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Ber [7]
2 years 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]2 years 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]2 years ago
4 0
<span><span>u2</span> – 11u + 24 = 0 where u = (x2 – 1) the first awnser, a.</span>
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We use the binomial approxiation to the normal to solve this question.

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

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Z = 1.06

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0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0
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