600

8 0
4 years ago
A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.31 ​lb, the mean of all of the weight
a_sh-v [17]

Answer:

a) The difference is of 3.222 lbs.

b) 1.64 standard deviations.

c) Z = 1.64

d) Not significant, as the z-score of 1.64 is between -2 and 2.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The mean of all of the weights is x=2.088 ​lb, and the standard deviation of the weights is s=1.968 lb.

This means that \mu = 2.088, \sigma = 1.968

a. What is the difference between the weight of 5.31 lb and the mean of the​ weights?

This is X - \mu = 5.31 - 2.088 = 3.222

The difference is of 3.222 lbs.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

This is the z-score. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.31 - 2.088}{1.968}

Z = 1.64

1.64 standard deviations.

c. Convert the weight of 5.31 lb to a z score.

Z = 1.64, as found above.

d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly​ high, is the weight of 5.31 lb​ significant?

Not significant, as the z-score of 1.64 is between -2 and 2.

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Answer:

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