Problem definition

Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati

Sedbober [7]

Answer:

n this question, we are asked to find the probability that

R1 is normally distributed with mean 65 and standard deviation 10

R2 is normally distributed with mean 75 and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

4 0

4 years ago

(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator

MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

$100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\$

Carrier frequency 100kHz

Upper side band

$100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz$

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

$100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz$

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

$100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz$

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

$100k, 99.9k, 99.8k\ \ and \ \99.6kHz$

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

$100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz$

3 0

3 years ago

Along with refining craft skills another way to increase the odds for career advancement is to

Xelga [282]

The acquisition of additional certifications with a personal refined craft skills can increase the odds for career advancemen.

<h3>What is a career advancement?</h3>

An advancement is achieved in a career if a professional use their skill sets, determination or perserverance to achieve new career height.

An example of a career advancement is when an employee progresses from entry-level position to management and transits from an occupation to another.

Therefore, the Option A is correct.

Read more about career advancement

<em>brainly.com/question/7053706</em>

7 0

2 years ago

Which of the following statements about pitot-static systems is FALSE? a). A pitot probe measures the Total Pressure of the free

Pavlova-9 [17]

Answer:

C

Explanation:

Pitot tube:

Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.

As we know that stagnation pressure is the summation of dynamic and static pressure.

Stagnation pressure = Static pressure + Dynamic pressure

So

Dynamic pressure = Stagnation pressure - Static pressure

We know that dynamic pressure

$P_{dynamic}=\dfrac{\rho V^2}{2}$

On the other hand Pitot tube measure the dynamic pressure.

So option C is correct.

5 0

3 years ago

7. An energy auditor is part of what career field?

Rashid [163]

An energy auditor is part of power operations

6 0

2 years ago