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2 years ago

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The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with mean μ=1497 and

standard deviation σ=322. A certain test-retake preparation course is designed for students whose SAT scores are in the lower 25%, percent of those who take the test in a given year. What is the maximum SAT score in 2014 that meets the course requirements?

1 answer:

Sav [38]2 years ago

4 0

Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

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The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

$\frac{w}{Q1} = \frac{35}{100}$

W = $1.2*\frac{35}{100}*1000kj/s$

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

$\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ$

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

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2 years ago

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Answer:

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Consider a 6-bit cyclic redundancy check (CRC) generator, G = 100101, and suppose that D = 1000100100. 1. What is the value of R

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Answer:

The value of R is 10101

Explanation:

As per the given data

D = 1000100100

G = 100101

Redundant bit = 6-bits - 1-bit = 5-bits

No add fice zero to D

D = 100010010000000

Now calculate R as follow

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Workings are attached with this question

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3 years ago

A girl operates a radio-controlled model car in a vacant parking lot. The girl's position is at the origin of the xy coordinate

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Answer:

a) 17.20

b) 11.31

c) 14.42

d) 12.65

Explanation:

(a)

The girl is at the origin of the x,y coordinates (i.e 0,0,0 )

the position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

at t=2s, the position vector is

$\vec{r}= 10, 14,0$

Therefore, the the distance between the car and the girl is

$s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}\$

s = 17.20

(b)

The position of the car at t = 0s is $\vec{r}_0 = 2,6,0$

The position of the car at t = 2s is $\vec{r}_2 = 10,14,0$

The distance of the car traveled in the interval from t=0s to t=2 s is as follows:

$s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02} = 11.31$

(c)

The position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

The velocity of the car is

$\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0$

the direction of the car's velocity at t = 2s is going to be

$\vec{v}\mid _t=2 8, 12,0$

Thus; The speed of the car is

$v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\ v_{t=2}= 14.42$

(d) the car's acceleration is:

$\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0$

The magnitude of car's acceleration at t=2s is

$\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65$

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3 years ago

A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°c. determine

Doss [256]

By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

<u>Given the following data:</u>

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

<h3>How to calculate the average heat transfer coefficient.</h3>

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

$T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45$

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

$\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}$

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

Kinematic viscosity, v = $1.750 \times 10^{-5}$ m²/s.

Thermal conductivity, k = 0.02699 W/mk.

Thermal diffusivity, α = $2.416 \times 10^{-5}$ m²/s.

Prandtl number, Pr = 0.7241.

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

$R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10} }\\\\R_{aL}=1.48 \times 10^{7}$

Also take note, g(Pr) is given by this equation:

$g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }$

g(Pr) = 0.430

For GrL, we have:

$G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7$

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

$N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times [\frac{4}{3} \times (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k$

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

$N_{uL}=\frac{h_{L}L}{k} =0.68 + \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19} } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 + \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19} })\\\\h_{L}=4.87\;W/m^2k$

Read more on heat transfer here: brainly.com/question/10119413

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