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Pepsi [2]
2 years ago
12

An igloo can be modeled as a hemisphere. Its radius measures 6.3 m. Find its volume in cubic meters. Round your answer to the ne

arest tenth if necessary.
Mathematics
1 answer:
Bas_tet [7]2 years ago
6 0

The volume of the igloo is 523.43 m^3

<h3></h3><h3>How to get the volume of the igloo?</h3>

We know that the volume of a sphere of radius R is:

V = (4/3)*pi*R^3

Where pi = 3.14

Then the volume of a hemisphere is half of that.

In this case, we have a hemisphere with a radius of 6.3m, then the volume will be:

V = (1/2)*(4/3)*3.14*(6.3m)^3 = 523.43 m^3

If you want to learn more about volume:

brainly.com/question/1972490

#SPJ1

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Answer:

Answer:

x=\frac{75}2

Step-by-step explanation:

Give letters, as in the attached image.

The triangles ABE and CDE are similars (AAA). In particular AE:CE=BE:DE \rightarrow 46:30=(x+20):x \rightarrow 46x=30(x+20) \rightarrow 23x=15x+300 \rightarrow 8x=300\rightarrow  x=\frac{75}2

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The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x
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Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

       P(a < X < b) =  \int_{a}^{b} \frac{2}{x^{3}} dx = 2\int_{a}^{b} x^{-3} dx

                            = 2[ \frac{x^{-3+1} }{-3+1}]^{b}_a   dx    { Because \int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a }

                            = 2[ \frac{x^{-2} }{-2}]^{b}_a = \frac{2}{-2} [ x^{-2} ]^{b}_a

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a) Now P(X < 5) = P(1 < X < 5)  {because x > 1 }

     Comparing with general probability we get,

     P(1 < X < 5) = \frac{1}{1^{2} } - \frac{1}{5^{2} } = 1 - \frac{1}{25} = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/8^{2} - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = \frac{1}{6^{2} } - \frac{1}{10^{2} } = \frac{1}{36} - \frac{1}{100 } = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

                                = (\frac{1}{1^{2} } - \frac{1}{6^{2} }) + (1/10^{2} - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

               ⇒  P(1 < X < x) = 0.75

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               ⇒  \frac{1} {x^{2} } = 1 - 0.75 = 0.25

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Therefore, value of x such that P(X < x) = 0.75 is 2.

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