F(x) = x3 + 6x2 – x f(x) = 30?
1 answer:
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We have to solve this equation:
Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:
Now we can find the roots of the quadratic polynomial as:
Then, the solutions to the equation are:
x = 0
x = 3 - √3
x = 3 + √3
Answer:
a) Figure attached
b)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 95%
c)
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%
d)
And using the z score we got:
And that correspond to approximately 0.15%
Step-by-step explanation:
Part a
For this case we can see the figure attached.
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part b
Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 95%
Part c
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%
Part d
We want this probability:
And using the z score we got:
And that correspond to approximately 0.15%
