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avanturin [10]
3 years ago
5

Write an equation for the polynomial function that has 2 and -3 + 5i as zero

Mathematics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

p(x) = {x}^{3}  + 4{x}^{2}  + 22x- 68

Step-by-step explanation:

We want to write an equation for the polynomial function with roots;

x = 2

and

x =  - 3 + 5i

Note that complex roots come in pairs and in conjugates.

Therefore

x =  - 3 - 5i

is also a root.

By the factor theorem:

(x-2), (x+3+5i), and (x+3-5i) are factors of the polynomial.

Let p(x) be the polynomial, then in factored form.

p(x) = (x - 2)(x + 3 + 5i)(x + 3 - 5i)

We expand now to get:

p(x) = (x - 2)( {x}^{2}  + 3x  -  5ix + 3x + 9 - 15i + 5ix + 15i + 25)

Simplify now

p(x) = (x - 2)( {x}^{2}  + 6x +34)

We expand further;

p(x) = x( {x}^{2}  + 6x + 34) - 2( {x}^{2}  + 6x + 34)

p(x) = {x}^{3}  + 6 {x}^{2}  + 34x - 2 {x}^{2}   - 12x  - 68

p(x) = {x}^{3}  + 4{x}^{2}  + 22x- 68

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A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
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On $500 at 6% for 18 months what is the interest
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Mike has a collection of 16 antique tin toys, including 2 airplanes. If Mike randomly selects a toy, what is the probability it
myrzilka [38]

Answer:

C. 1/8

Step-by-step explanation:

This is because the number two goes into sixteen eight times. This is the simplest form of the fraction.

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Find the slope of the following graph and write your result in the empty box.
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Answer:

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Step-by-step explanation:

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F(x) = 9x-7. f o f (2)
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Answer: 11

Step-by-step explanation: First, 9 times 2 is 18. Then subtract 18-7 which equals 11.

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