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lesya [120]
3 years ago
8

Polynomial Functions help

Mathematics
1 answer:
Studentka2010 [4]3 years ago
6 0
The key piece of information for these questions is the Fundamental Theorem of Algebra, which states that a degree n polynomial has n complex roots. A complex root can be either real or imaginary. 

First question, regarding the polynomial y = x^3 - 3x^2 + 16x - 48:
We know there is one real root, the x-intercept.
Since it's a third degree polynomial, there are three complex roots in total. 
Therefore, there is one real root and two imaginary roots.
Answer is B

Second question:
You probably can guess the answer, now that you know the Fundamental Theorem of Alegebra:
There are 3 real zeros, each with multiplicity one, meaning each root only happens once. It's a 5th degree polynomial, so there are a total of 5 roots, implying 2 imaginary roots. 

Answer is C) 3 real and 2 imaginary zeroes.

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Find The missing length
Reil [10]
C is the hypotenuse of a right triangle whose leg lengths are 7 and 9.  Thus, 

c^2 = 7^2 + 9^2 = 49 + 81 = 130.

Then c = sqrt(130) (answer)
7 0
3 years ago
Nikki, Kylie and Andy are arguing about who has the biggest candy bar. Below are their measurements of each candy bar:
saveliy_v [14]

Answer:

Andy.  

Step-by-step explanation:

10 is greater than 9.99 by 0.01.  10 is also greater than 1.999 by 8.001

6 0
3 years ago
Read 2 more answers
If g(x) = -25, then x =
statuscvo [17]
Answer :

-25

!!!!!!!!!!!!!!!!!
5 0
2 years ago
Someone help please, please be correct.
saveliy_v [14]

Answer:

Which question?

Step-by-step explanation:

5 0
3 years ago
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid \sum \limits ^{n}_{i=1} n^2 P(X=n)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

3 0
3 years ago
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