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3 years ago

Beach ball with diameter of 24 inches what is the volume of air in the beach ball to the nearest of a cubic foot?

Mathematics

1 answer:

Kamila [148]3 years ago

5 0

Answer:

4 ft³

Step-by-step explanation:

From the question;

Diameter of the beach ball is 24 Inches

We are required to determine the volume of the air in the beach ball;

We need to know that the volume of the air is the same as the volume of the beach ball.
The ball is in the shape of a sphere

$Volume of sphere=\frac{4}{3}pi*r^3$

Radius = 12 inches

But, 1 ft = 12 inches

Thus, radius = 1 ft

Therefore;

Taking pi as 3.142

Volume = 4/3 × 3.142 × 1³

= 4.189 ft³

= 4 ft³ (nearest cubic foot)

Therefore, the volume of the air in the beach ball is 4 ft³

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Answer:

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And the variance would be given by:

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And the deviation would be:

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Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

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