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Ber [7]
1 year ago
14

Which quadratic equation is equivalent to (x2 – 1)2 – 11(x2 – 1) + 24 = 0? u2 – 11u + 24 = 0 where u = (x2 – 1) (u2)2 – 11(u2) +

24 where u = (x2 – 1) u2 + 1 – 11u + 24 = 0 where u = (x2 – 1) (u2 – 1)2 – 11(u2 – 1) + 24 where u = (x2 – 1)
please answer quickly timed test
Mathematics
2 answers:
alex41 [277]1 year ago
6 0

we have

(x^{2}-1)^{2}-11(x^{2}-1)+24=0 --------> equation

Let

u=x^{2}-1

substitute the value of u in the equation

(u)^{2}-11(u)+24=0

u^{2}-11u+24=0

therefore

<u>the answer is the first option </u>

u^{2}-11u+24=0  where  u=x^{2}-1

Vinvika [58]1 year ago
4 0
<span><span>u2</span> – 11u + 24 = 0 where u = (x2 – 1) the first awnser, a.</span>
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And we can find this probability with this difference:

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And we can find the real value with the following excel code using the binomial distribution:

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And we got 0.118 a very close value from the value obtained using the normal approximation

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=400, p=0.2)

The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=400*0.2=80 \geq 10

n(1-p)=400*(1-0.2)=320 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

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And we want this probability:

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We can use the z score formula given by:

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And we got 0.118 a very close value from the value obtained using the normal approximation

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