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3 years ago

(3-4i)/(4-5i) HELP ME PLEASE --- algebra 2 question i = the square root of -1 (<<

Mathematics

1 answer:

posledela3 years ago

8 0

Answer:

Step-by-step explanation:

Multiply bu the conjuguate of the demonimator.

$\dfrac{3-4i}{4-5i} \\\\\\=\dfrac{(3-4i)(4+5i)}{(4-5i)(4+5i)} \\\\\\=\dfrac{12-16i+15i+20x}{16-i^2*25} \\\\\\=\dfrac{32-i}{41} \\\\\\=\dfrac{32}{41} -\dfrac{i}{41} \\\\\\$

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PLZ HELP MEEEE Four plus eight divided by two times (six minus three)

Maru [420]

The answer is 16
by PEMDAS

5 0

3 years ago

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Write the equation of the line that passes through the points (3, 1) and (-2, 6).

tia_tia [17]

Answer:

Equation is Y=mx+c (that's how I know it, it may differ)

Step-by-step explanation:

First get the slope which is M and you just gotta do the second y - the first y from the brackets and divide them by the second x - the first x

6-1 /-2-3 = 5/-5

= -1

So now u gotta find c, C= Y-mx

And u have got two brackets, which have an x and y, and u now have m.

Pick one of the brackets like (-2,6) and put them in the equation so it will be : C =6+1*-2

So C =4

The equation now is : Y = - 1x + 4

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2 years ago

The Pick 4 games in many state lotteries announce a four‑digit winning number each day. Each of the 10,000 possible numbers 0000

barxatty [35]

Answer:

a) P(a)= 0.0001

b) P(b)= 0.0024

Step-by-step explanation:

Given:

Chosen number = 3079

Possible numbers that have chance of winning from 0000 to 9999= 10,000

Winning condition= if made choice matches the winning digits

Now

a)What is the probability that the winning number matches your number exactly?

From 0000 to 9999, only one time can the number 3079 can be the winning number, so

P(a)= 1/10000

= 0.0001

b)What is the probability that the winning number has the same digits as your number in any order?

Winning numbers from 0000 to 10000 that have same digits as 3079 are

4 x 3 x 2 x 1 = 24 ways

Following are the 24 numbers from 0000 to 9999 that have matching digits to 3079:

0379

0397

0973

0937

0793

0739

3079

3097

3970

3907

3790

3709

7039

7093

7930

7903

7390

7309

9073

9037

9370

9307

9730

9703

P(b)= 24/10000

=0.0024 !

8 0

4 years ago

Isabel is running for president of the chess club, and she received 33 votes. There are 60 members in the club. What percentage

borishaifa [10]

Answer:

55%

Step-by-step explanation:

Create an equation.

p = votes received / total members

Solve

p = 33 / 60

p = .55

Multiply

Multiply the decimal by 100 to convert the decimal into a whole number.

.55 * 100 = 55

Answer

55 percent of the club members voted for Isabel.

4 0

3 years ago

Read 2 more answers

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

3 years ago