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3 years ago

Factor the expression over the complex numbers. x2+36 Enter your answer in the box.

Mathematics

2 answers:

Zina [86]3 years ago

8 0

hope this help Answer:( x - 3i√2)(x + 3i√2)

solve x² + 18 = 0

x² = - 18 ⇒ x = ±√- 18 = ±3i√2

factors are ( x - (3i√2))(x - (-3i√2))

x² + 18 = (x - 3i√2)(x + 3i√2

Step-by-step explanation:

mafiozo [28]3 years ago

4 0

Answer:

$x^{2}+36=(x-6i)(x+6i)$

Step-by-step explanation:

The given expression is

$x^{2}+36$

It's understood that this expression is equal to zero

$x^{2} +36=0$

Now, we isolate the variable and then apply a squared root to each side of the equation

$x^{2}=-36\\x=\±\sqrt{-36}$

At this point, you'll find that the equation doesn't have solution in the real numbers, that is, all solutions are in the complex numbers. We need to add the imaginary number $i=\sqrt{-1}$ to continue

$x=\±\sqrt{-36}\\x=\±\sqrt{36}i\\x=\±6i\\\\x_{1}=6i\\x_{2}=-6i$

Now, representing these solution as factors, we would have

$x-6i=0\\x+6i=0$

Finally,

$x^{2}+36=(x-6i)(x+6i)$

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

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3 years ago

Suppose a designer has a palette of 8 colors to work with, and wants to design a flag with 2 vertical stripes, all different col

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Answer:

For a flag with 2 different colours, there are 8 possible colours for the first colour to be chosen from and 7 possible colours (excluding the first colour already chosen) for the second colour to be chosen from, given that colours are different. Hence there are a total of 8 x 7 = 56 possible flags to be created.

Step-by-step explanation:

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3 years ago

Find the unit price for each option shown below. Round to the nearest cent when necessary. Option I: 10 candy bars for $6.75 Opt

natita [175]

Option I: 10 candy bars for $6.75

6.75/10 = 0.68 

Option II: 12 candy bars for $7.25 

7.25 / 12 = 0.60

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