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3 years ago

Factor 2x^2+x-3 please

2 answers:

8 0

1
Split the second term into two terms
2{x}^{2}+3x-2x-32x2+3x−2x−3

2 Factor out common terms in the first two terms, then in the last two termsx(2x+3)-(2x+3)x(2x+3)−(2x+3)
3 Factor out the common term 2x+32x+3(2x+3)(x-1)(2x+3)(x−1)

RoseWind [281]3 years ago

4 0

(2x + 3)(x - 1)

Is the factorisation.

3 * -1 = -3
3 + (2 * -1) = 1

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See attachment.

EXPLANATION

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3 years ago

Let the probability of success on a Bernoulli trial be 0.26. a. In five Bernoulli trials, what is the probability that there wil

andreyandreev [35.5K]

Answer:

0.3898 = 38.98% probability that there will be 4 failures

Step-by-step explanation:

A sequence of Bernoulli trials forms the binomial probability distribution.

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.

Let the probability of success on a Bernoulli trial be 0.26.

This means that $p = 0.26$

a. In five Bernoulli trials, what is the probability that there will be 4 failures?

Five trials means that $n = 5$

4 failures, so 1 success, and we have to find P(X = 1).

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

$P(X = 1) = C_{5,1}.(0.26)^{1}.(0.74)^{4} = 0.3898$

0.3898 = 38.98% probability that there will be 4 failures

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3 years ago