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kakasveta [241]
3 years ago
9

Algebra 2 please help

Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

1. Negative leading coefficient, fifth degree

2. Positive leading coefficient, fourth degree

Step-by-step explanation:

The graph of a polynomial eventually rises or falls depending on its degree and the sign of the leading coefficient (lc = coefficient of term with the highest exponent)

\begin{array}{ccl}\textbf{Degree} & \textbf{Lc} & \textbf{End Behaviour}\\\text{Even} & + & \text{Up on left and right}\\\text{Even} & - & \text{Down on left; up on right}\\\text{Odd} & + & \text{Down on left and right}\\\text{Odd} & - & \text{Up on left; down on right}\\\end{array}

In addition,  

  • the degree is at least  the number of zeros and  
  • at least 1 greater than the number of local extrema ("bumps"), and  
  • a flattened zero or bump (flex point) shows that shows that it is degenerate (occurs multiple times)

Graph 1

Up on left, down on right —  lc is negative; odd degree

Three zeros —  at least third degree

Two bumps —  at least third degree

Flex points — one at the origin: odd degree, so it occurs 3, 5, 7, or more times.

The polynomial has a positive leading coefficient and is probably fifth degree.

An example is the polynomial y = -x⁵ + x³ (Fig.1).

Graph 2

Up on left and right —  lc is positive; even degree

Four zeros —  at least fourth degree

Three bumps —  at least fourth degree

Flex points — none obvious (although they could be present)

The polynomial has a positive leading coefficient and is probably fourth degree.

An example is the polynomial y = x⁴ - x³ - 4x² - 4x (Fig.2).

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more
I am Lyosha [343]

Answer:

The probability is  P( X \le 4 ) = 0.0054

Step-by-step explanation:

From the question we are told that

   The percentage that are on time is  p =  0.76

   The  sample size is n =  11

   

Generally the percentage that are not on time is

     q =  1- p

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The  probability that no more than 4 of them were on time is mathematically represented as

        P( X \le 4 ) =  P(1 ) +  P(2) + P(3) +  P(4)

=>     P( X \le 4 ) =  \left n } \atop {}} \right.C_1 p^{1}  q^{n- 1} +   \left n } \atop {}} \right.C_2p^{2}  q^{n- 2} +  \left n } \atop {}} \right.C_3 p^{3}  q^{n- 3}  +  \left n } \atop {}} \right.C_4 p^{4}  q^{n- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{11- 1} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{11- 2} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{11- 3}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{11- 4}

P( X \le 4 ) =  \left 11 } \atop {}} \right.C_1 p^{1}  q^{10} +   \left 11 } \atop {}} \right.C_2p^{2}  q^{9} +  \left 11 } \atop {}} \right.C_3 p^{3}  q^{8}  +  \left 11 } \atop {}} \right.C_4 p^{4}  q^{7}

= \frac{11! }{ 10! 1!}  (0.76)^{1}  (0.24)^{10} +   \frac{11!}{9! 2!}  (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!}  (0.76)^{3}  (0.24)^{8}  + \frac{11!}{7!4!}  (0.76)^{4}  (0.24)^{7}

P( X \le 4 ) = 0.0054

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3 years ago
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