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asambeis [7]
3 years ago
15

How can we factorise the expression: 2x^6+17x^3+8?​

Mathematics
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

(a + 2)( {a}^{2}  - 2a +  4 )(2 {x}^{ 3}  + 1)

Step-by-step explanation:

2x^6+17x^3+8 \\  = 2x^6+16x^3 +  {x}^{3} +8 \\  =  2{x}^{3} ( {x}^{3}  + 8) + 1( {x}^{3}  + 8) \\  = ( {x}^{3}  + 8)(2 {x}^{ 3}  + 1) \\  = ( {x}^{3}  +  {2}^{3} )(2 {x}^{ 3}  + 1)  \\  = (a + 2)( {a}^{2}  - 2a +  {2}^{2} )(2 {x}^{ 3}  + 1)   \\  = (a + 2)( {a}^{2}  - 2a +  4 )(2 {x}^{ 3}  + 1)  \\

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Umm I need help with math
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Answer:

Step-by-step explanation:

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3 years ago
Bonnie withdrew $1,200 from her savings account. She spent 1/5 of the money on travel arrangements. She spent 2/3 of the remaini
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3 years ago
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Use the elimination method to solve the system of equations. Choose the correct ordered pair. 
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If there are a total of 25 questions, and I scored 56% on my test, how many answers did I get right, and how many answers did I
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Answer:

14 correct and 11 wrong

Step-by-step explanation:

If you got a 56% , we need to change this to decimal form 56% = .56

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8 0
3 years ago
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
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