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Neko [114]
3 years ago
10

Find all zeros of f(x)= x^3 + 3x^2 − 3x − 1

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
4 0
F(1) =  1^3 + 3(1)^1 - 3 - 1 = 1 + 3 - 3 - 1 = 0

so one zero  is x = 1, and x-1 is a factor

dividing f(x)  by x - 1 gives x^2 + 4x + 1

x = [-4 +/- sqrt (16 - 4)] /  2

x = -0.27, -3.73

zeroes are 1 , -0.27 and -3.73   Answer



  
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Answer:

There are no options to choose from, so I'll just give the solution

Step-by-step explanation:

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egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

  • The lower bound is of 0.484.
  • The upper bound is of 0.7292.
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In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

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Supposing that it increases on 37 out of 61 days:

n = 61, \pi = \frac{37}{61} = 0.6066

95% confidence level

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The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292

The ​95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

A similar problem is given at brainly.com/question/16807970

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