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sammy [17]
3 years ago
5

Find the quotient. (2x^4-3x^3-6x^2+11x+8)/(x-2)

Mathematics
2 answers:
meriva3 years ago
6 0

Answer:

2x^3+x^2-4x+3+14/x-3 // Option B on edge

Step-by-step explanation:

Took the test

stiv31 [10]3 years ago
4 0

Since the remander is 0 the answer is 2x²+5x+2

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Can someone please help me?
Lisa [10]

Answer:

B

Step-by-step explanation:

Using a graphing calculator, the graph shows an almost parabola like shape between 0 and 12

Plus if the width was equal or greater than twelve we would have a 0 or even negative volume which is impossible

7 0
3 years ago
Part I
Crazy boy [7]

Answer:

130

Step-by-step explanation:

10 x 13 = 130

13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 = 130

4 0
3 years ago
A taxi charges $3,plus $1.50 for every mile traveled. Mr.lewis rode in the taxi from his house to the airport and was charged $3
torisob [31]

Answer:

18 miles

Step-by-step explanation:

$30 - $3 = 27  initial charge

$27/1.50 = 18  miles driven

4 0
2 years ago
At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is
Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

81% of students own a wireless device.

This means that p = 0.81

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487

34.87% probability that all 5 have a wireless device

5 0
3 years ago
What is the prime factorization of 48?
babunello [35]

Answer:

2x2x2x2x3

Step-by-step explanation:

6 0
3 years ago
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