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AleksAgata [21]

3 years ago

6

A ball bearing is shaped like a sphere and has a diameter of 2.5 centimetres. What is the volume contained inside the ball beari

ng? Use 3.14 for pi. Round your answer to the nearest hundredth. 8.18 cubic centimeters 7.15 cubic centimeters 7.08 cubic centimeters 6.89 cubic centimeters

2 answers:

kvv77 [185]3 years ago

8 0

To solve:

The diameter is already given (2.5), and it can be divided by 2 to find the radius.

2.5/2=1.25

Plug 1.25 and Pi into the equation for a sphere to find the volume:

(4/3)(1.25^3)(3.14)=8.177

Answer: About 8.18 cm^3

grandymaker [24]3 years ago

5 0

Answer: 8.18 cubic centimeters

Step-by-step explanation:

Given : A ball bearing is shaped like a sphere .

The diameter of ball = 2.5 centimeters

Then , the radius of the ball = $\dfrac{2.5}{2}=1.25\text{ cm}$

We know that the volume of a sphere is given by :-

$\text{Volume}=\dfrac{4}{3}\pi r^3$

Then , the volume of ball is given by :-

$\text{Volume}=\dfrac{4}{3}(3.14) (1.25)^3\\\\\Rightarrow\ \text{Volume}=8.17708333333\approx8.18\text{ cubic centimeters}$

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2 years ago

Ashin has been working in UAE for a decade and saved AED 125,500 which he wants to transfer now to his parent's account in India

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3 years ago

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

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$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

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$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

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