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borishaifa [10]
2 years ago
3

A random experiment consists of selecting two balls in succession from an urn containing two black balls and and one white ball.

(a) Specify the sample space for this experiment. (b) Suppose that the experiment is modified so that the ball is immediately put back into the urn after the first selection. What is the sample space now? number of repetitions of the experiment in part a? In part b? outcome of the first draw in either of these experiments? (c) What is the relative frequency of the outcome (white, white) in a large (d) Does the outcome of the second draw fron the urn depend in any way on the outcome of the first draw in either of these experiments?
Engineering
1 answer:
Alex2 years ago
8 0

Answer:

(a) (B,B) (B,W) (W,B)

(b) (B,B) (B,W) (W,B) (WW)

(c) 1/9

(d) Yes it does in the experiment in part (a)

Explanation:

Black balls = 2

White Balls = 1

No. of balls selected = 2

(a) The balls are selected without replacement so, we can define the sample space as the kinds of possible outcomes:

1st ball is black and the 2nd ball is also black: (B, B)

1st ball is black and the 2nd ball is white: (B,W)

1st ball is white and the second ball is black: (W,B).

So, the sample space is: (B,B), (B,W), (W,B)

(b) Now, the balls are selected with replacement i.e. the first ball is taken out and then put back into the urn and then the second ball is taken out. So, the possibilities can be:

1st ball is black, 2nd ball is black: (B, B)

1st ball is black, 2nd ball is white: (B, W)

1st ball is white, 2nd ball is black: (W, B)

1st ball is white, 2nd ball is white: (W, W)

So, the sample space is: (B,B), (B,W), (W,B), (W,W).

No. of repetitions of the experiment in part (a) is 1. i.e. (B,B)

No. of repetitions of the experiment in part (b) is 2 i.e. (B,B) and (W,W).

The outcome of the first draw can be determined based on the probabilities for each ball.

P(Black) = No. of black balls/Total no. of balls

P(Black)= 2/3

P(White) = No. of white balls/Total no. of balls

P(White) = 1/3

Outcome of the first draw is most likely to be a black ball in either of these experiments because it has a higher probability.

(c) We can calculate the relative frequency of (W,W) as:

(W,W) = 1/3 x 1/3 = 1/9

(d) Yes, the outcome of the second draw depends on the first draw in the experiment in part (a) because the balls are drawn without replacement. If the first ball is white then the second ball can not be white hence, the outcome of the second draw depends on the first one.

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Alik [6]

Answer:

Given data

w1/w2=6.5/1

Power=5 KW

wp=1800 rpm

angle=14 degrees

Based on above values,the minimum diameter=30 mm

5 0
2 years ago
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
2 years ago
Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = \sqrt{400^2 + 650^2 - 2(400)(650)cos30}

      = \sqrt{160000 + 422500 - 80210}

      = 708.72 Ib

using the sine law to determine the design angle

\frac{sin}{400}  = \frac{sin 30}{Fac}

hence ∅ = sin^{-1} (\frac{sin 30 *400}{708.72} )

              = sin^{-1} 0.0871 =  4.9968 ≈ 5⁰

4 0
2 years ago
Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi
leva [86]

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

3 0
2 years ago
The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
Sonja [21]

Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, Wmec=0.95*6kW=5.7 kW.

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump \Delta P. So P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W.

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is \varepsilon=3.6kW/5.7kW=0.6315=63.15\%

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T[/tex]

Finally, the temperature rise is:

\Delta T=2100/75348 \ºC=0.028 \ºC

7 0
2 years ago
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