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Alborosie
1 year ago
14

Patrick is an electrical engineer who is testing the voltage of a circuit given a certain current and resistance. He uses the fo

llowing formula to calculate voltage: voltage=(current) × (resistance) The circuit he tests had a current of 4+j2 amps and a resistance of 2-j3 ohms. What is the voltage of the current?
Mathematics
1 answer:
morpeh [17]1 year ago
3 0

Based on the given current and the given resistance, the voltage of the current is 14 + j8

<h3>How to determine the voltage of the current?</h3>

From the question, we have the following parameters

current of 4+j2 amps and a resistance of 2-j3 ohms.

This can be rewritten as:

Current (I) = 4+j2 amps

Resistance (R) = 2-j3 ohms.

The formula to calculate voltage is given as

Voltage=(current) × (resistance)

Rewrite as:

V = I * R

Substitute the known values in the above equation

V = (4 + j2) * (2 - j3)

Expand the brackets

V = 8 - j4 + j12 + 6

Collect the like ters

V = 8 + 6 - j4 + j12

Evaluate the like term

V = 14 + j8

Based on the given current and the given resistance, the voltage of the current is 14 + j8

Read more about current and voltage at:

brainly.com/question/14883923

#SPJ1

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2 years ago
A glacier grew g feet over winter. It shrank 25 feet in summer.if the total change was -10 feet how much had the glacier grown i
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3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
What are the answers for both of these what are zeros<br> help
Archy [21]
Graph 5- e)
graph 6- b)
4 0
3 years ago
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