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Sav [38]
3 years ago
6

1. While solutions to the general quintic equation in terms of radicals are impossible to obtain, they do exist. Prove that the

given quintic equation has real solutions. x^5-3x+1=0
Mathematics
2 answers:
xenn [34]3 years ago
8 0

Answer with step-by-step explanation:

Quintic function :It is defined as the function of degree five.

It can be defined ax^5+bx^4+cx^3+dx^2+ex+f=0

Wherea\neq0,b,c,d,e and f are members of field , typically rational numbers,real numbers and complex numbers .

We are given that quintic equation

x^5-3x+1=0

We have to prove that given quintic equation has real solutions

Using Discarte's rule

f(x)=x^5-3x+1

When are going from coefficient of first term to coefficient of second term then sign change and when we are going from coefficient of second term to coefficient of third term then sigh also changes .Therefore,, there are two times sign change .Hence, possible positive real roots are 2 or zero.Then number of positive real roots are even in number .

f(-x)=-x^5+3x+1

When we are going from first term to second term then sign change and then from second term to third term then sign does not changes.

Hence, possible negative real root is   1.

Therefore, we can say given quintic equation has real root.

Hence, proved

Marta_Voda [28]3 years ago
3 0

Answer:

Step-by-step explanation:

Given is a function as x^5-3x+1

Equating to 0 we have equation

If the function f(x) has x intercepts then the solutions are real

Let us use remainder theorem and change of signs rule

f(0) = 1>0

f(-1) = -1+3+1=3

f(-2) = -32+6+1<0

This implies there is a real root between -1 and -2.

f(1) = -1

Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.

f(2) = 32-6+1>0

Since f(1) and f(2) have different signs there exists a real root between 1 and 2.

Thus there are definitely three real solutions as

one between -1 and 0, one between 0 and 1, and third between 1 and 2.

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Answer: V=0.0028278\ m^3

Step-by-step explanation:

Given

Length of the pipe l=4\ m

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Volume of the pipe

\Rightarrow V=\dfrac{\pi }{4}[d_o^2-d_i^2]\\\\\text{Insert the values}\\\\\Rightarrow V=\dfrac{\pi}{4}[5^2-4^2]\times 10^{-4}\times 4\\\\\Rightarrow V=28.278\times 10^{-4}\ m^3\\\\\Rightarrow V=0.0028278\ m^3

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Answer:

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Step-by-step explanation:

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A report on consumer financial literacy summarized data from a representative sample of 1,664 adult Americans. Based on data fro
daser333 [38]

Answer:

a. The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

b. Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Sample of 1,664 adult Americans, 939 people in the sample would have given themselves a grade of A or B in personal finance.

This means that n = 1664, \pi = \frac{939}{1664} = 0.5643

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 - 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.54

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 + 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.588

The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B?

Yes, because the confidence interval is entirely above 0.5.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

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Step-by-step explanation:

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