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photoshop1234 [79]

3 years ago

13

Write out the form of the partial fraction decomposition of the function (as in this example). Do not determine the numerical va

lues of the coefficients. (a) x4 − 2x3 + x2 + 3x − 2 x2 − 2x + 1

1 answer:

butalik [34]3 years ago

5 0

$\dfrac{x^4-2x^3+x^2+3x-2}{x^2-2x+1}$

The degree of the numerator exceeds the degree of the denominator, so first you have to divide:

$x^2+\dfrac{3x-2}{x^2-2x+1}$

Now, $x^2-2x+1=(x-1)^2$ , so the remainder term can be expanded to get

$\boxed{x^2+\dfrac a{x-1}+\dfrac b{(x-1)^2}}$

You might be interested in

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

3 years ago

Simplify the expressioh 6.9-(-5)+4.1

Vadim26 [7]

Answer:

16

Step-by-step explanation:

6.9-(-5)+4.1

When we subtract negatives, it is like adding positives

6.9 +(5)+4.1

11.9 + 4.1

16

8 0

3 years ago

It would mean a lot.

LUCKY_DIMON [66]

First, subtract y2 - y1 to find the vertical distance. Then, subtract x2 - x1 to find the horizontal distance.

Formula to find distance given two points.

Square root (X2 - X1)^2 + (Y2 - Y1)^2

Xa Ya Xb Yb
A = (3, -4) B = (-1, 3)

Xa goes into X2 and Xb goes into X1

(3 - (-1))^2

Ya goes into Y2 and Yb goes into Y1

(-4 - 3)^2

Square root (3 - (-1))^2 + (-4 - 3)^2

Square root (4)^2 + (-7)^2

Square root 16 + 49

Square root 65

= 8.06

The error was Drako had (3 - 4)^2 when it should have been (3 - (-4))^2 because a positive is subtracting a negative.

6 0

3 years ago

If x + 3y = 7 and x - 3y =7 , then find y<br><br> Please help me

telo118 [61]

Answer:

$x + 3y = 7 , x - 3y = 7 : y = , x = 7$

Step-by-step explanation:

$\left[\begin{array}{ccc}x + 3y = 7\\x - 3y = 7\end{array}\right]$

Isolate $x$ for $x + 3y = 7 : x = 7 - 3y$ :

Substitute $x = 7 - 3y$ :

$\begin{bmatrix}7-3y-3y=7\end{bmatrix}$

Isolate $y$ for $7 - 3y - 3y = 7 : y = 0$ :

For $x$ $=7-3y$

Substitute $y = 0$

$x=7-3 * 0$

$7 - 3 * 0 = 7$

$x = 7$

The solution to the system of equations are:

$y=0,\:x=7$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you. Have a good day! =)

3 0

3 years ago

Help with number 7!! You don’t have to do it but how do you do the problem??

musickatia [10]

80 * 5/8 will give you how many are thoroughbreds.

80 * 5/8 = 50

Now subtract this from all of the horses.

80 - 50 = 30.

There are 30 quarter horses.

OR If 5/8 of the horses are thoroughbreds, then 3/8 are quarter horses.

80 * 3/8 = 30.

6 0

3 years ago

Read 2 more answers