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3 years ago

Who knows how to answer this. It's Dividing complex numbers.

Mathematics

2 answers:

Andreas93 [3]3 years ago

7 0

Answer:

$\large\boxed{1+i}$

Step-by-step explanation:

$\frac{4+2i }{3-i}$

Factor the numerator

$\frac{2(2+i) }{3-i}$

Multiply both numerator and denominator by the conjugate of the denominator.

$\frac{2(2+i) }{3-i} \cdot \frac{3+i }{3+i} = \frac{2(2+i)(3+i) }{9-i^2}$

Once $i^2=-1$

$\frac{2(2+i)(3+i) }{10}$

Solving the numerator

$2(2+i)(3+i) = 2(6+2i+3i+i^2) = 2(6+5i-1)=2(5+5i)$

$\frac{2(5+5i)}{2 \cdot 5} =\frac{5+5i}{5}=\boxed{1+i}$

Elodia [21]3 years ago

4 0

Answer:

answe is 1+i

mark this as brainliest!!

mark this as brainliest!!follow me!

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Alex and bob are playing 5 chess games. Alex is 3 times more likely to win than bob. What is the probability that both of them w

givi [52]

Answer:

12.2%

Step-by-step explanation:

Lets firstly find out the probabilities for Alex and Bob. We know that the prb that Alex wins is 3 times the prob for Bob, so:

P(A) = 3P(B)

And, as these values are probabilities they must sum 1:

P(A) + P(B) = 1

Replacing the first equation in the second:

3P(B) + P(B) = 1

4P(B) = 1

Dividing both sided by 4:

4P(B)/4 = 1/4

P(B) =1/4

So, the probability for Bob is 1/4. Then the probability for Alex is 3*1/4 = 3/4

Now we have to find the probability that both win at least two games. Lets think about the possible combination of cases. As they play 5 games and both have to win at least to there is a remaining game that cane variate, this is, can be won by Alex or Bob. So, we can have:

AABBB

AABBA

Where A is "Alex win" and B is "Bob win" (here we do not pay attention to the order).

Thus, we have to find the probability that Alex wins 3 games and Bob 2 games, and the probability that Alex wins 2 games and Bob 3 games. As the games are independent, i.e., the result of one games does not affect the followings we can just multiply the probabilities:

P(3A and 2B) = (3/4)*(3/4)*(3/4)*(1/4)*(1/4) = 0.42*0.06 = 0.026

P(2A and 3B) = (3/4)*(3/4)*(1/4)*(1/4)*(1/4) = 0.56*0.016 = 0.096

So,

P (at least 2) = P(3A and 2B) + P(2A and 3B) = 0.026 + 0.096 = 0.122 = 12.2%

7 0

3 years ago

Im so lost im in algebra 1 and i feel so slow

Andrei [34K]

Answer:

Hey but you get better at in no time lol

Step-by-step explanation:

no need to worry

7 0

2 years ago

Please simplify 6-(2-4x)

umka2103 [35]

I'm pretty sure the answer would be 4+4x

6 0

3 years ago

Read 2 more answers

Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hear

Sedbober [7]

Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of $\mu = 1.35$ .

The standard deviation is of $\sigma = 0.33$ .

Item 1:

Considering his ratio, we have that X = 0.73, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.73 - 1.35}{0.33}$

$Z = -1.88$

His z-score was of Z = -1.88.

Item 2:

The probability is the p-value of Z = -1.88, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.96 - 1.35}{0.33}$

$Z = 1.85$

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

7 0

2 years ago

Solve for x. Enter your answer in the box. x = °

liq [111]

As we can notice the shape is a Pentagon. A Pentagon's angles should add up to 540. So we can use simple addition and subtraction to solve.

121 + 108 + 102 + 100 = 431

540 - 431 = 109

109 should be your answer

4 0

3 years ago

Read 2 more answers

Who knows how to answer this. It's Dividing complex numbers.​

Who knows how to answer this. It's Dividing complex numbers.