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3 years ago

What is the interval of convergence

Mathematics

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

The interval of converges of a power series is the interval of input values for which the series converges.

Step-by-step explanation:

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assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an avera

aleksley [76]

Let
n = number of data
s = standard deviation (sample)
S = standard deviation (population)

The working equations is

$\frac{(n-1) s^{2} }{ x^{2}_{right} } \ \textless \ S^{2} \ \textless \ \frac{(n-1) s^{2} }{ x^{2}_{left} }$

To find $x^{2}_{right}$ , : (1 - 0.90)/2 = 0.05

To find $x^{2}_{left}$ , : 1 - 0.05 = 0.95

Degrees of freedom = n-1 = 24 - 1 = 23

This is shown in the figure attached. Since there is no row for df=23, we interpolate. Thus,

$x^{2}_{left} = 13.093$

$x^{2}_{right} = 35.17$

Substitute all values,

$\frac{(24-1) 5.6^{2} }{ 35.17} \ \textless \ S^{2} \ \textless \ \frac{(24-1) 5.6^{2} }{ 13.093} }$

Thus the answer is,

$20.51\ \textless \ S^{2} \ \textless \ 55.09$

5 0

3 years ago

Mrs. Matthews is 24 years old and her daughter Nina is 6 years old. How old will Nina be when she is half as old as her mother?

Lena [83]

Answer:

Nina will be 18 years old when she is half as old as her mother.

Step-by-step explanation:

With the information provided, you can write an equation that indicates that Mrs. Matthews current age plus x which would be the number of years until Nina will be half as old as her mother would be equal to two times nina's age plus x, as you would have to multiply Nina's age at that time for 2 to be equal to her mother's age. The equation would be:

24+x=2(6+x)

Now, you can solve for x:

24+x=12+2x

24-12=2x-x

12=x

Finally, you have to add the value of x plus Nina's current age to determine her age when she is half as old as her mother:

12+6=18

Acording to this, the answer is that Nina will be 18 years old when she is half as old as her mother.

6 0

3 years ago

What is the factor of the trinomial 30x square+35x-15

Dafna11 [192]

I hope this helps you

5 (6x^2+7x-- 3)

5 (3x.2x+7-1.3)

5. (3x-1)(2x+3)

4 0

3 years ago

In rectangle QRST, diagonals QS and RT are drawn and intersect at point P. Which of the following st

natta225 [31]

Answer:

Step-by-step explanation:

8 0

3 years ago

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

8 0

3 years ago