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3 years ago

7. If n # 81, find the critical values x 2/l and x 2/h for a confidence level of 99%, (1 point)

Mathematics

2 answers:

Stels [109]3 years ago

7 0

Answer: 51.172, 116.321

GREYUIT [131]3 years ago

6 0

Answer: 51.172, 116.321

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In ABC, and intersect each other at point Q. According to a theorem on medians, Q divides in the ratio 2 : 1. What are the coord

Flauer [41]

How to find rapidly the coordinates of Q:
since Q is the center of gravity of the triangle ABC, so we have the following vector relationship
vecQA +vecQB +vecQC =vec0
vecQA=(x-3, y+2)
vecQB=(x-1, y+5)
vecQC=(x-7, y+5)
vec0=(0, 0)

so, vecQA +vecQB +vecQC =vec0 is equivalent to

x-3 +x-1+x-7 =0, and y+2+y+5+y+5=0 so 3x-11=0 implies x=11/3
and 3y+12=0 implies y=-12/3

finally the the coordinates of point Q is (11/3, -4)

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3 years ago

Help me out with this problem ?

3241004551 [841]

Answer:

(-4.5, 2.5)

Step-by-step explanation:

The solution for two graphed linear equations is the intersection (The point where two lines meet or cross).

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3 years ago

Which of the following rational numbers is equal to 3 point 1 with bar over 1?

Alecsey [184]

Answer:

28/9

Step-by-step explanation:

22/9 = 2.444444...

24/9 = 2.66666...

26/9 = 2.88888...

28/9 = 3.1111111...

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3 years ago

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A scientist planted seeds in 4 sections of soil for experiment. Not alll of the seeds grew into the plant .after 20 days, the sc

Karolina [17]

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3 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

4 years ago