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2 years ago

Find the value of x.

Mathematics

1 answer:

Gnom [1K]2 years ago

5 0

Answer:

x=28

Step-by-step explanation:

$x-8=20\\+8 +8\\x=28$

Hope this helps!

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Assume that we want to construct a confidence interval. Do one of the following, as appropriate:_________.

andreyandreev [35.5K]

Answer:

B. z Subscript alpha divided by 2 zα/2 = 1.96.

Step-by-step explanation:

We are given that we want to construct a confidence interval. For this, the summary statistics for randomly selected weights of newborn girls:

n = 236, $\bar x$ = 30.3 hg, s = 7.2 hg. The confidence level is 95%.

As we can clearly see here that the population standard deviation is unknown and the sample size is also very large.

It has been stated that when the population standard deviation is unknown, we should use t-distribution but since the sample size is very large so we can use z distribution also as it is stated that at very large samples; the t-distribution corresponds to the z-distribution.

Here, $\alpha$ = level of significance = 1 - 0.95 = 0.05 or 5%

$\frac{\alpha}{2}=\frac{0.05}{2}$ = 0.025 or 2.5%

So, the value of $Z_(_\frac{\alpha}{2} _)$ in the z table is given as 1.96 with a 2.5% level of significance.

3 0

3 years ago

Cameron needs to order some new supplies for the restaurant where he works. The

zvonat [6]

Answer:

598=194+8x

Step-by-step explanation:

To solve you would subtract 194 from its self and 598

598=194+8x

-194 -194

404=8x

then you would divide 8 from it's self and 404

which would make it 50.5

(but you cant buy half a package of spoons so realistically it'd be 50 or 51.

hope this helped :)

6 0

2 years ago

Read 2 more answers

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

3 years ago

I have asked so many times and no one will answer. :(

andrew-mc [135]

Answer:

1) $\frac{1}{2^7}$ ; 2) $11^7$