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11 months ago

Divide g(x) = x^4 - 2x^3 + X + 3 by x+3

Mathematics

1 answer:

SVEN [57.7K]11 months ago

8 0

Let's first get the coefficients of the numerator: x^4 - 2x^3 + x + 3 = 1, -2, 0, 1, 3

There is no x^2 in the expression, thus, the coefficient for x^2 = 0

Zero of the denominator: x + 3; x = -3

Using synthetic division,

-3 I 1 -2 0 1 3

I_________________

-3 I 1 -2 0 1 3

I_________________

-3 I 1 -2 0 1 3

I_____-3___________

1 -5

-3 I 1 -2 0 1 3

I_____-3__15_______

1 -5 15

-3 I 1 -2 0 1 3

I_____-3__15_-45____

1 -5 15 -44

-3 I 1 -2 0 1 3

I_____-3__15_-45____

1 -5 15 -44

-3 I 1 -2 0 1 3

I_____-3__15_-45_132__

1 -5 15 -44 135

The remainder is 135. Which transforms it into 135/x+3.

Thus, the quotient of x^4 - 2x^3 + x + 3 divided by x + 3 is:

$x-5x^2+15x\text{ - 44 + }\frac{135}{x+3}$

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Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One

Naily [24]

Answer:

The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

The expected value of Y is $E(Y)=\frac{73}{2} \approx 36.5$ and the variance of Y is $Var(Y)=\frac{179}{4} \approx 44.75$

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

The probability mass function $p_{X}(x)$ of X is given by

$p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}$

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function $p_{Y}(x)$ of Y is given by

$p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}$

The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

The variance of X is

$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

$E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}$

$Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45$

The variance of Y is

$E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)$

$E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377$

$Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75$

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zloy xaker [14]

Answer:

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Step-by-step explanation:

Hope It helps!

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Zolol [24]

Answer:(2,0) x=2,y=0

Step-by-step explanation:

Let's solve your system by elimination.

−x+y=−2;2x−3y=4

Multiply the first equation by 2,and multiply the second equation by 1.

2(−x+y=−2)

1(2x−3y=4)

Becomes:

−2x+2y=−4

2x−3y=4

Add these equations to eliminate x:

−y=0

Then solve−y=0 for y:

−y=0

−y/−1 =0/−1

(Divide both sides by -1)

y=0

Now that we've found y let's plug it back in to solve for x.

Write down an original equation:

−x+y=−2

Substitute0foryin−x+y=−2:

−x+0=−2

−x=−2(Simplify both sides of the equation)

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(Divide both sides by -1)

x=2

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mina [271]

Answer:

Hope this helps

Step-by-step explanation:

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