A pipe fitter would fabricate which one of the following systems?

University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, r

abruzzese [7]

Solution :

The percentage of the students who have a chance of repeating their current year = 3%

The drop out students for the first year and the sophomores = 6%

Drop out rate of first year and the seniors = 4%

Now for the state space :

S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }

Therefore

the first year students are indicated as '1'

Sophomores are indicated as '2'

Juniors are indicated as '3'

Seniors are indicated as '4

Graduates are indicated as 'G'

Dropouts are indicated as 'D'

The transition diagram is attached below.

The probability of the students who have the chance of repeating their current year = 3/100 = 0.03

Probability of first year dropouts and sophomores = 6/100 = 0.06

Probability of dropout rate of juniors and seniors = 4/100 = 0.04

Therefore, the probability matrix can be made as :

1 2 3 4 G D

$\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}$ $\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}$

Here, G represents 'graduates' and D represents 'Dropouts.'

5 0

3 years ago

PLS HELP

RSB [31]

Answer:

T=kg·m^2/s^2

Explanation:

T = kg (m^2/s^2) m^3 /m^3

Here I wrote down the unit for every dimension.

T=kg m^2 / s^2

m^3 is divided between m^3, this is equal to 1.

Result: T=kg·m^2/s^2

PD: I'm not sure if this is what you ask for. I hope it helps

4 0

3 years ago

If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ p ^ is approximately a normal distribution.

defon

Answer:

It is true that the sampling distribution of the proportions is approximately a normal distribution.

Explanation:

Solution

Given that:

N = 50

P= 0.8

Thus

p ^ =√p(1-p)/n

p ^=√0.8 (1-0.8)/50

p ^= 0.0566

Therefore,we conclude that the sampling distribution of the proportions is approximately a normal distribution.

4 0

3 years ago

An oil reservoir has a average porosity of 20%, an area of 100 acres, and a av- erage thickness of 10 feet. The connate water sa

hram777 [196]

Answer:a) recovery factor = 53.3% ,bi) volume of oil per acre-ft = 11090.64bbl/acre-ft

bii) bulk volume of the reservoir in acre-ft = 1000 acre-ft

c) 62214.74 cubic feet

Explanation: a) recovery factor is the percentage amount of oil that can recovered from a reservoir, it is the oil produced divided by oil initially in place

Recovery factor= 1 - (Soi/1-Swi)

= 1 - (0.35/1-0.25)

= 0.533 × 100%

= 53.3%

bi) volume of oil which may be recovered per acre-ft = 7758.porosity.(1- Swi-Soi)

= 7758 x 0.2 x (1-0.25-0.35)

= 620.64 bbl/acre-ft

bii) bulk volume of the reservoir in acre-ft= Area x thickness

= 100 acres x 10 ft

= 1000 acre-ft

c) total volume of oil in cubic feet

since we have gotten volume as 1000 acre-ft we simply multiply it by the volume of oil gotten in answer bi)

= 1000 acre-ft x 620.64 bbl/ acre-ft

= 620640bbl

So we convert from barrel(bbl) to cubic feet, and 1 barrel is equal to 5.609 cubic feet, so to convert the answer from barrel to cubic feet we multiply the answer 620640 bbl by 5.609

= 620640 bbl x 5.609

= 3481580.711 cubic feet

4 0

3 years ago

Isormophous phase diagram

shusha [124]

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0

3 years ago