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Nastasia [14]
2 years ago
5

A lampshade has a height of 12cm and upper and lower diameters of 10cm and 20cm A)what area of material is required to cover tha

t curved surface of the frustum B)what is the volume of the frustum (Give both answers in terms of pie)
Mathematics
1 answer:
netineya [11]2 years ago
6 0

Answer:

(a)\ Area = 195\pi

(b)\ Volume = 700\pi

Step-by-step explanation:

Given

h = 12

d_1 = 10 --- lower diameter

d_2 = 20 --- upper diameter

Solving (a): The curved surface area

This is calculated as:

Area = \pi l(r_1 + r_2)

Where

r_1 = 0.5 * d_1 = 0.5 * 10 = 5 --- lower radius

r_2 = 0.5 * d_2 = 0.5 * 20 = 10 --- upper radius

And

l = \sqrt{h^2 + (r_1 - r_2)^2 ---- l represents the slant height of the frustrum

l = \sqrt{12^2 + (5 - 10)^2

l = \sqrt{12^2 + (-5)^2

l = \sqrt{144 + 25

l = \sqrt{169

l = 13

So, we have:

Area = \pi l(r_1 + r_2)

Area = \pi * 13(5 + 10)

Area = \pi * 13(15)

Area = 195\pi

Solving (b): The volume

This is calculated as:

Volume = \frac{1}{3}\pi * h * (r_1^2 + r_2^2 + (r_1 * r_2))

This gives:

Volume = \frac{1}{3}\pi * 12 * (5^2 + 10^2 + (5 * 10))

Volume = \frac{1}{3}\pi * 12 * (25 + 100 + (50))

Volume = \frac{1}{3}\pi * 12 * (175)

Volume = \pi *4 * 175

Volume = 700\pi

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Answer:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

Step-by-step explanation:

We know the following info:

\bar X_1 = 21.9 sample mean for group 1

\bar X_2 = 19.8 sample mean for group 2

s_1 = 3.4 sample standard deviation for group 1

s_2 = 3.5 sample standard deviation for group 2

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n_2 = 200 sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

Now we just need to find the critical value. The confidence level is 0.95 then the significance is 1-0.95 =0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df= n_1 +n_2 -2= 200+200-2= 398

The critical value for this case would be :t_{\alpha/2}=1.966  

And replacing into the confidence interval formula we got:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

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Answer:

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Step-by-step explanation:

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12=3.46 rounded to 3.5

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