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3 years ago

In an international film festival, a penal of 11 judges is formed to judge the best film. At

Engineering

1 answer:

Julli [10]3 years ago

7 0

Answer:

International Film Festival

Judging the best best film:

a. The probability that out of five judges (random sample), three are in favor of film FA is:

= 33%.

b. The demerits of classical probability are:

1. Classical probability can only be used with events that have definite numbers of possible outcomes.

2. Classical probability can only handle events where each outcome is equally likely.

3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.

Explanation:

a) Number of judges = 11

Number of judges in favor of FA film = 6

Number of judges in favor of FB film = 5

Probability of judges in favor of FA film = 6/11

Probability of judges in favor of FB film = 5/11

Random sample of judges = 5

Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11

= 18/55

= 33%

b) Classical probability is the simple probability showing that each event has equal chance of happening. It can be contrasted with empirical probability that is obtained from experiments.

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A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp

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Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

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3 years ago

The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e

snow_lady [41]

The equivalent force-couple system at O is the force and couple experienced when at point O due to the applied force at point A

The equivalent force couple system at O due to force F are;

Force, F = (8.65·i - 4.6·j) KN

Couple, M₀ ≈ 40.9 k kN·m

The reason the above values are correct is as follows:

The known values for the cantilever are;

The height of the beam = 0.65 m

The magnitude of the applied force, F = 9.8 kN

The length of the beam = 4.9 m

The angle away from the vertical the force is applied = 26°

The required parameter:

The equivalent force-couple system at the centroid of the beam cross-section of the cantilever

Solution:

The equivalent force-couple system is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The equivalent force $\overset \longrightarrow F$ = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The equivalent force $\overset \longrightarrow F$ ≈ (8.65·i - 4.6·j) KN

The couple acting at point O due to the force F is given as follows;

The clockwise moment = 9.8 kN × cos(28°) × 4.9

The anticlockwise moment = 9.8 kN × sin(28°) 0.65/2

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The sum of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = 9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9 ≈ 40.9 kN·m

The couple acting at O, due to F, M₀ ≈ 40.9 kN·m

The equivalent force couple system acting at point O due the force, F, is as follows

F = (8.65·i - 4.6·j) KN

M₀ ≈ 40.9 k kN·m

Learn more about equivalent force systems here:

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