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12345 [234]
3 years ago
8

Help again im stuck on this problem

Mathematics
1 answer:
kap26 [50]3 years ago
4 0

Answer:

0.85,8/9,0.92

Step-by-step explanation:

8/9=0.8888

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The set of all Real numbers between 4 and 7 including 4
erik [133]
4,5,6,7 is your answer
8 0
3 years ago
Given the complex number z_1=3\big(\cos \frac{14\pi}{15} +i\sin \frac{14\pi}{15}\big)z 1 ​ =3(cos 15 14π ​ +isin 15 14π ​ ) and
mars1129 [50]

The product of <em>z₁ =</em> 3 · (cos 14π/15 + i · sin 14π/15) and <em>z₂ =</em> 3 √3 · (cos 11π/15 + i · sin 11π/15) in <em>rectangular</em> form with fully simplified expressions is <em>z₁ · z₂ =</em> 7.794 - i · 13.5.

<h3>How to determine the product of two complex numbers</h3>

Let be two numbers of the form <em>z = a + i · b</em>, where <em>i =</em> √-1, the product of two of these numbers in <em>rectangular</em> form is described by the following formula:

<em>z₁ · z₂ = (a + i · b) · (c + i · d) = (a · c - b · d) + i · (a · d + b · c)</em>   (1)

If we know that a = 3 · cos 14π/15, b = 3 · sin 14π/15, c = 3√3 · cos 11π/15, d = 3√3 · sin 11π/15, then the result in rectangular form is:

<em>z₁ · z₂ =</em> 7.794 - i · 13.5

The product of <em>z₁ =</em> 3 · (cos 14π/15 + i · sin 14π/15) and <em>z₂ =</em> 3 √3 · (cos 11π/15 + i · sin 11π/15) in <em>rectangular</em> form with fully simplified expressions is <em>z₁ · z₂ =</em> 7.794 - i · 13.5. \blacksquare

<h3>Remark</h3>

The statement presents typing mistakes and is poorly formatted, the correct form is introduced below:

<em>Given the complex number z₁ = 3 · (cos 14π/15 + i · sin 14π/15) and z₂ = 3 √3 · (cos 11π/15 + i · sin 11π/15), express the result of z₁ · z₂ in rectangular form with fully simplified fractions and radicals.</em>

<em />

To learn more on complex numbers, we kindly invite to check this verified question: brainly.com/question/10251853

8 0
2 years ago
!!!!!!!!! Helppppppp plsssss
Travka [436]

Answer:

(6 / 4) * (7 + 9)

Step-by-step explanation:

sry that took me so long lol

4 0
2 years ago
What fraction of the total # of students earned at least a C? Include small explanation on how you did it.
alisha [4.7K]
6/30 or 3/15.Add all the number together and then you get 6/30 because 6 of them were c.Then simplify it.Also, I like your picture |-/
5 0
4 years ago
Read 2 more answers
An experimenter flips a coin 100 times and gets 59 heads. Find the 98% confidence interval for the probability of flipping a hea
lara [203]

Answer:

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

An experimenter flips a coin 100 times and gets 59 heads.

This means that n = 100, \pi = \frac{59}{100} = 0.59

98% confidence level

So \alpha = 0.02, z is the value of Z that has a p-value of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 - 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.4756

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 + 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.7044

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).

5 0
3 years ago
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