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iVinArrow [24]
2 years ago
8

The output voltage of a power supply is normally distributed with mean 12 V and standard deviation 0.11 V. If the upper and lowe

r specifications for voltage are 12.15 V and 11.85 V respectively, what is the probability that the power supply selected at random will confirm to the specifications on voltage?
Engineering
1 answer:
podryga [215]2 years ago
4 0

Answer:

82.62%

Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation.\\\\Given \ that\ \mu=12V, \sigma=0.11V.\\\\For\ x11.85V:\\\\z=\frac{11.85-12}{0.11} =-1.36\\\\

From the normal distribution table, P(11.85 < x < 12.15) = P(-1.36 < z < 1.36) = P(z < 1.36) - P(z < -1.36) = 0.9131-0.0869 = 0.8262 = 82.62%

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Answer:

None of these is correct.

Explanation:

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3 0
2 years ago
This problem demonstrates aliasing. Generate a 512-point waveform consisting of 2 sinusoids at 200 and 400-Hz. Assume a sampling
aalyn [17]

Answer and Explanation:

clear all; close all;  

N=512;  

t=(1:N)/N;

fs=1000;  

f=(1:N)*fs/N;

x= sin(2*pi*200*t) + sin(2*pi*400*t);  

y= sin(2*pi*200*t) + sin(2*pi*900*t);

for n = 1:20  

a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))

b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))  

c(n) = sqrt(a(n).^2+b(n).^2)  

theta(n) =-(360/(2*pi))*atan(b(n)./a(n));  

end  

plot(f(1:20),c(1:20),'rd');

disp([a(1:4),b(1:4),c(1:4),theta(1:4)])

8 0
3 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
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Answer: true

Explanation:

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Sorry need.points I'm new

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