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atroni [7]
2 years ago
11

Cone A is divided into Cone B and Frustum C. B A 12cm C 20cm 20cm Express the volume of Frustum C as a fraction of the volume of

Cone A. Give your answer in its simplest form.​

Mathematics
1 answer:
zaharov [31]2 years ago
7 0

The volume of the frustum is volume of the whole cone(A) minus the smaller cone(B) which is would give the volume of frustum(C) = 256cm³

<h3>Calculation of a frustum</h3>

The volume of cone A V=πr²h/3

Where radius = 20cm

The volume of cone B = V=πr²h/3

Where radius = 12cm

Therefore volume of frustum =

V=π * 20² * h/3 - π * 12² *h/3

The variables will cancel out each other

V = 20² - 12²

V = 400- 144

V = 256cm³

Therefore, the volume of the frustum(C) = 256cm³

Learn more about cone here:

brainly.com/question/1082469

#SPJ1

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6 Bridesmaids want to make 330 decorations for a wedding. Each
Alex_Xolod [135]

Answer:

about  3.21 hours each

Step-by-step explanation:

First divide 330 by 6 since there are six people.

Every person needs to do 55 decorations.

I would take the average of 2-4 minutes and multiply it by 55.

55 x 3.5 = 192.5 minutes

to convert to hours divide by 60.

192.5 / 60 = 3.21

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An orchard sells 4 pounds of apples for one dollar how much would 5 pounds of apples cost
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1.25

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2 years ago
What is the vertex of the parabola?<br><br> A. (-1,0)<br> B. (0,-3)<br> C. (1,-4)<br> D. (3,0)
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5 0
2 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
3 years ago
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