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4 years ago

Find the remainder when x³+3x²+3x+1 is divided by x+1

Mathematics

1 answer:

creativ13 [48]4 years ago

7 0

The answer should be 0 as your remainder.

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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.

7 0

4 years ago

-9+-5x please help meee cause I'm not Finna fail

Westkost [7]

Answer:

-9-5x

Step-by-step explanation:

they aren't the same because of the variable on 5x. You can simplify the problem to -9-5x but that's about it unless it equals something.

7 0

3 years ago

Solve each question and check your solution -2(3x-4)=4(x+3)+6

agasfer [191]

Answer:

x = -1

Step-by-step explanation:

-2(3x-4)=4(x+3)+6

Distribute

-6x +8 = 4x +12 +6

Combine like terms

-6x +8 = 4x+18

Add 6x to each side

-6x+6x +8 = 4x+6x +18

8 = 10x+18

Subtract 18 from each side

8-18 =10x+18-18

-10 = 10x

Divide each side by 10

-10/10 = 10x/10

-1 =x

8 0

4 years ago

WIIL GIVE BRAINLIEST IF YOU HELP ME

cricket20 [7]

Answer:

What is the question?

Step-by-step explanation:

5 0

3 years ago

A minimum-phase continuous-time system has all its poles and zeros in the left-half s-plane. If a minimum-phase continuous-time

vlabodo [156]

Answer:

Yes, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.

Step-by-step explanation:

Bilinear Transform:

In digital signal processing, the bilinear transform is used to convert continuous time system into discrete time system representation.

Minimum-Phase:

We know that a system is considered to be minimum phase if the zeros are situated in the left half of the s-plane in continuous time system. In the same way, a system is minimum phase when its zeros are inside the unit circle of z-plane in discrete time system.

The bilinear transform is used to map the left half of the s-plane to the interior of the unit circle in the z-plane preserving the stability and minimum phase property of the system. Therefore, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.

8 0

4 years ago

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